A second pendulum is shifted from Delhi to London . If the accelereation due to gravity at london is `981cm/s^(2)` and increases in length of the pendulum is observed to be 0.2 cm , then the acceleration due to gravity at Delhi is

Options

A. `983cm/s^(2)`
B. `979cm/s^(2)`
C. `985cm/s^(2)`
D. `977cm//s^(2)`

Answer

B

Explanation

Time period of pendulum is given by
`t = 2pisqrt((l)/(g))`
`rArr l = (gT^(2))/(4pi^(2))`
For second.s pendulum, T =2 s
`therefore l = (g(2^(2)))/(4pi^(2)) = (g)/(pi^(2))`
At London,
`therefore l = (g)/(pi^(2)) = (981)/(10) = 98.1` cm `(pi^(2) = 9.8696 ~~ 10)`
At Delhi, `therefore l = 0 .2 = (g.)/(pi^(2)) = (g.)/(10)`
`rArr g. - 10l - 2 = 10 xx 98.1 - 2 = 979 cm//s^(2)`