JEE Main 2020 Physics Paper With Solutions Jan 8 Shift 1

13. If finding the electric field around a surface is given by $\left | \vec{E} \right | = \frac{q_{enclosed}}{\varepsilon_{0} \left | A \right |}$ is applicable. In the formula Ɛ₀ is permittivity of free space, A is area of Gaussian and 𝑞_𝑒𝑛𝑐 is charge enclosed by the Gaussian surface. This equation can be used in which of the following equation?

1. a. Only when the Gaussian surface is an equipotential surface.
2. b. Only when $\left | \vec{E} \right |$ = constant on the surface.
3. c. Equipotential surface and $\left | \vec{E} \right |$ is constant on the surface
4. d. for any choice of Gaussian surfaces.

14. The dimension of stopping potential 𝑉₀ in photoelectric effect in units of Planck’s constant (h), speed of light (c), and gravitational constant (G) and Ampere (A) is

1. a. h^2/3 c^5/3 G^1/3 A⁻¹
2. b. h²c^1/3G^3/2A⁻¹
3. c. h⁰ G ^-1 c ⁵ A⁻¹
4. d. h^−2/3 c^−1/3 G^4/3 A⁻¹

Solution:

1. V = K(h)^a(I)^b(G)^c(c)^d

   Unit of stopping potential is (V₀) Volt.

   We know [h] = ML²T⁻¹

   [I] = A

   [G] = M⁻¹L³T⁻²

   [C] = LT⁻¹

   [V] = ML²T⁻³A⁻¹

   ML²T⁻³A⁻¹ = (ML²T⁻¹)^a(A)^b (M⁻¹L³T⁻²)^c(LT⁻¹)^d

   ML²T^-3A^-1 = M^a-cL^2a+3c+dT^{−a−2c−d}A^b

   a - c = 1

   2a + 3c + d = 2

   -a - 2c - d = -3 b = -1

   On solving,

   c = -1

   a = 0

   d = 5

   b = -1

   V = K(h)⁰(A)⁻¹(G)⁻¹(c)⁵

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15. A leak proof cylinder of length 1 m, made of metal which has very low coefficient of expansion is floating in water at 0^o C such that its height above the water surface is 20 cm. When the temperature of water is increases to 4^o C, the height of the cylinder above the water surface becomes 21 cm. The density of water at T = 4^o C relative to the density at T = 0^o C is close to

1. a. 1.01
2. b. 1.03
3. c. 1.26
4. d. 1.04

Solution:

1. 

   Since the cylinder is in equilibrium, it’s weight is balanced by the Buoyant force.

   mg = A(80)(ρ_0oc)g

   mg = A(79)(ρ_4oc)g

   ρ₄^oc/ ρ_0oc = 80/79 = 1.01

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16. The graph which depicts the result of Rutherford gold foil experiment with α- particle is:

𝜃: Scattering angle

𝑁 ∶ Number of scattered 𝛼 − particles is detected

(Plots are schematic and not to scale)

Solution:

1. 
   

   N ∝ 1/sin⁴(θ/2)

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17. At time t = 0 magnetic field of 1000 Gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to 500 Gauss, in the next 5 s, then induced EMF in the loop is:

1. a. 56 μV
2. b. 28 μV
3. c. 30 μV
4. d. 48 μV

Solution:

1. $\epsilon =\left | \frac{-d\phi }{dt} \right | = \left | \frac{-A\: dB }{dt} \right |$

   $(16\times 4-4\times 2)\frac{(1000-500)}{5}\times 10^{-4}\times 10^{-4}$

   = 56×500×10^-8/5

   = 56× 10^-6 V

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18. Choose the correct Boolean expression for the given circuit diagram:

1. a. A.B
2. b. A + B
3. c. $\bar{A}+ \bar{B}$
4. d. $\bar{A}. \bar{B}$

Solution:

1. First part of figure shown OR gate and second part of figure shown NOT gate.

   So, Y = $\bar{A}. \bar{B}$

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19. Consider a solid sphere of density ρ(r) = ρ₀ (1- r²/R²), 0< r ≤ R. The minimum density of a liquid in which it float is just

1. a. (2/5) ρ₀
2. b. (2/3) ρ₀
3. c. ρ₀/5
4. d. ρ₀/3

Solution:

1. Let the mass of the sphere be m and the density of the liquid be ρ_L

   ρ_{= ρ0 ( 1 - r²/R²), 0< r ≤ R}

   Since the sphere is floating in the liquid, buoyancy force (F_B) due to liquid will balance the weight of the sphere.

   𝐹 = mg

   ρ_L (4π/3) R³g = ∫ρ(4πr² dr)g

   ρ_L (4π/3) R³ = ∫ ρ₀ ( 1 - r²/R²) 4πr²dr

   ρ_L (4/3)π R³ = $\int_{0}^{R}\rho _{0}4\pi (r^{2}-\frac{r^{4}}{R^{2}})dr= \rho _{0}4\pi\left ( \frac{r^{3}}{3} -\frac{r^{5}}{5R^{2}}\right )_{0}^{R}$

   ρ_L = (2/5) ρ₀

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20. The critical angle of a medium for a specific wavelength, if the medium has relative permittivity 3 and relative permeability 4/3 for this wavelength, will be

1. a. 15°
2. b. 30°
3. c. 45°
4. d. 60°

Solution:

1. If the speed of light in the given medium is V then,

   V = 1/√(μƐ)

   We know that, n = c/v

   n = √(μ_r Ɛ_r) = 2

   sin θ_c = ½

   θ_c = 30⁰

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21. A body of mass 𝑚=0.10 𝑘𝑔 has an initial velocity of $3\hat{i}$ 𝑚/𝑠. It collides elastically with another body, B of the mass which has an initial velocity of $5\hat{j}$ 𝑚/𝑠. After collision, A moves with a velocity $v= 4(\hat{i}+\hat{j})$ 𝑚/𝑠. The energy of B after collision is written as (𝑥/10) 𝐽, the value of 𝑥 is

Solution:

1. Mass of each object, 𝑚₁= 𝑚₂ = 0.1 𝑘𝑔

   Initial velocity of 1st object, 𝑢₁=5 𝑚/𝑠

   Initial velocity of 2nd object, 𝑢₂ =3 𝑚/𝑠

   Final velocity of 1st object, 𝑉₁= $v= 4\hat{i}+4\hat{j}$ 𝑚/𝑠 =√(4²+4² )= 16√2 𝑚/𝑠

   For elastic collision, kinetic energy remains conserved

   Initial kinetic energy (K_i) = final kinetic energy (K_f)

   ½ mu₁² + ½ mu₂^{2 =} ½ mV₁² + ½ mV₂²

   ½ m(5)² + ½ m(3)^{2 =} ½ m(16√2)² + ½ mV₂²

   V₂ = √2 m/s

   Kinetic energy of second object = ½ mV₂²

   = ½ × 0.1 × √2²

   = 0.1

   = 1/10 J

   x = 1

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22. A point object in air is in front of the curved surface of a plano-convex lens. The radius of curvature of the curved surface is 30 cm and the refractive index of lens material is 1.5, then the focal length of the lens (in cm) is

Solution:

1. Applying Lens makers’ formula,

   1/f = (μ - 1)[(1/R₁) - (1/R₂)]

   R₁ =∞

   R₂ = -30 cm

   μ = 1.5

   1/f = (1.5 - 1)[(1/∞) - (1/-30)]

   1/f = 0.5/30

   f = 60 cm

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23. A particle is moving along the x-axis with its coordinate with time t given by x(t) = -3t² + 8t + 10 𝑚. Another particle is moving along the y-axis with its coordinate as a function of time given by y = 5 - 8t³ m. At t = 1 s, the speed of the second particle as measured in the frame of the first particle is given as √v . Then v (m/s) is

Solution:

1. x = −3t² + 8t + 10

   $\vec{V_{A}}$ = (−6t + 8)$\hat{i}$

   = 2 $\hat{i}$

   y = 5 − 8t³

   $\vec{V_{B}}$ = −24t² $\hat{j}$

   $\left | \overrightarrow{V_{B/A}} \right | = \left | \overrightarrow{V_{B}} -\overrightarrow{V_{A}} \right | = \left | -24\hat{j} -2\hat{i}\right |$

   v = √(24² + 2²)

   v = 580 m/s

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24. A one metre long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP is 300 m/s, the frequency difference between the fundamental and second harmonic of this pipe is __Hz.

Solution:

1. V = √(B/ρ)

   V_pipe/ V_air = $\sqrt{\frac{\frac{B}{2\rho }}{\frac{B}{\rho }}}= \frac{1}{\sqrt{2}}$

   V_pipe = V_air/√2

   f_n = V_pipe(n+1)/2l

   f₁-f₀ = V_pipe/ 2l

   = 300/2√2

   = 106.06 Hz ( if √2 = 1.414) ≈ 106 Hz

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25. Four resistors of resistance 15 Ω, 12 Ω, 4Ω and 10Ω respectively in cyclic order to form a wheatstone’s network. The resistance that is to be connected in parallel with the resistance of 10 to balance the network is __.

Solution:

1. 

   [(10R)/(10+R) ]×12 = 15 ×4

   R = 10Ω

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