JEE Main 2021 Physics question paper for March 17, Shift 2 exam is provided here. We are offering accurate solutions which will further help students to perform well in the JEE Main exam. Students can have a glance at the question paper and get a better understanding of the question types, difficulty level, marking scheme. JEE Main 2021 Physics March 17, Shift 2 paper has been solved by our subject experts.
Section - A
Question 1. Two identical blocks A and B each of mass m resting on the smooth horizontal floor are connected by a light spring of natural length L and spring constant K. A third block C of mass m moving with a speed v along the line joining A and B collides elastically with A. The maximum compression in the spring is
Solution:
Answer: (d)
Let v_{1} be the common velocity of blocks A and B at the maximum compression of the spring
From the conservation of momentum
mv = mv_{1} + mv_{1}
v_{1} = V_{2 }
& from energy conservation
The maximum compression in the spring is
Question 2. A solid sphere of mass 2 kg radius 0.5m is rolling with an initial speed of 1 ms^{-1} goes up an inclined plane which makes an angle of 300 with the horizontal plane, without slipping. How long will the sphere take to return to the starting point A?
Solution:
Answer: (d)
For a solid sphere, c = ⅖
a = 3.5 m/sec^{2}
Time of ascent is given by
v = u + at
0 = 1 - 3.5 t
Time of decent
Total time,
Question 3. If one mole of a polyatomic gas has two vibrational modes and β is the ratio of molar specific heats for polyatomic gas β=C_{p} / C_{v} then the value of β is :
Solution:
Answer: (d)
Degree of freedom of polyatomic gas
Here, no of translational dof = 3,
no of rotational dof = 3
and no of vibrational dof = 2×2 = 4
Therefore, total no of dof
f = 3 + 3 + 4 = 10
β = C_{P }/C_{V} =1 + (2/f) = 1+ (2/10)
β=12/10=1.2
Question 4. Two cells of emf 2E and E with internal resistance r_{1} and r_{2 }respectively are connected in series to an external resistor R (see figure). The value of R, at which the potential difference across the terminals of the first cell becomes zero is
Solution:
Answer: (d)
If potential difference across terminals of first cell is zero
V_{A} = V_{B}
2 E = i r_{1}
2R + 2r_{1} + 2r_{2} = 3r_{1}
R = r_{1}/2- r_{2}
Question 5. A sound wave of frequency 245 Hz travels with a speed of 300 ms^{-1 }along the positive x-axis. Each point of the medium moves to and fro through a total distance of 6 cm. What will be the mathematical expression of the travelling wave?
Solution:
Answer: (d)
General expression for wave travelling along positive x axis is of the form
Y = A sin (kx - ωt)
Here
A = 6/2 = 3cm = 0.03 m
ω = 2π f = 2π ×245
ω = 1.54 ×10^{3}rad s-1
k = ω/v=(1.54×10^{3})/300
k = 5.1 m^{-1}
y = 0.03 sin (5.1x – 1.5 × 10^{3}t)
Question 6. A carrier signal C(t) = 25 sin (2.512×10^{10}t) is amplitude modulated by a message signal m(t)= 5 sin (1.57×10^{8}t) and transmitted through an antenna. What will be the bandwidth of the modulated signal?
Solution:
Answer: (c)
Bandwidth of modulated signal is given by,
β = 2f_{m(t)}
β = 2× {(1.57×10^{8})/2π}
β = 50 MHz
Question 7. Two particles A and B of equal masses are suspended from two massless springs of spring constants K_{1} and K_{2} respectively. If the maximum velocities during oscillations are equal, the ratio of the amplitude of A and B is :
Solution:
Answer: ( c)
Since, V_{max }= Aω
Given, A_{1}ω_{1}= A_{2}ω_{2}
We know that ω = √ k/m
Question 8. Match List I with List II
List I |
List II |
||
a) |
Phase difference between current and voltage in a purely resistive AC circuit |
i. |
𝜋/2, current leads voltage |
b) |
Phase difference between current and voltage in a pure inductive AC circuit |
ii. |
zero |
c) |
Phase difference between current and voltage in a pure capacitive AC circuit |
iii. |
𝜋/2, current lags voltage |
d) |
Phase difference between current and voltage in an LCR series circuit |
iv. |
tan-1 (Xc - XL/R) |
Choose the most appropriate answer from the options given below :
Solution:
Answer: (d)
(a) phase difference b/w current & voltage in a purely resistive AC circuit is zero
(b) phase difference b/w current & voltage in a pure inductive AC circuit is 𝜋/2; current lags voltage.
(c) phase difference b/w current & voltage in a pure capacitive AC circuit is 𝜋/2; current lead voltage.
(d) phase difference b/w current & voltage in an LCR series circuit is = tan^{-1} (X_{c }- X_{L}/R)
Question 9. A geostationary satellite is orbiting around an arbitrary planet ‘P’ at a height of 11R above the surface of ‘P’, R being the radius of ‘P’. The time period of another satellite in hours at a height of 2R from the surface of ‘P’ is _____. ‘P’ has a time period of rotation of 24 hours.
Solution:
Answer: (d)
From Kepler’s law
T^{2}∝ R^{3}
Since the time period of a geostationary satellite is equal to the time period of rotation of the planet i.e. 24 hours,
Using Kepler’s law,
(24/T)^{2}= (12R/3R)^{3}
T = 3 sec
Question 10. The velocity of a particle is v=v_{0}+gt+Ft^{2}. Its position is x=0 at t=0; then its displacement after time (t = 1) is :
Solution:
Answer: (d)
v = v_{0} + gt + Ft^{2}
dy/dt=v_{0}+gt+Ft^{2}
x=v_{0}+g/2+F/3
Question 11. A block of mass 1 kg attached to a spring is made to oscillate with an initial amplitude of 12 cm. After 2 minutes the amplitude decreases to 6 cm. Determine the value of the damping constant for this motion. (take ln2 = 0.693).
Solution:
Answer: (c)
Amplitude of a damped oscillation is given by;
6 = 12 e^{–b×60}
1/2=e^{-60b }
In(2) = 60b
b = ln(2)/60=1.16×10^{-2}Kg/s
Question 12. An object is located 2 km beneath the surface of the water. If the fractional compression ∆V/V is 1.36%, the ratio of hydraulic stress to the corresponding hydraulic strain will be ……..
[Given: density of water is 1000 kgm^{–3} and g= 9.8 ms^{–2}]
Solution:
Answer: (d)
Bulk modulus
β = Δp/(ΔV/V)
β = 1.44 × 10^{9} N/m^{2 }
Question 13. Two identical photocathodes receive the light of frequencies f_{1} and f_{2} respectively. If the velocities of the photo-electrons coming out are v_{1} and v_{2} respectively, then
Solution:
Answer: (d)
Using the expression for K.E of photoelectrons
[φ is the same for same material of photocathodes]
Subtracting equation (1) by equation (2)
Question 14. The atomic hydrogen emits a line spectrum consisting of various series. Which series of hydrogen atomic spectra lie in the visible region?
Solution:
Answer: (a)
Balmer series of hydrogen atomic spectrum is lying in the visible region when the electron jumps from a higher energy level to n = 2 orbit.
Question 15. Which one of the following will be the output of the given circuit?
Solution:
Answer: (c)
Output of the above circuit can be obtained as
which is XOR gate
Question 16. The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of 15 resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.
Solution:
Answer: (a)
Applying KCL for point B,
3V_{B}-30+20V_{B}-20V_{D}+30V_{B}=0
53V_{B }– 20V_{D} = 30 ____(1)
Similarly applying KCL for point D,
V_{D} – 10 + 4V_{D} – 4V_{B }+ 12V_{D }= 0
- 4V_{B} + 17V_{D} = 10 _____(2)
after solving equation (1) & (2)
V_{D} = 0.792 volt
V_{B} = 0.865 volt
Then the current through the galvanometer
= 4.87 mA
Question 17. Which one is the correct option depicting the two different thermodynamic processes?
Solution:
Answer: (a)
Isothermal process means constant temperature which is only possible in the graph (c) & (d)
for adiabatic process
PV^{γ} = constant -------(1)
Therefore, PV = nRT
P∝T/V
So, (T/V)V^{γ} = constant
TV^{γ-1} = constant -------(2)
Similarly,
V ∝ T/P
P(T/P)^{γ} = constant
P^{1- γ}T^{γ} = constant -------(3)
Therefore, differentiating equation (3) w.r.t temp
P^{1-γ}γT^{γ-1}dT+T^{γ}(1-γ)P^{1-γ-1}dP=0
It gives (+ve) slope.
Question 18. A hairpin-like shape as shown in the figure is made by bending a long current-carrying wire. What is the magnitude of a magnetic field at point P which lies on the centre of the semicircle?
Solution:
Answer: (b)
Magnetic field due to each of the straight wire =
Magnetic field due to semicircular arc =
Thus, the total magnetic field at point P
Question 19. A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to (81/100) of the height through which it falls. Find the average speed of the ball. (Take g=10 ms^{-2})
Solution:
Answer: (a)
Total distance d = h + 2e^{2}h + 2e^{4}h + 2e^{6}h + 2e^{8}h +….
d = h + 2e^{2}h (1+e^{2}+e^{4}+e^{6}+…..)
Total time = T + 2 eT + 2e2T + 2e3 T +…..
Total time = T + 2e T (1+e + e2 + e3+……)
Total time =
Average speed of the ball
Therefore, h’ = e^{2}h
(81/100) = e^{2}
e = 9/10 = 0.9
= 2.50 m/sec
Question 20. What happens to the inductive reactance and the current in a purely inductive circuit if the frequency is halved?
Solution:
Answer: (c)
X_{L} = ωL
If the frequency is halved, X'_{L}=(X_{L}/2)
[inductive reactance is halved]
Therefore, I = V/X_{L }
& I’ = 2V/X_{L}=2I [current will be doubled]
SECTION –B
Question 21. The electric field intensity produced by the radiation coming from a 100 W bulb at a distance of 3 m is E. The electric field intensity produced by the radiation coming from 60 W at the same distance is √(x/5)E. Where the value of x = ________
Solution:
Answer: (3)
The intensity of electromagnetic radiation,
I = 1/2C∈_{0}E^{2 }
where E is electric field intensity at a point
E^{2} ∝ I
I = Power/Area
E^{2} ∝ (P/A)
E ∝ √P
[at the same distance, A will be the same]
So the value of x = 3
Question 22. The image of an object placed in the air formed by a convex refracting surface is at a distance of 10 m behind the surface. The image is real and is at 2^{nd}/3 of the distance of the object from the surface. The wavelength of light inside the surface is 2/3 times the wavelength in air, The radius of the curved surface is x/13 m. The value of ‘x’ is _____
Solution:
Answer: (30)
Since wavelength of light inside the surface is 32 times the wavelength in air,
R =150/65)
R = (30/13)
Thus, the value of x = 30
Question 23. A 2μF capacitor C_{1} is first charged to a potential difference of 10V using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor C_{2} of 8 μF. The charge in C_{2} on equilibrium condition is ________ μC. (Round off to the Nearest Integer)
Solution:
Answer: (16)
After capacitor C_{1} is fully charged,
When battery is removed & the capacitor is connected
At equilibrium condition, let voltage across each capacitor be V.
Then, using conservation of charge
2V + 8V = 20
10V = 20
V = 2 volt
Q = CV
Q = 8×2 = 16μc
Question 24. A particle of mass m moves in a circular orbit in a central potential field U(r) =U_{0r}4. If Bohr's quantization conditions are applied, radii of possible orbital r_{n} vary with n^{1/α}, where α is _____
Solution:
Answer: (3)
Therefore, mv^{2}/r = 4U_{0}r^{3}
Then, v∝r^{2}
Therefore, mvr = nh/2𝜋
Thus, r^{3}α n
r α (n)^{1/3 }
So the value of α = 3
Question 25. The electric field in a region is given by
Solution:
Answer: (640)
From Gauss’ law
= (2/5) E_{0} x (0.4)
= (2/5) x 4 x 10^{3} x 0.4
Φ = 640 Nm^{2} c^{-1}
Question 26. A body of mass 1 kg rests on a horizontal floor with which it has a coefficient of static friction 1/√3. It is desired to make the body move by applying the minimum possible force F N. The value of F will be __________. (Round off to the Nearest Integer)
[Take g = 10 ms^{-2}]
Solution:
Answer: (5)
Minimum possible force ⇒
Fmin = 5 N
Question 27. Seawater at a frequency f = 9 × 10^{2} Hz, has permittivity ∈ = 80∈_{0} and resistivity ρ = 0.25 Ωm. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source V(t) = V_{0} sin(2πft). Then the conduction current density becomes 10x times the displacement current density after time t= 1800. The value of x is __________.
(Given : 1/4𝜋∈_{0} = 9 × 10^{9}Nm^{2}C^{-2})
Solution:
Answer: (6)
Given:-
f = 9 × 10^{2} Hz
∈=∈_{0}∈_{r}
∈=80∈_{0}
So, ∈_{r}=80
ρ = 0.25 Ωm
V(t) = V_{0} sin (2πft)
Displacement current, I_{d} = dq/dt=(cdv/dt)
Where d is the distance between plates & conduction current I_{c }= V/R
Divide equation (1) and (2)
I_{c} = 10^{6} I_{d}
So x = 6
Question 28. The disc of mass M with uniform surface mass density σ is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position x/3, R/𝜋,xR/3𝜋. x is ____(Round off to the Nearest Integer) [a is an area as shown in the figure]
Solution:
Answer: (4)
Each segment of disc subtending angle dθ can be considered as a triangle of height R and base Rdθ
Mass of a segment of disc subtending angle dθ
d = σ (½) R ×Rdθ
d = (σR^{2}dθ)/2
Therefore, x coordinate of COM of quarter disc
Here, x is the x-coordinate of COM of dm mass. And for a triangular section it’s COM is at a cross section of median which will be at a distance 2R/3 from vertex. That’s why they have taken x-coordinate as (2R/3)cosθ
= (2R/3)(2/𝜋)
= 4R/3π
So the value of x = 4
Question 29. Suppose you have taken a dilute solution of oleic acid in such a way that its concentration becomes 0.01 cm^{3 }of oleic acid per cm^{3} of the solution. Then you make a thin film of this solution (monomolecular thickness) of area 4 cm^{2} by considering 100 spherical drops of radius (3/4𝜋)^{⅓} x 10^{-3} cm. Then the thickness of the oleic acid layer will be x × 10^{-14} m where x is _________.
Solution:
Answer: (25)
Volume of film = Area of film × thickness
Also, we know: Volume of film = Volume of 100 spherical drops
4t_{T}=100×(4/3)πr^{3 }
=100×(4π/3) x (3/40π) x 10^{-9 }
=10^{-8} cm^{3}
Hence, the thickness of the film;
t_{T}=25×10^{-10}cm
=25×10^{-12} m
& Thickness of oleic acid layer
t_{0}=0.01
t_{T}=25×10^{-14}m
So, x = 25
Question 30. A boy of mass 4 kg is standing on a piece of wood having mass 5 kg. If the coefficient of friction between the wood and the floor is 0.5, the maximum force that the boy can exert on the rope so that the piece of wood does not move from its place is ________N. (Round off to the Nearest Integer)
[Take g = 10 ms^{-2}]
Solution:
Answer: (30)
[R is the reaction force between man and block while N from ground to block]
From FBD of [man + wooden block] system
Therefore, f=T [horizontal direction]
μN = T
μ(90-T) = T
[Because T + N = 90 (in vertical direction)]
0.5 (90–T) = T
90–T = 2T
3T = 90 T = 30 N