2017 Code-b-solution.pdf

Test Booklet Code

B Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.: 011-47623456 Fax : 011-47623472

Time : 3 hrs.

Answers & Solutions for

M.M. : 360

JEE (MAIN)-2017 (Physics, Mathematics and Chemistry)

Important Instructions : 1.

The test is of 3 hours duration.

2.

The Test Booklet consists of 90 questions. The maximum marks are 360.

3.

There are three parts in the question paper A, B, C consisting of Physics, Mathematics and Chemistry having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for each correct response.

4.

Candidates will be awarded marks as stated above in Instructions No. 3 for correct response of each question. ¼ (one-fourth) marks of the total marks allotted to the question (i.e. 1 mark) will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet.

5.

There is only one correct response for each question. Filling up more than one response in any question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 4 above.

6.

For writing particulars/marking responses on Side-1 and Side-2 of the Answer Sheet use only Black BallPoint Pen provided in the examination hall.

7.

No candidate is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any electronic device, etc. except the Admit Card inside the examination hall/room.

1

JEE (MAIN)-2017 (Code-B)

PART–A : PHYSICS 1.

A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like :

Answer (2) Sol. n1 = initial number of moles; 5

n1 =

(1)

PV 10  30 3 1 1   1.24  10 RT1 8.3  290

n2 = final number of moles

0

T 2

T

5

PV 10  30 3  1.20  10 = 2 2  RT2 8.3  300

Change of number of molecules (2)

0

T 4

T 2

nf – ni = (n2 – n1) × 6.023 × 1023

T

 – 2.5 × 1025 3.

(3)

0

(4)

0

T 2

T

Which of the following statements is false? (1) A rheostat can be used as a potential divider (2) Kirchhoff’s second law represents energy conservation (3) Wheatstone bridge is the most sensitive when all the four resistances are of the same order of magnitude (4) In a balanced Wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed Answer (4) Sol. In a balanced Wheatstone bridge, the null point remains unchanged even if cell and galvanometer are interchanged. 4. The following observations were taken for determining surface tension T of water by capillary method: diameter of capillary, D = 1.25 × 10–2 m rise of water, h = 1.45 × 10–2 m. Using g = 9.80 m/s 2 and the simplified relation

T

T

Answer (2) Sol. K.E 

1 m2 A2 cos2 t 2

T 4 2.

rhg  103 N / m , the possible error in surface 2 tension is closest to T 

T 2

The temperature of an open room of volume 30 m3 increases from 17°C to 27°C due to the sunshine. The atmospheric pressure in the room remains 1 × 105 Pa. If ni and nf are the number of molecules in the room before and after heating, then nf – ni will be

(1) (2) (3) (4) Answer

(1) 2.5 × 1025 Sol.

(2) –2.5 × 1025

T D h  100   100   100 T D h

(3) –1.61 × 1023 =

(4) 1.38 × 1023 2

2.4% 10% 0.15% 1.5% (4)

0.01 0.01  100   100 1.25 1.45

JEE (MAIN)-2017 (Code-B)

7.

= 1.489

The moment of inertia of a uniform cylinder of length  and radius R about its perpendicular bisector is I.  What is the ratio such that the moment of inertia R is minimum?

 1.5%

(1) 1

100 100  = 125 145

= 0.8 + 0.689

5.

In amplitude modulation, sinusoidal carrier frequency

3

(2)

used is denoted by c and the signal frequency is

2

denoted by m . The bandwidth ( m ) of the signal is such that m  c . Which of the following frequencies is not contained in the modulated wave? (1) m  c

(2) c – m

(3) m

(4) c

(3)

3 2

(4)

3 2

Answer (3)

Answer (3) Sol. Modulated wave has frequency range. c ± m

Sol. 2R

 Since c >> m

l

 m is excluded. 6.

A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is (1) Real and at a distance of 40 cm from the divergent lens (2) Real and at a distance of 6 cm from the convergent lens

I

mR 2 m2  4 12

I

m ⎡ 2 2 ⎤ ⎢R  ⎥ 4 ⎣⎢ 3 ⎥⎦



(3) Real and at a distance of 40 cm from convergent lens

m ⎡ v 2 ⎤ ⎢  ⎥ 4 ⎣⎢  3 ⎦⎥

dI m ⎡ v 2 ⎤   0 d  4 ⎢⎣ 2 3 ⎥⎦

(4) Virtual and at a distance of 40 cm from convergent lens

v

Answer (3) f1 = 25 cm



f2 = 20 cm

Sol.



2

v

2 3

23 3

I1

R 2   25 cm

15 cm

2 R

For converging lens u = –40 cm which is equal to 2f

2



  R

 Image will be real and at a distance of 40 cm from convergent lens. 3

3 2 3 2

23 3

JEE (MAIN)-2017 (Code-B)

8.

An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays. It produces continuous as well as characteristic X-rays. If λmin is the smallest possible wavelength of X-ray in the spectrum, the variation of log λmin with log V is correctly represented in

9.

A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by (1) t  T log 1.3  T (2) t  log 1.3 

(1) log min

log V

T log2 (3) t  2 log1.3

(2) log min

log1.3 (4) t  T log2

log V

Answer (4) (3) log min

Sol.

log V

N0  N0 e t  0.3 N0 e t

 et = 1.3  t = ln 1.3

(4) log min

⎛ ln 2 ⎞ ⎜ T ⎟ t  ln 1.3 ⎝ ⎠

log V Answer (3)

t  T.

Sol. In X-ray tube min 

hc eV

t T

⎛ hc ⎞ ln min  ln ⎜ ⎟  lnV ⎝ e ⎠

ln(1.3) ln 2

log(1.3) log2

 10. An electric dipole has a fixed dipole moment p , which makes angle  with respect to x-axis. When  subjected to an electric field E 1  Ei , it experiences  a torque T 1   k . When subjected to another electric    field E 2  3E1j it experiences a torque T 2  T 1 . The angle  is

Slope is negative Intercept on y-axis is positive

log min

(1) 60° (2) 90° (3) 30°

log V

(4) 45° 4

JEE (MAIN)-2017 (Code-B)

Answer (1) Sol. y

Answer (1) Sol. Let molar heat capacity at constant pressure = Xp and molar heat capacity at constant volume = Xv

p



Xp – Xv = R

x

MCp – MCv = R

R M

z  p  p cos i  p sin j

Cp – Cv =

 E 1  Ei

For hydrogen; a =

   T 1  p  E1

For N2; b =



 

a = 14b …(i)

13. A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by :

 E 2  3E1j  T2  ( p cos i  p sin j )  3E1j  k  3 pE1 cos  k

R 28

a = 14 b

= ( p cos i  p sin j )  E i

 k  pE sin   k

R 2

(Given : room temperature = 30°C, specific heat of copper = 0.1 cal/gm°C)

…(ii)

From (i) and (ii)

pE sin   3 pE cos 

(1) 1250°C

(2) 825°C

(3) 800°C

(4) 885°C

tan   3

Answer (4)

 = 60°

Sol. 100 × 0.1 × (t – 75) = 100 × 0.1 × 45 + 170 × 1 × 45

11. In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be (1) 135°

(2) 180°

(3) 45°

(4) 90°

10t – 750 = 450 + 7650 10t = 1200 + 7650 10t = 8850 t = 885°C 14. A body of mass m = 10–2 kg is moving in a medium and experiences a frictional force F = –kv2. Its initial speed is v0 = 10 ms–1. If, after 10 s, its energy is

Answer (2) Sol. In common emitter configuration for n-p-n transistor, phase difference between output and input voltage is 180°.

1 mv 02 , the value of k will be 8

12. Cp and Cv are specific heats at constant pressure and constant volume respectively. It is observed that Cp – Cv = a for hydrogen gas Cp – Cv = b for nitrogen gas

(1) 10–4 kg m–1

(2) 10–1 kg m–1 s–1

(3) 10–3 kg m–1

(4) 10–3 kg s–1

Answer (1)

The correct relation between a and b is : (1) a = 14b (3) a 

1 b 14

1 mv 02 kf 1 8   Sol. 1 ki 4 mv 02 2

(2) a = 28b (4) a = b 5

JEE (MAIN)-2017 (Code-B)

16. A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle  with the vertical is

vf 1  vi 2

vf 

v0 2

 kv 2  v0 2

dv

mdv dt t0

∫ v2 ∫ 

v0

0

z

 kdt m

 x

v0

k ⎡ 1⎤ 2 ⎢  v ⎥  m t0 ⎣ ⎦v 0

1 2 k    t0 v0 v0 m 

1 k   t0 v0 m

k



m v 0 t0

(1)

3g cos  2

(2)

2g cos  3

(3)

3g sin  2

(4)

2g sin  3

Answer (3) Sol. Torque at angle 

102 10  10

  Mg sin  

= 10–4 kg m–1 15. When a current of 5 mA is passed through a galvanometer having a coil of resistance 15 Ω, it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0-10 V is (1) 2.535 ×

103

 2

, 

(M, l) 

Ω

Mg

(2) 4.005 × 103 Ω (3) 1.985 × 103 Ω

= I

(4) 2.045 × 103 Ω

I = Mg sin 

Answer (3)

 2

M 2     Mg sin  3 2

Sol. ig = 5 × 10–3 A G = 15 Ω

 sin  g 3 2

Let series resistance be R. V = ig (R + G) 10 = 5 × 10–3 (R + 15)



R = 2000 – 15 = 1985 = 1.985 × 103 Ω 6

3g sin  2

 I 

M 2 3

JEE (MAIN)-2017 (Code-B)

17. Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths r = 1/2, is given by

–E 

4 E 3

Stress =

2 ⎛ mf ⎞ ⎛ Ai ⎞  1 ⎜⎝ mi ⎟⎠ ⎜⎝ Af ⎟⎠

2



1

–3 E 3 4

4 (3) r  3

93 92

(2) r 

1 3

10 Current (amp.)

2 (4) r  3

Answer (2) (1) 250 Wb

0.5 s Time (2) 275 Wb

(3) 200 Wb

(4) 225 Wb

Sol. From energy level diagram hc E

1 



Answer (1)

hc ⎛E ⎞ ⎜3⎟ ⎝ ⎠

2 

Sol. ε =

1 1  2 3

(3) 9

(2) (4)

d dt

∫ d   R ∫ idt Magnitude of change in flux = R × area under current vs time graph

1 81

= 100 ×

1 9

1 1 × × 10 2 2

= 250 Wb

Answer (3) Sol.

d dt

iR =

18. A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor of (1) 81

9

19. In a coil of resistance 100 Ω, a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is

–2 E

(1) r 

(Mass)  g Area

20. In a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is

vf  93 vi

∵ Density remains same So, mass  Volume

mf  93 mi (Area)f  92 (Area)i 7

(1) 9.75 mm

(2) 15.6 mm

(3) 1.56 mm

(4) 7.8 mm

JEE (MAIN)-2017 (Code-B)

Answer (4) Sol. For 1

g

For 2

y

m1D d



m 2 4   n 1 5

y

(3)

n 2 D d

d O

g (4)

For 1

d R

O

m1D y , 1 = 650 nm d

Answer (2)

= 7.8 mm

g

21. A magnetic needle of magnetic moment 6.7 × 10–2 Am2 and moment of inertia 7.5 × 10–6 kg m2 is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is (1) 6.98 s

(2) 8.76 s

(3) 6.65 s

(4) 8.89 s

Sol. gs

O

d

d=R

Variation of g inside earth surface

Answer (3) d Rg 

I Sol. T = 2π MB

Gm R2

d  R  gs 

2  1.06 = 2π = –2 10 6.7  10  0.01

7.5  10 –6

d Rg 

For 10 oscillations,

d

Gm R2

Gm d2

t = 10T = 2π × 1.06 23.

= 6.6568 ≈ 6.65 s

2V

22. The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth's radius) :

2V

R

g

2V

2V

(1) 0.5 A

(2) 0 A

(3) 1 A

(4) 0.25 A

Answer (2)

(2)

Sol. The potential difference in each loop is zero.

d O

1

In the above circuit the current in each resistance is

d O

2V 1

1

g (1)

2V

∴ No current will flow.

R 8

JEE (MAIN)-2017 (Code-B)

24. A particle A of mass m and initial velocity v collides m with a particle B of mass which is at rest. The 2 collision is head on, and elastic. The ratio of the de-Broglie wavelengths A to B after the collision is A 2  B 3

(1)

(2)

V P = V K  V = V0 (1 + t) V V0 = t

A 1  B 2

 A 1  B 3

(3)

(4)

A 2 B m1  m

(m1  m2 )v 0 m1  m2

=

m2 

v 3

m 2

v2 

(3) 4.5 J

(4) 22 J

dv dt

v

∫ dv  ∫ 6t dt

2m1v 0 m1  m2

0

1

⎡t2 ⎤ v  6⎢ ⎥ ⎣2⎦

0

= 3 ms–1

m ⎡ 4v ⎤ 2mv  2 ⎢⎣ 3 ⎥⎦ 3

 de-Broglie wavelength

W = KE 

 A p2   2 :1 B p1

1  1  9  4.5 J 2

27. An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light = 3 × 108 ms–1)

25. An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and  is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by :

(1) 17.3 GHz

(2) 15.3 GHz

(3) 10.1 GHz

(4) 12.1 GHz

Answer (1)

(1)

3 PK

(2) 3PK

(3)

P 3 K

(4)

Sol. For relativistic motion

P K

f = f0

Answer (3) Sol. K =

(2) 18 J

Sol. 6t  1

4v = 3

p2 

(1) 9 J

Answer (3)

⎡v ⎤ p1  m. ⎢ ⎥ ⎣3 ⎦



P P P = t  t = = K 3 K K

26. A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 second will be

Answer (4) Sol. v1 

 = 3

c v c v

;

v = relative speed of approach

c 2  10 3  17.3 GHz f = 10 c c 2 c

P ⎛ V ⎞ ⎜– V ⎟ ⎝ ⎠

9

JEE (MAIN)-2017 (Code-B)

28. In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be :

E

1 2 3 8

r

1 2 3 8

r1

8 F 8 F 8 F 8 F

r2

250 V 250 V 250 V 250 V

r1 (2) CE (r1  r )

(4) CE

(3) CE

1000 V 30. A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?

r1 (r2  r )

v

Answer (1) Sol. In steady state, flow of current through capacitor will be zero.

E

t

(1)

r

i

r1

v

C

E r  r2

v

Er2C VC = i r2C = r  r 2

VC = CE

t

(2)

r2 i=

1 2 3 8

1000 V

C

r2 (1) CE (r  r2 )

1 2 3 8

(3)

r2 r  r2

t

v

29. A capacitance of 2 F is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 F capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is (1) 24

(2) 32

(3) 2

(4) 16

t

(4)

Answer (1) Sol. Acceleration is constant and negative

v

Answer (2)

t

Sol. Following arrangement will do the needful : 8 capacitors of 1F in parallel with four such branches in series. 10

JEE (MAIN)-2017 (Code-B)

PART–B : MATHEMATICS 31. Let k be an integer such that the triangle with vertices (k, –3k), (5, k) and (–k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point 1⎞ ⎛ (1) ⎜ 2, ⎟ ⎝ 2⎠

1⎞ ⎛ (2) ⎜ 2,  ⎟ ⎝ 2⎠

⎛ 3⎞ (3) ⎜⎝ 1, ⎟⎠ 4

3⎞ ⎛ (4) ⎜ 1,  ⎟ ⎝ 4⎠

2y  4  x  5 x  2y  1  0

Solving (i) & (ii), 2y = 1

y

Sol. Area =

1 5 2 k

32. If, for a positive integer n, the quadratic equation,

3k 1

x( x  1)  ( x  1)( x  2)  ...  ( x  n  1)( x  n )  10n

1  28 1

k 2

1 2

⎛ 1⎞ Orthocentre is ⎜ 2, ⎟ ⎝ 2⎠

Answer (1) k

...(ii)

has two consecutive integral solutions, then n is equal to

k  5 4k 0 5  k k  2 0   56 k 2 1

Answer (1)

(k 2  7k  10)  4k 2  20k   56

Sol. Rearranging equation, we get

(2) 12

(3) 9

(4) 10

nx 2  1  3  5  ....  (2n  1) x

5k 2  13k  10   56 5k 2  13k  46  0

 1 2  2  3  ...  ( n  1)n  10n

5K 2  13K  66  0 ⇒ nx 2  n 2 x 

5k 2  13k  46  0

k =

(1) 11

13  169  920 10

(n  1)n(n  1)  10n 3

⎛ n 2  31 ⎞ ⇒ x 2  nx  ⎜ ⎟0 ⎝ 3 ⎠

= 2, –4.6 reject

Given difference of roots = 1  || = 1

For k = 2

 D1

A (2, –6) =

8

 n2 

m

E

(5, 2) B

D

m= –2

So, n = 11 33. The

C (–2, 2)

m=0

f (x) 

Equation of AD, x=2

4 2 (n  31)  1 3

function x 1 x2

⎡ 1 1⎤ f : R  ⎢ , ⎥ ⎣ 2 2⎦

, is

(1) Neither injective nor surjective

...(i)

(2) Invertible

Also equation of BE,

(3) Injective but not surjective

1 y  2  ( x  5) 2

(4) Surjective but not injective 11

defined

as

JEE (MAIN)-2017 (Code-B)

Answer (4) Sol. f ( x ) 

f ( x ) 

Answer (1) Sol.

x 1 x

2

1 1 1

(1  x 2 )  1  x  2 x 2 2



1 x2

1 a 10

2 2

(1  x ) (1  x ) f (x) changes sign in different intervals.

a b 1  –(1 – a)2 = 0

 Not injective. y

 a=1

x

For a = 1

1 x2

Eq. (1) & (2) are identical i.e.,x + y + z = 1

yx 2  x  y  0

To have no solution with x + by + z = 0

For y  0

b=1 36. The area (in sq. units) of the region

⎡ 1 1⎤ D  1  4 y  0 ⇒ y  ⎢  , ⎥  {0} ⎣ 2 2⎦ 2

{(x, y) : x  0, x + y  3, x2  4y and y  1  x }

For, y = 0 x = 0  Part of range

is

⎡ 1 1⎤  Range : ⎢  , ⎥ ⎣ 2 2⎦  Surjective but not injective.

(1)

5 2

(2)

59 12

(3)

3 2

(4)

7 3

34. The following statement (p q) [(~ p q) q] is

Answer (1)

(1) A fallacy

y

(2) A tautology (3) Equivalent to ~ p q

) ,2 (1

(4) Equivalent to p ~ q Answer (2) Sol.

p

q

T T F

T F T

F

F

(2, 1)

(0, 1)

Sol. x 

p  q (~p  q) (~p  q)  q (p  q)  [(~p  q)  q] T F T

T T T

T F T

T T T

T

F

T

T

O

x=1 x=2

x+

x y=

3

x=0 y

(a tautology)

Area of shaded region 35. If S is the set of distinct values of b for which the following system of linear equations

1 2 ⎛ ⎛ x2 ⎞ x2 ⎞  ∫ ⎜ x  1 ⎟ dx  ∫ ⎜ (3  x )  ⎟ dx ⎝ 4 ⎠ ⎝ 4 ⎠ 0 1

x y z 1

5 sq. unit 2 37. For any three positive real numbers a, b and c, 

x  ay  z  1 ax  by  z  0

has no solution, then S is

9(25a2  b2 )  25(c 2  3ac )  15b(3a  c ).

(1) A singleton

Then (1) a, b and c are in G.P.

(2) An empty set

(2) b, c and a are in G.P.

(3) An infinite set

(3) b, c and a are in A.P.

(4) A finite set containing two or more elements

(4) a, b and c are in A.P. 12

JEE (MAIN)-2017 (Code-B)

Answer (3) 2

2

dy ( x  2)( x  3)  y (2 x  5)  1 dx

2

Sol. 9(25a  b )  25(c  3ac )  15b (3a  c )  (15a )2  (3b)2  (5c )2  45ab  15bc  75ac  0

dy (6)  1( 5)  1 dx

 (15a  3b)2  (3b  5c )2  (15a  5c )2  0

dy 6  1 dx 6

It is possible when

Now slope of normal = –1

15a  3b  0 and 3b  5c  0 and 15a  5c  0

Equation of normal y – 1 = –1(x – 0)

15a  3b  5c

y+x–1=0

⎛ 1 1⎞ Line (i) passes through ⎜ , ⎟ ⎝2 2⎠

a b c   1 5 3

40. A hyperbola passes through the point P ( 2, 3) and has foci at (±2, 0). Then the tangent to this hyperbola at P also passes through the point

 b, c, a are in A.P. 38. A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is (1) 484

Sol.

(4) 469

Answer (2) Sol. X(4 L 3 G)

(1) (  2,  3)

(2) (3 2, 2 3)

(3) (2 2, 3 3)

(4) ( 3, 2)

Answer (3)

(2) 485

(3) 468

x2 y 2  1 a2 b2 a2  b2  4

Y(3 L 4 G)

3L0G

0L3G

2L1G

1L2G

1L2G

2L1G

0L3G

and

4b

3L0G



4

C2  3C1

  2

2 3  2 1 2 a b

2

Required number of ways = 4C3  4C3 

... (i)

4

C1  3C2

  C  2

3

2

 x2 

39. The normal to the curve y ( x  2)( x  3)  x  6 at the point where the curve intersects the y-axis passes through the point

⎛ 1 1⎞ (3) ⎜ , ⎟ ⎝2 2⎠

⎛ 1 1⎞ (4) ⎜ ,  ⎟ ⎝2 3⎠

3 b2

1

 a2  1

3

= 485

⎛ 1 1⎞ (2) ⎜  ,  ⎟ ⎝ 2 2⎠



 b2  3

= 16 + 324 + 144 + 1

⎛ 1 1⎞ (1) ⎜ , ⎟ ⎝2 3⎠

2

y2 1 3

 Tangent at P ( 2, 3) is

2x 

y 3

1

Clearly it passes through (2 2, 3 3) 41. Let a, b, c  R. If f(x) = ax2 + bx + c is such that a + b + c = 3 and f ( x  y )  f ( x )  f ( y )  xy ,  x, y  R, 10

Answer (3) then

Sol. y ( x  2)( x  3)  x  6

∑ f (n )

is equal to

n 1

At y-axis, x = 0, y = 1

(1) 255

(2) 330

Now, on differentiation.

(3) 165

(4) 190

13

JEE (MAIN)-2017 (Code-B)

  |c a | 3

Answer (2)

    ⇒ | c |2  | a |2 2(a  c )  9

Sol. As, f ( x  y )  f ( x )  f ( y )  xy Given, f (1)  3

  932 a c  2 2 43. Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If BPC =  then tan  is

Putting, x  y  1  f (2)  2f (1)  1  7 Similarly, x  1, y  2  f (3)  f (1)  f (2)  2  12 10

Now,

∑ f (n ) =

n 1

f (1)  f (2)  f (3)  ...  f (10)

(1)

4 9

(2)

6 7

(3)

1 4

(4)

2 9

= 3 + 7 + 12 + 18 + ... = S (let) Now, S  3  7  12  18  ...  t n Again, S  3  7  12  ...  t n 1  t n

Answer (4)

We get, t n  3  4  5  ... n terms

Sol. tan  

n (n  5) = 2

=

=

So, S10 =

∑ tn

n 1

1 2

∑ n



 5∑n

Solving tan  

n (n  1)(n  8) 6 10  11 18  330 6

P

4x

A

2 9

(2) 12.5 (3) 10 (4) 25 Answer (4)

1 8

r

Sol.

25 8 (3) 2

r

(2)



r

2r  r  20

(4) 5

   ⇒ | a  b | | c | sin 30  3

 

(1) 30

     such that | c  a | 3, (a  b )  c  3 and the angle      between c and a  b be 30°. Then a  c is equal to

Answer (3)    Sol. | (a  b )  c |  3

x C x

44. Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is

   42. Let a  2iˆ  ˆj  2kˆ and b  iˆ  ˆj . Let c be a vector

(1)

1 2

1  tan  1 4  1 2 1  tan  4

 2

B

tan     

n

i.e., Sn =

1 4

  a  b  2iˆ  2 ˆj  kˆ

A = area =

   |a |  3  ab

A

 ⇒ |c | 2

14

... (i)

 r 2  r 2  2 2

r 2 ⎛ 20  2r ⎞ ⎜ ⎟ 2⎝ r ⎠

... (ii)

JEE (MAIN)-2017 (Code-B)

⎛ 20r  2r 2 ⎞ 2 A⎜ ⎟  10r  r ⎝ 2 ⎠

46. If  2  sin x 

⎛ ⎞ y ⎜ ⎟ is equal to ⎝ 2⎠

A to be maximum dA  10  2r  0 ⇒ r  5 dr

(1) d2A dr 2

(3) 

Hence for r = 5, A is maximum Now, 10 + ·5 = 20  = 2 (radian)

(2) 2 3

(4) 

Sol. (2  sin x )

is equal to

(1) –1

(2) –2

1 cos x dy  dx  0 y 1 2  sin x

(3) 2

(4) 4

dx

∫ 1  cos x  4

ln| y  1|  ln(2  sin x )  ln C

Answer (3) 3 4

∫  4

( y  1)(2  sin x )  C

1 dx  x 2 2cos2 2 dx

3 4

∫  4

Put x = 0, y = 1

x sec dx 2 2

(1  1)  2  C  C = 4 Now, ( y  1)(2  sin x )  4

3 ⎤4

x ⎡ tan ⎥ 1⎢ 2  ⎢ ⎥ 2⎢ 1 ⎥ ⎣ 2 ⎦

For, x 

 2

( y  1)(2  1)  4

4

 tan

1 3

dy  ( y  1)cos x  0 dx

⎛⎞ y (0)  1, y ⎜ ⎟  ? ⎝2⎠

45. The integral

1 3

Answer (2)

2  5 2  25 sq m 2 3 4

Sol.

4 3

 2  0

Area =

dy   y  1 cos x  0 and y(0) = 1, then dx

y 1

3   tan 8 8

⎡  1  cos ⎢  4  ⎢ tan   8 ⎢ 1  cos ⎢⎣ 4

3 1  cos 3 4  tan  3 8 1  cos 4

y

2 1 2 1



2 1 1

4 3

4 1 1 3 3

47. Let In  ∫ tann xdx,(n  1) . If

I4  I6  a tan5 x  bx 5  C, where C is a constant of integration, then the ordered pair (a, b) is equal to

⎤ ⎥ 2 1  2  1⎥ 2 1 ⎥ ⎥⎦

 ( 2  1)  ( 2  1) 2

15

⎛ 1 ⎞ (1) ⎜  ,0 ⎟ ⎝ 5 ⎠

⎛ 1 ⎞ (2) ⎜  ,1⎟ ⎝ 5 ⎠

⎛1 ⎞ (3) ⎜ ,0 ⎟ ⎝5 ⎠

⎛1 ⎞ (4) ⎜ , 1⎟ ⎝5 ⎠

JEE (MAIN)-2017 (Code-B)

49. The value of

Answer (3)

( 21C1  10C1 )  ( 21C2  10C2 )  ( 21C3  10C3 ) 

n Sol. In  ∫ tan xdx, n  1

( 21C4  10C4 )  ...  ( 21C10  10C10 ) is

I4  I6  ∫ (tan4 x  tan6 x )dx

(1) 220 – 210

 ∫ tan4 x sec 2 xdx

(2) 221 – 211 (3) 221 – 210

Let tanx = t

(4) 220 – 29

sec2x dx = dt

Answer (1)

 ∫ t 4 dt

Sol.

5

t  C 5  a

1 tan5 x  C 5



50.

1 2  1 2  3k , then k is equal to

(2) –z

(3) z

(4) –1

Answer (2) Sol. 2 + 1 = z , z  3 i

1  3i Cube root of unity. 2

1

1 2

1 1   1

2



1 1

2

1

3

1

(1)

1 4

(2)

1 24

(3)

1 16

(4)

1 8



1 



2

0 

(   2 x )3

 x 2

1

2

2

equals

cot x  cos x

Sol. lim

  1    0  2 7

(   2)3

 x 2

Answer (3)

C1  C1 + C2 + C3 1

cot x  cos x

lim

Put,



= 3 (2 – 4)

lim

 x t 2

tan t  sin t 8t 3

t 0

⎡⎛ 1  3i ⎞ ⎛ 1  3i ⎞ ⎤ ⎟⎜ ⎟⎥ = 3 ⎢⎜ 2 ⎠ ⎝ 2 ⎠⎦ ⎣⎝

= lim

sin t  2 sin2

t 0

= 3 3i = –3z

=

 k = –z 16

1 . 16

8t 3

t 2



C0  21C1  ...  21C21  1

C1  10C2  ...  10C10  210  1

7

(1) 1



21



10

1

2



= 220 – 210

where z  3 . If

1

1 2

 Required sum = (220 – 1) – (210 – 1)

48. Let  be a complex number such that 2 + 1 = z

1

C1  21C2  ...  21C10 

= 220 – 1

1 ,b 0 5

1

21

JEE (MAIN)-2017 (Code-B)

For Q,  = 2

51. If 5 (tan2x – cos2x) = 2cos 2x + 9, then the value of cos 4x is (1)  (3)

7 9

(2) 

1 3

(4)

Distance PQ  2 12  42  52  2 42

3 5

53. The distance of the point (1, 3, –7) from the plane passing through the point (1, –1, –1), having normal x 1 y  2 z  4   perpendicular to both the lines 1 2 3 x  2 y 1 z  7   and , is 1 1 2

2 9

Answer (1) Sol. 5 tan2x = 9 cos2x + 7 5 sec2x – 5 = 9 cos2x + 7

(1)

Let cos2x = t

5  9t  12 t 9t2 + 12t – 5 = 0 t

1 3

as

t

(3)

(2)

74 10

(4)

83

20 74 5 83

Answer (3)

5 3

Sol. Let the plane be a( x  1)  b( y  1)  c( z  1)  0

1 cos x  , cos 2x = 2cos2x – 1 3 2

It is perpendicular to the given lines a – 2b + 3c = 0

1 3 cos4x = 2 cos2 2x – 1

= 

=

2a – b – c = 0 Solving, a : b : c = 5 : 7 : 3

2 1 9

 The plane is 5x + 7y + 3z + 5 = 0

7 9

Distance of (1, 3, –7) from this plane =

= 

(1) 6 5

(2) 3 5

(3) 2 42

(4)

10 83

1 ⎛ 6 x x ⎞ 1 ⎟ is 54. If for x  ⎛⎜ 0, ⎞⎟ , the derivative of tan ⎜ ⎝ 1  9x 3 ⎠ ⎝ 4⎠

52. If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to the line, x y z   is Q, then PQ is equal to 1 4 5

x  g ( x ) , then g(x) equals (1)

3 1  9x 3

42

Answer (3) Sol. Equation of PQ,

10

x 1 y  2 z  3   1 4 5

1  9x 3

(3)

1 9x3

Let M be (  1, 4  2, 5  3)

P

9

(2)

(4)

3x x

3x 1 9x3

Answer (2)

M

Sol. f ( x )  2 tan1(3 x x ) f ( x ) 

Q As it lies on 2x + 3y – 4z + 22 = 0

g( x ) 

=1 17

9 x 1 9x3

9 1 9x3

1⎞ ⎛ For x  ⎜ 0, ⎟ ⎝ 4⎠

JEE (MAIN)-2017 (Code-B)

55. The radius of a circle, having minimum area, which touches the curve y = 4 – x2 and the lines, y = |x| is

57. The eccentricity of an ellipse whose centre is at the origin is

(1) 4  2  1

1 . If one of its directrices is x = – 4, then 2

⎛ 3⎞ the equation of the normal to it at ⎜ 1, ⎟ is ⎝ 2⎠

(2) 2  2  1

(1) x + 2y = 4

(3) 2  2  1

(2) 2y – x = 2

(4) 4  2  1

(3) 4x – 2y = 1

Answer (4) (4) 4x + 2y = 7

Sol. Centre : (0, 4 – r)  Length of perpendicular from (0, 4 – r) to x – y = 0 is r 

Answer (3)

04r

Sol.

2

 r  4   2r

r  

4 1 2

, r 

4

x = –4

1 2

r  4( 2  1)

e

56. A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with replacement, then the variance of the number of green balls drawn is (1)

a  4 e a  4  e

6 25

a2

12 (2) 5

Now, b2  a2 (1  e2 )  3

(3) 6

Equation to ellipse

(4) 4 Answer (2)

x2 y 2  1 4 3

Sol. n = 10 p(Probability of drawing a green ball) =

 p

1 2

15 25

Equation of normal is

3 y x 1 2  1 3 4 23

3 2 , q 5 5

var(X) = n.p.q = 10 

6 12  25 5

 4 x  2y  1  0 18

JEE (MAIN)-2017 (Code-B)

58. If two different numbers are taken from the set {0, 1, 2, 3, ......, 10}; then the probability that their sum as well as absolute difference are both multiple of 4, is (1)

(2)

(3)

(4)

∵ P( A  B  C ) 



7 55

P( A  B  C ) 

3 1 7   8 16 16

⎡ 2 3 ⎤ 2 60. If A  ⎢ ⎥ , then adj (3A + 12A) is equal to ⎣ 4 1 ⎦

6 55

⎡ 72 63 ⎤ (1) ⎢ ⎥ ⎣ 84 51 ⎦

12 55

⎡ 72 84 ⎤ (2) ⎢ ⎥ ⎣ 63 51 ⎦

14 45

⎡ 51 63 ⎤ (3) ⎢ ⎥ ⎣84 72 ⎦

Answer (2) Sol. Total number of ways =

11C 2

⎡ 51 84 ⎤ (4) ⎢ ⎥ ⎣ 63 72 ⎦

= 55 Favourable ways are (0, 4), (0, 8), (4, 8), (2, 6), (2, 10), (6, 10) Probability =

1 16

Answer (3)

6 55

⎡ 2 3 ⎤ Sol. A  ⎢ ⎥ ⎣ 4 1 ⎦

59. For three events A, B and C, P (Exactly one of A or B occurs) = P(Exactly one of B or C occurs) A  I 

1 = P (Exactly one of C or A occurs) = and 4

2

3

4

1 

= (2 – 2– + 2) – 12

1 . 16 Then the probability that at least one of the events occurs, is

f ( )   2  3  10

3 (1) 16

7 (2) 32

 A2 – 3A –10I = 0

7 (3) 16

7 (4) 64

P(All the three events occur simultaneously) =

∵ A satisfies f ( )

A2 – 3A = 10I 3A2 – 9A = 30I

Answer (3)

3A2 + 12A = 30I + 21A

Sol. P ( A)  P (B )  P ( A  B ) 

1 4

P (B )  P (C )  P (B  C ) 

1 4

P (C )  P ( A)  P ( A  C ) 

1 4

⎡30 0 ⎤ ⎡ 42 63 ⎤ ⎢ ⎥⎢ ⎥ ⎢⎣ 0 30 ⎥⎦ ⎢⎣ 84 21 ⎥⎦ ⎡ 72 63 ⎤ ⎢ ⎥ ⎢⎣ 84 51 ⎥⎦

P ( A)  P (B )  P (C )  P ( A  B )  P (B  C )

3  P( A  C )  8



 19



⎡ 51 63 ⎤ adj(3 A2  12 A)  ⎢ ⎥ ⎢⎣84 72 ⎥⎦

JEE (MAIN)-2017 (Code-B)

PART–C : CHEMISTRY 63. The increasing order of the reactivity of the following halides for the SN1 reaction is

61. Which of the following compounds will form significant amount of meta product during mono-nitration reaction?

CH 3CHCH 2CH 3

I.

OH

Cl

II.

(1)

CH3CH2CH2Cl

III. p–H3CO – C6H4 – CH2Cl

OCOCH3 (2)

(1) (III) < (II) < (I)

(2) (II) < (I) < (III)

(3) (I) < (III) < (II)

(4) (II) < (III) < (I)

Answer (2) Sol. Rate of SN1 reaction  stability of carbocation

NH2

CH3 – CH – CH2 – CH3

I. (3)

CH3 – CH – CH2 – CH3

Cl

NHCOCH3

II.

(4) Answer (3)

CH3 – CH2 – CH2

CH2 – Cl

CH2

OCH3

OCH3

III.

NH2

NH3 H+

Sol.

CH3 – CH2 – CH2 – Cl

So, II < I < III

NO2

Increase stability of carbocation and hence increase reactivity of halides.

NH3

(51%)

NH3

NO2

+

64. The radius of the second Bohr orbit for hydrogen atom is

NH3 NO2

+ NO2

(Planck's Const. h = 6.6262 × 10–34 Js; mass of electron = 9.1091 × 10–31 kg;

(2%)

charge of electron e = 1.60210 × 10–19 C;

(47%)

permittivity of vacuum 62. U is equal to

0 = 8.854185 × 10–12 kg–1 m–3 A2)

(1) Isochoric work

(1) 1.65 Å

(2) Isobaric work

(2) 4.76 Å

(3) Adiabatic work

(3) 0.529 Å

(4) Isothermal work

(4) 2.12 Å Answer (4)

Answer (3) Sol. For adiabatic process, q = 0

Sol. r = a0

 As per 1st law of thermodynamics, U = W

n2 = 0.529 ×4 Z

= 2.12 Å 20

JEE (MAIN)-2017 (Code-B)

Answer (3)

65. pKa of a weak acid (HA) and pKb of a weak base (BOH) are 3.2 and 3.4, respectively, The pH of their salt (AB) solution is

Sol. Mass of hydrogen =

(1) 7.2

10  75 = 7.5 kg 100

Replacing 1H by 2H would replace 7.5 kg with 15 kg

(2) 6.9

 Net gain = 7.5 kg

(3) 7.0

68. Which of the following, upon treatment with tert-BuONa followed by addition of bromine water, fails to decolourize the colour of bromine?

(4) 1.0 Answer (2)

O

Sol. pH = 7 +

(1)

1  pK a – pKb  2

Br

C6H5 = 7+

1  3.2 – 3.4 2

(2)

Br O

= 6.9 (3)

Br

66. The formation of which of the following polymers involves hydrolysis reaction?

O

(1) Nylon 6 (4)

(2) Bakelite

Br

(3) Nylon 6, 6

Answer (1)

(4) Terylene

CH3

Answer (1)

Na+O–

Sol. Caprolactam is hydrolysed to produce caproic acid which undergoes condensation to produce Nylon-6.

O Sol.

O

O NH

CH3

CH3

Br

C

H3O+

HO (Caprolactam)

C

CH3 (CH2)5 – NH2

O

(Caproic acid)

O

C

CH3

CH3 67. The most abundant elements by mass in the body of a healthy human adult are :

(Product)

The above product does not have any C = C or C  C bond, so, it will not give Br2-water test.

Oxygen (61.4%); Carbon (22.9%); Hydrogen (10.0%) and Nitrogen (2.6%).

69. In the following reactions, ZnO is respectively acting as a/an

The weight which a 75 kg person would gain if all 1H atoms are replaced by 2H atoms is

(a) ZnO + Na2O  Na2ZnO2 (b) ZnO + CO2  ZnCO3

(1) 15 kg

(1) Base and acid (2) 37.5 kg

(2) Base and base

(3) 7.5 kg

(3) Acid and acid

(4) 10 kg

(4) Acid and base 21

JEE (MAIN)-2017 (Code-B)

Answer (4)

72. A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is 'a', the closest approach between two atoms in metallic crystal will be

Sol. In (a), ZnO acts as acidic oxide as Na2O is basic oxide. In (b), ZnO acts as basic oxide as CO2 is acidic oxide.

(2) 2 2a

(1) 2a 2a

(3)

70. Both lithium and magnesium display several similar properties due to the diagonal relationship, however, the one which is incorrect, is

(4)

a 2

Answer (4) Sol. In FCC, one of the face is like

(1) Both form basic carbonates

A

(2) Both form soluble bicarbonates

a

(3) Both form nitrides (4) Nitrates of both Li and Mg yield NO2 and O2 on heating

C

B

By ABC,

Answer (1)

2a2 = 16r2

Sol. Mg forms basic carbonate

 r2 

3MgCO3 · Mg  OH2 ·3H2O but no such basic

carbonate is formed by Li.

 r

71. 3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is

1 2 a 8

1 2 2

a

Distance of closest approach = 2r =

a 2

73. Two reactions R 1 and R 2 have identical preexponential factors. Activation energy of R1 exceeds that of R2 by 10 kJ mol–1. If k 1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then ln(k2/k1) is equal to

(1) Six (2) Zero (3) Two

(R = 8.314 J mole–1 K–1)

(4) Four Answer (4)

CH3 Sol. CH3 – CH = C – CH2 – CH3 3-methyl pent-2-ene

(2) 12

(3) 6

(4) 4

Answer (4)

HBr R2O2

Sol. k1 = Ae k2 = Ae

CH3 CH3 – *CH –*C – CH2 – CH3 Br

(1) 8

–Ea /RT 1

–Ea /RT 2

1 k2 Ea – Ea2  e RT 1 = k1

H Product (X)

ln

k 2 Ea1 – Ea2 = k1 RT

=

10  103 4 8.314  300

Since product (X) contains two chiral centres and it is unsymmetrical. So, its total stereoisomers = 22 = 4. 22

JEE (MAIN)-2017 (Code-B)

74. The correct sequence of reagents for the following conversion will be

O

HO

HO

CHO

Answer (2) Sol. For Tyndall effect refractive index of dispersion phase and dispersion medium must differ significantly. Secondly, size of dispersed phase should not differ much from wavelength used.

CH3

76. Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution?

CH3 CH3

(1)

(1) [Ag(NH3)2]+OH–, H+/CH3OH, CH3MgBr

OH

(3) CH3MgBr, [Ag(NH3)2]+OH–, H+/CH3OH

O

HOH2C

(4) [Ag(NH3)2]+OH–, CH3MgBr, H+/CH3OH (2)

Answer (1)

O

OH

–

[Ag(NH3)2] OH

CH3 – OH/H

Sol.

+

HO

O

HOH2C

Esterification

C=O

CH2OH

HO

O +

CH2OH

HO OCOCH 3

(2) CH3MgBr, H+/CH3OH, [Ag(NH3)2]+OH–

H

O

HOH2C

C=O

CH2OH

HO OCH 3

(3)

OH O

O

C – OCH3

(i) CH3MgBr (3 moles) (ii) H2O

HO

HO – C

CH3

HOH2C

CH3

(4)

CH3

OH

O CH OCH 2 3

OH OH

75. The Tyndall effect is observed only when following conditions are satisfied

Answer (1) Sol. Sugars in which there is free anomeric –OH group are reducing sugars

(a) The diameter of the dispersed particles is much smaller than the wavelength of the light used.

OH

(b) The diameter of the dispersed particle is not much smaller than the wavelength of the light used

CH2 O

CH2 – OH O

KOH(aq)

HO O – C– CH 3

(c) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude

OH

(d) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude

OH CH2 O

(1) (a) and (d)

HO

(2) (b) and (d) (3) (a) and (c)

OH

(4) (b) and (c) 23

CH2 – OH OH

+ CH3COOK

Free anomeric group

JEE (MAIN)-2017 (Code-B)

79. The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are

77. Given C(graphite) + O2(g)  CO2(g);

(1) ClO– and ClO3–

rHº = –393.5 kJ mol–1 H2(g) +

(2) ClO2– and ClO3–

1 O2 (g)  H2O(l); 2

rHº = –285.8 kJ

(3) Cl– and ClO– (4) Cl– and ClO2–

mol–1

Answer (3)

CO2(g) + 2H2O(l)  CH4(g) + 2O2(g);

Sol. Cl2  2NaOH  NaCl  NaOCl  H2 O Cold & dilute

rHº = +890.3 kJ mol–1 Based on the above thermochemical equations, the value of rHº at 298 K for the reaction

80. The major product obtained in the following reaction is

C(graphite) + 2H2(g)  CH4(g) will be (1) +74.8 kJ

Br

mol–1

C6H5

(2) +144.0 kJ mol–1 (3) –74.8 kJ

mol–1

 CO2(g);

Answer (2)

Br

 H2O(l); mol–1

Sol.

...(ii)

 CH4(g) + 2O2(g); rH° = 890.3 kJ mol–1

(i) + 2 × (ii) + (iii), we get  CH4(g);

C6H5

(+)

t-BuOK  (E-2)

C6H5 C6H5

(1) C6H5COONa

rH°= –393.5 –285.8 × 2 + 890.3 = –74.8 kJ

C6H5

H

81. Sodium salt of an organic acid 'X' produces effervescence with conc. H2SO4. 'X' reacts with the acidified aqueous CaCl2 solution to give a white precipitate which decolourises acidic solution of KMnO4. 'X' is

...(iii)

By applying the operation

C(graphite) + 2H2(g)

(+)

(4) (–)C6H5CH(OtBu)CH2C6H5

rH° = –285.8 kJ CO2(g) + 2H2O(l)

t-BuOK 

(3) (+)C6H5CH(OtBu)CH2C6H5

rH° = –393.5 kJ mol–1 ...(i) 1 O2 (g) 2

C6H5

(2) C6H5CH = CHC6H5

Answer (3) Sol. C(graphite) + O2(g)

H

(1) (±)C6H5CH(OtBu)CH2C6H5

(4) –144.0 kJ mol–1

H2 (g) 

Sodium hypochlorite

(2) HCOONa

mol–1

(3) CH3COONa (4) Na2C2O4

78. Which of the following reactions is an example of a redox reaction?

Answer (4)

(1) XeF4 + O2F2  XeF6 + O2

Sol. Na2C2O4  H2SO4   Na2SO4  H2C2O4

(2) XeF2 + PF5  [XeF]+ PF6–

( X)

(3) XeF6 + H2O  XeOF4 + 2HF

Conc.

oxalic acid

Conc. H SO

2 4 H2C2O4   CO   CO2   

(4) XeF6 + 2H2O  XeO2F2 + 4HF

–H2O

Answer (1)

effervescence

Na2 C2 O 4  CaCl2  CaC2 O 4   2NaCl ( X)

Sol. Xe is oxidised from +4(in XeF4) to +6(in XeF6)

white ppt.

 2 2MnO4–  5C2O2–  10CO2  8H2O 4  16H  2Mn

Oxygen is reduced from +1 (in O2F2) to zero (in O2) 24

JEE (MAIN)-2017 (Code-B)

82. Which of the following species is not paramagnetic?

84. Which of the following molecules is least resonance stabilized?

(1) NO (2) CO

(1)

(3) O2 (4) B2

(2)

Answer (2)

O

Sol. CO has 14 electrons (even)  it is diamagnetic NO has 15e–(odd)  it is paramagnetic and has 1 unpaired electron in 2p molecular orbital.

(3)

N

10e–

B2 has (even) but still paramagnetic and has two unpaired electrons in 2p x and 2p y (s-p mixing). (4)

O

O2 has 16 e– (even) but still paramagnetic and has two unpaired electrons in *2px and *2py molecular orbitals.

Answer (4)

83. The freezing point of benzene decreases by 0.45ºC when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be

Sol. However, all molecules given in options are stabilised by resonance but compound given in option (2) is least resonance stabilised (other three are aromatic)

O

(Kf for benzene = 5.12 K kg mol–1) (1) 64.6%

(2) 80.4%

(3) 74.6%

(4) 94.6%

O–

Answer (4) 85. On treatment of 100 mL of 0.1 M solution of CoCl3  6H2O with excess AgNO3; 1.2 × 1022 ions are precipitated. The complex is

0.2 / 60  1000 Sol. 0.45 = i(5.12) 20

 i = 0.527

(1) [Co(H2O)4Cl2]Cl  2H2O

  2CH3 COOH    CH3COOH2 1– 

(2) [Co(H2O)3Cl3]  3H2O

 2

(3) [Co(H2O)6]Cl3 (4) [Co(H2O)5Cl]Cl2  H2O

  i  1– 2  0.527 = 1 –



Answer (4)  2

Sol. Millimoles of AgNO3 =

 = 0.473 2

1.2  1022 6  1023

 1000 = 20

Millimoles of CoCl3·6H2O = 0.1 × 100 = 10

  = 0.946

 Each mole of CoCl3·6H2O gives two chloride ions.

 % association = 94.6%

 [Co(H2O)5Cl]Cl2·H2O 25

JEE (MAIN)-2017 (Code-B)

86. The major product obtained in the following reaction is

Answer (3) –

Sol. Permissible limit of F in drinking water is upto – 1 ppm. Excess concentration of F > 10 ppm causes decay of bones.

O

O

88. 1 gram of a carbonate (M2CO3) on treatment with excess HCl produces 0.01186 mole of CO2. The molar mass of M2CO3 in g mol–1 is

DIBAL-H

COOH OH

(1) 1186

(2) 84.3

(3) 118.6

(4) 11.86

Answer (2) Sol. M2CO3 + 2HCl  2MCl + H2O + CO2

CHO

(1)

nM2CO3  nCO2

COOH

1

OH

MM2CO3

CHO

(2)

MM2CO3 =

CHO 89. Given º ECl

2 /Cl

COOH

2

CHO Answer (2) Sol. DIBAL — H reduces esters and carboxylic acids into aldehydes

OH H

7

4

(1) Cr

(2) Mn2+

(3) Cr3+

(4) Cl–

Answer (1) Sol. For Cr3+ ,

O

DIBAL-H

COOH

º  1.36 V, ECr  –0.74 V 3 /Cr

Among the following, the strongest reducing agent is

(4)

O

–

º º ECr  1.33 V, EMnO  1.51 V – O2– /Cr 3  /Mn2 

CHO

O

1 0.01186

= 84.3 g/mol

CHO

(3)

 0.01186

E°Cr 3+ /Cr O2– = – 1.33 V 2 7

For Cl–,

E°Cl– /Cl = – 1.36 V 2

For Cr,

E°Cr/Cr 3 = 0.74 V

For Mn2 , E°Mn2 /MnO – = – 1.51 V

CHO

4

Positive E° is for Cr, hence it is strongest reducing agent.

87. A water sample has ppm level concentration of following anions

90. The group having isoelectronic species is

F– = 10; SO42– = 100; NO3– = 50

(1) O2–, F–, Na+, Mg2+

The anion/anions that make/makes the water sample unsuitable for drinking is/are

(2) O–, F–, Na, Mg+

(1) Only NO3–

(3) O2–, F–, Na, Mg2+

(2) Both SO42– and NO3–

(4) O–, F–, Na+, Mg2+

(3) Only F–

Answer (1)

(4) Only SO42–

Sol. Mg2+, Na+, O2– and F– all have 10 electrons each.

   26