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: Direct proportion
For a fixed period, $l\propto g$.

Step 2: Compare
The planet has $g=\frac{7}{2}g_{e}$, so the length is $\frac{7}{2}$ times 1 m, which is $3.5$ m.

Step 3: Final answer
The length scales with g, giving 3.5 m.

Final Answer:
The length scales with g, giving 3.5 m. \[ \boxed{\text{(C) }\dfrac{7}{2}\ \text{m}} \]