Solutions to Mock IIT Advanced/Test -...
Transcript of Solutions to Mock IIT Advanced/Test -...
Vidyamandir Classes
VMC/2013/Solutions 1 Mock IIT Advanced/Test - 3/Paper-1
Solutions to Mock IIT Advanced/Test - 3[Paper-1]/2013
[CHEMISTRY]
Vidyamandir Classes
VMC/2013/Solutions 2 Mock IIT Advanced/Test - 3/Paper-1
Vidyamandir Classes
VMC/2013/Solutions 3 Mock IIT Advanced/Test - 3/Paper-1
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VMC/2013/Solutions 4 Mock IIT Advanced/Test - 3/Paper-1
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VMC/2013/Solutions 5 Mock IIT Advanced/Test - 3/Paper-1
Vidyamandir Classes
VMC/2013/Solutions 6 Mock IIT Advanced/Test - 3/Paper-1
Vidyamandir Classes
VMC/2013/Solutions 7 Mock IIT Advanced/Test - 3/Paper-1
Solutions to Mock IIT Advanced/Test - 3[Paper-1]/2013
[PHYSICS]
24.(A)
1 1 1
20 10V− =−
20 1V m= = −
Co-ordinate is (20 cm, 0.2 cm).
25.(C) For inductor di
V Ldt
= −
Or LV
di dtL
= −∫ ∫
V = (10 t − 20) where 0 4t ms≤ < ( )2
0
1 1 1010 20 20
2
t
Lt
i t dt tL L
⇒ = − + = − −
∫
25 20
Lt t
iL L
= − +
26.(B) Force of buoyancy
( )0 0BF A x L x gρ ρ = + − Weight sW LA gρ=
For equilibrium BF W=
( )0 0SLA g A x L x gρ ρ ρ ⇒ = + − ( )0s wL x L xρ ρ ρ⇒ = + −
( )00
1000 800 2
1000 300 7
w ss w w
w
x x
L L
ρ ρρ ρ ρ ρ
ρ ρ− −
⇒ = − + ⇒ = = =− −
27.(C) x a cos tω=
v a sin tω ω= −
1 11
2 2
aa cos t cos tω ω= ⇒ =
1 13 3 2 6
T Tt t
π πω
π⇒ = ⇒ = =
×
( )6
0
6
0
3
T /
T /
a sin t dt
av
Tdt
ω ω−
< >= =∫
∫
28.(A) Given 2 4y x= −
When 0 2x y= , = ±
The separation between open ends = 4.
Magnetic force on a current carrying element placed in a uniform field B is ( ) ɵ ɵ( )3 4 5 60i L B j k i× = × =�� ��
ɵ
29.(B) Here at the instant of sliding limiting friction f Nµ= will act on the two blocks for limiting equilibrium, we use:
F = T Nµ+ …(i)
and T = Nµ …(ii)
and 2T + N = 50 …(iii)
On solving (ii) and (iii), we get
50
252 1
N Nµ
= =+
Now from (i) and (ii)
F = 2 Nµ = 2 0 5 25 25. N× × =
Vidyamandir Classes
VMC/2013/Solutions 8 Mock IIT Advanced/Test - 3/Paper-1
30.(C)
31.(B) After 2 sec speed of boy will be 2 2 4v m / s= × =
At this moment centripetal force on boy is 2 30 16
806
rmv
F NR
×= = = .
Tangential force on boy is 30 2 60tF ma N= = × = .
Total friction acting on boy is 2 2 100r tF F F N= + = .
At the time of slipping F mgµ=
or 100 30 10µ= × × ⇒ 1
3µ = .
32.(B)
33.(C) By conserving angular momentum and linear momentum and also using concept of coefficient restitution, we get
1
21cmv v
ev cos
ω
θ
+ += =
ℓ
. . . . . .(i)
1cmmv cos mv mvθ = − . . . . . .(ii)
2
12 12 2
mmv cos mvθ ω= −
ℓ ℓ ℓ . . . . . .(iii)
Now solving, we get : 2
5cmv v cosθ= ,
12ω
5v cosθ=ℓ
, 13
5v v cosθ= −
34.(B) Change in angular momentum of the rod is 2 12 1
Δ12 5 5
cmm
L I v cos m v cosω θ θ= = × =ℓ
ℓℓ
35.(B) 13 2
Δ5 5
v v cos v v cos v cos v cosθ θ θ θ= + = − = 36.(AB)
37.(AB) ( )
( )11 1
2020 1
R
F R
µ
µ
−= = ⇒
− . . . . .(i)
If equiconvex lens of f = 20
Then ( )1 21
20 Rµ= −
From eq. (1) and (2) we conclude.
∴ R = 40 ( )1µ − . . . . .(ii)
That option (A) is correct
After Silvering :
10Feq cm= − (Combination will behave like a concave mirror)
(B) 15u = −
1 1 1 1 1 1
15 10eqv u F v+ = ⇒ − = −
1 1 1
30 So Bis correct15 10
v cmv= − ⇒ = −
(C) 20u = − 20v = − So C is wrong
Alternative :
For planoconvex ( )1 2
1 1 11
f R Rµ
= − −
1 2R R R= ∞ = r = 20 ( )1µ −
For equiconvex ( )1 1 11
20 R Rµ
= − +
( )40 1R µ= −
Vidyamandir Classes
VMC/2013/Solutions 9 Mock IIT Advanced/Test - 3/Paper-1
38.(ABD)
Speed will be maximum when the block passes through the equilibrium position.
springF mg=
(B) From conservation of energy,
( ) 2 21 1
2 2maxmg l x Kx mv+ = + . . . .(i)
where 4 4
mg mg lx l
K mg= = =
Putting 4
lx = in equation (1), we get
3
2maxv gL=
(C) is incorrect as in equilibrium compression is 4l / which is more than 8l / .
(D) Time taken is 1
24 4 4
T m L
K g
ππ= =
39.(BD) Equivalent circuit :
Induced emf in a rotating rod with constant angular
Velocity 2ω
ω2
B ae; = , (here a = Radius)
Now use Kirchoff’s Law.
40.(3) 1v
mu n
= = −
( )0
1L
L
Xm
– X X n= = −
−
Lens formula ⇒( )0
1 1 1
L LX X X f− =− −
( )0
2 20
1 10L L
L L
dXdX dX.
dt dt dtX X X
− − − =
−
( ) ( )0
2 2 20 0
1 1 1L
L L L
dXdX
dt dtX X X X X
− + = − −
0
0; LdX dXV V
dt dt= =
⇒
2 2
01 1
1V Vn n
− + + = +
⇒ [ ]0 04
1 4 43
VV V V− + = ⇒ = ⇒ 03
4
VV =
41.(1) In first case 10
30 10 11
KR
× = − ×+
In second case 5
10 5 22
KR
× = − ×+
Solving above two equations we get : 1R =
42.(1) Work done ΔQ= (heat dissipated) Let e : induced emf
=2 2( )e N B v
t tR R
×= ×
ℓ=
2 2 32100 (0 4) 2 5 10
110 000
. .v
,
−× × ×× ×
2 32 5 10. −= ×ℓ
4 2
25 10 5 10 m− −= × = ×ℓ
25 10
1v
−×=
⇒ 86 25 10 16 1W . J Jµ−= × × =
Vidyamandir Classes
VMC/2013/Solutions 10 Mock IIT Advanced/Test - 3/Paper-1
43.(9) 1
22 3
2 03
2
P
RkQ
kQE
RR= − =
⇒ 12
4
9 2
QQ =
2
1
9=
8
Q
Q
29 8
98
Q Cµ×
= =
44.(8) Initially 0100 2x g=
0 20x cm=
Finally, equilibrium position
0100 1x g′ =
0 10′ =x cm
1
2100 10
Tπ
π2
= =
2 10
Tt
π= =
21
10 502 100
S cmπ
= × × =
Total distance = 50 + 10 + 20 = 80cm.
So, x = 8
45.(4) Paschen series 7 6
2
1 1 11 097 10 10
9.
nλ
= × − ≤
2
1 1 1
9 10 97.n− ≤
2
1 1 97
9 10 97
.
.x≥
×
29 1097
197x
×≥
31097
7197
n≤ ≈ ⇒ 7 3 6 3 5 3 4 3 4lines− , → , − , − ⇒
46.(1) 2 2
3
24(ω )
10 10z L R
−= = +
×
22 2 2 1
(ω ) ωω
R L R LC
+ = + −
( ) 1ω ω
ωL L
C= − +
2 6
1 15
2ω 2 100 100 10L H
C π π −= = =
× × ×
( ) ( )2 2 22400 500 Rπ= +
R = 2 2(2400) (5 100)π− × = 2 210 (24) 25π−
= 10 326 180× ≈
I = 1A
Vidyamandir Classes
VMC/2013/Solutions 11 Mock IIT Advanced/Test - 3/Paper-1
Solutions to Mock IIT Advanced/Test - 3[Paper-1]/2013
[MATHEMATICS]
47.(B) Let ( )22P t , t and
2
1 2Q ,
tt
−
Circle with OP as diameter is 2 2 21 2 0S x y t x ty≡ + − − =
Circle with OQ as diameter is 2 22 2
1 20S x y x y
tt≡ + − + =
R lies on 1 2 0S S− = i.e. 1
2
= −
y t xt
∴ 1
1tt
− = −
Point of intersection of tangents is 1
1, tt
− −
i.e. ( )1 1,− −
48.(D) ( ) nA Adj A A I=
Clearly, 4 3A , n= =
( ) ( )21 44 256n
Adj Adj A A−
= = =
1 24 16
nAdj A A
−= = =
∴ ( ) 256
1616
Adj Adj A
Adj A= =
49.(C) 2 1
1 2 3x . . .x
+ + +
=
2 11
1
2
xx
x
+
Now using the property that 1 1 1
1x x x
− < ≤
(i.e. using sandwich theorem)
We get : ( ) ( )
2 11
1 1 11 1
2 2 2
+ − < ≤ +
xx
x xx
Now applying sandwich theorem the required limit is 1
2
50.(C) 2a c b+ = and 22ac b bc− = .
Eliminating ‘b’ from these two equations, we get ( )2 24 3 0 3a ac c a c a c− + = ⇒ = ≠∵
51.(B)
2
5 2 3 2 43 2
x xa b
x x x x
ln a ln bI x dx
a b a b
= + ∫
2 3
2 36
x x
x x
ln a bdx
a b= ∫
But 2 3x xa b t=
2 3t ln a b dx dt=
2 3 2 2 3 2
1 1 1
6 6
lnt lntI dt dt
tln a b t ln a b t
− − = = −
∫ ∫
2 3
1
6
lnetk
tln a b
= − +
2 3
2 3 2 3
1
6
= − +
x x
x x
lnea bk
ln a b a b
Vidyamandir Classes
VMC/2013/Solutions 12 Mock IIT Advanced/Test - 3/Paper-1
52.(C) P(unequal no. of heads and tails) = 1 P− (equal no. of heads and tails) = ( )
2
2
1 1 2 !1 1
2 2 ! 4
n nn
n n
nC
n
− = −
53.(A) ( ) ( )
93
257
53 64
625 3
3 37 5
409
loglog
loglog
N+
= −
33 325 6
3 325 6 6
409
log log
N+ = −
( ) ( )25 6 6 25 6 6
409N
+ −=
N = 1
2 2 1 0log N log= =
54.(B) The auxiliary equation is 3 27 16 12 0D D D− + − =
or ( ) ( )23 4 4 0D D D− − + = or D = 3, 2, 2
∴ by rule (ii), ( ) 2 31 2 3
x xy c x c e c e= + +
55.(C) The general solution tells that the auxiliary equation has roots 1 1 1i, i,− + − − . So, the auxiliary equation is
( ) ( ) ( )1 1 1 0D i D i D+ − + + − = or ( ){ } ( )21 1 1 0D D+ + − =
or ( )( )2 2 2 1 0D D D+ + − = or 3 22 0D D+ − =
∴ the differential equation is 3 2
3 22 0
d y d yy
dx dx+ − =
56.(D) Statement 1 : Locus of point of intersection of only perpendicular lines is a circle and other vertices B and C do
not form a circle
Statement 2 : Obvious (standard Definition)
57.(C) 1+
−
−→∞ →
+=
+
n n
n nn
| sin x | | cos x |a lim lim
α
α α
α α
2
21 1
n
nn
| sin x | | cos x |lim lim | sin x|
α
α
α+
−
−→∞ →
+= =
+
1−
−
−→∞ →
+=
+
n n
n nn
| sin x | | cos x |b lim lim
α
α α
α α
2
21 1−→∞ →
+= =
+
n
nn
| sin x | | cos x |lim lim | cos x |
α
α
α
( )2 1
14 2 2 2n
nc lim cos cos . . . . . . cos
n n n n
π π π π
→∞
−= + + + +
( )114 4
4 4 4 2
4
n
nsin cos
nlim sin cosn
sinn
π ππ π π
π→∞
− = = =
OR alternatively
11
0 0
1 1
4 2 4 2 2
−
→∞=
= = =∑ ∫n
nr
r xc lim cos cos dx
n n
π π π π
( )1
2f x max | sin x |,| cos x |,
=
∴ range of f (x) is
11
2,
58.(D) If circum centre is ( )S α
1 8| z | SA Rα− = = = And 1 2 3 4r r r r R+ + − =
Vidyamandir Classes
VMC/2013/Solutions 13 Mock IIT Advanced/Test - 3/Paper-1
59.(ABD) 5 21 1 10p C C= × =
. . . .i.e. select one match out of 5 which is wrongly forcast in 5C1 ways. Suppose its true results is win,
then it can be wrongly forcast as loss or draw in 2C1 ways.
Similarly ( )35 23 1 80q C C= =
Similarly ( )55 25 1 32r C C= = ⇒ 2 5 8q r, p q= = and ( )2 p r q+ >
60.(AC)
61.(ACD)
(A) ( )2 2 2x iy i x y i xy− = − + ⇒ 2x xy= − and 2 2x y y− = −
0x = or 1
2y
−= ;
For x = 0, y = 0 or 1
For1
2y
−=
3
2x = ±
(B)
71
1 0+
+ =
z
z ⇒
( ) ( )1 2 1 2 11
7 7
k kcos i sin
z
π π+ ++ = +
⇒ 1
1 cos i sinz
θ θ− = − − where ( )2 1
7
k πθ
+=
⇒ 1
1− =
− −z
co s i sinθ θ
1
1z
cos i sinθ θ
−⇒ =
− −
1
22 2 2
sin sin icosθ θ θ
−=
−
( )12 2
22
2
sin i cos
Re z
sin
θ θ
θ
− + −
= ⇒ = ⇒ ( )6
0
7
2k
k
Re z
=
−=∑
(C) 21 1
−= i nz z e π / ⇒ ( ) ( )2 2
x iy x iy cos i sinn n
π π − = + −
2 2
xcos y sin xn n
π π+ = and
2 2x sin y cos y
n n
π π− + = −
2 2
1x cos y sinn n
π π − =
and 2 2
1x sin y cosn n
π π = +
and also
2 1y
x= − (given) ⇒ ( )2 1
8tan tan
n
π π= − = ⇒ n = 8
(D) 3 2
1 2
i sin
i sin
θ
θ
+
− is purely real
⇒ 3 2
01 2
i sinIm
i sin
θ
θ
+ = −
( ) ( )1 2 3 2 0sin sinθ θ+ =
0sin ; nθ θ π= =
( ) 0, ;θ π π θ∈ − ∴ =
Vidyamandir Classes
VMC/2013/Solutions 14 Mock IIT Advanced/Test - 3/Paper-1
62.(AC) As 0 1b a a c c b , x− + − + − = = is a root. So, the other root is 2 or 1
2
∴ 1 2c b
b a
−⋅ =
− or
11
2
c b
b a
−⋅ =
−
∴ 2 2b a c b− = − or 2 2b a c b− = −
∴ 3 2b a c= + or 3 2b a c= +
∴ 2
2 1
a cb
+=
+ or
2
1 2
a cb
+=
+
∴ B divides AC in the ratio 2 : 1 or 1 : 2 internally.
63.(0) ∵ | z | = real and positive, imaginary part is zero
∴ arg | z | = 0 ⇒ [ ] 0arg | z | =
∴ [ ]100 100
0 0
0 0
x x
arg | z | dx .dx
= =
= =∫ ∫
64.(6) Since, 2 3 0a b c+ + = . . . .(i)
∴ 2 3 0a c b c c c× + × + × = taking cross product with c
⇒ ( )2 0 0c a b c− × + × + =
∴ ( )2c a b c× = × . . . .(ii)
Again from eq. (i)
2 3 0× + × + × =a b b b c b taking cross product with b
∴ ( )3a b b c× = × . . . .(iii)
Given ( )a b b c c a b cλ× + × + × = ×
⇒ ( ) ( ) ( ) ( )3 2b c b c b c b cλ× + × + × = × [from equations (ii), (iii)]
⇒ ( ) ( )6 b c b cλ× = ×
∴ 6λ =
65.(1) { }
1
1 1
1
− −
→∞=
+∑
a a a aan
ank
k n . n k . klim
n n
/ /
1
1
1
→∞=
= +
∑
a an
nk
k klim .
n n n
/
( )1
1
0
= +∫ a ax x dx
/
( )1
1 1 1
0
1 11
+ +
= + + +
a ax x
a
a
/1
11 1
a
a a= + =
+ +
66.(1) ( ) ( ) ( )2 2 0t f x tf x f '' x− + =′ , has equal roots
Discriminant = ( )( ) ( ) ( )24 4 0f x f x f x− =′ ′′
( )( )
( )( )
f x f x
f x f x
′′ ′=
′ . . . . integrating both side.
( )( ) ( )ln f x ln f x ln c= −′
⇒ ( ) ( )f x cf x= ′
Vidyamandir Classes
VMC/2013/Solutions 15 Mock IIT Advanced/Test - 3/Paper-1
( ) ( )1
0 02
f cf c= ⇒ =′
( )( )
2f x
f x
′=
ln f (x) = 2x + k
⇒ (∵ f (0) = 1) ⇒ k = 0
⇒ ln f (x) = 2x
∴ ( ) 2xf x e=
2 2 2 24 4 0x x xt e te e− + =
⇒ 2 4 4 0t t− + =
t = 2
( ) 2
0 0
1 1 22 2 1 1
2 2 2
x
x x
f x t elim lim
x x→ →
− −∴ − = × − = − =
67.(4) ( ) ( ) ( )2 22 1f x kx x x= − +′ ∵ f (x) is divisible by x3, so it does not have the terms of x
2, x
1, x
0
( )5 4 3 22 2k x x x x= − + −
⇒ ( ) ( )3
3 210 24 15 4060
kxf x x x x= − + −
( )( )
( )3 2 3 8 264
4 1 4 39
f
f
−= × =
−
68.(1) x > 0, y > 0 & 8 2| x iy | r+ + = & ( )84 8 4
yarg x iy tan
x
π π− + = ⇒ =
−
⇒ (x + 8)2 + y
2 = 2r
2, 8y x= − ⇒ x > 8 ∵ y is also positive
⇒ ( ) ( )2 2 28 8 2x x r+ + − = ⇒ 2 22 2 64 2x r+ × =
⇒ 2 264x r+ = ⇒
2 2 64r x− =
⇒ ( ) ( ) 64r x r x− + =
∵ r N∈ and 8 0x , y> > , the only possibility is
⇒ 2 32r x , r x− = + = ⇒ 15 17x , r= =
69.(5) ( ) ( ) ( ) ( ) ( )5 5 51 1 1 1 1
++ = + + = + +
n n nx x x x x
( )5 5 5 4 5 3 5 2 5 1 5 00 1 2 3 4 5C x C x C x C x C x C x= + + + + +
2 1 5
0 1 2 1 5n n n n r n r n r n n
r r r nC C x C x . . . C x C x . . . C x . . . C x+ ++ +
× + + + + + +
Comparing coefficient of xr + 5
⇒ 5 5 5 55 0 1 1 5 5
n n n nr r r rC C C C C . . . C C++ + += + + +
1 2 3 4 55 10 10 5n n n n n nr r r r r rC C C C C C+ + + + += + + + + +
⇒ LHS =( )( )
55 55
4 444
coeff. of in 1 5
5coeff. of in 1
++ ++
+ +++
+ + += = =
+ ++
nr nr
n nrr
x x C n n k
r r kCx x (Given)
⇒ k = 5