Question

A solid sphere of radius 18 cm is rolling down from rest from the top of an inclined plane of length 14 m and angle of inclination $30^\circ$. The time taken by the sphere to reach the bottom of the inclined plane is (Acceleration due to gravity = $10~\text{m s}^{-2}$)

• A. 2.8 s
• B. 4.2 s
• C. 3.5 s
• D. 1.4 s

Hint:

For rolling bodies, use $a = \frac{g \sin \theta}{1 + k}$ where $k$ depends on shape: sphere = 2/5, cylinder = 1/2. Use $s = \frac{1}{2} a t^2$ for distance along plane.

Answer 1

The Correct Option is C

1. For rolling sphere, $a = \frac{g \sin \theta}{1 + k}$, $k = \frac{2}{5}$ for solid sphere.
2. $a = \frac{10 \sin 30^\circ}{1 + 2/5} = \frac{5}{1.4} \approx 3.57~\text{m s}^{-2}$.
3. Using $s = \frac{1}{2} a t^2 \implies 14 = \frac{1}{2} \cdot 3.57 \cdot t^2 \implies t^2 \approx 7.84 \implies t \approx 2.8~\text{s}$.