A small sphere carrying a charge `q` is hanging in between two parallel plates by a string of length `L`. Time period of pendulum is `T_(0)`. When parallel plates are charged, the time period changes to `T`. The ratio `T//T_(0)` is equal to

Options

A. `((g+(qE)/(m))/(g))^(1//2)`
B. `((g)/(g+(qE)/(m)))^(3//2)`
C. `((g)/(g+(qE)/(m)))^(1//2)`
D. None of these

Answer

c

Explanation

Net downward force `mg'= mg+QE`
`implies` Effect acceleration `g' (g+(QE)/(m))`
Hence time period `T= 2pi sqrt(l/(g'))= 2pisqrt((1)/((g+(QE)/(m))))`
Required ratio`= (T)/(T_(0))=[(g)/(g+(QE)/(m))]^(1//2)`