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

Concept: Acceleration of rolling body: \[ a=\frac{g\sin\theta}{1+\frac IK} \] For solid sphere \[ I=\frac25mr^2 \] Hence \[ a=\frac{g\sin\theta}{1+\frac25} \] \[ a=\frac57g\sin\theta \]

Step 1: Initial acceleration At \(30^\circ\) \[ a_1=\frac57g\sin30 \] \[ a_1=\frac57g\left(\frac12\right) \] \[ a_1=\frac{5g}{14} \]

Step 2: Final acceleration At \(45^\circ\) \[ a_2=\frac57g\sin45 \] \[ a_2=\frac57g\left(\frac1{\sqrt2}\right) \]

Step 3: Percentage increase \[ \%\ increase=\frac{a_2-a_1}{a_1}\times100 \] \[ =\left(\frac{\sin45-\sin30}{\sin30}\right)\times100 \] \[ =\left(\frac{0.707-0.5}{0.5}\right)\times100 \] \[ =41.4\% \] Thus \[ \boxed{41.4} \]