Question

A solid sphere rolls down from the top of an inclined plane. On reaching the bottom of the plane, its velocity is '\(V_1\)'. When the same sphere slides down from the top of the same plane of same height, its velocity on reaching the bottom is '\(V_2\)'. The ratio \(V_1:V_2\) is (neglect friction)

• A. \(\sqrt{7}:\sqrt{5}\)
• B. \(\sqrt{7}:\sqrt{3}\)
• C. \(\sqrt{3}:\sqrt{5}\)
• D. \(\sqrt{5}:\sqrt{7}\)

Hint:

Rolling converts part of the energy into rotation; sliding without friction does not.

Answer 1

The Correct Option is D

Step 1: Use the fraction of energy in translation:
For rolling, the ratio of translational KE to total KE is $\frac{1}{1 + I/(mR^2)} = \frac{1}{1 + 2/5} = \frac57$.

Step 2: Link to speed:
Total KE is $mgh$ in both cases. Translational KE $\frac12 mV_1^2 = \frac57mgh$, while $\frac12 mV_2^2 = mgh$.

Step 3: Divide:
$\frac{V_1^2}{V_2^2} = \frac57$, so $V_1 : V_2 = \sqrt5 : \sqrt7$.

Final Answer:
The ratio is root 5 to root 7, option (D). \[ \boxed{\sqrt{5}:\sqrt{7}} \]