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Q.

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
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