Question

29. The length of a seconds pendulum on the surface
of the Earth is 100 cm. Find the length of the
seconds pendulum on the surface of the moon.
Take, &M=78E
(a) 1.66 m
(c) 33.2 cm
(b) 16.6 cm
(d) 3.32 m​

Answer 1

Answer:

The time period of a simple pendulum on earth is given as

T = 2π(l/g)1/2

now,

for moon g' = g/6

this would mean that the gravitational force exerted by moon would be six times less compared to that of earth.

so, the time period would be :-

T' = 2π(l/g/6)1/2

or,

T' = √6T

T' > T

Time period of simple pendulum would increase on moon.