AIEEE 2012 solution (26th may)

Detailed solution given below Q1

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ans 1 Q2

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ans2 NOTE- actual ans is 0125m Q10

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ans3

Q11

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ans1 Q12

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ans2 (actual ans is 8888)

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ANS (ans is 4) Q83

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ANS2 By Saurav gupta Electronics amp tele comm Engg (2nd year) Jadavpur university

AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012)

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG(2ND YEAR)

JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1

Q2 q1+q2 = Q ---eq(1) given


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given

or q14πrsup2=q24πRsup2

q1rsup2= q2Rsup2-------- eq(2)

solving eq (1) amp (2) we get

q1 =Q(1+rsup2Rsup2)

now

V=14πϵ˳ (q1r+q2R)

=14πϵ˳ q1(1r+ q1 Rrsup2)

=Q4πϵ˳ (r+R)(Rsup2+rsup2)

Ans -4

Q3 Sinα ≃ tanα≃α


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4

Q4 E=[Ecosωt+Ecos(ωt) cos(ωt)]icirc

=[Ecosωt+E2cos(ω+ωt)

+E2cos(ω-ωt)]icirc

Ans 4

Q5 N MSD = (N+1) VSD

1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD

=[1 ndash N(N+1)]MSD

a(N+1)

Ans 2

Q6 Case1

Tension unit length = 100gl

Case2

Tension unit length= 2mgl

2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g

Putting T=1 amp g≃πsup2

R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)

Solving 1 amp 2 we get

50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

υ(fundamentaloutsidewater)=vλ

=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1

Q12 Mu = mv + mvrsquo


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

frac12 musup2 = frac12 mvsup2 + frac12 mvrsquosup2

Solving eq (1) amp(2)

V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳

IfυisdoubledKEismorethandoubled

Also

Photocurrent depends on intensity not on frequency

Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v

E=E˳cos(wt-2πλy)ẑ

Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16 I Esup3= 136 zsup2nsup2

= 136 x 9

= 1224

TE = IEsup1 + IEsup2 + IEsup3

=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3

Q19 vsup2 = usup2 + 2al

Vrsquosup2 = usup2 + 2a(l2)

=usup2 + frac12 (vsup2-usup2)

Vrsquosup2 = (vsup2 + usup2)2

Vrsquo = radic(vsup2+usup2)2

Ans -2

Q20 Statement 1 ndashtrue


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true

Bnearthecentreisμ˳nibutdecresesaswemovetowardsitrsquosendandbecomesμ˳ni2atthe extreme end which explains statement 1

Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4

Q22 Y=Asin(ωt+2π3)

At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ

Fκ = μ(mg- Fsinθ)

Now

Fcosθ = Fκ

F = μW(cosθ + μsinθ)---(1)

For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ

or cosθ= 1radic1+μsup2

or sinθ = μradic1 +μsup2

substituting these values in eq ndash1

F = μWradic1 +μsup2

Ans -2

Q24

As we pull the connector by say x amount there will be growth in current in the circuit which will

be opposed by the inductor(due to self inductance)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Here inductor is analogous to spring(mechanical analog of inductor)

Opposing force prop -x

Hence the connector will execute SHM

Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv

= α2 (vfsup2- vᵢsup2)

= αvsup2 (msup2-1)2

Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant

fMr = (1-f)Mvrsquorrsquo

frsup2ω = (1-f)rrsquosup2ωrsquo

ωωrsquo = (1-f)f x (rrsquor)sup2---(1)

now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)

putting this in eq ---(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680

Hence 30 efficiency is not possible by carnotrsquos theorem

Ans 2

Q29

Vol(initially) = vol(finally)

n 43 πrsup3 = 43 π Rsup3

n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs

= n 4πrsup2T - 4πRsup2T

Solving we get

Vsup2 = 8πTRsup2m [Rr -1]

Put m = ρ x 43 πRsup3

V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2

AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)

BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61

Lt x-gt 2ˉ x[x] le limx-gt 2ˉ f(x) le lim x-gt2ˉ radic6-x

2 le limx-gt2ˉ f(x) le 2

Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given

Limx-gt2ˉ f(x) ne f(2)

Hence f(x) is not continuous at x=2

Statement 2- ----false

Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63

g(x) = axsup33 + bxsup2 + cx

g(x) = axsup2 + bx + c

now

g(1)=0 g(x)= 0 and g(x) is obviously continuous

by rollersquos theorem

at some point between (01)

grsquo(x)=0

hence

axsup2 +bx +c = 0

has itrsquos root between (01)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1

= asup2bsup2csup2 1 -1 1

1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65

2secθ = sec(θ+ф) + sec(θ-ф)

2cos(θ+ф)cos(θ-ф)

cos(θ+ф) +cos(θ-ф)

= (cossup2θ ndashsinsup2θ) cosθcosф

Cossup2θ(cosф-1) = -sinsup2ф

Cossup2θ = 2 cossup2ф

K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66

M = L + (N2 ndash c)x hf

25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2

Secsup2x (1+xsup2) -1

1+xsup2

intSecsup2x dx - int dx1+xsup2

F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

f(x)= tanx ndash tanˉsup1x + c

f(0) = 0

C = 0

f(1) = tan1 ndash tanˉsup11

= tan1 ndash π4

Ans 4

Q68

dr of normal to plane 1 ---(100)

dr of normal to plane 2 ---(2-5a3)

dr of normal to plane 3 ---(3b1-3)

let dr of common line (pqr)

now

line is perpendicular to plane1

P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

line is perpendicular to plane 2

2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)

Solving these eq we get

a =15

Ans 3

Q69

lim f(1-α) ndashf(1)

α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)


Ysup2(1 +msup2) - y(3m + 2msup2) +1msup2-3m = 0

Since two points are equidistant from and opp sides of x-axis

Sum of the roots of this eqᶯ must be 0

ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z

Statement 1mdashtrue

Statement 2---false

As |z|=|w| only implies that they are equidistant from centre of argand diagram

Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib

αβ = asup2 + bsup2

|α| +|β| = radicasup2+bsup2 + radicasup2+bsup2

=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a

Slope of this line m1= (2a-1a)

Slope of given line m2= 2

Since these lines are perpendicular

m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now (a2a) are middle points of (10) and (αβ)

we get

α = 45 β = 35

Ans 4

Q74

sup2C1 x ⁴C2 x 3 + sup2C2 x ⁴C1 x 3

= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3

no of one one relation is ⁸P3 = 336

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy

=˳int⁸ ysup1 sup3 dy

= 12

Ans 2

Q77

P q = ~p v q

Basic formula

Can check it by truth table

Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(Y ndash 2)= -dxdy|ˣ ⁴ (x-4)

y-2 =4-x

x +y = 6

Ans 4

Q79

|a+b| = radic asup2+bsup2+2abcosθ

|c| = radic 9+25+2x3x5cosθ

30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09

˳int⁰⁹ o dx = 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81

I F = exp( int 2x dx)

=xsup2

Now

yxsup2 = int xsup2xsup2 dx + c

y = xsup35 + cxˉsup2

Ans 4

Q82

bf-ec -ae -ab

Aˉsup1 = -1bd cd -ad 0

-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get

c=0 e=0 f=0 a=1a=bsup2a=dsup2

or bsup2 =1 dsup2 =1

or b = plusmn1 d= plusmn 1

hence there are 4 possible matrices

Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2

dwdt = n dwdt + w dndt

= 8tsup3-6tsup2+14t-3

dwdt|at (t=1) = 13

Ans 3

Q84

radic (x-1)sup2 + ysup2

radic (x+1)sup2 + ysup2

Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)

Hence the circumcentre of the triangle is (53 0)

Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4

These 2 lines are parallel

p(111)

line 1

q(2λ-λ2λ) line2

line pq and line(1 or2) are perpendicular

hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86

1 + 43 + 109 + helliphellip

1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)

n + frac12[1- 13ᶯˉsup1]

Ans 1

Q87

Foci of ellipse= radicasup2-bsup2

=radic16-bsup2

Foci of hyperbola = radicasup2+bsup2

=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88

Tanˉsup1(sin(cosˉsup1radic23))

Tanˉsup1(1radic3)

π6

Ans 3

Q89

c is perpendicular to a x b

c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

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Q82

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ANS (ans is 4) Q83

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ANS3 Q84

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ANS4

Q85

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ANS1 Q86

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ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


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LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

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JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


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-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q7

1)

2)

3)

4)

ans1 Q8

1)

2)

3)

4)

ans4

Q9

1)

2)

3)

4)


1)

2)

3)

4)

ans3

Q11

1)

2)

3)

4)

ans1 Q12

1)

2)

3)

4)


Q13

1)

2)

3)

4)

ans3 Q14

1)

2)

3)

4)

ans1

Q15

1)

2)

3)

4)

ans3 Q16

1)

2)

3)

4)

ans4

Q17

1)

2)

3)

4)

ans4 Q18

1)

2)

3)

4)

ans3

Q19 1)

2)

3)

4)

ans2 Q20

1)

2)

3)

4)

ans3

Q21

1)

2)

3)

4)

ans4 Q22

1)

2)

3)

4)

ans3 Q23

1)

2)

3)

4)

ans2

Q24

1)

2)

3)

4)

ans4 Q25

1)

2)

3)

4)

ans4

Q26 1)

2)

3)

4)

ans3 Q27

1)

2)

3)

4)

ans2

Q28

1)

2)

3)

4)

ans2 Q29

1)

2)

3)

4)

ans3

Q30 1)

2)

3)

4)

ans2

Q61

1)

2)

3)

4)

ANS1 Q62

1)

2)

3)

4)

ANS3

Q63

1)

2)

3)

4)

ANS 4 Q64

1)

2)

3)

4)

ANS2 Q65

1)

2)

3)

4)

ANS4

Q66

1)

2)

3)

4)

ANS2 Q67

1)

2)

3)

4)

ANS4 Q68

1)

2)

3)

4)

ANS3

Q69 1)

2)

3)

4)

ANS1 Q70

1)

2)

3)

4)

ANS3 Q71

1)

2)

3)

4)

ANS2

Q72 1)

2)

3)

4)

ANS2 Q73

1)

2)

3)

4)

ANS4 Q74

1)

2)

3)

4)

ANS1

Q75 1)

2)

3)

4)

ANS1 Q76

1)

2)

3)

4)

ANS2 Q77

1)

2)

3)

4)

ANS1 Q78

1)

2)

3)

4)

ANS4

Q79

1)

2)

3)

4)

ANS3 Q80

1)

2)

3)

4)

ANS2 Q81

1)

2)

3)

4)

ANS4

Q82

1)

2)

3)

4)

ANS (ans is 4) Q83

1)

2)

3)

4)

ANS3 Q84

1)

2)

3)

4)

ANS4

Q85

1)

2)

3)

4)

ANS1 Q86

1)

2)

3)

4)

ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q9

1)

2)

3)

4)


1)

2)

3)

4)

ans3

Q11

1)

2)

3)

4)

ans1 Q12

1)

2)

3)

4)


Q13

1)

2)

3)

4)

ans3 Q14

1)

2)

3)

4)

ans1

Q15

1)

2)

3)

4)

ans3 Q16

1)

2)

3)

4)

ans4

Q17

1)

2)

3)

4)

ans4 Q18

1)

2)

3)

4)

ans3

Q19 1)

2)

3)

4)

ans2 Q20

1)

2)

3)

4)

ans3

Q21

1)

2)

3)

4)

ans4 Q22

1)

2)

3)

4)

ans3 Q23

1)

2)

3)

4)

ans2

Q24

1)

2)

3)

4)

ans4 Q25

1)

2)

3)

4)

ans4

Q26 1)

2)

3)

4)

ans3 Q27

1)

2)

3)

4)

ans2

Q28

1)

2)

3)

4)

ans2 Q29

1)

2)

3)

4)

ans3

Q30 1)

2)

3)

4)

ans2

Q61

1)

2)

3)

4)

ANS1 Q62

1)

2)

3)

4)

ANS3

Q63

1)

2)

3)

4)

ANS 4 Q64

1)

2)

3)

4)

ANS2 Q65

1)

2)

3)

4)

ANS4

Q66

1)

2)

3)

4)

ANS2 Q67

1)

2)

3)

4)

ANS4 Q68

1)

2)

3)

4)

ANS3

Q69 1)

2)

3)

4)

ANS1 Q70

1)

2)

3)

4)

ANS3 Q71

1)

2)

3)

4)

ANS2

Q72 1)

2)

3)

4)

ANS2 Q73

1)

2)

3)

4)

ANS4 Q74

1)

2)

3)

4)

ANS1

Q75 1)

2)

3)

4)

ANS1 Q76

1)

2)

3)

4)

ANS2 Q77

1)

2)

3)

4)

ANS1 Q78

1)

2)

3)

4)

ANS4

Q79

1)

2)

3)

4)

ANS3 Q80

1)

2)

3)

4)

ANS2 Q81

1)

2)

3)

4)

ANS4

Q82

1)

2)

3)

4)

ANS (ans is 4) Q83

1)

2)

3)

4)

ANS3 Q84

1)

2)

3)

4)

ANS4

Q85

1)

2)

3)

4)

ANS1 Q86

1)

2)

3)

4)

ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q11

1)

2)

3)

4)

ans1 Q12

1)

2)

3)

4)


Q13

1)

2)

3)

4)

ans3 Q14

1)

2)

3)

4)

ans1

Q15

1)

2)

3)

4)

ans3 Q16

1)

2)

3)

4)

ans4

Q17

1)

2)

3)

4)

ans4 Q18

1)

2)

3)

4)

ans3

Q19 1)

2)

3)

4)

ans2 Q20

1)

2)

3)

4)

ans3

Q21

1)

2)

3)

4)

ans4 Q22

1)

2)

3)

4)

ans3 Q23

1)

2)

3)

4)

ans2

Q24

1)

2)

3)

4)

ans4 Q25

1)

2)

3)

4)

ans4

Q26 1)

2)

3)

4)

ans3 Q27

1)

2)

3)

4)

ans2

Q28

1)

2)

3)

4)

ans2 Q29

1)

2)

3)

4)

ans3

Q30 1)

2)

3)

4)

ans2

Q61

1)

2)

3)

4)

ANS1 Q62

1)

2)

3)

4)

ANS3

Q63

1)

2)

3)

4)

ANS 4 Q64

1)

2)

3)

4)

ANS2 Q65

1)

2)

3)

4)

ANS4

Q66

1)

2)

3)

4)

ANS2 Q67

1)

2)

3)

4)

ANS4 Q68

1)

2)

3)

4)

ANS3

Q69 1)

2)

3)

4)

ANS1 Q70

1)

2)

3)

4)

ANS3 Q71

1)

2)

3)

4)

ANS2

Q72 1)

2)

3)

4)

ANS2 Q73

1)

2)

3)

4)

ANS4 Q74

1)

2)

3)

4)

ANS1

Q75 1)

2)

3)

4)

ANS1 Q76

1)

2)

3)

4)

ANS2 Q77

1)

2)

3)

4)

ANS1 Q78

1)

2)

3)

4)

ANS4

Q79

1)

2)

3)

4)

ANS3 Q80

1)

2)

3)

4)

ANS2 Q81

1)

2)

3)

4)

ANS4

Q82

1)

2)

3)

4)

ANS (ans is 4) Q83

1)

2)

3)

4)

ANS3 Q84

1)

2)

3)

4)

ANS4

Q85

1)

2)

3)

4)

ANS1 Q86

1)

2)

3)

4)

ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


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= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

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Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

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Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

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Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

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we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

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Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

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y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

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JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

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JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

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N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

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Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

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statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

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bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

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r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q13

1)

2)

3)

4)

ans3 Q14

1)

2)

3)

4)

ans1

Q15

1)

2)

3)

4)

ans3 Q16

1)

2)

3)

4)

ans4

Q17

1)

2)

3)

4)

ans4 Q18

1)

2)

3)

4)

ans3

Q19 1)

2)

3)

4)

ans2 Q20

1)

2)

3)

4)

ans3

Q21

1)

2)

3)

4)

ans4 Q22

1)

2)

3)

4)

ans3 Q23

1)

2)

3)

4)

ans2

Q24

1)

2)

3)

4)

ans4 Q25

1)

2)

3)

4)

ans4

Q26 1)

2)

3)

4)

ans3 Q27

1)

2)

3)

4)

ans2

Q28

1)

2)

3)

4)

ans2 Q29

1)

2)

3)

4)

ans3

Q30 1)

2)

3)

4)

ans2

Q61

1)

2)

3)

4)

ANS1 Q62

1)

2)

3)

4)

ANS3

Q63

1)

2)

3)

4)

ANS 4 Q64

1)

2)

3)

4)

ANS2 Q65

1)

2)

3)

4)

ANS4

Q66

1)

2)

3)

4)

ANS2 Q67

1)

2)

3)

4)

ANS4 Q68

1)

2)

3)

4)

ANS3

Q69 1)

2)

3)

4)

ANS1 Q70

1)

2)

3)

4)

ANS3 Q71

1)

2)

3)

4)

ANS2

Q72 1)

2)

3)

4)

ANS2 Q73

1)

2)

3)

4)

ANS4 Q74

1)

2)

3)

4)

ANS1

Q75 1)

2)

3)

4)

ANS1 Q76

1)

2)

3)

4)

ANS2 Q77

1)

2)

3)

4)

ANS1 Q78

1)

2)

3)

4)

ANS4

Q79

1)

2)

3)

4)

ANS3 Q80

1)

2)

3)

4)

ANS2 Q81

1)

2)

3)

4)

ANS4

Q82

1)

2)

3)

4)

ANS (ans is 4) Q83

1)

2)

3)

4)

ANS3 Q84

1)

2)

3)

4)

ANS4

Q85

1)

2)

3)

4)

ANS1 Q86

1)

2)

3)

4)

ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

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Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

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σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

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α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

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LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

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dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

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α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

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Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

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=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

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V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

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KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

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= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

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Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

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JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

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JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

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JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

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Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

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Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

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ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


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now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

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Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

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p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

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Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

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-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

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Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

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f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

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2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

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JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

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Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

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we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

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Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


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N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

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bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q15

1)

2)

3)

4)

ans3 Q16

1)

2)

3)

4)

ans4

Q17

1)

2)

3)

4)

ans4 Q18

1)

2)

3)

4)

ans3

Q19 1)

2)

3)

4)

ans2 Q20

1)

2)

3)

4)

ans3

Q21

1)

2)

3)

4)

ans4 Q22

1)

2)

3)

4)

ans3 Q23

1)

2)

3)

4)

ans2

Q24

1)

2)

3)

4)

ans4 Q25

1)

2)

3)

4)

ans4

Q26 1)

2)

3)

4)

ans3 Q27

1)

2)

3)

4)

ans2

Q28

1)

2)

3)

4)

ans2 Q29

1)

2)

3)

4)

ans3

Q30 1)

2)

3)

4)

ans2

Q61

1)

2)

3)

4)

ANS1 Q62

1)

2)

3)

4)

ANS3

Q63

1)

2)

3)

4)

ANS 4 Q64

1)

2)

3)

4)

ANS2 Q65

1)

2)

3)

4)

ANS4

Q66

1)

2)

3)

4)

ANS2 Q67

1)

2)

3)

4)

ANS4 Q68

1)

2)

3)

4)

ANS3

Q69 1)

2)

3)

4)

ANS1 Q70

1)

2)

3)

4)

ANS3 Q71

1)

2)

3)

4)

ANS2

Q72 1)

2)

3)

4)

ANS2 Q73

1)

2)

3)

4)

ANS4 Q74

1)

2)

3)

4)

ANS1

Q75 1)

2)

3)

4)

ANS1 Q76

1)

2)

3)

4)

ANS2 Q77

1)

2)

3)

4)

ANS1 Q78

1)

2)

3)

4)

ANS4

Q79

1)

2)

3)

4)

ANS3 Q80

1)

2)

3)

4)

ANS2 Q81

1)

2)

3)

4)

ANS4

Q82

1)

2)

3)

4)

ANS (ans is 4) Q83

1)

2)

3)

4)

ANS3 Q84

1)

2)

3)

4)

ANS4

Q85

1)

2)

3)

4)

ANS1 Q86

1)

2)

3)

4)

ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

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α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


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LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

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JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

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JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

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JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

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Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


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N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

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bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q17

1)

2)

3)

4)

ans4 Q18

1)

2)

3)

4)

ans3

Q19 1)

2)

3)

4)

ans2 Q20

1)

2)

3)

4)

ans3

Q21

1)

2)

3)

4)

ans4 Q22

1)

2)

3)

4)

ans3 Q23

1)

2)

3)

4)

ans2

Q24

1)

2)

3)

4)

ans4 Q25

1)

2)

3)

4)

ans4

Q26 1)

2)

3)

4)

ans3 Q27

1)

2)

3)

4)

ans2

Q28

1)

2)

3)

4)

ans2 Q29

1)

2)

3)

4)

ans3

Q30 1)

2)

3)

4)

ans2

Q61

1)

2)

3)

4)

ANS1 Q62

1)

2)

3)

4)

ANS3

Q63

1)

2)

3)

4)

ANS 4 Q64

1)

2)

3)

4)

ANS2 Q65

1)

2)

3)

4)

ANS4

Q66

1)

2)

3)

4)

ANS2 Q67

1)

2)

3)

4)

ANS4 Q68

1)

2)

3)

4)

ANS3

Q69 1)

2)

3)

4)

ANS1 Q70

1)

2)

3)

4)

ANS3 Q71

1)

2)

3)

4)

ANS2

Q72 1)

2)

3)

4)

ANS2 Q73

1)

2)

3)

4)

ANS4 Q74

1)

2)

3)

4)

ANS1

Q75 1)

2)

3)

4)

ANS1 Q76

1)

2)

3)

4)

ANS2 Q77

1)

2)

3)

4)

ANS1 Q78

1)

2)

3)

4)

ANS4

Q79

1)

2)

3)

4)

ANS3 Q80

1)

2)

3)

4)

ANS2 Q81

1)

2)

3)

4)

ANS4

Q82

1)

2)

3)

4)

ANS (ans is 4) Q83

1)

2)

3)

4)

ANS3 Q84

1)

2)

3)

4)

ANS4

Q85

1)

2)

3)

4)

ANS1 Q86

1)

2)

3)

4)

ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

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α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

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LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

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dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

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KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q19 1)

2)

3)

4)

ans2 Q20

1)

2)

3)

4)

ans3

Q21

1)

2)

3)

4)

ans4 Q22

1)

2)

3)

4)

ans3 Q23

1)

2)

3)

4)

ans2

Q24

1)

2)

3)

4)

ans4 Q25

1)

2)

3)

4)

ans4

Q26 1)

2)

3)

4)

ans3 Q27

1)

2)

3)

4)

ans2

Q28

1)

2)

3)

4)

ans2 Q29

1)

2)

3)

4)

ans3

Q30 1)

2)

3)

4)

ans2

Q61

1)

2)

3)

4)

ANS1 Q62

1)

2)

3)

4)

ANS3

Q63

1)

2)

3)

4)

ANS 4 Q64

1)

2)

3)

4)

ANS2 Q65

1)

2)

3)

4)

ANS4

Q66

1)

2)

3)

4)

ANS2 Q67

1)

2)

3)

4)

ANS4 Q68

1)

2)

3)

4)

ANS3

Q69 1)

2)

3)

4)

ANS1 Q70

1)

2)

3)

4)

ANS3 Q71

1)

2)

3)

4)

ANS2

Q72 1)

2)

3)

4)

ANS2 Q73

1)

2)

3)

4)

ANS4 Q74

1)

2)

3)

4)

ANS1

Q75 1)

2)

3)

4)

ANS1 Q76

1)

2)

3)

4)

ANS2 Q77

1)

2)

3)

4)

ANS1 Q78

1)

2)

3)

4)

ANS4

Q79

1)

2)

3)

4)

ANS3 Q80

1)

2)

3)

4)

ANS2 Q81

1)

2)

3)

4)

ANS4

Q82

1)

2)

3)

4)

ANS (ans is 4) Q83

1)

2)

3)

4)

ANS3 Q84

1)

2)

3)

4)

ANS4

Q85

1)

2)

3)

4)

ANS1 Q86

1)

2)

3)

4)

ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q21

1)

2)

3)

4)

ans4 Q22

1)

2)

3)

4)

ans3 Q23

1)

2)

3)

4)

ans2

Q24

1)

2)

3)

4)

ans4 Q25

1)

2)

3)

4)

ans4

Q26 1)

2)

3)

4)

ans3 Q27

1)

2)

3)

4)

ans2

Q28

1)

2)

3)

4)

ans2 Q29

1)

2)

3)

4)

ans3

Q30 1)

2)

3)

4)

ans2

Q61

1)

2)

3)

4)

ANS1 Q62

1)

2)

3)

4)

ANS3

Q63

1)

2)

3)

4)

ANS 4 Q64

1)

2)

3)

4)

ANS2 Q65

1)

2)

3)

4)

ANS4

Q66

1)

2)

3)

4)

ANS2 Q67

1)

2)

3)

4)

ANS4 Q68

1)

2)

3)

4)

ANS3

Q69 1)

2)

3)

4)

ANS1 Q70

1)

2)

3)

4)

ANS3 Q71

1)

2)

3)

4)

ANS2

Q72 1)

2)

3)

4)

ANS2 Q73

1)

2)

3)

4)

ANS4 Q74

1)

2)

3)

4)

ANS1

Q75 1)

2)

3)

4)

ANS1 Q76

1)

2)

3)

4)

ANS2 Q77

1)

2)

3)

4)

ANS1 Q78

1)

2)

3)

4)

ANS4

Q79

1)

2)

3)

4)

ANS3 Q80

1)

2)

3)

4)

ANS2 Q81

1)

2)

3)

4)

ANS4

Q82

1)

2)

3)

4)

ANS (ans is 4) Q83

1)

2)

3)

4)

ANS3 Q84

1)

2)

3)

4)

ANS4

Q85

1)

2)

3)

4)

ANS1 Q86

1)

2)

3)

4)

ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q24

1)

2)

3)

4)

ans4 Q25

1)

2)

3)

4)

ans4

Q26 1)

2)

3)

4)

ans3 Q27

1)

2)

3)

4)

ans2

Q28

1)

2)

3)

4)

ans2 Q29

1)

2)

3)

4)

ans3

Q30 1)

2)

3)

4)

ans2

Q61

1)

2)

3)

4)

ANS1 Q62

1)

2)

3)

4)

ANS3

Q63

1)

2)

3)

4)

ANS 4 Q64

1)

2)

3)

4)

ANS2 Q65

1)

2)

3)

4)

ANS4

Q66

1)

2)

3)

4)

ANS2 Q67

1)

2)

3)

4)

ANS4 Q68

1)

2)

3)

4)

ANS3

Q69 1)

2)

3)

4)

ANS1 Q70

1)

2)

3)

4)

ANS3 Q71

1)

2)

3)

4)

ANS2

Q72 1)

2)

3)

4)

ANS2 Q73

1)

2)

3)

4)

ANS4 Q74

1)

2)

3)

4)

ANS1

Q75 1)

2)

3)

4)

ANS1 Q76

1)

2)

3)

4)

ANS2 Q77

1)

2)

3)

4)

ANS1 Q78

1)

2)

3)

4)

ANS4

Q79

1)

2)

3)

4)

ANS3 Q80

1)

2)

3)

4)

ANS2 Q81

1)

2)

3)

4)

ANS4

Q82

1)

2)

3)

4)

ANS (ans is 4) Q83

1)

2)

3)

4)

ANS3 Q84

1)

2)

3)

4)

ANS4

Q85

1)

2)

3)

4)

ANS1 Q86

1)

2)

3)

4)

ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q26 1)

2)

3)

4)

ans3 Q27

1)

2)

3)

4)

ans2

Q28

1)

2)

3)

4)

ans2 Q29

1)

2)

3)

4)

ans3

Q30 1)

2)

3)

4)

ans2

Q61

1)

2)

3)

4)

ANS1 Q62

1)

2)

3)

4)

ANS3

Q63

1)

2)

3)

4)

ANS 4 Q64

1)

2)

3)

4)

ANS2 Q65

1)

2)

3)

4)

ANS4

Q66

1)

2)

3)

4)

ANS2 Q67

1)

2)

3)

4)

ANS4 Q68

1)

2)

3)

4)

ANS3

Q69 1)

2)

3)

4)

ANS1 Q70

1)

2)

3)

4)

ANS3 Q71

1)

2)

3)

4)

ANS2

Q72 1)

2)

3)

4)

ANS2 Q73

1)

2)

3)

4)

ANS4 Q74

1)

2)

3)

4)

ANS1

Q75 1)

2)

3)

4)

ANS1 Q76

1)

2)

3)

4)

ANS2 Q77

1)

2)

3)

4)

ANS1 Q78

1)

2)

3)

4)

ANS4

Q79

1)

2)

3)

4)

ANS3 Q80

1)

2)

3)

4)

ANS2 Q81

1)

2)

3)

4)

ANS4

Q82

1)

2)

3)

4)

ANS (ans is 4) Q83

1)

2)

3)

4)

ANS3 Q84

1)

2)

3)

4)

ANS4

Q85

1)

2)

3)

4)

ANS1 Q86

1)

2)

3)

4)

ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q28

1)

2)

3)

4)

ans2 Q29

1)

2)

3)

4)

ans3

Q30 1)

2)

3)

4)

ans2

Q61

1)

2)

3)

4)

ANS1 Q62

1)

2)

3)

4)

ANS3

Q63

1)

2)

3)

4)

ANS 4 Q64

1)

2)

3)

4)

ANS2 Q65

1)

2)

3)

4)

ANS4

Q66

1)

2)

3)

4)

ANS2 Q67

1)

2)

3)

4)

ANS4 Q68

1)

2)

3)

4)

ANS3

Q69 1)

2)

3)

4)

ANS1 Q70

1)

2)

3)

4)

ANS3 Q71

1)

2)

3)

4)

ANS2

Q72 1)

2)

3)

4)

ANS2 Q73

1)

2)

3)

4)

ANS4 Q74

1)

2)

3)

4)

ANS1

Q75 1)

2)

3)

4)

ANS1 Q76

1)

2)

3)

4)

ANS2 Q77

1)

2)

3)

4)

ANS1 Q78

1)

2)

3)

4)

ANS4

Q79

1)

2)

3)

4)

ANS3 Q80

1)

2)

3)

4)

ANS2 Q81

1)

2)

3)

4)

ANS4

Q82

1)

2)

3)

4)

ANS (ans is 4) Q83

1)

2)

3)

4)

ANS3 Q84

1)

2)

3)

4)

ANS4

Q85

1)

2)

3)

4)

ANS1 Q86

1)

2)

3)

4)

ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q30 1)

2)

3)

4)

ans2

Q61

1)

2)

3)

4)

ANS1 Q62

1)

2)

3)

4)

ANS3

Q63

1)

2)

3)

4)

ANS 4 Q64

1)

2)

3)

4)

ANS2 Q65

1)

2)

3)

4)

ANS4

Q66

1)

2)

3)

4)

ANS2 Q67

1)

2)

3)

4)

ANS4 Q68

1)

2)

3)

4)

ANS3

Q69 1)

2)

3)

4)

ANS1 Q70

1)

2)

3)

4)

ANS3 Q71

1)

2)

3)

4)

ANS2

Q72 1)

2)

3)

4)

ANS2 Q73

1)

2)

3)

4)

ANS4 Q74

1)

2)

3)

4)

ANS1

Q75 1)

2)

3)

4)

ANS1 Q76

1)

2)

3)

4)

ANS2 Q77

1)

2)

3)

4)

ANS1 Q78

1)

2)

3)

4)

ANS4

Q79

1)

2)

3)

4)

ANS3 Q80

1)

2)

3)

4)

ANS2 Q81

1)

2)

3)

4)

ANS4

Q82

1)

2)

3)

4)

ANS (ans is 4) Q83

1)

2)

3)

4)

ANS3 Q84

1)

2)

3)

4)

ANS4

Q85

1)

2)

3)

4)

ANS1 Q86

1)

2)

3)

4)

ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q61

1)

2)

3)

4)

ANS1 Q62

1)

2)

3)

4)

ANS3

Q63

1)

2)

3)

4)

ANS 4 Q64

1)

2)

3)

4)

ANS2 Q65

1)

2)

3)

4)

ANS4

Q66

1)

2)

3)

4)

ANS2 Q67

1)

2)

3)

4)

ANS4 Q68

1)

2)

3)

4)

ANS3

Q69 1)

2)

3)

4)

ANS1 Q70

1)

2)

3)

4)

ANS3 Q71

1)

2)

3)

4)

ANS2

Q72 1)

2)

3)

4)

ANS2 Q73

1)

2)

3)

4)

ANS4 Q74

1)

2)

3)

4)

ANS1

Q75 1)

2)

3)

4)

ANS1 Q76

1)

2)

3)

4)

ANS2 Q77

1)

2)

3)

4)

ANS1 Q78

1)

2)

3)

4)

ANS4

Q79

1)

2)

3)

4)

ANS3 Q80

1)

2)

3)

4)

ANS2 Q81

1)

2)

3)

4)

ANS4

Q82

1)

2)

3)

4)

ANS (ans is 4) Q83

1)

2)

3)

4)

ANS3 Q84

1)

2)

3)

4)

ANS4

Q85

1)

2)

3)

4)

ANS1 Q86

1)

2)

3)

4)

ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q63

1)

2)

3)

4)

ANS 4 Q64

1)

2)

3)

4)

ANS2 Q65

1)

2)

3)

4)

ANS4

Q66

1)

2)

3)

4)

ANS2 Q67

1)

2)

3)

4)

ANS4 Q68

1)

2)

3)

4)

ANS3

Q69 1)

2)

3)

4)

ANS1 Q70

1)

2)

3)

4)

ANS3 Q71

1)

2)

3)

4)

ANS2

Q72 1)

2)

3)

4)

ANS2 Q73

1)

2)

3)

4)

ANS4 Q74

1)

2)

3)

4)

ANS1

Q75 1)

2)

3)

4)

ANS1 Q76

1)

2)

3)

4)

ANS2 Q77

1)

2)

3)

4)

ANS1 Q78

1)

2)

3)

4)

ANS4

Q79

1)

2)

3)

4)

ANS3 Q80

1)

2)

3)

4)

ANS2 Q81

1)

2)

3)

4)

ANS4

Q82

1)

2)

3)

4)

ANS (ans is 4) Q83

1)

2)

3)

4)

ANS3 Q84

1)

2)

3)

4)

ANS4

Q85

1)

2)

3)

4)

ANS1 Q86

1)

2)

3)

4)

ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q66

1)

2)

3)

4)

ANS2 Q67

1)

2)

3)

4)

ANS4 Q68

1)

2)

3)

4)

ANS3

Q69 1)

2)

3)

4)

ANS1 Q70

1)

2)

3)

4)

ANS3 Q71

1)

2)

3)

4)

ANS2

Q72 1)

2)

3)

4)

ANS2 Q73

1)

2)

3)

4)

ANS4 Q74

1)

2)

3)

4)

ANS1

Q75 1)

2)

3)

4)

ANS1 Q76

1)

2)

3)

4)

ANS2 Q77

1)

2)

3)

4)

ANS1 Q78

1)

2)

3)

4)

ANS4

Q79

1)

2)

3)

4)

ANS3 Q80

1)

2)

3)

4)

ANS2 Q81

1)

2)

3)

4)

ANS4

Q82

1)

2)

3)

4)

ANS (ans is 4) Q83

1)

2)

3)

4)

ANS3 Q84

1)

2)

3)

4)

ANS4

Q85

1)

2)

3)

4)

ANS1 Q86

1)

2)

3)

4)

ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

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α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

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JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

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JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

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JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

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JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

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JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

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KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

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= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

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JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

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JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

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Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

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Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


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now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

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Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

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p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

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JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

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-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

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JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

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f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

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JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

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JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

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Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

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N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q69 1)

2)

3)

4)

ANS1 Q70

1)

2)

3)

4)

ANS3 Q71

1)

2)

3)

4)

ANS2

Q72 1)

2)

3)

4)

ANS2 Q73

1)

2)

3)

4)

ANS4 Q74

1)

2)

3)

4)

ANS1

Q75 1)

2)

3)

4)

ANS1 Q76

1)

2)

3)

4)

ANS2 Q77

1)

2)

3)

4)

ANS1 Q78

1)

2)

3)

4)

ANS4

Q79

1)

2)

3)

4)

ANS3 Q80

1)

2)

3)

4)

ANS2 Q81

1)

2)

3)

4)

ANS4

Q82

1)

2)

3)

4)

ANS (ans is 4) Q83

1)

2)

3)

4)

ANS3 Q84

1)

2)

3)

4)

ANS4

Q85

1)

2)

3)

4)

ANS1 Q86

1)

2)

3)

4)

ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q72 1)

2)

3)

4)

ANS2 Q73

1)

2)

3)

4)

ANS4 Q74

1)

2)

3)

4)

ANS1

Q75 1)

2)

3)

4)

ANS1 Q76

1)

2)

3)

4)

ANS2 Q77

1)

2)

3)

4)

ANS1 Q78

1)

2)

3)

4)

ANS4

Q79

1)

2)

3)

4)

ANS3 Q80

1)

2)

3)

4)

ANS2 Q81

1)

2)

3)

4)

ANS4

Q82

1)

2)

3)

4)

ANS (ans is 4) Q83

1)

2)

3)

4)

ANS3 Q84

1)

2)

3)

4)

ANS4

Q85

1)

2)

3)

4)

ANS1 Q86

1)

2)

3)

4)

ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q75 1)

2)

3)

4)

ANS1 Q76

1)

2)

3)

4)

ANS2 Q77

1)

2)

3)

4)

ANS1 Q78

1)

2)

3)

4)

ANS4

Q79

1)

2)

3)

4)

ANS3 Q80

1)

2)

3)

4)

ANS2 Q81

1)

2)

3)

4)

ANS4

Q82

1)

2)

3)

4)

ANS (ans is 4) Q83

1)

2)

3)

4)

ANS3 Q84

1)

2)

3)

4)

ANS4

Q85

1)

2)

3)

4)

ANS1 Q86

1)

2)

3)

4)

ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q79

1)

2)

3)

4)

ANS3 Q80

1)

2)

3)

4)

ANS2 Q81

1)

2)

3)

4)

ANS4

Q82

1)

2)

3)

4)

ANS (ans is 4) Q83

1)

2)

3)

4)

ANS3 Q84

1)

2)

3)

4)

ANS4

Q85

1)

2)

3)

4)

ANS1 Q86

1)

2)

3)

4)

ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q82

1)

2)

3)

4)

ANS (ans is 4) Q83

1)

2)

3)

4)

ANS3 Q84

1)

2)

3)

4)

ANS4

Q85

1)

2)

3)

4)

ANS1 Q86

1)

2)

3)

4)

ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q85

1)

2)

3)

4)

ANS1 Q86

1)

2)

3)

4)

ANS1 Q87

1)

2)

3)

4)

ANS4

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

Q88 1)

2)

3)

4)

ANS3 Q89

1)

2)

3)

4)

ANS4 Q90

1)

2)

3)

4)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q1

Bqv= mvsup2R

R= mvBq

Now

eV= frac12 mvsup2

or mv= radic2emV

R= radic2emVBq

Rprop radicm

Rrsquo = radic2R

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

σ1=σ2--- given




q1 =Q(1+rsup2Rsup2)

now




Ans -4



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=dD

Δx=λdD dα

=λα D

Ans 4




Ans 4


1 VSD = N(N+1) MSD

NOW


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


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LC = 1MSD -1VSD


a(N+1)

Ans 2

Q6 Case1


Case2


2m=100

m=50kg

Ans -1

Q7


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

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JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

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JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

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= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

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p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

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JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

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JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

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JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

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we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dqdt = dqdtˢ

9KA(100-θ)(L2)=KAθ(L2)

θ=90⁰

Ans -1

Q8 F = dPdt

=ddt(mv)

= v dmdt

= 6

Ans 4

Q9 s

τ=-mgsinθR

≃-mgθR

Iα=-mgθR

mgsinθ

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

α=-mgRθI

ωsup2=mgRI

now

I = mRsup2 + mRsup2

=2mRsup2

ωsup2=g2R

T=2πradic(2R)g


R= 18

=0125

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 υn= (2n-1)v4L

case1

500 = (2n ndash 1)v4 x16 ----(1)

Case2

500 = (2nrsquo-1)v4 x 50-----(2)


50n -16nrsquo = 17

n≃1nrsquo≃2

now solving for v

v = 500 x4 x 503

v = 3333333ms

now


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

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JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


=v2l

≃276 Hz

υ(2nd lowest)=2υ

=552 Hz

Ans -3

Q11 λav= λα+λβ

now

Tfrac12=1λ

Tav=1λα+1λβ

Ans -1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



V = -13u

Vrsquo= 23u

(ΔEE)x100= 8888

Ans 2

Q13 Sinθ gt1μ

μgμw=15(43)

= 98

Sinθgt89

Ans 3

Q14


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

KE=hυ-hυ˳


Also


Statement 1- false

Statement 2-true

Ans -1

Q15 q(E + v x B)= 0

E = - v x B

E = B x v


Ans -3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 136 x 9

= 1224


=54 +756 +1224

= 2034

Ans 4

Q17 P=Vsup2R

R1ltR2

P1gtP2

Ans -4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q18 H=Isup2Rt

ΔH=2ΔI+ΔR+Δt

= 6

Ans -3






Ans -2



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

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JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

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JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

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JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Statement 2- true


Ans -3

Q21 V = e ndash iR

120 ndash 50 x 01

= 115

Ans 4


At t=0

Y = A sin 120⁰

Checking option


BY

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JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

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Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans -3

Q23 Fsinθ

Fκ Fcosθ


Now

Fcosθ = Fκ


For minimum force

mg


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dFdθ = 0

solving we get

tanθ = μ





Ans -2

Q24




BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY




Ans 4

Q25

P =αv

Vf= m Vᵢ

W = int p dv

= αintvdv



Ans -4

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ф = EA cos 45⁰

= E πasup2radic2

Ans -3

Q27

L = constant

Mvr = constant




now

mr = mrsquorrsquo

fMr =(1-f)Mrrsquo

rrsquor = f(1-f)



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

ωωrsquo = f(1-f)

Ans -2

Q28

ᶯ(max) = 1- T2T1

= 1 ndash 273373

= 2680


Ans 2

Q29



n = Rsup3rsup3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

12mvsup2=Δs


Solving we get



V = 6Tρ[1r -1R]⁰⁵

Ans- 3

Q30

Self explanatory

Ans -2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61



Limx-gt2ˉ f(x) = 2

Statement 1---true

f(2)=1 ----given




Ans 1

Q62

n(f) = ⁴C1 x 2 x 4

= 32

n(s) = 5 =120


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

p = 32120

= 415

Ans 3

Q63



now




grsquo(x)=0

hence

axsup2 +bx +c = 0



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q64

bsup2+csup2 ab ac

ab csup2+asup2 bc

ac bc asup2+bsup2

-asup2 ab ac

= ab -bsup2 bc

ac bc -csup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

-1 1 1


1 1 -1

= 4asup2bsup2csup2

Ans 2

Q65







K= plusmnradic2

Cosθ =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q66


25 = 20 + (50-45) x 10f

F =10

Ans 2

Q67

Xsup2 + secsup2x -1

1+xsup2


1+xsup2


F(x) = int

F(x) = dx

dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


f(0) = 0

C = 0


= tan1 ndash π4

Ans 4

Q68





now


P=0-----(1)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


2p -5aq + 3r = 0----(2)


3pb + q ndash 3r = 0----(3)


a =15

Ans 3

Q69


α - 0 αsup3 + 3α

frsquo(1-α)

3αsup2 + 3

= frsquo(1)3

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= -533

Ans 1

Q70

(x + 32)sup2 + ysup2 = (32)sup2---(1)

Y = mx+1---(2)





ie

3m + 2msup2 = 0

3m + 2= 0

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q71

θ π-θ

w = -a ndash ib

-w =a +ib

= z


Statement 2---false


Z=a+ib W= -a+ib


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q72

Pxsup2 + qx + r = 0

αβ = rp

let α= a+ib

then β=a-ib



=2radicasup2+bsup2

=2radicrp

Since rgtp

|α|+|β|gt2

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q73

(01)

(a2a) Y=2x

(αβ)

(Y-1)x= (2a-1)a




m1 x m2= - 1

we get

a = 25


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


we get

α = 45 β = 35

Ans 4

Q74


= 96

Ans 1

Q75

n p(s) = 2sup3

n(s) = 3


Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76

Area= int x dy


= 12

Ans 2

Q77

P q = ~p v q

Basic formula


Ans 1

Q78

Xsup2 = 8y


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


y-2 =4-x

x +y = 6

Ans 4

Q79



30cosθ = 15

θ= 60⁰

Ans 3

Q80

˳int⁰⁹ [x -2[x]] dx

[x] = 0 0lexle09



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q81


=xsup2

Now



Ans 4

Q82

bf-ec -ae -ab


-bd 0 0


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 0 d

A = 0 b e

a c f

now

Amacrsup1 = A

comparing we get





Ans

Q83


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

N = 2tsup2 + 3 w = tsup2 -t +2



dwdt|at (t=1) = 13

Ans 3

Q84



Solving we get

Xsup2+ysup2-103 x+1 = 0

Centre = (53 0)


Ans 4

= frac12


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q85

Given lines

x2 = y-1 = z2

(x-1)4 = (y-1)-2 = (z-1) 4


p(111)

line 1

q(2λ-λ2λ) line2


hence

(2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0

λ = 13

q = (23-1323)

length pq = radic2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

statement 1----true

Ans 1

Q86


1 +(1+13)+ (1 + 19) + hellip

n + (13 +19 + 127 +hellip)


Ans 1

Q87


=radic16-bsup2


=3

radic16-bsup2 = 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

bsup2 =7

Ans 4

Q88


Tanˉsup1(1radic3)

π6

Ans 3

Q89


c(a x b) = 0

1 -2 3

2 3 -1 =0

r 1 2r-1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

r= 0

Ans 4

Q90

n = m(m-1)2

n-1 = m(m-1)2 - 1

ᶯC2 = n(n-1)2

Solving we get

ᶯC2 = 3 ˉ sup1C4

Ans 2

AIEEE 2012 solution (26th may)

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Transcript of AIEEE 2012 solution (26th may)