Question

If the length of seconds pendulum on the surface of the earth is 1 m. The length of the seconds pendulum on the surface of the planet is \(\left\{g_{planet}=\frac{7}{2}g_{earth}\right\}\) {periodic time of the seconds pendulum is same on both planets}

• A. \(\frac{1}{7}\ m\)
• B. \(\frac{2}{7}\ m\)
• C. \(\frac{7}{2}\ m\)
• D. \(7\ m\)

Hint:

A seconds pendulum has the same period everywhere, so the ratio of length to gravitational acceleration stays the same.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
\(T=2\pi\sqrt{\dfrac{l}{g}}\). A seconds pendulum has period \(2\ \text{s}\) on both planets, so \(\dfrac{l}{g}\) is the same.

Step 2: Calculation
\(\dfrac{l_{p}}{g_{p}}=\dfrac{l_{e}}{g_{e}}\)
\(l_{p}=1\times\dfrac{7}{2}=\dfrac{7}{2}\ \text{m}\)

Step 3: Final answer
The seconds pendulum on the planet is \(\frac{7}{2}\) m long.

Final Answer:
The seconds pendulum on the planet is \(\frac{7}{2}\) m long. \[ \boxed{\text{(C) }\dfrac{7}{2}\ \text{m}} \]