JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2023
- Hard
- Numerical
A hollow spherical ball of uniform density rolls up a curved surface with an initial velocity \(3\, m / s\) (as shown in figure). Maximum height with respect to the initial position covered by it will be \(...........cm\).

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Answer & Solution
Correct Answer
(A) \(75\)
Step-by-step Solution
Detailed explanation
At highest point \(KE _{ f }=0\)
Initial \(KE =\) Translational \(KE +\) Rotational \(KE\)
\(=\frac{1}{2} mv ^2+\frac{1}{2} I \omega^2\)
In case of rolling \(v = R \omega\)
\(=\frac{1}{2} m v^2+\frac{1}{2} \times \frac{2}{3} m R^2 \times \frac{v^2}{R^2}\)
\(=\frac{5}{6} m v^2\)
Apply energy conservation
\(KE _{ i }+ PE _{ i }= KE _{ f }+ PE _{ f }\)
\(\frac{5}{6} mv ^2= mgh\)
\(h =\frac{5}{6 \times 10} \times 9\,m =\frac{15}{20} m =75\,cm\)
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