An infinite nonconducting sheet has surface charge density σ. There is a small hole in the sheet as shown in the figure. A uniform rod of length l having linear charge density λ is hinged in the hole as shown. If the mass of the rod is m, then the time period of oscillation forsmall angular displacement is
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a
πmε03σλ
b
2π2mε03σλ
c
π2mε03σλ
d
4πmε03σλ
answer is B.
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Detailed Solution
τ=2(σ2ε0λdx)xsinθ=σε0λ∫0ℓ/2x.dxsinθ=σ2ε0λl24sinθτ=−Iα or α=−τ/I=σ2ε0λl2412ml2sinθα=−(3σλ2mε0θ) (for small angle)T=2πω=2π2mε03σλ