Q. The mass and diameter of a planet are twice those of earth. The period of oscillation of pendulum on this planet will be (If it is a second’s pendulum on earth)
A) $\frac{1}{\sqrt{2}}s$
B) $2\sqrt{2}\ s$
C) $2\ s$
D) $\frac{1}{2}s$
Explanation
(b) As we know$\ g = \frac{GM}{R^{2}}$
$\Rightarrow \frac{g_{earth}}{g_{planet}} = \frac{M_{e}}{M_{p}} \times \frac{R_{\rho}^{2}}{R_{e}^{2}} \Rightarrow \frac{g_{e}}{g_{p}} = \frac{2}{1}$
$Also\ T \propto \frac{1}{\sqrt{g}} \Rightarrow \frac{T_{e}}{T_{p}} = \sqrt{\frac{g_{p}}{g_{e}}} \Rightarrow \frac{2}{T_{p}} = \sqrt{\frac{1}{2}}$
$\Rightarrow T_{p} = 2\sqrt{2}\ s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.