Question

A small object of uniform density rolls up a curved surface with an initial velocity \( v \). It reaches upto a maximum height of \( \frac{7v^2}{10g} \) with respect to initial position. Then the object is

• A. ring
• B. solid sphere
• C. hollow sphere
• D. disc

Hint:

Always remember standard \( \frac{I}{mR^2} \) values: Ring = 1, Disc = \( \frac{1}{2} \), Solid sphere = \( \frac{2}{5} \), Hollow sphere = \( \frac{2}{3} \).

Answer 1

The Correct Option is B

Concept: For rolling motion: \[ \text{Total K.E.} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \] Using \( v = \omega R \): \[ \text{Total K.E.} = \frac{1}{2}mv^2 \left(1 + \frac{I}{mR^2}\right) \] At maximum height: \[ mgh = \frac{1}{2}mv^2 \left(1 + \frac{I}{mR^2}\right) \]

Step 1: Compare with given height.
\[ h = \frac{7v^2}{10g} \] So, \[ \frac{1}{2}\left(1 + \frac{I}{mR^2}\right) = \frac{7}{10} \]

Step 2: Solve for moment of inertia term.
\[ 1 + \frac{I}{mR^2} = \frac{7}{5} \Rightarrow \frac{I}{mR^2} = \frac{2}{5} \]

Step 3: Identify object.
\[ I = \frac{2}{5}mR^2 \] This corresponds to a solid sphere.