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: Understanding the Concept:
For a solid sphere of mass \(m\), radius \(R\), \(I = \frac25mR^2\). Using energy conservation from the top of the incline at height \(h\).

Step 2: Rolling case:
\(mgh = \frac12mV_1^2 + \frac12I\omega^2\) with \(\omega = V_1/R\):
\[ mgh = \frac12mV_1^2\left(1 + \frac25\right) = \frac12mV_1^2\cdot\frac75 \Rightarrow V_1^2 = \frac{10}{7}gh \]

Step 3: Sliding case:
No friction, so no rotation: \(mgh = \frac12mV_2^2\), giving \(V_2^2 = 2gh\).

Step 4: Ratio:
\[ \frac{V_1^2}{V_2^2} = \frac{10/7}{2} = \frac57 \Rightarrow V_1 : V_2 = \sqrt5 : \sqrt7 \]

Step 5: Why the other options are wrong.
Options (A) and (B) put the larger number first, which would make the rolling sphere faster. Rolling is slower because some energy goes into rotation. Option (C) uses \(\sqrt3\), which does not match the moment of inertia of a sphere.

Final Answer:
The ratio is \(\sqrt5 : \sqrt7\), option (D). \[ \boxed{\sqrt{5}:\sqrt{7}} \]