Solution_JEEAdvanced2016Paper-1

Vidyamandir Classes

JEE ENTRANCE EXAM-2016/ADVANCED

P1-16-3-6 PAPER-1 Code : 6

Time : 3 Hours Maximum Marks : 186

Read the following Instructions very carefully before you proceed.

GENERAL
1. This sealed booklet is your Question Paper. Do not break the seal till you are told to do so.
2. The paper CODE is printed on the right hand top corner of this sheet and the right hand top corner of the back

cover of this booklet.
3. Use the Optical Response Sheet (ORS) provided separately for answering the questions.
4. The paper CODE is printed on the left part as well as the right part of the ORS. Ensure that both these codes are

identical and same as that on the question paper booklet. If not, contact the invigilator for change of ORS.
5. Blank spaces are provided within this booklet for enough work.
6. Write your name, roll number and sign in the space provided on the back cover of this booklet.
7. After breaking the seal of the booklet at 9.00 am, verify that the booklet contains 36 pages and that all the

54 questions along with the options are legible. If not, contact the invigilator for replacement of the booklet.
8. You are allowed to take away the Question Paper at the end of the examination.

OPTICAL RESPONSE SHEET
9. The ORS (top sheet) will be provided with an attached Candidate’s Sheet (bottom sheet).

The Candidate’s Sheet is a carbon-less copy of the ORS.
10. Darken the appropriate bubbles on the ORS by applying sufficient pressure. This will leave an impression at the

corresponding place on the Candidate’s Sheet.
11. The ORS will be collected by the invigilator at the end of the examination.
12. Your will be allowed to take away the Candidate’s Sheet at the end of the examination.
13. Do not tamper with or mutilate the ORS. Do not use the ORS for rough work.
14. Write your name, roll number and code of the examination centre, and sign with pen in the space provided for

this purpose on the ORS. Do not write any of these details anywhere else on the ORS. Darken the appropriate
bubble under each digit of your roll number.

DARKING THE BUBBLE ON THE ORS
15. Use a BLACK BALL POINT PEN to darken the bubble on the ORS.
16. Darken the bubble COMPLETELY.
17. The correct way of darkening a bubble is as:
18. The ORS is machine-gradable. Ensure that the bubbles are darkened in the correct way.
19. Darken the bubble ONLY IF you are sure of the answer. There is NO WAY to erase or “un-darken” a darkened

bubble.

VMC/Paper-1 1 JEE Entrance Exam-2016/Advanced

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PART-I PHYSICS

SECTION I (Maximum Marks : 15)

 This section contains Five questions.

 Each question has Four options (A), (B), (C) and (D). ONLY ONE of these four options is correct.

 For each question, darken the bubble corresponding to the correct option in the ORS.

 For each question, marks will be awarded in one of the following categories :

Full Marks : +3 If, only the bubble corresponding to the correct option is darkened.

Zero Marks : 0 If none of the bubbles is darkened.

Negative Marks : –1 In all other cases.

1. A parallel beam of light is incident from air at an angle  on the side

PQ of a right angled triangular prism of refractive index n  2.
Light undergoes total internal reflection in the prism at the face PR
when  has a minimum value of 45 . The angle  of the prism is :

(A) 15 (B) 22.5

(C) 30 (D) 45

1.(A)

For face PQ, 1.0sin 450  2 sin 
   300
Also,     90  90  450  1800
     450
   150
Hence (A)

2. In a historical experiment to determine Planck’s constant, a metal surface was irradiated with light of different
wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant
data for the wavelength () of incident light and the corresponding stopping potential (V0 ) are given below:

m V0 (volt)

0.3 2.0
0.4 1.0
0.5 0.4

Given that c  3108 ms1 and e  1.61019C, Planck’s constant (in units of Js) found from such an
experiment is :

VMC/Paper-1 2 JEE Entrance Exam-2016/Advanced

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(A) 6.0 1034 (B) 6.4 1034 (C) 6.6 1034 (D) 6.81034

2.(B) eV0  hc  w  V0   hc   1  w
3.   e   e

i.e. Slope of V0 us 1 gives hc . i.e. m  y  1 2 106  0.4 1.0 106
 e x 1  1 11

0.4 0.3 0.5 0.4

This gives : m  1.2 106

m  hc  1.2 106 1.61019  h
e 3 108

This gives h  6.4 1034. Hence (B)

A water cooler of storage capacity 120 litres can cool water at a
constant rate of P watts. In a closed circulation system (as shown
schematically in the figure), the water from the cooler is used to cool
an external device that generates constantly 3 kW of heat (thermal
load). The temperature of water fed into the device cannot exceed
30C and the entire stored 120 litres of water is initially cooled to
10C.
The entire system is thermally insulated. The minimum value of P (in watts) for which the device can be
operated for 3 hours is : (Specific heat of water is 4.2 kJ kg 1K 1 and the density of water is 1000 kg m3 )

(A) 1600 (B) 2067 (C) 2533 (D) 3933

3.(B) Overall water receives heat at the rate of 3000  P watts.
4.
Temperature of water increases from 10° C to 30°C

Let time be t. Then, 3000  P t  120 4200  20

t  3 60 60 s .

Solving gives : P  2067 watt.
Hence (B)
An infinite line charge of uniform electric charge density  lies along the axis of an electrically conducting
infinite cylindrical shell of radius R. At time t = 0, the space inside the cylinder is filled with a material of
permittivity  and electrical conductivity . The electrical conduction in the material follows Ohm’s law.
Which one of the following graphs best describes the subsequent variation of the magnitude of current density
j(t) at any point in the material?

(A) (B) (C) (D)

4.(A) Let charge density on wire be  ' at time t. JEE Entrance Exam-2016/Advanced
Then, E  2k

r
Using j  E (differential form of ohm's law)

j   .2k'
r

VMC/Paper-1 3

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i   . 2K .Q (Q is charge on length L of wire)
2 rL rL

 1 dQ   .2k Q    dQ  4k dt
2 dt Q

 n  Q   4k.t  t
 Q0 
   Q  Q0 e 0

 j  2k . Q0 et 0  j  2k0 e t 0
rL r

5. A uniform wooden stick of mass 1.6 kg and length l rests in an inclined manner on a smooth, vertical wall of

height h (<l) such that a small portion of the stick extends beyond the wall. The reaction force of the wall on the

stick is perpendicular to the stick. The stick makes an angle of 300 with the wall and the bottom of the stick in

on a rough floor. The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the

stick. The ratio h/l and the frictional force f at the bottom of the stick are : (g  10m s2 )

(A) h  3 , f  16 3 N (B) h  3 , f  16 3 N
l 16 3 l 16 3
(C)
5.(D) h 3 3,f 8 3N (D) h  3 3 , f  16 3 N
l 16 3 l 16 3

N sin 30° + N = 1.6 × 10
N = 32

3

f = N cos 30° = 32  3  16 3
32 3

0 must be zero
mg × L cos60  N  h

2 sin 60
16 L  1  32  2h

22 3 3

h   33 
l  16 

VMC/Paper-1 4 JEE Entrance Exam-2016/Advanced

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SECTION II (Maximum Marks : 32)

 This section contains EIGHT questions.

 Each question has FOUR options options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s)

is(are) correct.

 For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS.

 For each question, marks will be awarded in one of the following categories:

Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened.

Partial Marks : +1 For darkening a bubble corresponding to each correct option, provided NO

incorrect option is darkened.

Zero Marks : 0 If none of the bubbles is darkened.

Negative Marks : –2 In all other cases.

For example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will result in +4

marks; darkening only (A) and (D) will result in +2 marks; and darkening (A) and (B) will result in -2 marks, as a

wrong option is also darkened.

6.   )   t 3iˆ   t 2 ˆj,
The position vector r of a particle of mass m is given by the following equation : r (t

where   10 / 3 m s3,   5 m s2 and m  0.1 kg. At t  1 s, which of the following statement(s) is(are) true

about the particle?

(A) The velocity  is given by   (10iˆ  10 ˆj) m s 1
v v

(B) The angular momentum  with respect to the origin is given by   (5 / 3)kˆ N m s
L L

(C) The force F is given by F  (iˆ  2 ˆj)N

(D) The torque  with respect to the origin is given by   (20 / 3)kˆ N m
 

6.(ABD) 
r t3ˆi  t2 ˆj

x t3, y  t2

 1  10 ˆi  5ˆj
r
3

dx  3t2  10  dy  2t  10
dt dt

ax  6t  10  6 1  20
3

ay  2

 (A) V  10ˆi 10ˆj True
   

(B) L  m r  v

  10
3
L
 0.1 iˆ  5ˆj  10iˆ 10ˆj

 5 kˆ True
L 3


F  0.1
   (C)
20iˆ 10ˆj  2ˆi  ˆj Not true

 
(D)   r  F

   10 
 3 

 ˆi  50ˆj  2iˆ  ˆj

VMC/Paper-1 5 JEE Entrance Exam-2016/Advanced

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   20 kˆ True

3

7. A transparent slab of thickness d has a refractive index n (z) that

increases with z. Here z is the vertical distance inside the slab,

measured from the top. The slab is placed between two media with

uniform refractive indices n1 and n2 ( n1), as shown in the figure. A

ray of light is incident with angle i from medium 1 and emerges in

medium 2 with refraction angle  f with a lateral displacement l.

Which of the following statement(s) is (are) true?

(A) l is independent of n2 (B) n1 sini  n2 sin f

(C) l is dependent on n (z) (D) n1 sini  (n2  n1)sin f

7.(ABC) For a slab with variable refractive index, n sin remains constant. Hence (B)

Also, l depends on equation of curve and hence on how refractive index varies with z. Hence (C)
As the equation of the curve is independent of n2 hence  is independent of n2. Hence (A).
8. A plano-convex lens is made of a material of refractive index n. When a small object is placed 30 cm away in

front of the curved surface of the lens, an image of double the size of the object is produced. Due to reflection

from the convex surface of the lens, another faint image is observed at a distance of 10 cm away from the lens.

Which of the following statement(s) is(are) true?

(A) The refractive index of the lens is 2.5

(B) The radius of curvature of the convex surface is 45 cm

(C) The faint image is erect and real

(D) The focal length of the lens is 20 cm

8.(AD) For refraction from lens,

u  30 , v  60

Hence, 1  1  1
60 30 F

 F  20 cm ................(i)

Hence, (D)
For reflection from spherical surface,

u  30 , v  10 cm

1 1 2
10 30 R
 R  30 cm .............(ii)

By lens maker's formula: 1  (   1 )  1  1     2.5 Hence, (A)
20  30  

9. Highly excited states for hydrogen-like atoms (also called Rydberg states) with nuclear charge Ze are defined by

their principal quantum number n, where n >> 1. Which of the following statement(s) is(are) true ?

(A) Relative change in the radii of two consecutive orbitals does not depend on Z

(B) Relative change in the radii of two consecutive orbitals varies as 1/n
(C) Relative change in the energy of two consecutive orbitals varies as 1/n3

(D) Relative change in the angular momenta of two consecutive orbitals varies as 1/n

9.(ABD) For (AB) Rn1  Rn  n 12  n2  2n 1  21
Rn z n2 nn
z
2

VMC/Paper-1 6 JEE Entrance Exam-2016/Advanced

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z2  z2
(n 1)2 n2
For (C) Un1 Un   2n 1  1
Un z2 (n 1)2 n

n2

For (D) Ln1  Ln  (n 1)  (n)  1
Ln n n

10. A length-scale (  ) depends on the permittivity (  ) of a dielectric material, Botzmann constant kB  , the

absolute temperature (T), the number per unit volume (n) of certain charged particles, and the charge (q) carried
by each of the particles. Which of the following expression(s) for  is(are) dimensionally correct ?

(A)   nq2  (B)    kBT 
  kBT   nq2 

(C)   q2  (D)   q2 
  n2 3kBT    n1 3kBT 

10.(BD) [n]  [L3]

[q]  [AT ]

[ ]  [M 1t3 A2T 4 ]
[T ]  [K ]

[l]  [L]

[kB ]  [M 1L2T 2K 1]

(A) RHS.  [L3A2T 2 ]  [L3 A2T 2 ]  [L2 ]  [L1] Wrong
[M 1L3T 4 A2][M1L2T 2K 1][K ] [L1T 2 A2]

(B) RHS [M 1L3T 4 A2][M1L2T 2K 1][K ]  [L1T 2 A2 ]  [L] Correct
[L3][ A2T 2] [L3T 2 A2 ]

(C) RHS [ A2T 2 ]  L3 Wrong
1L3T 4 A2][L2 ][M 1L2T 2K 1][K ]
[M

(D) RHS [ A2T 2]  [ A2T 2 ]  [L] Correct
[M 1L3T 4 A2 ][L1][M 1L2T 2K 1] [L2T 2 A2 ]

11. Two loudspeakers M and N are located 20 m apart and emit sound at frequencies 118 Hz and 121 Hz,
respectively. A car is initially at a point P, 1800 m away from the midpoint Q of the line MN and moves towards
Q constantly at 60 km/hr along the perpendicular bisector of MN. It crosses Q and eventually reaches a point R ,
1800 m away from Q. Let v(t) represent the beat frequency measured by a person sitting in the car at time t. Let
vP, vQ and vR be the beat frequencies measured at locations P, Q and R, respectively. The speed of sound in air

is 330 m s1 . Which of the following statements(s) is(are) true regarding the sound heard by the person?

(A) The plot below represents schematically the variation of beat frequency with time

(B) vP  vR  2vQ
(C) The plot below represents schematically the variation of beat frequency with time

VMC/Paper-1 7 JEE Entrance Exam-2016/Advanced

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(D) The rate of change in beat frequency is maximum when the car passes through Q
11.(ABD)

Speed of car, V = 60 km/h
500 m/s
3

At a point S, between P & Q

v'M  vM  C  V cos  ; v'N  vN  C  V cos 
 C   C 

 v  (vN  vM ) 1  V cos  
C 

similarly, between Q & R

v  (vN  vM ) 1  V cos  
C 

d (v)  (vN  vM )V sin  d
dt C dt

  0º at P & R as they are large distance apart.

 slope of graph is zero.

at Q,   90º

 sin  is max.
also value of d is max
dt

as d  V , where V is its velocity and r is length of line joining P & S.
dt r

and r is minimum at Q.
 slope is maximum at Q.
 (A) & (B) are correct.

At P, v  (vN  vM ) 1  V  (  0º )
C 

At R, v  (vN  vM ) 1  V  (  0º )
C 

At Q, v  (vN  vM ) (  90º )

VMC/Paper-1 8 JEE Entrance Exam-2016/Advanced

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 vP  vR  2(vN  VM )  2vQ Hence (B) is correct.

12. A conducting loop in the shape of a right angled isosceles triangle of

height 10 cm is kept such that the 90° vertex is very close to an

infinitely long conducting wire (see the figure). The wire is electrically

insulated from the loop. The hypotenuse of the triangle is parallel to

the wire.

The current in the triangular loop is in counterclockwise direction and increased at a constant rate of 10 As1 .

Which of the following statement(s) is (are) true?

(A) The induced current in the wire is in opposite direction to the current along the hypotenuse

(B) The magnitude of induced emf in the wire is  0  volt
  

(C) There is a repulsive force between the wire and the loop

(D) If the loop is rotated at a constant angular speed about the wire, an additional emf of  0  volt is
  

induced in the wire

12.(BC) By reciprocity theorem of mutual induction it can be assumed that current in infinite wire is varying at 10 A/S

and EMF is induced in triangular loop.

Flux of magnetic field through triangle loop, if current in infinite wire is  , can be calculated as follows:

d  0i  2 y dy
2 y

d  0i dy


  0i  l  EMF  0  l   di  0 (10cm) 10 A  0 volts.
     2  dt  S  
 2   

Hence, (B) is correct.

If we assume the current in the wire towards right then as the flux in the loop increases we know that the induced

current in the wire is counter clockwise. Hence the current in the wire is towards right.

Field due to triangular loop at the location of infinite wire is into the paper. Hence, force on infinite wire is away

from the loop. Hence (C)

By cylindrical symmetry about infinite wire, rotation of triangular loop will not cause any additional EMF.

13. An incandescent bulb has a thin filament of tungsten that is heated to high temperature by passing an electric

current. The hot filament emits black-body radiation. The filament is observed to break up at random locations

after a sufficiently long time of operation due to non-uniform evaporation of tungsten from the filament. If the

bulb is powered at constant voltage, which of the following statement(s) is(are) true?

(A) The temperature distribution over the filament is uniform

(B) The resistance over small sections of the filament decreases with time

(C) The filament emits more light at higher band of frequencies before it breaks up

(D) The filament consumes less electrical power towards the end of the life of the bulb

13.(CD) (A) Since evaporation is given to be non-uniform, temperature must be non-uniform.

(B) Due to evaporation, cross sectional area of wire decreases. Hence resistance of any given section

increases.

(C) Consider a small section where area has been reduced.

VMC/Paper-1 9 JEE Entrance Exam-2016/Advanced

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Power generated per unit volume will determine the temperature of this section:

dP  i2dR

dP  i2 dR  i2dx
dV dV A(dx A)

 dP  i2   T4  i
dV A2 A3

Power radiated =  AT 4

Assuming that effect of A3 dominates i, (C) will be correct.
(D) P  V 2 . Overall R decreases, V remains constant. Hence (D)

R

SECTION III (Maximum Marks : 15)
 This section contains FIVE questions.
 The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive.
 For each question, darken the bubble corresponding to the correct integer in the ORS.
 For each question, marks will be awarded in one of the following categories:

Full Marks : +3 If only the bubble corresponding to the correct answer is darkened.
Zero Marks : 0 In all other cases.

14. Two inductors L1 (inductance 1 mH, internal resistance 3  ) and L2 (inductance 2 mH, internal resistance 4
14.(8)  ), and a resistor R (resistance 12  ) are all connected in parallel across a 5 V battery. The circuit is switched

on at time t  0 . The ratio of the maximum to the minimum current  Imax / Imin  drawn from the battery is

________.

At t  0
Imin at t = 0 as current in both inductors = 0
 Using KVL, 5 – 12 I = 0

VMC/Paper-1 10 JEE Entrance Exam-2016/Advanced

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 imin  5 A
12

I max is at t   as current in both inductors will be max.

t 

Inductors will behave like resistors only 1  1  1  1  2
Req 3 4 12 3

 I max  5  10  Imax  10 / 3  8
3/ 2 3 Imin 5 /12

15. A metal is heated in a furnace where a sensor is kept above the metal surface to read the power radiated (P) by
15.(9)
the metal. The sensor has a scale that displays log2  P / P0  , where P0 is a constant. When the metal surface is

at a temperature of 487°C, the sensor shows a value 1. Assume that the emissivity of the metallic surface remains

constant. What is the value displayed by the sensor when the temperature of the metal surface is raised to 2767

°C?

Power radiated P  A4

1st reading, R1  log2   A  760 4  ; 2nd reading, R2  log2  A 30404 
   
 P0   P0 
  

R2  R1  log2  A  3040 4  P0   log2  3040 4  4 log2 4  8
   760 
 P0  A  7604 


 R2  8  R1  9

16. A hydrogen atom in its ground state is irradiated by light of wavelength 970 Å . Taking
hc / e  1.237 106 eV m and the ground state energy of hydrogen atom as 13.6 eV , the number of lines

present in the emission spectrum is________.

16.(6) Incident energy Ei  hc  1.237 106  12.75eV
17.  970 1010

 e jumps to state of E = –13.6+12.75= –0.85 eV  n  4

 Number of lines in emission spectrum is 6.

Consider two solid spheres P and Q each of density 8 gm cm3 and diameters 1 cm and 0.5 cm, respectively.

Sphere P is dropped into a liquid of density 0.8gm cm3 and viscosity   3 poiseulles. Sphere Q is dropped into

a liquid of density 1.6 gm cm3 and viscosity   2 poiseulles. The ratio of the terminal velocities of P

and Q is________.

17.(3) Terminal velocity VT  2 r2 (d  p)g
9 

VMC/Paper-1 11 JEE Entrance Exam-2016/Advanced

Vidyamandir Classes

VP  rp2  Q  d  P    1 2   2   8 – 0.8  4 2  7.2 =3
VQ  0.5   3  – 3 6.4
rQ2 P (d  Q ) (8 1.6)

18. The isotope 12 B having a mass 12.014 u undergoes  -decay to 12 C. 12 C. has an excited state of the nucleus
5 6 6

(162 C*) at 4.041 MeV above its ground state. If 12 B decays to 12 C * , the maximum kinetic energy of the
5 6

 -particle in units of MeV is________. (1u  931.5 MeV / c2 , where c is the speed of light in vacuum)

18.(9) 12B  162C  0 e  v
–i
5

Mass of 12 C = 12. 000 u (by definition if 1 a.m.u.)
6

Q – value of reaction, Q = (MB – MC) × c2 = (12.014 – 12.000) × 931.5 = 13.041 MeV

4.041 MeV of energy is taken by 12 C*  max. K.E. of  -particle is (13.041 – 4.041) = 9 MeV
6

PART-II CHEMISTRY

SECTION I (Maximum Marks : 15)

 This section contains Five questions.

 Each question has Four options (A), (B), (C) and (D). ONLY ONE of these four options is correct.

 For each question, darken the bubble corresponding to the correct option in the ORS.

 For each question, marks will be awarded in one of the following categories :

Full Marks : +3 If, only the bubble corresponding to the correct option is darkened.

Zero Marks : 0 If none of the bubbles is darkened.

Negative Marks : –1 In all other cases.

19. P is the probability of finding the 1s electron of hydrogen atom in a spherical shell of infinitesimal thickness, dr,
at a distance r from the nucleus. The volume of this shell is 4 r2dr . The qualitative sketch of the dependence of
P on r is :

(A) (B) (C) (D)

19.(C) r = 0 at nucleus
For 1s orbital probability of finding electron can’t be zero for any value of r.

20. One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0

L against a constant pressure of 3.0 atm. In this process, the change in entropy of surroundings Ssurr  in

J K–1 is (1 L atm = 101.3 J)

(A) 5.763 (B) 1.013 (C) –1.013 (D) –5.763

20.(C) For isothermal expansion : U  0  qirrev.  wirrev.  (PV)  3(2 1)  3 L atm

SSurrounding  qirrev.  3101 3   1 013 J K1
T 300

VMC/Paper-1 12 JEE Entrance Exam-2016/Advanced

Vidyamandir Classes

21. Among [Ni(CO)4 ], [NiCl4 ]2 , [Co(NH3)4 Cl2 ]Cl, Na3[CoF6 ], Na 2O2 and CsO2, the total number of

paramagnetic compounds is

(A) 2 (B) 3 (C) 4 (D) 5

21.(B) [NiCl4 ]2 , Na3[CoF6 ],CsO2 are paramagnetic.
Cl is a weak field ligand with Ni2 in [NiCl4]2 hence no pairing of electrons
F is a weak field ligand with Co3 in Na3[CoF6 ] hence no pairing of electrons
O2 is superoxide anion in CsO2.

22. The increasing order of atomic radii of the following Group 13 elements is
22.(B)
(A) Al < Ga < In < Tl (B) Ga < Al < In < Tl

(C) Al < In < Ga < Tl (D) Al < Ga < Tl < In

Increasing order of atomic radii: Ga < Al < In < Tl

Ga is smaller due to poor shielding of d-orbital

23. On complete hydrogenation, natural rubber produces

(A) ethylene-propylene copolymer (B) vulcanised rubber

(C) polypropylene (D) polybutylene

SECTION II (Maximum Marks : 32)

 This section contains EIGHT questions.

 Each question has FOUR options options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s)

is(are) correct.

 For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS.

 For each question, marks will be awarded in one of the following categories:

Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened.

Partial Marks : +1 For darkening a bubble corresponding to each correct option, provided NO

incorrect option is darkened.

Zero Marks : 0 If none of the bubbles is darkened.

Negative Marks : –2 In all other cases.

For example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will result in +4

marks; darkening only (A) and (D) will result in +2 marks; and darkening (A) and (B) will result in -2 marks, as a

VMC/Paper-1 13 JEE Entrance Exam-2016/Advanced

Vidyamandir Classes

wrong option is also darkened.
24. The product(s) of the following reaction sequence is(are)

(A) (B) (C) (D)

25. The correct statement(s) about the following reaction sequence is(are)

Cumene C9H12  i HO32O P CHCl3/ NaOH Q  major   R  minor 
ii

NaOH

Q S

PhCH2 Br

(A) R is steam volatile

(B) Q gives dark violet coloration with 1% aqueous FeCl3 solution
(C) S gives yellow precipitate with 2,4-dinitrophenylhydrazine

(D) S gives dark violet coloration with 1% aqueous FeCl3 solution.

26. The crystalline form of borax has 14 JEE Entrance Exam-2016/Advanced

(A) tetranuclear B4O5 OH4 2 unit

(B) all boron atoms in the same plane

VMC/Paper-1

Vidyamandir Classes

(C) equal number of sp2 and sp3 hybridized boron atoms
(D) one terminal hydroxide per boron atom.

27. The reagent(s) that can selectively precipitate S 2 from a mixture of S 2 and SO42 in aqueous solution is (are)

(A) CuCl2 (B) BaCl2

(C) Pb(OOCCH3)2 (D) Na2 FeCN5 NO

27.(A) By using CuCl2, S 2 can be precipitated selectively from a mixture of S2 and SO42

28. A plot of the number of neutrons (N) against the number of protons (P) of stable nuclei exhibits upward deviation

from linearity for atomic number, Z > 20. For an unstable nucleus having N/P ratio less than 1, the possible

mode(s) of decay is(are) :

(A)   decay  emission (B) orbital or K-electron capture

(C) neutron emission (D)   decay positron emission

28.(BD) N/P can be increased by  decay and K electron capture. In both processes number of protons decreases and
number of protons increases.

29. Positive Tollen’s test is observed for

(A) (B) (C) (D)

29.(ABC) Aldehyde gives positive Tollen's test. JEE Entrance Exam-2016/Advanced

VMC/Paper-1 15

Vidyamandir Classes

 -Hydroxy ketones are also oxidized by mild oxidizing agent including Tollen's reagent.

30. The compound(s) with TWO lone pairs of electrons on the central atom is(are) :

(A) BrF5 (B) ClF3 (C) XeF4 (D) SF4

31. According to the Arrhenius equation,

(A) a high activation energy usually implies a fast reaction.

(B) rate constant increases with increase in temperature. This is due to a greater `

number of collisions whose energy exceeds the activation energy.

(C) higher the magnitude of activation energy, stronger is the temperature dependence of the rate

constant.

(D) the pre-exponential factor is a measure of the rate at which collisions occur, irrespective of

their energy

31.(BCD) Ea


k  Ae RT

High activation energy means lower value of k and slow reaction.

On increasing temperature energy of particles increases hence greater number of collisions occur whose energy

exceeds the activation energy.

Rate of increase of k with temperature is higher when Ea has a large value

Pre-exponential factor is a measure of frequency of collisions.

SECTION III (Maximum Marks : 15)
 This section contains FIVE questions.
 The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive.
 For each question, darken the bubble corresponding to the correct integer in the ORS.
 For each question, marks will be awarded in one of the following categories:

Full Marks : +3 If only the bubble corresponding to the correct answer is darkened.
Zero Marks : 0 In all other cases.

32. In the following monobromination reaction, the number of possible chiral products is _________.

32.(5) Total five products are formed. 16 JEE Entrance Exam-2016/Advanced

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33. The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its
molality. Density of this solution at 298 K is 2.0 g cm–3. The ratio of the molecular weights of the solute

and solvent,  MWsolute  is_________.
 MWsolvent 
 

33.(9) m  B 1000  m  1000 . . . .(i)
34. A  MA 9M A

Also using,

m  1000M  MM B  1000 . . . .(ii)
1000d  MM B

1000
1

1000 2  MMB

Using (i) and (ii) : M B  9 (Molality = molarity)
MA

The number of geometric isomers possible for the complex  CoL2Cl2  L  H2 NCH2CH2O

is_________.
34.(5) Total five isomers are possible

35. In neutral or faintly alkaline solution, 8 moles of permanganate anion quantitatively oxidize thiosulphate
anions to produce X moles of a sulphur containing product. The magnitude of X is_________.

VMC/Paper-1 17 JEE Entrance Exam-2016/Advanced

Vidyamandir Classes

36. The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The
36.(4) absolute temperature of an ideal gas is increased 4 times and its pressure is increased 2 times. As a result,
the diffusion coefficient of this gas increases x times. The value of x is _________.
Diffusion co-efficient  .Cmean
T

P
Cmean  T

Diffusion co-efficient  T
Diffusion co-efficient  T T

P
Diffusion co-efficient  T 3/ 2

P
If T is increased four time and pressure is increased two times diffusion co-efficient should become 4 times.

PART-III MATHEMATICS

SECTION I (Maximum Marks : 15)

 This section contains Five questions.

 Each question has Four options (A), (B), (C) and (D). ONLY ONE of these four options is correct.

 For each question, darken the bubble corresponding to the correct option in the ORS.

 For each question, marks will be awarded in one of the following categories :

Full Marks : +3 If, only the bubble corresponding to the correct option is darkened.

Zero Marks : 0 If none of the bubbles is darkened.

Negative Marks : –1 In all other cases.

37. A computer producing factory has only two plants T1 and T2 . Plant T1 produces 20% and plant T2 produces

80% of the total computers produced. 7% of computers produced in the factory turn out to be defective. It is

known that P (computer turns out to be defective given that it is produced in plant T1 ) = 10 P (computer turns

out to be defective given that it is produced in plant T2 ).

Where P(E) denotes the probability of an event E. A computer produced in the factory is randomly selected and it

does not turn out to be defective. Then the probability that it is produced in plant T2 is

(A) 36 (B) 47 (C) 78 (D) 75
73 79 93 83

VMC/Paper-1 18 JEE Entrance Exam-2016/Advanced

Vidyamandir Classes

37.(C)

P [Defective] = 0.07

 [Non – defective] = 0.93

Let P[Defective from T1] = x

 P [Defective from T2] = x
10

Figure

A.T.P. 0.2(1  x)  0.8 1 x   0.93
10 

 0.07 = 0.28 x  x 1
4

0.8  1 x  8  39 78
10  10 40 93
 P [Non-defective from T2] =  =
0.93 93

100

38. A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the

selection of a captain (from among these 4 members) for the team. If the team has to include at most one boy,

then the number of ways of selecting the team is

(A) 380 (B) 320 (C) 260 (D) 95

38.(A) 6 Girls 4 Boys

Case-I 4 Girls

Select 4 Girls from 6 in 6 C4 ways.

Select captain in 4 C1 ways

 by FPC, 6 C4  4C1  60

Case-II 3 Girls, 1 Boy

Select 3 Girls from 6 in 6 C3 & 1 Boy from 4 in 4 C 1

Select captain in 4 C1

 by FPC, 6 C3  4C1  4C1  320
 Total = 320 + 60 = 380.

39. The least value of   R for which 4x2  1  1 , for all x > 0, is :
x

(A) 1 (B) 1 (C) 1 (D) 1
64 32 27 25

39.(C) 4h3  h 1  0 x0

f  x  12n2 1  0

VMC/Paper-1 19 JEE Entrance Exam-2016/Advanced

Vidyamandir Classes

x2  1
12 

x 1
12

1 31
 12  2 1  2
4      12 
1  0

4 4  4 1  0
31 11
12 2 12 2 12 2  2

31

4 12 12 2  2
0

33

12 2  2

31

8 12 2  2  0

31

12 2  2  0

1 8

 2 3

12 2

  64 12
144

 4  1
9 12 27

40. Let       . Suppose 1 and 1 are the roots of the equation x2  2xsec   1  0 and 2 and 2 are the
6 12

roots of the equation x2  2x tan  1  0 . If 1 > 1 and 2 > 2 , then 1 + 2 equals :

(A) 2(sec   tan ) (B) 2 sec 

(C) 2 tan  (D) 0

40.(C) For 1 & 1

Roots are 2 sec   2 tan   sec   tan 
2

   sec   0
6 12

tan   0

 1  1

 1  sec   tan  ; 1  sec   tan 

For 2 & 2

Roots are 2 tan   2sec   sec   tan  &  sec   tan 
2

 2  2  2  sec   tan  ; 2   sec   tan 

 1  2  2 tan 

41. Let S  x   ,  : x  0,   . The sum of all distinct solutions of the equation
 
2 

3 sec x  cos ec x  2tan x  cot x  0 in the set S is equal to :

(A)  7 (B)  2 (C) 0 (D) 5
9 9 9

VMC/Paper-1 20 JEE Entrance Exam-2016/Advanced

Vidyamandir Classes

41.(C) 3 1 2 cos 2x

 0
cos x sin x sin x cos x

 3 sin x  cos x  2 cos 2x  cos  x     cos 2x
 3 

 x    2n  2x x    2n  2x
3 3

 x     2n x  2n  
3 39

Possible values    , 7 , 5
3 99 9

 sum  3    7  5  0
9

SECTION II (Maximum Marks : 32)

 This section contains EIGHT questions.

 Each question has FOUR options options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s)

is(are) correct.

 For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS.

 For each question, marks will be awarded in one of the following categories:

Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened.

Partial Marks : +1 For darkening a bubble corresponding to each correct option, provided NO

incorrect option is darkened.

Zero Marks : 0 If none of the bubbles is darkened.

Negative Marks : –2 In all other cases.

For example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will result in +4

marks; darkening only (A) and (D) will result in +2 marks; and darkening (A) and (B) will result in -2 marks, as a

wrong option is also darkened.

42. Let f : 0,   R be a differential function such that f  x  2  f  x for all x  0,  and f 1  1 .
Then:
x

(A) lim f   1   1 (B) lim xf  1   2
x   x 
x  0 x  0

(C) lim x2 f  x  0 (D) f  x  2 for all x  0, 2
x  0

42.(A) f : 0,   R

f ' x  2  f  x  x 0,, f 1  1

x

dy  1 y  2
dx x

e 1 dx
x
If   x

xy  2x dx  c  x2  c

y  x c
x

x  1, y  1  c  0

VMC/Paper-1 21 JEE Entrance Exam-2016/Advanced

Vidyamandir Classes

(i) f 'x 1 c ; f '  1   1 cx2
x2  x 

lim f '  1  1
x 
x0

(ii) xf  1   x  1  cx  1 cx 2
 x   x 

lim xf  1   1
 x 
x  0

(iii) x2 f ' x  x2  c

lim x2 f ' x  c  0

x0

(iv) | f (x) |  x  c , x  0  x  c
xx

Let c = 2 f (x)  x  2
x

Let x = 1 f (x)  3 contradicts f  x  2

43. The circle C1 : x2  y2  3 , with centre at O, intersects the parabola x2  2 y at the point P in the first quadrant.

Let the tangent to the circle C1 at P touches other two circles C2 and C3 at R2 and R3 , respectively. Suppose

C2 and C3 have equal radii 2 3 and centres Q2 and Q3 respectively. If Q2 and Q3 lie on the y-axis, then :

(A) Q2Q3  12 (B) R2R3  4 6

(C) Area of the triangle OR2R3 is 6 2 (D) Area of the triangle PQ2Q3 is 4 2
43.(ABC)

2y  y2  3 22 JEE Entrance Exam-2016/Advanced
y2  2y 3  0
(y  3)( y 1)  0
y  1,  3
P( 2, 1)
Tangent at P is 2x  y  3
Let Q2 or Q3 (0, k)

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k3  2 3
3

x  3  6  9 or  3

Q2 0,  3, Q3 0,9

  Q2Q3  12 2
 R2R3  122  4 144  48  96
3

Area of OR2 R3  1 4 6 3 6 2 squints
2 3

Area of PQ2Q3  1 12  2 6 2
2

 44. A solution curve of the differential equation x2  xy  4x  2y  4 dy  y2  0 x  0 , passes through the point
dx

(1, 3). Then the solution curve :

(A) intersects y = x + 2 exactly at one point (B) intersects y= x + 2 exactly at two points

(C) Intersects y   x  22 (D) does NOT intersect y   x  32

44.(AD) Let x+2=X x>0

y=Y  X>2

(x  2)2  y (x  2)  dx  X 2  YX  dX   X 2   X   dX
y2 dy Y 2 dY  Y   Y  dY

Let X  tY  dX  t  Y dt
dY dY

 t2  t  t  Y dt  Ydt  t2  dt  dX
dY dY t2 Y

 1  log Y  c   Y  log Y  c …… (i)
tX

(i) passes through (3, 3)

 1  log 3  c  c  log  1   log 3  c  log  1 
 e   3e 

 curve is

Y  log Y  log  1 
X  3e 

or Y  log  Y  …… (ii)
X  3e 

Now option (A)

Y=X using in (ii)

 1  log  Y   Y 3  x=3
 3e 

 at one point.
Option (C) is y = x2

Replace in (ii)

 X2  log  X2   X 2  3e . e X
X  3e 

But x > 2 LHS > 4

RHS  3
e

 No solution.

VMC/Paper-1 23 JEE Entrance Exam-2016/Advanced

Vidyamandir Classes

Option (D) is Y  ( X  1)2

Replace in (ii)

 ( X 1)2  ( X 1)2  1)2  3e . e X 1  2 
X  3e  X 
 log  (X

X>2
LHS > 9
RHS < 3 . e7 / 2

 Not intersects.
45. In a triangle XYZ , let x, y, z be the lengths of sides opposite to the angles X, Y, Z, respectively, and

2s  x  y  z . If s  x  s  y  s  z and area of incircle of the triangle XYZ is 8 , then :
43 2 3

(A) area of the triangle XYZ is 6 6 (B) the radius of circumcircle of the triangle XYZ is 35 6
6

(C) sin X sin Y sin Z  4 (D) sin2  X  Y  3
2 2 2 35  2  5

45.(ACD) r2  8
3

s  x  s  y  s  z  3s   x  y  z   s
432 99

9s  9x  4s,3s  3y  s,9s  9z  2s

5s  9x, 2s  3y,7s  9z

 s  s  5s   s  2s   s  7s   s  4s  s  2s  2 2 s2
 9   3   9  9 3 9 93

r2 2 2 2s
3 93

s  9  x  5, y  6, z  7

  2 2 81  6 6
93

R  xyz  5 6 7  35  35 6
4 4 6 6 4 6 24

r  4R sin x sin y sin z
222

22

 sin X sin Y sin Z  3  2 2  6  4
2 2 2 35 6 3 35 35

6

2 sin2  X Y   1  cos  X Y  1 cos Z 1 x2  y2  z2  1 25  36  49  1 12  6
 2  2xy 60 60 5

sin 2  X Y   3
 2  5

46. Let RS be the diameter of the circle x2  y2 1 , where S is the point (1, 0). Let P be a variable point (other than

R and S) on the circle and tangents to the circle at S and P meet at the point Q. The normal to the circle at P
intersects a line drawn through Q parallel to RS at point E. Then the locus of E passes through the point(s).

(A)  1  1 (B)  1 1 (C)  1  1  (D)  1   1 
 3   4 2   3   4 2 
 3   3 

VMC/Paper-1 24 JEE Entrance Exam-2016/Advanced

Vidyamandir Classes

46.(AC) S  1 0 R  1 0

P  cos sin 

Tangent at P:
x cos   y sin  1

Tangent at S : x = 1

Solving Q  1 tan 
2 

Normal at P :

x sin   y cos   0

Line through Q and  to RS

y  tan 
2

Solving E  1  1  sec2  tan 2   h k
2 2 

 Locus : h  1 1 k2  1
2

 or x  1 1 y2  1 or x  y2  1  2x  y2  1
2 22

Options (A) and (C) satisfy the locus

3 1 2

47. Let P  2 0   , where    . Suppose Q  qij  is a matrix such that PQ = kI, where k   k  0 and


3 5 0 

I the identity matrix of order 3. If q23  k and det Q  k2 , then :
8 2

(A)   0 k  8 (B) 4  k  8  0 (C) det  P adj Q  29 (D) det Q adj P   213

47.(BC) Q  KP1

 5 10  

 3 6 (3  4)


P1  10 12 2 

20 12

q23   k
8

k.(3  4)  k
20 12 8

  1

|Q|  k2 |P| 8 |Q| 8
2

| kP1 |  k2
2

k3  k2
|P| 2

k3 k2


20 12 2

k=4

4  k  8  0 Ans. B

| Padj  Q |  | P || Q |2  23  26  29 Ans. C

VMC/Paper-1 25 JEE Entrance Exam-2016/Advanced

Vidyamandir Classes

48. Let f     g     and h     be differentiable functions such that f  x  x3  3x  2 ,

g  f  x  x and h g  g  x  x for all x  . Then :

(A) g2  1 (B) h1  666 (C) h 0  16 (D) h  g 3  36

15

48.(BC) f  x  3x2  3 0 x

g  x  f 1  x

 g  g  x  f 1 f 1  x  f 1 of 1

gx  1 1
3
f 0

3

2 3
   hx  fof x 
x3  3x  x3  3x  2 2

     h x  3 x3  3x  2 2 3x2  3  3 3x2  3

h1  3366  18  666
h0  8  6  2  16

  h g 3  f f f 1 3  f 3  27  9  2  38

49. Consider a pyramid OPQRS located in the first octant (x  0, y  0, z  0) with O as origin, and OP and OR

along the x-axis and the y-axis, respectively. The base OPQR of the pyramid is a square with OP = 3. The point
S is directly above the mid-point T of diagonal OQ such that TS = 3. Then :
(A) the acute angle between OQ and OS is 

3
(B) the equation of the plane containing the triangle OQS is x  y  0

(C) the length of the perpendicular from P to the plane containing the triangle OQS is 3
2

(D) the perpendicular distance from O to the straight line containing RS is 15
49.(BCD) 2

(A) cos   OQ  OS 26 JEE Entrance Exam-2016/Advanced
OQ OS

cos   1
3

  cos1 1
3

VMC/Paper-1

Vidyamandir Classes

x yz
(B) 3 3 0  0  x  y  0

3 33
22
(C) 3  0  3

22
(D)

RM  OR  RS
RS

RM  3 …… (i)
2 …… (ii)

OR  3

 OM  OR2  RM 2

OM  15
2

SECTION III (Maximum Marks : 15)
 This section contains FIVE questions.
 The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive.
 For each question, darken the bubble corresponding to the correct integer in the ORS.
 For each question, marks will be awarded in one of the following categories:

Full Marks : +3 If only the bubble corresponding to the correct answer is darkened.
Zero Marks : 0 In all other cases.

50. Let z  1  3 i , where i  1, and r, s  1, 2, 3 . Let P (z)r z2s 
  and I be the identity matrix of order
2  z2s zr 

2. Then the total number of ordered pairs (r, s) for which P2  I is __________.

50.(1) z   1 3i
2

z 

 r  2 s
 P   

 s 

2 r 


P2  I

VMC/Paper-1 27 JEE Entrance Exam-2016/Advanced

Vidyamandir Classes

 r  2 s  r  2 s  1 0
  1

 2 s  r  2 s r   0
   
   


 r  4s  r  2s  1  1r  1 0
  1

 r  2s 1  1r   0
4s  2r  



 2r  4s  1 and r  2s 1  1r  0 r can be 1 or 3

If r  1

2  4s  1

4s  1 2

s    s  1

If r  3

6  4s   1

4s   2

No “s” possible

Ordered pair of r, s is 1, 1

51. Let m be the smallest positive integer such that the coefficient of x2 in the expansion of
51.(5)
(1  x)2  (1  x)3  ...  (1  x )49  (1  mx)50 is (3n  1) 51C3 for some positive integer n. Then the value of n

is __________.
S = (1 + x)2 + (1 + x)3 + ….. + (1 + x)49 + (1 + mx)50

S  1  x2 (1  x)48  1  (1  mx)50

x

S  1  x50  1  x2  (1  mx)50

xx
Coefficient of x2

 Coeff . of x3 in (1  x)50  0  coeff . of x2 in (1  mx)50

 50C3  50C2 m2  (3n 1) 51C3 .

50 C3  50 C2 m2  3n  1
51 C3 51C3

50  49  48  50  49 6 m2  3n 1
51  50  49 2 51  50 49

16  1 m2  3n  1
17 17
m2 1

 3n 
17 17
m2 1  51n

m2  51n 1

n5

x 0, 1 x t2
52. 0 1 t4  2x 1
The total number of distinct for which dt is __________.

VMC/Paper-1 28 JEE Entrance Exam-2016/Advanced

Vidyamandir Classes

x x t2
52.(1) Letf   t4 dt  2x  1 . . . . .(i)
01

f x  x2 2
 x4
1

Now as x2  1  2  1 1
x2 1 2
x2  x2

 f x  0  f  x is decreasing function

Now f 01 . . . . (ii)

1  1 t2
andf 0  t4 1 . . . . (iii)

1

t2  1   1 t2 1
1 t4 2 01  t4
1
2

0

 1 t2 1  1  1 . . . . .(iv)
0 1 t4 2

 f 1  0 . . . . . (v)

Therefore f  x will cross X-axis exactly at one point between [0, 1]

x x2 1 x3
53. The total number of distinct x  R for which 2x 4x2 1 8x3  10 is __________.

3x 9x2 1 27x3

53.(2) x x2 1 x3
2x 4x2 1 8x3  10
3x 9x 1  27x3

x x2 1 x x2 x3
2x 4x2 1  2x 4x2 8x3  10
3x 9x2 1 3x 9x2 27x3

1 x x2 1 x x2

1 2x 4x2  x  2x  3x 1 2x 4x2  10

1 3x 9x2 1 3x 9x2

 x  x  2x 1  6x3  10
 2x3 1  6x3  10
 x3 1 6x3  5

x3 t
t  6t 2 5
6t2  t 5  0
6t2  6t  5t  5  0

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Vidyamandir Classes

6t t 1  5t  1 0

t  1t  5 / 6

x3   1 x3  5 / 6

54. x  1 and x  5 / 61/ 3
54.(7)
Let ,   be such that lim x2 sin(x)  1 . Then 6(  ) equals __________.
x0 x  sin x

x2  sin x

lim  1
x0 x  sin x

x3
lim  1

x0 x  sin x

 1

6  1

 1
6

6    6 1  1   7
6 

VMC/Paper-1 30 JEE Entrance Exam-2016/Advanced