Question

The length of a seconds pendulum on the Earth to be decreased to make it as a seconds pendulum on the surface of the Moon is _________ times its length on the Earth.

• A. $\frac{1}{6}$
• B. $\frac{11}{6}$
• C. $\frac{5}{6}$
• D. $\frac{7}{6}$

Answer 1

The correct option is A

$\frac{1}{6}$

EXPLANATION:

We know that,

The acceleration due to gravity on the moon is only $\frac{1}{6}th$ that of earth

$g_m=\frac{g_e}{6}$--------------1

The time period of seconds pendulum is 2 s,

At Earth

$T=2\pi \sqrt{\frac{l_e}{g_e}}$

$2=2\pi \sqrt{\frac{l_e}{g_e}}$

$g_e={\pi }^2{l}_e$---------------2

At moon,

$T=2\pi \sqrt{\frac{l_m}{g_m}}$

$2=2\pi \sqrt{\frac{l_m}{g_m}}$

$2=2\pi \sqrt{\frac{l_m}{{\displaystyle \raisebox{1ex}{$g_e$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}}from1$

$2=2\pi \sqrt{\frac{l_m}{{\displaystyle \raisebox{1ex}{${\pi }^2{l}_e$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}}from2$

$6{\pi }^2{l}_m={\pi }^2{l}_e$

$l_m=\frac{1}{6}{l}_e$

The length of seconds pendulum on the surface of the Moon is $\frac{1}{6}$ times its length on the Earth.

option A is the correct answer