1.

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

Answer»

`pisqrt((mepsilon_0)/(3sigmalambda))`
`2pisqrt((2mepsilon_0)/(3sigmalambda))`
`pi/2 sqrt((mepsilon_0)/(3sigmalambda))`
`4pi sqrt((mepsilon_0)/(3sigmalambda))`

Solution :`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))`


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