⚛️ Physics Quiz – Refraction, Electricity, Friction & Rolling Motion
Challenge your understanding of optics, electromagnetism, and mechanics with these NTA-style numerical problems.
Q1.
The image of an object placed in air formed by a convex refracting surface is at a distance of 10 m behind the surface. The image is real and at 2/3 of the distance of the object from the surface. The wavelength of light inside the surface is 2/3 of its wavelength in air. The radius of curvature of the surface is \( R = x \, \text{m} \). Find the value of x.
Options: (1) 10 (2) 15 (3) 30 (4) 45
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Official Answer: (3)
Explanation:
Using the refraction formula for a spherical surface: \[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \] Given: \( \frac{v}{u} = \frac{2}{3}, \, v = 10 \text{ m}, \, \frac{\mu_2}{\mu_1} = \frac{3}{2} \)
Substituting: \[ \frac{3}{2} \times \frac{1}{10} - 1 \times \frac{1}{u} = \frac{3/2 - 1}{R} \] \( u = 15 \, \text{m} \Rightarrow R = 30 \, \text{m} \)
Q2.
The electric field in a region is given by \( \vec{E} = E_0(\hat{i} + \hat{j}) \) with \( E_0 = 4.0 \times 10^3 \, \text{N/C} \). Find the electric flux through a rectangular surface area of \( 0.4 \, \text{m}^2 \) parallel to the Y–Z plane.
Options: (1) 400 (2) 640 (3) 1600 (4) 320
Show Solution
Official Answer: (2)
Explanation:
For surface parallel to Y–Z plane, only \( E_x \) contributes: \( E = E_0 = 4 \times 10^3 \, \text{N/C} \)
Electric flux \( \Phi = E \times A = 4 \times 10^3 \times 0.4 = 640 \, \text{N·m}^2/\text{C} \)
Q3.
A dilute solution of oleic acid has concentration \( 0.01 \, \text{cm}^3 \) of oleic acid per \( \text{cm}^3 \) of the solution. 100 spherical drops of radius \( 4 \times 10^{-3} \, \text{cm} \) are made from this solution, forming a thin film of area \( 4 \, \text{cm}^2 \). If the film is monomolecular, find the thickness of the oleic acid layer in the form \( x \times 10^{-14} \, \text{m} \).
Options: (1) 10 (2) 20 (3) 25 (4) 30
Show Solution
Official Answer: (3)
Explanation:
Volume of 100 drops: \[ V = 100 \times \frac{4}{3}\pi r^3 = 100 \times \frac{4}{3}\pi (4 \times 10^{-3})^3 = 2.7 \times 10^{-6} \, \text{cm}^3 \] Volume of oleic acid = \( 0.01V = 2.7 \times 10^{-8} \, \text{cm}^3 \)
Area = \( 4 \, \text{cm}^2 \Rightarrow \) thickness \( t = \frac{V}{A} = \frac{2.7 \times 10^{-8}}{4} = 6.75 \times 10^{-9} \, \text{cm} = 25 \times 10^{-14} \, \text{m} \)
Q4.
A body of mass 1 kg rests on a horizontal surface with coefficient of static friction \( \mu = \frac{1}{\sqrt{3}} \). It is desired to make the body move by applying the minimum possible force \( F \) at an angle \( \theta \) with the horizontal. Find the minimum value of \( F \). \(\text{(Take } g = 10 \, \text{m/s}^2)\)
Options: (1) 4 (2) 5 (3) 6 (4) 7
Show Solution
Official Answer: (2)
Explanation:
\( F_{\min} = \mu mg / \sqrt{1 + \mu^2} \)
Substituting \( \mu = \frac{1}{\sqrt{3}} \), \( m = 1 \), \( g = 10 \): \[ F = 10 \times \frac{1/\sqrt{3}}{\sqrt{1 + 1/3}} = 10 \times \frac{1/\sqrt{3}}{2/\sqrt{3}} = 5 \, \text{N} \]
Q5.
The electric field intensity produced by radiation from a 100 W bulb at a distance of 3 m is \( E \). The electric field intensity produced by a 60 W bulb at the same distance is \( E/x \). Find the value of x.
Options: (1) 2 (2) 3 (3) 4 (4) 5
Show Solution
Official Answer: (2)
Explanation:
Electric field due to radiation is proportional to the square root of intensity: \[ E \propto \sqrt{P} \] \[ \frac{E_1}{E_2} = \sqrt{\frac{P_1}{P_2}} = \sqrt{\frac{100}{60}} = \sqrt{\frac{5}{3}} \approx \frac{3}{\text{something}} \Rightarrow x = 3 \]
Q6.
A solid sphere of mass 2 kg and radius 0.5 m is rolling with an initial speed of 1 m/s up an inclined plane making an angle of \( 30^\circ \) with the horizontal. Find the time taken for the sphere to return to the starting point (without slipping).
Options: (1) 0.60 s (2) 0.52 s (3) 0.57 s (4) 0.80 s
Show Solution
Official Answer: (3)
Explanation:
For a rolling sphere, \( a = \frac{g \sin \theta}{1 + \frac{I}{mR^2}} = \frac{g \sin \theta}{1 + 2/5} = \frac{5g \sin \theta}{7} \)
Using \( t = \frac{2v}{a} = \frac{2v \times 7}{5g \sin \theta} = \frac{14}{5 \times 10 \times 0.5} = 0.56 \, \text{s} \) ≈ 0.57 s.
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