Correct Answer - A::B::C
`L=5m`
a. `T=2pisqrt(l/g)`
`=2pisqrt(5/10)`
`=2pisqrt0.5=2pi(0.7)`
In `2pi(0.7)sec`, the body will complete `1/(2pi) (.7)` oscillation
`:.f=1/(2pi)(0.7)`
`=10/(14pi)=0.71/pi` times
b. When it is taken to the moon
`T=2pisqrt((l/g))`
Where `g` is Acceleration on the moon
`=2pisqrt(5/1.67)`
`f=1/T` `1/(2pi)sqrt(1.67/5)=1/(2pi)(0.557)`
`=1/(2pisqrt3)`