The driver of a car record a period of `(pi)/(3)` seconds for a pendulum of `1m` hung form the roof. The acceleration of the car is `(g = 10 ms^(-2))`

Options

A. `10ms^(-2)`
B. `15ms^(-2)`
C. `17.2 ms^(-2)`
D. `34.5ms^(-2)`

Answer

D

Explanation

Given: `T = (pi)/(2)second, l = 1m`As the car accelerates it will given a pseudoforce on bob and the bob will till itself form the vertical by an angle
`theta =tan^(-1)((a)/(g))` and the time period of oscillation is `T = 2pi sqrt((l)/(sqrt(g^(2)+a^(2))))` squaring both sides with `T` and `L ((pi)/(3))^(2) = 4pi^(2) (1)/(sqrt(g^(2)+a^(2))) rArr sqrt(g^(2)+a^(2)) = 36`.
`a^(2) = 36^(2) -10^(2), a = sqrt(36^(2) -10^(2)) = 34.5 ms^(-2)`
so, choice (d) is correct.