Q. The diameter of a flywheel is increased by 1%. Increase in its moment of inertia about the central axis is
A) 1%
B) 0.5%
C) 2%
D) 4%
Explanation
(c) $I = MR^{2}\ \ \therefore\ \log I = \log M + 2\log R$
Differentiating, we get $\frac{dI}{I} = 0 + 2\frac{dR}{R}$
$\therefore\ \ \frac{dI}{I} \times 100 = 2\left( \frac{dR}{R} \right) \times 100$
$= 2 \times \frac{1}{100} \times 100 = 2\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.