Question

The acceleration due to gravity on the surface of the moon is 1.7 m s–2. The time period of a simple pendulum on the moon if its item period on the earth is 3.5 s is (Given, g = 9.8 m s–2)

• A. 2.2 s
• B. 4.4 s
• C. 8.4 s
• D. 16.8 s

Answer 1

Answer: (C)

8.4 s

Answer 2

For moon,

For earth

But

$\therefore \frac { \mathrm { T } _ { \mathrm { m } } } { \mathrm { T } _ { \mathrm { e } } } = \sqrt { \frac { \mathrm { g } _ { \mathrm { e } } } { \mathrm { g } _ { \mathrm { m } } } }$

Or

$= \sqrt { \frac { 9.8 } { 1.7 } } \times 3.5 = 8.4 \mathrm {~s}$