Calculate the time period of a simple pendulum whose length is equal to radius of earth.
Hint `: L =R_(e) = 6.4 xx 10^(6) m , g = 9.8 ms^(-2)`
`T ' = 2pi sqrt(( R_(e))/(2g))`

Explanation

`L = R_(e) = 6.4 xx 10^(6) m , g= 9.8 ms^(-2)`
If `l gt gt R_(e)`
Effective length `l` can begiven by `(l)/(l')=(1)/(l) + (1)/(R_(e))= (l+R_(e))/( lR_(e))`
`l'= (R_(e))/( 1+(R_(e))/(l))`
`T ' = 2pi sqrt((R_(e))/((1+(R_(e))/(l))g))`
When `R_(e) = l`
`T ' = 2pi sqrt((R_(e))/(2g))`
`= 2pi sqrt((6.4 xx10^(6))/ ( 2 xx 9.8 ))`
`=60 `min