Question

An infinite nonconducting sheet has surface charge density s. There is a small hole in the sheet as shown in the figure. A uniform rod of length l having linear charge density `lambda` is hinged in the hole as shown. If the mass of the rod is m, then the time period of oscillation for small angular displacement is

A. `pisqrt((mepsilon_0)/(3sigmalambda))`
B. `2pisqrt((2mepsilon_0)/(3sigmalambda))`
C. `pi/2 sqrt((mepsilon_0)/(3sigmalambda))`
D. `4pi sqrt((mepsilon_0)/(3sigmalambda))`

Answer 1

Correct Answer - B
`r-2((sigma)/(2epsilon_(0))lambdadx)x sin theta=(sigma)/(epsilon_(0))lambdaint_(o)^(l//2)xdxsin theta=(sigma)/(2epsilon_(0))lambda(l^(2))/(4)sin theta`
`r=Ia` or `a=-tau//I=(sigma)/(2epsilon_(0))lambda(l^(2))/(4)(12)/(ml^(2))sin theta`
`a=-((3sigmalambda)/(2mepsilon_(0))theta)` (for small angle)
`tau=(2pi)/(omega)=2pisqrt((2mepsilon_(0))/(3sigmalambda))`