A pendulum has time period `T` is air when it is made to oscillates in water it acquired a time period `T = sqrt(27)` The density of the pendulum bob is equal to (density) of water `= 1)`

Options

A. `sqrt(2)`
B. 2
C. `2 sqrt(2)`
D. None of these

Answer

B

Explanation

The effictive acceleration of a bob in water `= g' = g (1 - (sigma)/(rho))` where `sigma` and `rho` are the density of water and the bob respectively. since the period of oscillation of the bob in air in all and water are given as `T = 2pi sqrt((l)/(g))` and `T'= 2pisqrt((l)/(g'))`
`:. T/(T') = sqrt((g')/(g)) = sqrt((g(1-sigma//rho))/(g)) = sqrt(1-(sigma)/(rho)) = sqrt(1-(l)/(rho))`
`:. T/(T') = sqrt((g')/(g)) = sqrt((g(1-sigma//rho))/(g)) = sqrt(1-(sigma)/(rho)) = sqrt(1-(l)/(rho))`
Putting `(T)/(T') = (1)/(sqrt(2))`, We obtain, (1)/(2) = 1 rArr rho = 2`