Question

The length of a second’s pendulum on the surface of the Earth is 0.9 m. The length of the same pendulum on surface of planet X such that the acceleration of the planet X is n times greater than the Earth

(a) 0.9 n

(b) \(\frac{0.9}{n}\) m

(c) 0.9 n²m

(d) \(\frac{0.9}{n^2}\)

Answer 1

(a) 0.9 n

Second’s pendulum on the surface of planet, 

Time period, T = 2 sec

T = \(2π \sqrt {\frac{l}{g}}\)