Question

What is the time period of the seconds pendulum, on a planet similar to earth, but mass and radius three times as that of the earth?

• A. $2 \, s$
• B. $6 \, s$
• C. $\frac{2}{\sqrt{3 }}s$
• D. $2\sqrt{3 }s$

Answer 1

Answer: (D)

$g=\frac{G M}{R^{2}}$
$\frac{g_{p}}{g_{e}}=\frac{M_{p}}{M_{e}}\times \left(\frac{R_{e}}{R_{p}}\right)^{2}=3\times \frac{1}{9}=\frac{1}{3}$
$T=2\pi \sqrt{\frac{l}{g}}$
$\frac{T_{p}}{T_{e}}=\sqrt{\frac{g_{e}}{g_{p}}}=\sqrt{3}$
$T_{p}=\sqrt{3}T_{e}$
Time period at earth for seconds pendulum $=2 \, s$
$\therefore T_{p}=2\sqrt{3 }s$