If `'L'` is length of sample pendulum and `'g'` is acceleration due to gravity then the dimensional formula for `((l)/(g))^(1/2)` is same that for

Options

A. frequency
B. velocity
C. time period
D. wavelength

Answer

C

Explanation

Hence `[(l)/(g)]^(1//2) = [(M^(0) L^(1) T^(0))/(M^(0) L^(1) T^(-2))]^(1//2) = M^(o) L^(0) T^(1)`