A pendulum clock calibrated at earth’s surface will read on the surface of the moon (acceleration due to gravity on the moon is 1/6th of that on earth)

Question

Question

This question was previously asked in

ISRO VSSC Technical Assistant Mechanical 9 June 2019 Official Paper

Answer 1

Concept:

\(Time~ Period= T =2\pi √{\frac Lg}\)

where L = Length of the pendulum, g = acceleration due to gravity

Calculation:

Given:

at moon

\(g'=\frac g6\)

where g' = acceleration due to gravity at the moon

\(New~Time~ Period= T' =2\pi √ {\frac L{g'}}=2\pi √ {\frac {6L}g}\)

\(T'=√ {6}T\)

As time period for oscillation increases √6 times, therefore speed becomes √6 times slower.