Question

A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities when reached at the bottom of the plane \(\sqrt{\frac{x}{5}}\) where the value of \(x\) is

• A. \(3\)
• B. \(3.5\)
• C. \(4\)
• D. \(2.5\)

Hint:

For rolling without slipping from height h, speed depends on the moment of inertia factor k through v squared equal to 2gh over 1 plus k.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For rolling without slipping from rest, \(v=\sqrt{\dfrac{2gh}{1+\dfrac{I}{mR^2}}}\).

Step 2: Ring and sphere
Ring: \(I=mR^2\), so \(v_1^2=\dfrac{2gh}{2}=gh\).
Solid sphere: \(I=\dfrac{2}{5}mR^2\), so \(v_2^2=\dfrac{2gh}{7/5}=\dfrac{10gh}{7}\).

Step 3: Ratio
\(\dfrac{v_1}{v_2}=\sqrt{\dfrac{gh}{10gh/7}}=\sqrt{\dfrac{7}{10}}\).

Step 4: Find x
Write \(\sqrt{\dfrac{7}{10}}=\sqrt{\dfrac{3.5}{5}}\), so \(x=3.5\), which is option (B).

Final Answer:
The value of x is 3.5. \[ \boxed{\text{(B) }3.5} \]