Question:
Time period of a simple pendulum is \(4\,\text{s}\) at a place on the earth where the acceleration due to gravity is
\[
\pi^2\,\text{m/s}^2.
\]
Then the length of the pendulum in meters is:
Time period of a simple pendulum is \(4\,\text{s}\) at a place on the earth where the acceleration due to gravity is
\[
\pi^2\,\text{m/s}^2.
\]
Then the length of the pendulum in meters is:
Show Hint
For a simple pendulum,
\[
T=2\pi\sqrt{\frac{l}{g}}.
\]
Always square the equation carefully after substitution.
Updated On: Jun 24, 2026
- \(4\)
- \(2\)
- \(\pi\)
- \(\dfrac{\pi}{2}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Use the formula for time period of a simple pendulum.
For a simple pendulum, \[ T=2\pi\sqrt{\frac{l}{g}} \] Given, \[ T=4\,\text{s} \] and \[ g=\pi^2\,\text{m/s}^2 \]
Step 2: Substitute the values.
\[ 4=2\pi\sqrt{\frac{l}{\pi^2}} \] Divide both sides by \(2\pi\): \[ \frac{2}{\pi}=\sqrt{\frac{l}{\pi^2}} \] Squaring both sides: \[ \frac{4}{\pi^2}=\frac{l}{\pi^2} \] Multiplying by \(\pi^2\): \[ l=4 \]
Step 3: Final conclusion.
Hence, the length of the pendulum is \[ \boxed{4\,\text{m}} \]
For a simple pendulum, \[ T=2\pi\sqrt{\frac{l}{g}} \] Given, \[ T=4\,\text{s} \] and \[ g=\pi^2\,\text{m/s}^2 \]
Step 2: Substitute the values.
\[ 4=2\pi\sqrt{\frac{l}{\pi^2}} \] Divide both sides by \(2\pi\): \[ \frac{2}{\pi}=\sqrt{\frac{l}{\pi^2}} \] Squaring both sides: \[ \frac{4}{\pi^2}=\frac{l}{\pi^2} \] Multiplying by \(\pi^2\): \[ l=4 \]
Step 3: Final conclusion.
Hence, the length of the pendulum is \[ \boxed{4\,\text{m}} \]
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