**Question 1:**Consider two plane convex lens of same radius of curvature and refractive index n

_{1}and n

_{2}respectively. Now consider two cases:

Case â€“ I: When n_{1} = n_{2} = n, then equivalent focal length of lens is f_{0}.

Case â€“ II: When n_{1} = n, n_{2} = n + Î”n, then equivalent focal length of lens is f= f_{0} + Î”f_{0} . Then correct options are:

a) If Î”n/n > 0, then Î”f_{0}/f_{0} < 0

b) |Î”f_{0}/f_{0}| < |Î”n/n|

c) If n = 1.5, Î”n = 10^{-3} and f_{0} = 20 cm then |Î”f_{0}= 0.02 cm

d) If Î”n/n < 0, then Î”f_{0}/f_{0} > 0

=

**Answer:** A and C

**Question 2:**In YDSE monochromatic light of wavelength 600 nm incident of slits as shown in figure.

If S_{1}S_{2} = 3mm, OP = 11 mm then

a) If Î± = 0.36/Ï€ degree then destructive interfaces at point P.

b) If Î± = 0.36/Ï€ degree then constructive interfaces at point O.

c) If Î± = 0 then constructive interfaces at O

d) Fringe width depends an Î±

Î”x = d sin Î± + d sin Î¸

= dÎ± + dy/D

(A)

3900 = (2n â€“ 1) Î»/2 => n = 7 dest

(B) Î”x = 3mm (.36/Ï€) (Ï€/180) = 600 nm

600nm = n600nm

=>n = 1 const

(C) Î± = 0, Î”x = 0

So constructive interference

(D) Fringe width does not depend on alpha;.

Answer: a::b::c

**Question 3:** A uniform rigid rod of mass m & length l is released from vertical position on rough surface with sufficient friction for lower end not to slip as shown in figure. When rod makes angle 60Â° with vertical then find correct alternatives

a)

b)

c) N = mg/16

d) a

_{radical}= 3g/4

mg â€“ N = ma_{v}

N = mg/16

Answer: b::c::d

**Question 4:**Monoatomic gas A having 5 mole is mixed with diatomic gas B having 1 mole in container of volume V0. Now the volume of mixture is compressed to V0/4 by adiabatic process. Initial pressure and temperature of gas mixture is P

_{0}and T

_{0}. [given 2

^{3.2}= 9.2] Choose correct option:

a)

b) Final pressure is between 9P

_{0}and 10P

_{0}

c) |W.D.| = 13 RT

_{0}

d) Average Translational kinetic energy

W = [P_{1}V_{1} â€“ P_{2}V_{2}]/[V-1]

P_{0}V_{0}^{8/5} = P_{2} (V_{0}/9)^{8/5}

P_{2} = 9.2 P_{0}

Answer: a::b::c

**Question 5:**The given arrangement is released from rest when spring is in natural length. Maximum extension in spring during the motion is x

_{0}, a

_{1}, a

_{2}and a

_{3}are accelerations of the blocks.

Make the correct options

a) a_{2} â€“ a_{1} = a_{1} â€“ a_{3}

b) x_{0} = 4mg/3k

c)

d) Acceleration a

_{1}at x

_{0}/4 is 3kx

_{0}/42m

2a_{1} = a_{2} + a_{3}

a_{1} – a_{3} = a_{2} – a_{1}

When we use m equivalent (for other opitions)

T/g’ = 2(2m)m/(2m+m) = 4m/3

2T/g’ = 8m/3

m_{eq.} = 8m^{2}/[m+2m] = 8m/3

(1/2) kx_{0}^{2} = 8mgx_{0}/3

x_{0} = 16mg/3k

V_{x0/2} = V_{max} = x_{0}w/2

=

a_{x0/4} = x_{0}w^{2}/4 = x_{0}/4 * 3k/14m = 3kx_{0}/42m

Answer: a::c::d

**Question 6:**A dipole of Dipole moment

_{0}is applied along direction of dipole. Two points A and B are lying on a equipotential surface of radius R centered at origin. A is along axial position of dipole and B is along equatorial position. There correct option are:

a) Net electric field at point A is 3E

_{0}

b) Net electric field at point B is Zero

c) Radius of equatorial surface R = (kp

_{0}/E

_{0})

^{1/3}

d) Radius of equatorial surface R = (âˆš2kp

_{0}/E

_{0})

^{1/3}

P = (P_{0}/âˆš2) (x + 1)

KP_{0}/r^{3} = E_{0}

(E_{A})_{net} = 2KP_{0}/r^{3} + E_{0} = 3E_{0}

(E_{A})_{net} = 0

Answer: b

**Question 7:**A free hydrogen atom after absorbing a photon of wavelength Î»

_{a}gets excited from state n = 1 to n = 4. Immediately after electron jumps to n = m state by emitting a photon of wavelength Î»

_{e}. Let change in momentum of atom due to the absorption and the emission are Î”P

_{a}and Î”p

_{e}respectively. If Î»

_{a}/ Î»

_{e }= 1/5.

Which of the following is correct

a) m = 2

b) Î”P_{a}/P_{e} = 1/2

c) Î»_{e} = 418 nm

d) Ratio of K.E. of electron in the state n = m to n = 1 is Â¼.

Answer: a::d

**Question 8:**In a cylinder a heavy piston is moving with speed v as shown diagram and gas is filled inside it. A gas molecule is moving with speed v

_{0}towards moving piston. Then which of the following is

correct (Assume v <<<< v_{0} Î”l/l ) and collision is elastic)

a) change in speed after collision is 2V

b) change is speed after collision is 2 v_{0} Î”l/l

c) rate of collision is V/l

d) When piston is at l/2 its kinetic energy will be four times.

Change in speed is (2V + V_{0} â€“ V_{0}) = 2V

Answer: a

**Question 9:** Calculate the work done if a particle is moved along a path AB−BC−CD−DE−EF−FA, as shown in the figure in presence of a force

dÏ‰ = Î±y dx + 2Î±x dx

E → F, y = 0, dy = 0, W<sub>EF</sub> = 0

F → A, x = 0, dx = 0, W<sub>F → A</sub> = 0

∴ w = α − α − α / 4 − α / 2 = -3α / 4

Given α = −1

⇒ W = + 3/4 J = 0.75 J.

Answer: 0.75 J

**Question 10:**In a given circuit inductor of L = 1mH and resistance R = 1Î© are connected in series to ends of two parallel conducting rods as shown. Now a rod of length 10 cm is moved with constant velocity of 1 cm/s in magnetic field B = 1T. If rod starts moving at t = 0 then current in circuit after 1 millisecond is x . 10

^{-3}A. Then value of x is: (given e

^{-1}= 0.37)

e = (V x B) dl = 10^{-3} v df

i = 10^{-3}(1 â€“ e^{-1})

i = 0.63 mA

Answer: 0.63 mA

**Question 11:**A prism is shown in the figure with prism angle 75Â° and refractive index âˆš3. A light ray incidents on a surface at incident angle Î¸. Other face is coated with a medium of refractive index n. For Î¸ â‰¤ 60

^{0}ray suffers total internal reflection find value of n

^{2}.

Answer: 1.5

**Question 12:**Perfectly reflecting mirror of mass M mounted on a spring constitute a spring mass system of angular frequency Î© such that [4Ï€MÎ©]/h = 10

^{24}m

^{-2}where h is plank constant. N photons of wavelength Î» = 8Ï€ x 10

^{-6}m strikes the mirror simultaneously at normal incidence such that the mirror gets displaced by 1 Î¼m. If the value of N is x * 10

^{12}, then find the value of x.

Photons are reflected

MV = 2Nh/Î» [mean]

V_{mean} = Î±A

A = 1 min

N =

N = 4MÏ€Î±/h x 10^{-12}

10^{24} x 10^{-12}

1 x 10^{-12}

Therefore, X = 1

**Question 13:**The sample of monoatomic gas undergoes a process as represented by P â€“ V graph (if P

_{0}V

_{0}= 1/3 RT

_{0}) then

(P) W_{1->2} = 1/3 RT_{0 } (Q) Q_{1->2->3} = 11/6 RT_{0 } (R) U_{1->2} = RT_{0}/2 (S) W_{1->2->3} = 1/3 RT_{0}

Which of the following options are correct

a) P, Q, R, S are correct

b) Only P, Q are correct

c) Only R, S are correct

d) Only P, R, S correct

Answer: a

**Question 14:**The sample of monoatomic gas undergoes a process as represented by T â€“ V graph (if P

_{0}V

_{0}= 1/3 RT

_{0}) then

(P) W_{1->2} = 1/3 RT_{0 } ln 2

(Q) Q_{1->2->3} = 1/6 RT_{0} (2 ln (2) + 3 )_{ }

(R) U_{1->2} = 0

(S) W_{1->2->3} = [RT_{0}/3] ln 2

Which of the following options are correct

a) P, Q are incorrect

b) R, S are incorrect

c) P, Q, S are incorrect

d) None of these

Ï‰_{1-2} = nRT_{0} ln 2

Q_{1-2-3} = Q_{12} + Q_{23}

= d Ï‰_{1-2} + dU_{2-3}

= [RT_{0}/3] ln 2 + n (f/2) RT_{0}

= [RT_{0}/3] ln 2 + (1/3) (3/2) RT_{0}

U_{1-2}=0

Ï‰_{1-2-3} = (1/3) R_{0}T_{0} ln 2

Answer: d

**Question 15:**Length of string of a musical instrument is varied from L

_{o}to 2L

_{o}in 4 different cases. Wire is made of different materials of mass per unit length Î¼, 2Î¼, 3Î¼, 4Î¼ respectively. For first case (string â€“ 1) length is L

_{o}, Tension is T

_{o}then fundamental frequency is f

_{o}, for second case length of the string is 3L

_{0/}2 (3rd Harmonic), for third case length of the string is 5L

_{0}/4 (5th Harmonic) and for the fourth case length of the string is 7L

_{0}/4 (14th harmonic). If frequency of all is same then tension in strings in terms of T

_{o}will be:

a) String – 1 (P) T

_{0}

b) String â€“ 2 (Q) T

_{0}/âˆš2

c) String – 3 (R) T

_{0}/2

d) String – 4 (S) T

_{0}/16

(T) 3T

_{0}/16

Answer:

(A) → P, (B) → R, (C) → T, (D) → S

**Question 16:**The free length of all four string is varied from L0 to 2L0. Find the maximum fundamental frequency of 1, 2, 3, 4 in terms of f0 (tension is same in all strings)

a) String â€“ 1 (P) 1

b) String â€“ 2 (Q) 1/2

c) String â€“ 3 (R) 1/âˆš2

d) String â€“ 4 (S) 1/âˆš3

(T) 1/16

(U) 3/16

(A) → P, (B) → R, (C) → S, (D) → Q

## Video Lessons – Paper 2 Physics

## JEE Advanced 2019 Physics Paper 2 Solutions

## Comments