Mtg-nov Issue-physics for You-iit Jee Paper

8/13/2019 Mtg-nov Issue-physics for You-iit Jee Paper

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MTG-NOV ISSUE-PHYSICS FOR YOU-IIT JEE PAPER

Answer key & Solutions

C H E M I S T R Y

P H Y S I C S

M A T H E M A T I C S

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

c c d c c c abd abc ac a b b a d c

16

a

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35

b a b a a b bd bd d bc d a d a b

36

d

41 42 43 44 45 46 47 48 49 50 51 52 53 54 55

c c a c d a bd ad bc c d c d a c a

56

d

17.Amax= 1s xk

mg

= 3 18.N= 14 N

19.(A) p,s, (B) q,r, (C) p,s, (D) q,r 20. (A) p, (B) p , (C) p , (D) p,s

37.

15538.9.6 102 =10

39. (A) p, (B) r, (C) s, (D) q 40. (A) q,r ; (B) p,s ; (C) p, r ; (D) q,r

57.a + b = 23 58.f (1) = 3

59.(A) r, (B) s, (C) p,q,r,s , (D) p 60.(A) q, (B) s, (C) r , (D) p


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A

C

B

P H Y S I C S

1.(c) Consider the motion of the rod as pure rotational motion about an axis passing through A and perpendicular

to the rod

Then by conservation of mechanical energy

2 21 1( ) ( ) 2

2 2B C B CI I m g l m gl + = +

22 2 21 1

( ) 32 2 4

lm l m mgl + =

2524

8l g =

24

5

g

l=

Minimum horizontal speed to be imparted to (2 )B l= 24

5

gl= . (c) is correct.

2.(c) Since collision is elastic, kinetic energy is conserved.

Before collision rod has only translational kinetic energy after collision rod will have translational as well as

rotational kinetic energy.

1H H< (c) is also correct.

3.(d) (Pressure) force due to liquid acts normal to the surface everywhere producing no net torque about O.

(d)

4.(c) 2 2 22I f a v =

, max .2

10p

vv a a f = = =

22 2

2

400

va f

= (c)

5.(c) PV nRT= ;dV nR

dT P=

n

Pis same in both cases. Therefore straight line B is both cases. (c)

6.(c) The velocity of the particle will first increase then decrease. Thus total potential energy will f irst decrease and

then increase.

7.(a,b,d)Since only six dif ferent wavelengths are emitted, therefore highest excited state is 4n = , therefore (a) iscorrect.

In the emitted radiations two wavelengths are shorter than o it means that initially atoms were in theexcited state therefore (b) is also correct.

Therefore (b) is also correct.

Transitions corresponding to

4 1, 3 1, 2 1 belong to Lyman series.

(d) is also correct.


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pressure node

/ 8 mid pt.

15/32

8.(a,b,c)Capacitive resistance

1159.2

2C

XfC

= =

Current in the capacitive branch

200.126

159.2

SC

C

VI A

X= = = (b) is correct.

Current in the resistive branch

200.2

100

SR

VI A

R= = = (a) is correct.

Total current T R CI I jI= +

| | 0.24TI A (c) is also correct.

9.(a,c) DX =Dyd for central maxima DX = 0, hence y = 0. (Does not depend upon l)

B =d

D

d

Dm

d

Dn 21 =

n 4500 = m 6000

3n = 4m. n = 4, m = 3.

10.(a)12420

0.6220000

cutoff A A = = =62pm

11.(b) At the middle of the column (i.e. at C).

distance of C from the closest node N, 9.3758

NC cm

= =

At C, pressure ampl.

0 0 0

2 1sin sin ( )

8 2P P kx P P

= = =

12.(b) For II overtone

5 5 5 330 1593.75

4 4 4 440 16

vL m cm

v= = = = =

13.(a)At closed end, Max pressure 0 0P P= + . Minimum pressure 0 0P P=

14.(d) The distance travelled by the atom between two consecutive collisions (2 1) 2m m= =

Time interval between two consecutive collisions2

rmsv=


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15.(c)Average kinetic energy3

2kT= (for monoatomic gas)

23 2131.38 10 / 160 3.31 10

2J K J

= =

16.(a)Pressure of gas,

21

3 rms

m

P vV= m = mass of gas, V= volume of box.

4

2 2

3 3 100 13 10

(1000)rms

PVm kg

v

= = =

17.[Ans. 3 ] Suppose origin is at the equilibrium position and the direction of increasing x is toward the right. Ifthe blocks are at the origin, the net force on them is zero. If the blocks are a small distance x to the right ofthe origin, value of the net force on them is 4kx. Applying Newton's second law to the two-block systemgives

4kx = 2maApplying Newton's second law to the lower block gives

k(x1 x) f= mawhere x1= initial stretch and fis the magnitude of the frict ional force.

f= k(x1+ x)The maximum value for x is the amplitude A and the maximum value for f is smg. Thus,

smg = k(x1+ Amax). Solving for Amaxgives

Amax= 1s xk

mg

= 3

18:[Ans: 14] Br

is uniform

Fr

= I Lr

Br

here Lr

=

OA = 0.25 i + j

Br

= 4(cos60 i + sin60j )

F

r

= 12(0.25 i + j ) 4(cos60 i + sin60j ) = 4.8

k

60cosk

60sin4

1

= 4.8 k

2

1

8

3

= 4.8 k

8

43

= 6 )k43 = 13.8 N= 14 N

19.(A) p,s, (B) q,r, (C) p,s, (D) q,r20. (A) p, (B) p , (C) p , (D) p,s


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C H E M I S T R Y

21.(b) 50N = 30 x 0.2 = 6

25

3

50

6N ==

50 ml HA + 20 ml NaOH HA + NaA6 meq 4 meq 2 meq 4 meq

pH = pKa + log2

4

5.8 = pKa + log 2pKa = 5.8 - log 2

= 5.8 - 0.3 = 5.5 (b) ]

22.(a) A(s) + 2B (aq) 2C (s) + D (aq) K = 1020

2022 10]B[

6

1

]B[]D[K ===

[B] = 4 x 1011M (A) ]

23.(b)

e.g. of Anionotropy.

+OH3

3NH RMgX

CH3NH

2+

+OH3


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24.(a)25.(a)NaNO2+HCl converts NH2into OH ]26.(b)

27.(bd)

more substituted carbon get H, D or T from BH3 or BD3 or

BT3.

Less substituted get, H, D or T from or ]

28.(bd)29.(d) Acetal hydrolyse

(d)]

30.(bc)

31.(d) bottle 2 + bottle 3 colourless gas

carbon dioxide sodium carbonate + acidIt suggests bottle 2 and 3 contain sodium carbonate and HCl.

Bottle 3 4+ blue precipitate confirms the 2Cu + in either of bottle considering available sulphate, chloride

nitrate and carbonate.Precipitate must be carbonate.

Hence, 3 carbonte sodium carbonate

4 2Cu +

2 HCl (from above)

32.(a) As bottle 3 sodium carbonate


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4 copper sulphate

2 hydrochloric acidHence, bottle 1 lead nitrate

Now, 3 2 4 3 2 4

white ppt.soluble

( ) ( )Pb NO CuSO Cu NO PbSO+ +

33.(d) Bottle 4 4CuSO

2 2

3 3 4( ) 4 ( ) ( ) ( )distinctive blue colour

Cu aq NH aq Cu NH aq+ ++

34.(a) 3 2 22 6 2 3 3HBrO HI HBr I H O+ + +

3HBrO gets reduced,

35.(b)1 x

Kat a x

=

3

300 2.92 10K = 3600 3.057 10K

= 31200 2.99 10K =

Hence, order 2n =

36.(d)18

log 4 61.5 / 2.303 8.314 301 3.19

EkJ mol

= =

37(155) A CrO2Cl

2B H

2CrO

4CHCl

)C()B(422

)A(22 HClCrOHOHClCrO ++

)E()D(42

)A(22 NaClCrONaNaOHClCrO ++

)yellow(42

)DorB(

24 BaCrOBaClCrO +

++ 3HNO.dilinlelubInso

3)EorC(

NOAgClAgNOCl3

Molecular weight of A(CrO2Cl2) = 155 ]

38.

0.1 M AgNO3

0.01 M AgNO3

+ + e)NH(AgM1.0

23Ag +

1.03 )NH(2

Ag +2.0

3 )NH(2 + + e)NH(Ag

01.023


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1.023 )NH(Ag

+ + Ag +2.0

3NH2 Ag +1.0

3 )NH(2 +01.0

23 )NH(Ag +

Ecell

= 0.0 ( )

( ) ( )

2

2

2.01.0

1.001.0log

1

06.0

= 9.6 102 ]

39. (A) p, (B) r, (C) s, (D) q

(A) CH3CH=CH2 )P(HBr33 CHCHCH

|Br

(Markownikoff addition)

(B) CH3CH=CH2 )peroxide(

)R(HBr CH3CH2CH2Br (Peroxide effect)

(C) CH3CH-CH2 onsubstitutiAlkylic

)S(NBS 22 CHCHCH|

Br

=

(D) CH3CH=CH2 )Q(Br2 23 CHCHCH||BrBr

40. (A) q,r ; (B) p,s ; (C) p, r ; (D) q,r

Complex Oxidation state Type Magnetic property

4

6[ ( ) ]Fe CN

+2 2 3d sp inner orbital complex Diamagnetic

2

2 6[ ( ) ]Fe H O +

+2 3 2sp d outer orbital complex Paramagnetic

2

3 4[ ( ) ]Cu NH +

+2 2d sp inner orbital complex Paramagnetic

2

4[ ( ) ]Ni CN +

+2 2d sp inner orbital complex Diamagnetic


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M A T H E M A T I C S

41.(c) Z = i + 2i2+ 3i3+ + 2007i2007

Use method of dif ference to get Z = 1004 1004i\ Re(Z) + Im(Z) = 2008 ]

42.(c) 10C4=

123478910

123478910 = 210]

43.(a) y = 2

x 2

tan 30 = 2t

t2

3

1=

t

2 t = 32

l(OP) = 34 Ans. ]

44.(c) f (0) = f (1) = 0 (obviously) ]45.(d) Using LMVT in [2, 5]

4 7

)2(f)5(f 3

28 f (5) f (2) 21 ]46.(a) y = ln x x = ey

A = bn

an

dyx

l

l

= bn

an

ydye

l

l

= (eln b) (eln a) = b a Ans. ]47.(bd) Let the roots are a d, a, a + d

\ a d + a + a + d = 6 a = 2again the product of roots taken 2 at a time

a(a d) + a(a + d) + (a + d)(a d) = 24a2 ad + a2+ ad + a2 d2= 243a2 d2= 24

d2

= 36 d = 6 or 6Hence the AP's are 4, 2, 8, ............

or 8, 2, 4, .........

now Sn= 2

n[ 8 + (n 1) 6] = n[ 4 + 3(n 1)] = n(3n 7) Ans.

or Sn= 2

n[16 + (n 1)(6)] = n[8 3(n 1)] = n(11 3n) Ans. ]

48.(a,d)We must have ( )222 10 16 4 2 4x x+ = + + .

49.(b,c)If PON = then as P is ( )acos ,bsin ( )NP cos tan a sin = = .

which is not true since

NP b sin= is given Choice (a) is false. QON = then ON acos= ( )OQ a=Q

and putting x a cos= in2 2

2 2 1

x y

a b+ =

We get y DN b sin = Choice (b) is correct.

Now normal at q will pass through origin since Q lies on circle 2 2 2x y a+ =

Q

P

NO


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50.(c) differentiating )x(g

0

dt)t(f = x2

\ ( ))x(gf g ' (x) = 2x ....(1)

since f and g inverse of each other hence ( ))x(gf = x.Hence (1) becomes x g'(x) = 2x or g'(x) = 2integrating g (x) = 2x + Cbut the initial condition shows C = 0hence , g (x) = 2x

its inverse f (x) =2

x C, D ]

51.(c) Let the focus of parabola be ( )1 1x , y

SP = distance of P f rom directrixHere P are (0,2) and (0,-2)Equation of directrix si

3xcos y sin + =

( ) ( )2 22

1 1 2 2 3x y sin+ = .....(i)

and ( ) ( )2 22

1 2 2 3x y sin+ + = + .....(ii)

Subtract, we get

18 24y sin=

1

3

ysin=

Adding, (i) and (ii),

( ) ( )2 2 21 12 4 2 4 9x y sin + + = +

2

2 2 11 1 4 4 99

y

x y+ + = +

2 2 21 1 19 9 36 4 81x y y+ + = +

2 21 19 5 45x y+ =

Locus of focus is

2 2

15 9

x y+ =

52.(d)

( ) ( ) ( )22 2

0 0 2x y + =

53.(a)

214 45 0 =

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