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Question: Answered & Verified by Expert
A solid cube of wood of side \( 2 a \) and mass \( M \) is resting on a horizontal surface as shown in the figure. The cube is free to rotate about a fixed axis \( A B \). A bullet of mass \( m(< < M) \) and speed \( v \) is shot horizontally at the face opposite to \( A B C D \) at a height of \( \frac{4 a}{3} \) from the surface to impart the cube and angular speed \( \omega \). It strikes the face and embeds in the cube. Then \( \omega \) is close to (note: the moment of inertia of the cube about an axis perpendicular to the face and passing through the center of mass is \( \frac{2 M a^{2}}{3} . \) )
PhysicsRotational MotionJEE Main
Options:
  • A \( \frac{M v}{m a} \)
  • B \( \frac{M v}{2 m a} \)
  • C \( \frac{m v}{M a} \)
  • D \( \frac{m v}{2 M a} \)
Solution:
2784 Upvotes Verified Answer
The correct answer is: \( \frac{m v}{2 M a} \)

Applying the parallel axis theorem, I=Icm+Ma2, here Icm is the moment of inertia passing through centre of mass, a is the distance between two axes.

The moment of inertia of solid cube about AB is I=23Ma2+Ma22

I=83Ma2

The cube is free to rotate about AB axis. Thus, the angular momentum should be conserved.

Angular momentum mvr=Iω, here r=4a3.

So, mv×4a3=Iω

ω=mv4a383Ma2=mv2Ma.

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