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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 for small angular displacement is

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$π \sqrt{\frac{{mε}_{0}}{3 σλ}}$

$2 π \sqrt{\frac{2 {mε}_{0}}{3 σλ}}$

$\frac{π}{2} \sqrt{\frac{{mε}_{0}}{3 σλ}}$

$4 π \sqrt{\frac{{mε}_{0}}{3 σλ}}$

answer is B.

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Detailed Solution

$τ = 2 \int_{0}^{ℓ / 2} (\frac{σ}{2 ε_{0}} λdx) xsin θ = \frac{σ}{ε_{0}} λ \int_{0}^{ℓ / 2} x \cdot dxsin θ = \frac{σ}{2 ε_{0}} λ \frac{ℓ^{2}}{4} \sin θ$

$τ = - Iα or α = - τ / I = \frac{σ}{2 ε_{0}} λ \frac{ℓ^{2}}{4} \frac{12}{{mℓ}^{2}} \sin θ$

$α = - (\frac{3 σλ}{2 {mε}_{0}} θ) (for small angle)$

$T = \frac{2 π}{ω} = 2 π \sqrt{\frac{2 {mε}_{0}}{3 σλ}}$

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