Question

The moment of inertia of a solid cylinder of mass 2.5 kg and radius 10 cm about its axis is

• A. 0.0725 kg m$^2$
• B. 12500 kg m$^2$
• C. 0.0125 kg m$^2$
• D. 72500 kg m$^2$

Hint:

Use the formula \(I = \frac{1}{2} m r^2\) for moment of inertia of a solid cylinder about its central axis.

Answer 1

The Correct Option is C

Moment of inertia \(I\) of a solid cylinder about its central axis is given by: \[ I = \frac{1}{2} m r^2 \] Given: \[ m = 2.5\, kg, \quad r = 10\, cm = 0.1\, m \] Calculate: \[ I = \frac{1}{2} \times 2.5 \times (0.1)^2 = \frac{1}{2} \times 2.5 \times 0.01 = 0.0125\, kg\, m^2 \]