Question

A solid sphere rolls down without slipping on an inclined plane of angle \(30^\circ\). If angle increases to \(45^\circ\), percentage increase in acceleration is nearly

• A. \(73.2\)
• B. \(41.4\)
• C. \(21.2\)
• D. \(36.6\)

Hint:

For rolling bodies acceleration depends on rotational inertia. First write correct moment of inertia and substitute into rolling acceleration formula.

Answer 1

The Correct Option is B

Step 1: Acceleration of a rolling body.
$a=\dfrac{g\sin\theta}{1+\dfrac{I}{mr^2}}$. For a solid sphere $\dfrac{I}{mr^2}=\dfrac25$, so $a=\dfrac{g\sin\theta}{1+\tfrac25}=\dfrac57 g\sin\theta$.
Step 2: Note the proportionality.
Since the $\dfrac57 g$ factor is the same at both angles, $a\propto\sin\theta$.
Step 3: Acceleration at $30^\circ$.
$a_1\propto\sin30^\circ=0.5$.
Step 4: Acceleration at $45^\circ$.
$a_2\propto\sin45^\circ\approx0.707$.
Step 5: Percentage increase.
$\dfrac{a_2-a_1}{a_1}\times100=\dfrac{0.707-0.5}{0.5}\times100=\dfrac{0.207}{0.5}\times100$.
Step 6: Evaluate.
$=0.414\times100=41.4\%$.
\[ \boxed{41.4} \]