What would the length of a seconds pendulum on the surface of the Earth be if the mass of the Earth remains constant but its volume shrinks to `(1)/(8)`th of its original volume. ltBrgt (Take orignal value of acceleration due to gravity as `9.8 m s^(-2)`)

Explanation

(i) The time period of a seconds pendulum = 2 s `= 2 pi sqrt((l)/(g))`
When the volume `(V_(2))` of the Earth shrinks to `(1)/(8)`th of its original volume `(V_(1))`, then the radius of the Earth becomes half its orginal radius.
`rArr V_(2) = (1)/(8) V_(1)`
`rArr (4)/(3)pi R_(2)^(3) = (1)/(8)[(4)/(3)pi R_(1)^(3)]`
`rArr R_(2) = (R_(1))/(2)`
We know, `g = (GM)/(R^(2))`
where 'g' is acceleration due to gravity and 'M' is the mass of the Earth.
`rArr g_(1) = (GM)/(R_(1)^(2)) "and" g_(2) = (GM)/(R_(2)^(2))`
Then, `2 = 2pi sqrt((l_(1))/(g_(1))) = 2pi((l_(2))/(g_(2)))`
Substitute, the value of `'g_(2)'` and obtain the value of `'l_(2)'`.
(ii) 3.975 m