Question

The mass and diameter of a planet are twice those of earth. What will be the period of oscillation of a pendulum on this planet. It is a second's pendulum on earth?

• A. $\sqrt{2}$ s
• B. $2\sqrt{2}$ s
• C. $\frac{1}{\sqrt{2}}$ s
• D. $\frac{1}{2\sqrt{2}}$ s

Answer 1

The correct option is B $2\sqrt{2}$ s

$T\propto \frac{1}{\sqrt{g}}\propto \frac{1}{\sqrt{M/R^2}}$
or $T\propto \frac{R}{\sqrt{M}}$
$\frac{T_1}{T_2}=\frac{R_1}{R_2}\sqrt{\frac{M_2}{M_1}}$
$\frac{R_1}{R_2}=\frac{1}{2},\frac{M_2}{M_1}=2,T_1=2s$
$\frac{2}{T_2}=\frac{1}{2}\sqrt{2}\Rightarrow T_2=\frac{4}{\sqrt{2}}=2\sqrt{2}s$