JEE Advanced · Physics · 6. Work Power Energy
A small object of uniform density rolls up a curved surface with an initial velocity \(v\). It reaches up to a maximum height of \(\frac{3 v^2}{4 g}\) with respect to the initial\(
\text { position. The object is }
\)
- A ring
- B solid sphere
- C hollow sphere
- D disc
Answer & Solution
Correct Answer
(D) disc
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
\frac{1}{2} m v^2+\frac{1}{2} I\left(\frac{v}{R}\right)^2 & =m g\left(\frac{3 v^2}{4 g}\right) \\
I & =\frac{1}{2} m R^2
\end{aligned}
\)
So, body is disc.
The correct option is (d).
\begin{aligned}
\frac{1}{2} m v^2+\frac{1}{2} I\left(\frac{v}{R}\right)^2 & =m g\left(\frac{3 v^2}{4 g}\right) \\
I & =\frac{1}{2} m R^2
\end{aligned}
\)
So, body is disc.
The correct option is (d).
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