Q. A uniform rod of length $2.0\ m$ is suspended through an end is set into oscillation with small amplitude under gravity. The time period of oscillation is approximately
A) $1.60\ s$
B) $1.80\ s$
C) $2.0\ s$
D) $2.40\ s$
Explanation
(d) This is the special case of physical pendulum and in this case
$T = 2\pi\sqrt{\frac{2l}{3g}}$
$\Rightarrow T = 2 \times 3.14\sqrt{\frac{2 \times 2}{3 \times 9.8}} = 2.31s \approx 2.4s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.