JEE Main 2021 Physics Paper With Solutions July 20 Shift 2

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July 20 Shift 2 – Physics Question Paper and Solutions

Question 1: If the kinetic energy of a particle becomes four times, then % change in momentum will be:

a. 200
b. 100
c. 150
d. 50

Answer: (b)

K.E. ⇒ K = P² / 2m

P ∝ √K

\(\begin{array}{l}\frac{P_{2}}{P_{1}} = \sqrt{\frac{K_{2}}{K_{1}}} \Rightarrow \frac{P_{2}}{P_{1}} = \sqrt{\frac{4K}{K}}\end{array} \)

⇒
\(\begin{array}{l}\frac{P_{2}}{P_{1}} = 2\end{array} \)

⇒
\(\begin{array}{l}\frac{P_{2} – P_{1}}{P_{1}}% = \frac{P_{2}}{P_{1}-1}\times 100 = (2-1)\times 100 = 100\end{array} \)

⇒
\(\begin{array}{l}\frac{\Delta P}{P_{1}}% = 100%\end{array} \)

Question 2: A RLC circuit is in its resonance condition. Its circuit components have value R = 5Ω, L = 2H, C = 0.5mF, V= 250V. Then find the power in the circuit.

a. 6 kW
b. 10 kW
c. 12 kW
d. 12.5 kW

Answer: (d)

As the circuit is in resonance. Thus,

X_L = X_C

∴ Z = R so, i_rms = V / Z = V / R

\(\begin{array}{l}P = i_{rms}^{2}R\end{array} \)

\(\begin{array}{l}P = \frac{V^{2}}{R} = \frac{250\times 250}{5} = 12500\frac{J}{S} = 12.5\: kW\end{array} \)

Question 3: A satellite is revolving around a planet in an orbit of radius R. Suddenly radius of orbit becomes 1.02 R, then what will be the percentage change in its time period of revolution?

a. 2%
b. 3%
c. 5%
d. 7%

Answer: (b)

\(\begin{array}{l}T \alpha R^{\frac{3}{2}}\end{array} \)

\(\begin{array}{l}T = kR^{\frac{3}{2}}\end{array} \)

\(\begin{array}{l}\frac{\Delta T}{T} = \frac{3}{2}\frac{\Delta R}{R} = 3%\end{array} \)
%

Question 4: A person walks up a stationary escalator in the time 𝑡₁. If he remains stationary on the escalator, then it can take him up in time 𝑡₂. Determine the time it would take to walk upon the moving escalator?

a.
\(\begin{array}{l}\frac{t_{1}t_{2}}{t_{1}+t_{2}}\end{array} \)

b.
\(\begin{array}{l}\frac{t_{1}t_{2}}{t_{1}-t_{2}}\end{array} \)

c.
\(\begin{array}{l}\frac{2t_{1}t_{2}}{t_{1}+t_{2}}\end{array} \)

d.
\(\begin{array}{l}\frac{2t_{1}t_{2}}{t_{1}-t_{2}}\end{array} \)

Answer: (a)

Suppose length of escalator = L

Speed of man w.r.t escalator L / t₁

Speed of escalator = L / t₂

Speed of man w.r.t ground when escalator is moving = L / t₁ + L / t₂

Time taken by the man to walk on the moving escalator =
\(\begin{array}{l}\frac{L}{\frac{L}{t_{1}}+\frac{L}{t_{2}}} = \frac{t_{1}t_{2}}{t_{1}+t_{2}}\end{array} \)

Question 5: For the given graph between decay rate and time find half-life (where R = decay rate).

a. 10 / 3 ln⁡2
b. 20 / 3 ln⁡2
c. 3 / 20 ln⁡2
d. 3 / 10 ln⁡2

Answer: (a)

\(\begin{array}{l}R = R_{o}e^{-\lambda t}\end{array} \)

ln R = ln ⁡R_o – λt

Slope = -λ = – 6 / 40

λ = 3 / 20

\(\begin{array}{l}t_{1/2} = \frac{ln2}{\lambda} = \frac{ln2}{3}\times 20 = \frac{20}{3}ln2\end{array} \)

Question 6. A wheel rotating with an angular speed of 600 rpm is given a constant acceleration of 1800 rpm² for 10 sec. The number of revolutions revolved by the wheel is.

a. 125
b. 100
c. 75
d. 50

Answer: (a)

ω_o = 600 rpm

α = 1800 rpm²

t = 10 sec = 1/6 min

θ = ω_ot + ½ αt²

θ = 600 ×10 / 60 + ½ × 1800 × 1 / 36

θ = 100 + 25 = 125 revolutions

Question 7:
\(\begin{array}{l}\left | \vec{P} \right |=\left | \vec{Q} \right |,\left | \vec{P}+\vec{Q} \right |=\left | \vec{P}-\vec{Q} \right |\end{array} \)
. Find the angle between
\(\begin{array}{l}\left | \vec{P} \right |\end{array} \)
and
\(\begin{array}{l}\left | \vec{Q} \right |\end{array} \)

a. 45°
b. 90°
c. 135°
d. 150°

Answer: (b)

\(\begin{array}{l}\left | \vec{P} + \vec{Q}\right | = \left | \vec{P} – \vec{Q}\right |\end{array} \)

\(\begin{array}{l}\left | \vec{P} \right |^{2} + \left | \vec{Q}\right |^{2} + 2\left | \vec{P} \right |\left | \vec{Q}\right | \: cos\Theta = \left | \vec{P} \right |^{2} + \left | \vec{Q}\right |^{2} – 2\left | \vec{P} \right |\left | \vec{Q}\right | \: cos\Theta\end{array} \)

\(\begin{array}{l}\left | \vec{P} \right |\left | \vec{Q}\right | \: cos\Theta = 0\end{array} \)

θ = 90°

Question 8: Time (T), Velocity (C) and angular momentum (h) is chosen as fundamental quantities instead of mass, length and time. In terms of these, dimension of mass would be

a. [M] = [T^-1 C^-2 h]
b. [M] = [T^-1 C² h]
c. [M] = [T^-1 C^-2 h^-1 ]
d. [M] = [TC^-2 h]

Answer: (a)

M ∝ T^x c^y h^z
[ M¹ L⁰ T⁰] = [T^x] [LT^-1]^y [ML2T^-1]^z

On comparing the powers, we get:

Z = 1 …………. (1)

X – y + z = 0. ……….(2)

Y +2z = 0 …………. (3)

So, y = -2

X = -1
[M] = [T^-1 C^-2 h]

Question 9: Find the relation between 𝛾 (adiabatic constant) and degree of freedom (𝑓).

a. f = 2 / γ – 1
b. f = γ / γ – 1
c. f = γ – 1 /2
d. f = γ-1 / γ

Answer: (a)

We know that,

C_v = f R / 2

C_p = (f / 2 + 1)R R

γ = C_p / C_v = 1 + 2 / f

F = 2 / γ – 1

Question 10: Two identical drops of Hg coalesce to form a bigger drop. Find the ratio of the surface energy of a bigger drop to a smaller drop.

a. 2 ^3/2
b. 3^2/5
c. 2 ^2/3
d. 5 ^2/3

Answer: (c)

\(\begin{array}{l}2\times \frac{4}{3}\pi r^{3} = \frac{4}{3}\pi R^{3}\end{array} \)

\(\begin{array}{l}\frac{R}{r} = 2^{\frac{1}{3}}\end{array} \)

\(\begin{array}{l}\frac{U_{bigger}}{U_{smaller}}=\frac{S\times4\pi R^{2}}{S\times4\pi r^{2}}=(\frac{R}{r})^{2} = 2^{\frac{2}{3}}\end{array} \)

Question 11: The velocities of a particle performing SHM at a distance of x₁ and x₂ from mean position are v₁ and v₂, find the time period of oscillation?

a.
\(\begin{array}{l}2\pi \sqrt{\frac{x^{\frac{2}{2}}+x^{\frac{2}{1}}}{v^{\frac{2}{1}}-v^{\frac{2}{2}}}}\end{array} \)

b.
\(\begin{array}{l}2\pi \sqrt{\frac{x^{\frac{2}{2}}-x^{\frac{2}{1}}}{v^{\frac{2}{1}}+v^{\frac{2}{2}}}}\end{array} \)

c.
\(\begin{array}{l}2\pi \sqrt{\frac{x^{\frac{2}{2}}-x^{\frac{2}{1}}}{v^{\frac{2}{1}}-v^{\frac{2}{2}}}}\end{array} \)

d.
\(\begin{array}{l}2\pi \sqrt{\frac{x^{\frac{2}{2}}+x^{\frac{2}{1}}}{v^{\frac{2}{1}}+v^{\frac{2}{2}}}}\end{array} \)

Answer: (c)

\(\begin{array}{l}v = \omega \sqrt{A^{2} – x^{2}}\end{array} \)

\(\begin{array}{l}v_{1} = \omega \sqrt{A^{2} – x_{1}^{2}}\end{array} \)

\(\begin{array}{l}v_{2} = \omega \sqrt{A^{2} – x_{2}^{2}}\end{array} \)

\(\begin{array}{l}(\frac{v_{1}}{\omega})^{2} – (\frac{v_{2}}{\omega})^{2} = x_{2}^{2} – x_{1}^{2}\end{array} \)

\(\begin{array}{l}\omega^{2} = \frac{v_{1}^{2}-v_{2}^{2}}{x_{2}^{2} – x_{1}^{2}}\end{array} \)

\(\begin{array}{l}\omega = \sqrt{\frac{v_{1}^{2}-v_{2}^{2}}{x_{2}^{2} – x_{1}^{2}}}\end{array} \)

\(\begin{array}{l}T =2\pi \sqrt{\frac{x_{2}^{2} – x_{1}^{2}}{v_{1}^{2}-v_{2}^{2}}}\end{array} \)

Question 12: Identify the correct graph between PV and T for an ideal gas.

Answer: (c)

PV = nRT ⇒ PV=CT

Therefore, PV v/s T graph is a straight line.

Question 13: In photoelectric effect stopping potential is 3V₀ for incident wavelength λ₀ and stopping potential for incident wavelength 2λ₀. Find threshold wavelength.

a. 3λ₀
b. 2λ₀
c. 4λ₀
d. 8λ

Answer: (c)

KE = hv – W

eV = hc / λ – W

For first case

e(3V_o) = hc /λ_o – W ……..(i)

For second case:

e(V_o)=hc / 2λ_o – W ……(ii)

From equation (i) and (ii), we get,

W = hc / (4λ_o)

For λ_th

w = hc / λ_th

⇒ hc / (4λ_o) = hc/λ_th

⇒ λ_th = 4λ_o

Question 14: A plane electromagnetic wave travels in free space. Electric field is
\(\begin{array}{l}\left | \vec{E} \right |=E_{0}\hat{i}\end{array} \)
and magnetic field is represented by
\(\begin{array}{l}\left | \vec{B} \right |=B_{0}\hat{k}\end{array} \)
. What is the unit vector along the direction of propagation of electromagnetic waves?

a.
\(\begin{array}{l}\hat{j}\end{array} \)

b.
\(\begin{array}{l}-\hat{k}\end{array} \)

c.
\(\begin{array}{l}-\hat{j}\end{array} \)

d.
\(\begin{array}{l}\hat{k}\end{array} \)

Answer: (c)

Direction of the EM wave is given by direction of
\(\begin{array}{l}\vec{E}\times \vec{B}\end{array} \)
.

Unit vector in direction
\(\begin{array}{l}\vec{E}\times \vec{B}\end{array} \)
, =
\(\begin{array}{l}\frac{\vec{E}\times \vec{B}}{\left | \vec{E}\times \vec{B} \right |}\end{array} \)

\(\begin{array}{l}\frac{E_{o}\hat{i}\times B_{o}\hat{k}}{E_{o}B_{o}Sin\: 90^{o}}\end{array} \)

\(\begin{array}{l}\hat{i}\times \hat{k}\end{array} \)

\(\begin{array}{l}-\hat{J}\end{array} \)

Question 15: Two satellites of mass M_A and M_B are revolving around a planet of mass M in radius R_A and R_B respectively. Then

a. T_A> T_Bif R_A > R_B
b. T_A> T_B if M_A> M_B
c. T_A= T_B if M_A> M_B
d. T_A > T_B if R_A< R_B

Answer: (a)

According to Kepler’s law of planetary motion,

T ∝ R^3/2

\(\begin{array}{l}\frac{T_{A}}{T_{B}} = \left ( \frac{R_{A}}{R_{B}} \right )^{\frac{3}{2}}\end{array} \)

So, if R_A > R_B then T_A > T_B.

Hence, option (a) is the correct answer.

Question 16: At 45° of the magnetic meridian angle of dip is 30° then find the angle of dip in a vertical plane at 45°?

a. tan⁻¹(1/√6)
b. tan⁻¹(1/√2)
c. tan⁻¹ (1/√4)
d. tan⁻¹(1/√3)

Answer: (a)

Let the horizontal and vertical components of the earth’s magnetic field at the meridian be V and H.

Angle of dip, tan⁡ θ = V / H……(i)

At angle of 45° from magnetic meridian, angle of dip = 30°

tan⁡ 30° = V / H cos⁡ 45°

⇒1 / √3 = V / H cos⁡ 45°

V / H = 1 / √6

tan ⁡θ = V / H = 1 / √6

⇒θ = tan^-1 ⁡(1 / √6)

Question 17: A sodium lamp in space was emitting waves of wavelength 2880 Å. When observed from a planet, its wavelength was recorded at 2886 Å. Find the speed of the planet?

a. 4.25 × 10⁵ m/s
b. 6.25 × 10⁵ m/s
c. 2.75 × 10⁵ m/s
d. 3.75 × 10⁵ m/s

Answer: (b)

\(\begin{array}{l}\frac{V_{rel}}{C} = \frac{\Delta \lambda}{\lambda}\end{array} \)

\(\begin{array}{l}V_{rel} = \frac{6}{2880}\times3 \times 10^{8}\end{array} \)

\(\begin{array}{l}V_{rel} = 6.25 \times 10^{5} m/s\end{array} \)

Question 18: For a body in pure rolling, its rotational kinetic energy is 1/2 times of its translational kinetic energy. The body should be?

a. Solid cylinder
b. Ring
c. Solid sphere
d. Hollow sphere

Answer: (a)

Given,

Rotational K.E = ½ Translational K.E

½ Iω² = ½ × ½ mv²

In pure rolling, v = Rω

½ Iω² = ¼ mR² ω²

I = ½ mR²

Hence, it is a solid cylinder.

Question 19: Magnetic susceptibility of a material is 499 & μ₀ = 4π × 10⁻⁷ SI unit. Then find μ_r

a. 500
b. 400
c. 300
d. 200

Answer: (a)

Given, 𝜒 = 499

The relative permeability, μ_r = 1 + 𝜒 = 500

Question 20: An electron having the de Broglie wavelength λ falls on an X-ray tube. The cut-off wavelength of emitted X-ray is

a. 2mcλ²/ h
b. 2h / mc
c. h / mc
d. 2mcλ²/ 3h

Answer: (a)

De-broglie wavelength, λ_B = h / P

P = h / (λ_B)

Kinetic energy of electron, E = P² / (2m_e) = h² / 2m_e λ_B²

For cutoff wavelength of emitted X-ray: E = hc / λ

h² / 2m_e λ_B²= hc / λ

λ = 2m_e λ_B² / h = 2mc λ²/ h where λ_B = λ and m_e = m

Question 21: A body is moved from rest along a straight line by a machine delivering constant power. Time taken by the body to travel a distance 𝑆 is proportional to:

a. 𝑆^1/3
b. 𝑆^2/3
c. 𝑆^1/2
d. 𝑆^1/4

Answer: (b)

Energy supplied in time t sec is,

E = P × t

Here P represents the power delivered by the machine.

⇒ Pt = ½ mv²

v ∝ √t

Writing the velocity in terms of the displacement of the body,

⇒ dS/dt = C√t

Here C is a constant.

⇒
\(\begin{array}{l}\int_{O}^{S}ds = C \int_{O}^{t} t^{\frac{1}{2}}dt\end{array} \)

\(\begin{array}{l}t^{\frac{3}{2}} = \frac{3S}{2C} \Rightarrow t = S^{\frac{2}{3}}\left ( \frac{3}{2C} \right ) \Rightarrow T\: \alpha \: S^{\frac{2}{3}}\end{array} \)

Question 22: A uniform rod of Young’s modulus Y is stretched by two tension forces T₁ and T₂ such that the rods get expanded to length L₁ and L₂ respectively. Find the initial length of the rod?

a.
\(\begin{array}{l}\frac{L_{1}T_{1}-L_{2}T_{2}}{T_{1}-T_{2}}\end{array} \)

b.
\(\begin{array}{l}\frac{L_{2}T_{1}-L_{1}T_{2}}{T_{2}-T_{1}}\end{array} \)

c.
\(\begin{array}{l}\frac{L_{1}T_{2}-L_{2}T_{2}}{T_{2}-T_{1}}\end{array} \)

d.
\(\begin{array}{l}\frac{L_{1}}{T_{1}}\times \frac{T_{2}}{L_{2}}\end{array} \)

Answer: (c)

Let the initial length of the rod to be L_o and area A

In this case the external force on the rod is tension hence from Hook’s law,

\(\begin{array}{l}\frac{F}{A} = Y\frac{\Delta l}{l}\end{array} \)

⇒
\(\begin{array}{l}\frac{T}{A} = Y\frac{\Delta l}{l}\end{array} \)

For both the cases we can relate the tension and elongation produced in rod as,

\(\begin{array}{l}\frac{T_{1}}{A} = \frac{Y(L_{1}-L_{o})}{L_{o}}\end{array} \)
……(i)

Here (L₁ – L_o) is the elongation in rod when T₁ is applied.

Similarly,

\(\begin{array}{l}\frac{T_{2}}{A} = \frac{Y(L_{2}-L_{o})}{L_{o}}\end{array} \)
……(ii)

Here (L₂ – L_o) is the elongation in rod when T₂ is applied.

On dividing Eq.(i) and (ii) we get,

\(\begin{array}{l}\frac{T_{1}}{T_{2}} = \frac{L_{1}-L_{o}}{L_{2}-L_{o}}\end{array} \)

⇒ T₁ L₂ – T₁ L_o = T₂ L₁ – T₂ L_o

Or,

\(\begin{array}{l}L_{o} = \frac{L_{1}T_{2}-L_{2}T_{1}}{T_{2}-T_{1}}\end{array} \)

Question 23: A block is projected up to a rough plane of inclination 30°. If time of ascending is half the time for descending and the coefficient of friction is μ =3/5√𝑛. Then 𝑛 =

Answer: (n = 3)

S = ½ a_At_A² ….. (1)

S = ½ a_Dt_D² ….. (2)

From equation (1) and (2)

\(\begin{array}{l}\frac{t_{A}^{2}}{t_{D}^{2}} = \frac{a_{D}}{a_{A}}\end{array} \)

⇒
\(\begin{array}{l}\frac{t_{A}^{2}}{t_{D}^{2}} = \frac{g \: sin \: \Theta – \mu g \: cos \Theta}{g \: sin \: \Theta + \mu g \: cos \Theta}\end{array} \)

⇒
\(\begin{array}{l}\frac{t_{A}}{t_{D}} = \sqrt{\frac{g \: sin \: \Theta – \mu g \: cos \Theta}{g \: sin \: \Theta + \mu g \: cos \Theta}}\end{array} \)

⇒
\(\begin{array}{l}\frac{1}{2} = \sqrt{\frac{1-\sqrt{3\mu }}{1+\sqrt{3\mu }}}\end{array} \)

⇒ 1 + √3μ = 4 – 4√3 μ

⇒ 5√3μ = 3

⇒ μ = 3 / 5√3

Question 24: I-V characteristic curve of a diode in a forward bias is given in fig. Find out dynamic resistance

a. 212.3 Ω
b. 205.3 Ω
c. 245.3 Ω
d. 233.3 Ω

Answer: (d)

Dynamic resistance = ΔV / ΔI

= 0.7 V / 3 mA = 233.3 Ω

Question 25: An electron is accelerated through a voltage of 40 kV. What will be its wavelength?

a. 0.061 Å
b. 0.011 Å
c. 0.021 Å
d. 0.161 Å

Answer: (a)

λ_B = h / P

= h / √2meV

=
\(\begin{array}{l}\frac{12.27}{\sqrt{V}}\end{array} \)
Å

=
\(\begin{array}{l}\frac{12.27}{\sqrt{40 \times 10^{3}}}\end{array} \)
= 0.061 Å

Question 26: Find the value of R_p in the given circuit? (V_Z = 8V)

a. 4Ω
b. 6Ω
c. 3Ω
d. 5Ω

Answer: (b)

Applying KVL

20 – 8 – 2R_P = 0

R_P = 6 Ω

Question 27: Two stars of masses m₁ and m₂ are in mutual interaction and revolving in orbits of radii r₁ and r₂ respectively. The time period of revolution for this system will be?

a.
\(\begin{array}{l}2\pi \sqrt{\frac{(r_{1}-r_{2})^{3}}{G(m_{1}+m_{2})}}\end{array} \)

b.
\(\begin{array}{l}2\pi \sqrt{\frac{(r_{1}+r_{2})^{3}}{G(m_{1}+m_{2})}}\end{array} \)

c.
\(\begin{array}{l}2\pi \sqrt{\frac{(r_{1}-r_{2})^{3}}{G(m_{1}-m_{2})}}\end{array} \)

d.
\(\begin{array}{l}2\pi \sqrt{\frac{(r_{1}+r_{2})^{3}}{G(m_{1}-m_{2})}}\end{array} \)

Answer: (b)

Let angular velocity will be ω

For mass m₁,

\(\begin{array}{l}\frac{Gm_{1}m_{2}}{(r_{1}+r_{2})^{2}} = m_{1}r_{1}\omega ^{2} = m_{1}\times \frac{m_{2}(r_{1}+r_{2})}{m_{1}+m_{2}}\omega ^{2}\end{array} \)

\(\begin{array}{l}\omega = \frac{\sqrt{G(m_{1}+m_{2})}}{(r_{1}+r_{2})^{\frac{3}{2}}}\end{array} \)

\(\begin{array}{l}T = \frac{2\pi }{\omega }\end{array} \)

\(\begin{array}{l}T = 2\pi\frac{(r_{1}+r_{2})^{3}}{G(m_{1}+m_{2})}\end{array} \)

Question 28: If N₀ active nuclei become N₀/16 in 80 days. Find half-life of nuclei?

a. 40 days
b. 20 days
c. 60 days
d. 30 days

Answer: (b)

Question 29: A gas is undergoing a change in state by an isothermal process AB as follow. Work done by gas in process AB is

a. 100 ln2 Joule
b. −100 ln2 Joule
c. 200 ln2 Joule
d. −200 ln2 Joule

Answer: (c)

W_isothermal = P₁V₁ ln⁡ V₂ / V₁

V₁ = 100 m³

V₂ = 200 m³

P₁ = 2 N/m²

W_isothermal = 2 × 100 ln 200 / 100

W = 200 ln 2 Joule

JEE Main 2021 July 20 Shift 2 Physics Question Paper

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