Question

A simple pendulum of length $30$ cm makes $20$ oscillations in $10$ s on a certain planet. Another pendulum makes $40$ oscillations in $10$ s on the same planet. Find the length of the second pendulum.

• A. $10$ cm
• B. $14$ cm
• C. $7.5$ cm
• D. $25$ cm

Hint:

On the same planet, $g$ is constant, so always relate pendulum lengths using $T^2 \propto \ell$.

Answer 1

The Correct Option is C

Concept: For a simple pendulum: \[ T = 2\pi\sqrt{\frac{\ell}{g}} \] On the same planet, $g$ remains constant. Hence, \[ T \propto \sqrt{\ell} \quad \text{or} \quad \ell \propto T^2 \]
Step 1: Calculate time periods First pendulum: \[ T_1=\frac{10}{20}=0.5\ \text{s} \] Second pendulum: \[ T_2=\frac{10}{40}=0.25\ \text{s} \]
Step 2: Use proportionality \[ \frac{\ell_1}{\ell_2}=\left(\frac{T_1}{T_2}\right)^2 \]
Step 3: Substitute values \[ \frac{30}{\ell_2}=\left(\frac{0.5}{0.25}\right)^2 =4 \] Step 4: Solve for $\ell_2$ \[ \ell_2=\frac{30}{4}=7.5\ \text{cm} \]
Step 5: Hence, the length of the second pendulum is: \[ \boxed{7.5\ \text{cm}} \]