Question

The period of seconds pendulum on a planet, whose mass and radius are three times that of earth, is

• A. $3\sqrt{2}$ second
• B. $\sqrt{3}$ second
• C. $2\sqrt{3}$ second
• D. $2\sqrt{2}$ second

Hint:

If gravity decreases, the time period of a pendulum increases.

Answer 1

The Correct Option is C

Step 1: Expression for acceleration due to gravity.
\[ g = \frac{GM}{R^2} \]

Step 2: Substitute planetary values.
\[ g' = \frac{G(3M)}{(3R)^2} = \frac{GM}{3R^2} = \frac{g}{3} \]

Step 3: Relation between time period and gravity.
\[ T \propto \frac{1}{\sqrt{g}} \]

Step 4: Calculate new time period.
\[ T' = T \sqrt{\frac{g}{g'}} = 2 \sqrt{3} \]

Step 5: Conclusion.
The time period of the seconds pendulum on the planet is $2\sqrt{3}$ seconds.