JEEADV2017P1 Solutions
JEEADV2017P1 Solutions
JEEADV2017P1 Solutions
SOLUTIONS
2017JEEEntranceExaminationAdvanced/Paper1Code7
PARTI PHYSICS
1.(AD) Net external force acting on the system along the x-axis is zero.
Along the x-axis
Momentum is conserved
mv = MV
From conservation of energy
Loss in GPE of particle of mass m = Gain in kinetic energy of both the masses.
2
1 2 1 1 2 1 mv
mgh = mv + MV 2 mgh = mv + M
2 2 2 2 M
m 2 2 gh
2 gh = 1 + v v= hence option (A) is correct.
M 1+
m
M
m m 2 gh
V= v= along ve x-axis
M M (m + M )
Hence option [B] is incorrect.
Since the location of center of mass does not change along the x-axis
mxm / G + M xM / G = 0 m ( R x ) Mx = 0 f
mR
x= .
m+M
mR
xM G =
m+M
Hence option (D) is correct.
MR
xm G = R x =
M +m
mR
Final position of m = x =
m+M
Hence option (C) is incorrect.
2.(AC) At 0 , X L ~ 0 and X C
Current will be nearly zero.
At >> 106 , XL >>1 and X C 0
Hence circuit does not behave like a capacitor.
At resonance frequency ( 0 ) i.e. when X C = X L current will be in phase with voltage and frequency is
independent of R.
1
Resonant frequency 0 = = 106 rad / s
LC
3.(ABD/BD)
Velocity of the wave depends upon the tension in the rope and mass per unit length of the rope.
Velocity is independent of frequency and wavelength
Option (B) is correct
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Since tension at the mid point is same, therefore speeds at that point will also be same. Option (A) is correct.
Since the direction of motion of pulses is opposite, velocities will be equal and opposite. With this point of
view A will be incorrect.
Since the velocities at position is dependent only on tension at every point ( = constant) therefore time taken
for the wave the reach A from O and O from A will be same.
TOA = TAO
Option (D) is correct
=
Since depends on source
T
Tension decreases as we move from O to A, therefore becomes shorter. Option (C) is incorrect.
4.(ACD) Molecules hitting the forward and rear surfaces will bounce back with speeds given above. Let mass of one
molecule be m0 . Then,
P0 ( forward ) = 2m0 (u + v)
P0 ( Rear ) = 2m0 (u v)
Let the rates of collision with front and rear surfaces be R1 and R2 respectively
So, R1 (u + v)
R2 (u v)
Force = P. R.
So, F1 = R1. 2m0 ( u + v ) , F2 = R2 . 2m0 ( u v )
F1 F2 2m0 ( u + v ) 2m0 (u v) 2
2
So,
2m0 [ 4uv ]
uv so, (A) is correct
Clearly, the net force due to gas is proportion to v, i.e. it is variable hence acceleration of plate is variable.
Finally the plate will start moving with terminal velocity.
Hence (C) is correct.
Resistive force = P. A V
Hence (D) is correct.
5.(A or ACD) (
Net power radiated = a T 4 T04 )
For small temperature difference, T = T T0
T 4 4T
P ( Net ) = A (T0 + T ) T04 = AT04 1 +
4
1 AT04
T0 T0
P ( Net ) = 4AT03 (T T0 )
(A) From the above result, P(Net) decreases with A
(B) Peak shits to shorter wave lengths for rise in temperature.
P(Radiated) = AT 4 = A (T0 + 10 ) > 460 watt
4
(C)
Hence (C) is incorrect
If P(Net) is taken, then (C) will be correct
(D) Energy radiated by a body is dependent only on its own temperature, not the temperature of
surroundings. Hence (D) is incorrect. If P(Net) is taken, then from the calculation shown above,
(D) will also be correct.
6.(ACD) Angular deviation : = i1 + i2 A
For min deviation : i1 = i2 so, i1 = A ...(i)
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Also, r1 + r2 = A
A
i.e. r1 = ...(ii)
2
i
r1 = 1 (A is correct)
2
By Snell's law, sin i1 = sin r1
A
i.e. sin A = sin
2
A
2 cos =
2
A = 2 cos 1
2
Hence (B) is incorrect
(C) is obviously correct
For tangential emergence, i2 90 .
So, sin r2 = 1
1
cos r2 = 1
2
Also, r1 = A r2
sin r1 = sin A cos r2 cos A sin r2
2 1 1
= ( sin A ) ( cos A )
By Snell's law on 1st surface,
sin i1 = sin r1
( )
1/ 2 A
sin i1 = sin A 2 1 cos A i1 = sin 1 sin A 4cos 2 1 cos A
2
i.e. (D) is correct
7.(AB) The angle that area vector makes with B at time t is = t.
1 = BA cos t 1 = BA sin (t )
2 = 2 BA cos(t ) 2 = 2 BA sin (t )
Due to orientation of loops, the two EMFs will work against each other.
So, (net) = BA sin(t )
So, (A) is correct.
We can see that net is maximum when = . So, (B) is correct. Obviously, (C) is incorrect. (D) is incorrect as
2
the EMF is proportional to difference in areas.
1
8.(5) V
n2
2
Vi n f nf 5
So, = = 6.25 = 2.5 =
V f ni ni 2
Minimum integral value of n f is 5.
9.(8) We know that for the given case,
sin = constant
So, 1.6 sin ( 30 ) = ( n mn ) sin 90
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330 + 2 332
Frequency received by car = 492 = 492 495 Hz
330 330
330
Frequency received by source = 495 498 Hz
330 2
Original frequency f1 = 492 Hz
Final frequency f 2 = 498 Hz
Beat frequency = | f1 f 2 | = 6 Hz
4 3 4 3
12.(6) R = r K (Conservation of volume)
3 3
R = radius of bigger drop
r = radius of smaller drop
R3 = r 3 K ...(i)
U i = S 4R 2 ( )
U f = KS 4r 2 ( )
KS 4r 2 ( ) S 4R 2
= 103
4S Kr 2 R 2 = 103
1
K 3 R 2 R 2 = 102
10 / 3 1 = 100
10 / 3 = 101
6
13.(D) For constant velocity, acceleration of particle should be zero, Hence net force should be zero.
E
qVB = qE V=
(For a = 0)
B
Electric field and magnetic field should be perpendicular.
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E0
Option (D) : V= y , E = E0 x , B = B0 z (for electron)
B0
E
( )
F B = q V B = e 0 y B0 z = E0 x
0
B
FE = E0 x
14.(B) V = 0, E = E0 y , B = B0 y (Proton)
FB = 0 only force along -y axis is acting due to electric field alone.
E0
15.(D) V =2 x ; E = E0 z ; B = B0 z (Proton)
B0
FB = along y axis
FE = along +z axis
So condition for helical path is satisfied
17.(A) Laplace correction is done is correction in the determination of the speed of sound in an ideal gas, which states
that the process assumed was not Isothermal but it is Adiabatic .
1
Wadiabatic 1 2 = ( P2V2 PV
1 1)
1
Corresponding graph is
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PARTII CHEMISTRY
19.(BC)
* Z represent vapour pressure of pure liquid L and as L 1 , the given graph is merging with ideal
graph of p L . Hence Option (C) is correct
* Since, vapour pressure of liquid L(p L ) is higher than the ideal values so L-M interactions are weaker
than L-L & M-M interactions. Hence Option (B) is correct choice.
20.(ABC)
In oxoacids acidic strength increases with increase in number of double bonded oxygen atoms because of
greater resonance stabilization of conjugate base. Hence HClO 4 is more acidic than HClO.
This equilibrium remain shifted in forward direction hence ClO 4 is weaker base than H 2O. Because such
equilibrium remain shifted towards weaker acid or weaker base side.
Cl2 + H 2O
HCl + HOCl
If M is Co then X : Co ( H 2 O )6 Cl2
Y : Co ( NH3 )6 Cl3
2
Z : [ CoCl4 ]
x = 3.87 due to presence of three unpaired electron in X
Co 2+ ;3d 7
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Co 2+ ;3d 7
Cl is weak a field ligand and coordination number is four hence hybridization state is sp3 .
3+
M ( NH3 ) Cl3 + 3AgNO3 M ( NH3 ) 6 + 3NO3 + 3AgCl(s)
6
2+
Co ( H 2 O )6
4Cl [CoCl4 ]2+ + 6H 2 O : H > 0
Pink Blue
(As octahedral complex changes to tetrahedral complex with H2O ligands replaced by Cl )
2+
At 0C, equilibrium shifts in reverse direction hence colour of the solution is pink due to Co ( H 2 O )6
23.(CD)
2 *2 2
25.(BC) Electronic configuration of F2 molecule is 1s , 1s , 2s , *2 2 2 2 *2 *2 *
2s , 2p , 2p 2p , 2px 2p , 2 p
z x y y z
Similar electronic configuration for other X 2 molecules.
*p x and *p y are highest energy occupied molecular orbital (HUMO) and *pz is lowest energy
unoccupied molecular orbital (LUMO).
Colour of X 2 molecules of group 17 elements is due to transition of electrons from * to * .
V+ M C
26.(6) Number of electrons around central atom (N) =
2
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6+6+2
[TeBr6 ]2 : N= =7
2
Number of bp = 6
Number of p = 1
7 + 2 1
[BrF2 ]+ : N= =4
2
Number of bp = 2
Number of p = 2
6 1 + 3
SNF3 : N= =4
2
Number of bp = 4
Number of p = 0
8+ 3+1
[XeF3 ] : N= =6
2
Number of bp = 3
Number of p = 3
The sum of the number of lone pairs of electrons on each central atom = 1 + 2 + 0 + 3 = 6
1000 K
27.(6) cm =
M
HA H + + A
c = 0.0015 = 104
1 l
= R =
15 A
1 1 l 120
= K = 5 107
R
A 1
K = 6.0 105
1000 6.0 105
And cm =
1.5 103
cm
cm = 40 = 0m = 40 15 = 6.0 102 = z 102
0m
So z = 6.0
28.(5) Compound having a close loop of (4n + 2) electrons is aromatic compound.
1.
2.
3.
4.
5.
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6.
7.
8.
9.
2 *2 2
N 2 Diamagnetic 1s , 1s , 2s , *2 2
2s , 2 2 2
, 2p
P x 2p y z
O 2 Paramagnetic 2
1s , *2
1s , 2
2s , *2 2
2s , 2 p z , 22p , 22p , *2 *1
2 p x , 2 p y
x y
2 *2 2
F2 Diamagnetic 1s , 1s , 2s , *2 2 2 2 *2 *2
2s , 2p z , 2p x 2p y , 2px 2py
The number of diamagnetic species is 6 i.e. H 2 , Li 2 , Be2 , C2 , N 2 and F2 are diamagnetic species
30.(2) a = 400pm
d = 8g cm 3
w = 256 g
z M0
d=
a3 NA
( )
3
d a3 NA 8 400 1010 6.023 1023
M0 = = = 77.09 g/mol
z 4
256
Number of atoms = 6.023 1023 = N 1024
77.09
2 1024 = N 1024
N=2
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zr
a0
31.(A) Incorrect combination is (I) (iii) (R) because in the expression of n, ,m , the exponential part must have e .
[A]
zr
a0
32.(D) The correction combination is (I) (i) (s) as n, ,m in the coloum 2 (i) has exponential part as e .
2 2
13.6z 3 13.6z
(4 1)
E4 E2 27
= 16 = 16 =
E6 E 2 13.6z 2
8 13.6z 2 32
(9 1)
36 16
[D]
33.(C) The correct combination for any hydrogen like species is [II] (ii) (P).
2s-orbital has one radial node and n, , m v/s r plot will start from a finite value and sign changes once from
+ve to ve.
[C]
34.(B)
35.(A)
36.(B)
PARTIII MATHEMATICS
az + b az + b
37.(AC) = 2iy
z +1 z +1
azz + az + bz + b (azz + bz + az + b)
= 2iy
( z + 1) ( z + 1)
( a b ) z z ( a b)
= 2iy
( z + 1) ( z + 1)
(a b) 2iy
= 2iy
( z + 1) ( z + 1)
( z + 1) ( z + 1) = 1
x + y2 + 2x + 1 = 1
x2 + 2 x + 1 = 1 y2
( x + 1) = 1 y 2
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x + 1 = 1 y2 x + 1 = 1 y2
x = 1 + 1 y2 x = 1 1 y2 .
x
38.(CD) (A) f 1 ( x) = ex f ( t ) sin t dt
0
f1 ( 0 ) = 1
0 < f ( x) < 1
0 < f ( x ) sin x < 1
f ' 1 ( x ) = e x f ( x ) sin x > 0
/ 2
(B) f ( x) + f ( t ) sin t dt > 0 x (0, 1)
0
x
2
(C) g ( x) = x f ( t ) cos t dt
0
/2
g (0) = f ( t ) cos t dt < 0
0
1
2
g (1) = 1 f ( t ) cos t dt > 0
0
(D) g ( x ) = x9 f ( x )
g (0) = f (0) < 0
g (1) = 1 f (1) > 0
39.(AC) P ( X ) = 1/ 3 . . .(i)
P( X Y )
P ( X / Y ) = 1/ 2 = . . .(ii)
P ( 4)
P (Y X )
P (Y / X ) = 2 / 5 = . . .(iii)
P(X )
2 P ( X Y )
= .3
5 1
2
P( X Y ) =
15
2
1 15 2.2 4
(ii) = = P (Y ) = =
2 P (Y ) 15.1 15
1 4 2 5+42 7
P ( X Y ) = P( X ) + P (Y ) P ( X Y ) = + = =
3 15 15 15 15
(
P X Y ) = P (Y ) P ( X Y ) = 1 2 15 = 1/ 2
(
P X /Y = ) P (Y ) P (Y ) 15 4
40.(ABD)
Lt = ( 1 + h ) cos ( ( 1 + h ) + [ 1 + h ])
x 1+
( 1 + h ) cos ( 1 + h 1) = ( 1 + h ) cos ( 2 + h )
( 1 + h ) cos [ 2 h ] ( 1 + h ) cos h 1
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Lt = ( h ) cos ( h 1) = h cos [ h + ] 0
x 0
Lt = h cos [ h ] = 0
x 0+
41.(ABD) 2x y + 1 = 0
y = 2x + 1
x2 y2
=1
a2 16
c2 = a 2 m2 b2
1 = a 2 .4 16
17
a2 =
4
17
a= .
2
42.(AB) For (A) and (B) det ( A2 ) = 1 ,
and for (C)
1 0 0 1 0 0
0 1 0 = 1 = 0 1 0
0 0 1 0 0 1
1 0 0 1 0 0 1 0 0
0 1 2 0 1 2 = 0 1 0
0 1 1 0 1 1 0 0 1
1 0 0
For (C): I 2 = 0 1 0 .
0 0 1
43.(B) y 2 = 16 x
Equation of a chord with a given middle point ( h, k ) is T = S 1
ky 8 ( x + h ) = k 2 16h
ky 8 x + 8h k 2 = 0
2x + y p = 0
8 k 8h k 2
= = k = 4
2 1 p
4 8h 16
=
1 p
4 p = 8h 16 p = 2h 4
h=3 p=2
/2
44.(2) g ( x ) = f (( t ) cos ect = g ( x) = 3 f ( x) cosec ( x)
x
f ( x)
lim g ( x ) = lim 3 =3 1 = 2 = 2
x 0
x 0 sin x
45.(5) x = 10!
10 10!
y= C9 9C1
2!
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46.(6) a d , a, a + d a>0
2 2
(a + d ) = a2 + ( a d )
a 2 + d 2 + 2ad = 2a 2 2ad + d 2
a 2 = 4ad a0
a = 4d
3d 4d 5d
1
A = 3d 4d = 24
2
6d 2 = 24
d2 = 4 d = 2 Rejected
d =2
a=8
6, 8, 10
y2 + 4 y + 1 = 0
D > 0 (Two real root)
p = 1
Case II : Touching y-axis
( x + 1)2 + ( y + 2 )2 =1
x2 + y2 + 2 x + 4 y + 4 = 0
y=0
x2 + 2x + 4 = 0
D<0 (None real root) Rejected
2 2
Case III : x + y + 2x + 4 y = 0
p=0
1 2
( )
2
48.(1) 1 1 = 2 1 = 0
1 1
2 = 1
= 1
= 1 (two planes are parallel) (Rejected)
= 1 (two planes are coincident)
49.(D) y = x+8
a a 2a
y 2 = 32 x, y = mx + , ,
m m2 m
50.(A) a= 2
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ma 2 a 2
x 2 + y 2 = a 2 , y = mx a 1 + m 2 , ,
c m
x 2 + y 2 = 2 tangent at ( 1, 1) x + y = 2 or y = x + 2
a2
51.(C) 3+ = a2
4
a2 = 4
x2 + y2 = 4
3x + 2 y = 4
1 1 1 1 x ln x
f ( x) = 1 + x ln x = ln x = =0
x x x x
1
ln x =
x
f (1) = f ( x ) = 1
f (0+ ) =
f (e 2 ) < 0
1 1
f ( x) = <0
x2 x
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