JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2016
- Hard
- MCQ
A cubical block of side \(30\,cm\) is moving with velocity \(2\,\,ms^{-1}\) on a smooth horizontal surface. The surface has a bump at a point \(O\) as shown in figure. The angular velocity (in rad/s) of the block immediately after it hits the bump, is

Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(5.0\)
Step-by-step Solution
Detailed explanation
Angular momentum, \(mvr = I\omega \)
momrnt of Inertia \((I)\) of cubical block is given by
\(\begin{array}{l}
I = m\left( {\frac{{{R^2}}}{6} + {{\left( {\frac{R}{{\sqrt 2 }}} \right)}^2}} \right)\\
\therefore \,\,\omega = \frac{{m.2\frac{R}{2}}}{{m\left( {\frac{{{R^2}}}{6} + {{\left( {\frac{R}{{\sqrt 2 }}} \right)}^2}} \right)}}\\
\Rightarrow \omega = \frac{{12}}{{8R}} = \frac{3}{{2 \times 0.3}} = \frac{{10}}{2} = 5\,rad/s.
\end{array}\)
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