JEE Advanced 2019 Question Paper 2 with solutions for Physics is available here. All questions are solved by subject experts and in a detailed manner. Listed solutions can also be downloaded in a PDF format for offline use. Practising JEE Advanced 2019 Physics Question Paper 2 will help the aspirants to study productively and at the same time understand the coverage of topics as well as the overall exam pattern for exam. To succeed in the JEE Advanced upcoming exams candidates are advised to practice sample papers to prepare well and effectively.
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 – I: When n_{1} = n_{2} = n, then equivalent focal length of lens is f_{0}.
Case – II: When n_{1} = n, n_{2} = n + Δn, then equivalent focal length of lens is f= f_{0} + Δf_{0} . Then correct options are:
Solution:
=
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
Solution:
Δx = d sin α + d sin θ
= dα + dy/D
(A)
3900 = (2n – 1) λ/2 => n = 7 dest
(B) Δx = 3mm (.36/π) (π/180) = 600 nm
600nm = n600nm
=>n = 1 const
(C) α = 0, Δx = 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 60° with vertical then find correct alternatives
Solution:
mg – 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:
Solution:
W = [P_{1}V_{1} – 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
Solution:
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
Solution:
P = (P_{0}/√2) (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 λ_{a} gets excited from state n = 1 to n = 4. Immediately after electron jumps to n = m state by emitting a photon of wavelength λ_{e}. Let change in momentum of atom due to the absorption and the emission are ΔP_{a} and Δp_{e} respectively. If λ_{a}/ λ_{e }= 1/5.
Which of the following is correct
Solution:
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} Δl/l ) and collision is elastic)
Solution:
Change in speed is (2V + V_{0} – 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
Solution:
dω = αy dx + 2αx 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 = 1Ω 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)
Solution:
e = (V x B) dl = 10^{-3} v df
i = 10^{-3}(1 – e^{-1})
i = 0.63 mA
Answer: 0.63 mA
Question 11: A prism is shown in the figure with prism angle 75° and refractive index √3. A light ray incidents on a surface at incident angle θ. Other face is coated with a medium of refractive index n. For θ ≤ 60^{0} ray suffers total internal reflection find value of n^{2}.
Solution:
Answer: 1.5
Question 12: Perfectly reflecting mirror of mass M mounted on a spring constitute a spring mass system of angular frequency Ω such that [4πMΩ]/h = 10^{24} m^{-2} where h is plank constant. N photons of wavelength λ = 8π x 10^{-6} m strikes the mirror simultaneously at normal incidence such that the mirror gets displaced by 1 μm. If the value of N is x * 10^{12}, then find the value of x.
Solution:
Photons are reflected
MV = 2Nh/λ [mean]
V_{mean} = αA
A = 1 min
N =
N = 4Mπα/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 – 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
Solution:
Answer: a
Question 14: The sample of monoatomic gas undergoes a process as represented by T – 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
Solution:
ω_{1-2} = nRT_{0} ln 2
Q_{1-2-3} = Q_{12} + Q_{23}
= d ω_{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
ω_{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 μ, 2μ, 3μ, 4μ respectively. For first case (string – 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:
Solution:
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)
Solution:
(A) → P, (B) → R, (C) → S, (D) → Q