Question

The mass and the radius of a planet are twice that of earth. Then, period of oscillation of a second pendulum on that planet will be ___________

(a) 1/√2s

(b) 2√2s

(c) 2s

(d) 1/2s

Answer 1

Right answer is (b) 2√2s

To explain: On earth,

g=GM/R^2

On planet,

g^‘=G2M/(4R)^2 = g/2

T^‘/T=√(g/g^‘) = √((g/g)/2) = √2

T^‘=√2×T=2√2 s.