MHT CET202613 April 2026Evening ShiftPhysicsRotational MotionActual
A thin uniform rod of length 2 m cross-sectional area 'A' and density 'd' is rotated about an axis passing through the centre and perpendicular to its length with angular velocity . If the value of in terms of its rotational kinetic energy E is ( E/Ad)^ 1/2 , then the value of is
Options
- A2
- B3
- C4
- D5
Correct answer
B. 3
Step-by-step solution
Mass of the rod, M = Volume density = A L d Given L = 2 , M = 2Ad Moment of inertia of the rod about an axis passing through its centre and perpendicular to its length is I = ML^2 12 Substituting the values of M and L , we get I = 2Ad 2^2 12 = 2Ad 3 Rotational kinetic energy is given by E = 1 2 I ^2 Substituting the value of I , E = 1 2 ( 2Ad 3 ) ^2 = Ad 3 ^2 Rearranging for , we get = ( 3E Ad )^ 1/2 Comparing this with the given expression = ( E Ad )^ 1/2 , we find = 3 . Answer: 3