A simple pendulum of length $l$ has a bob of mass $m$ , with a charge $q$ on it. A vertical sheet of charge, with charge $σ$ per unit area, passes through the point of suspension of the pendulum. At equilibrium, the string makes an angle $θ$ with vertical. Its time period of oscillation is $T$ in this position. Then

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$T < 2 π \sqrt{\frac{I}{g}}$

$\tan θ = \frac{σ q}{ε_{0} m g}$

$T > 2 π \sqrt{\frac{l}{g}}$

$\tan θ = \frac{σ q}{2 ε_{0} m g}$

answer is A, C.

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Detailed Solution

Since, $ros \cos θ$

Since, $T_{0} \cos θ = m g$

$T_{0} \sin θ = q E \Rightarrow \tan θ = \frac{q}{m g} (\frac{σ}{2 ε_{0}}) \Rightarrow T_{0} = \sqrt{(m g)^{2} + (q E)^{2}} \Rightarrow T_{0} > m g$

i.e. the effective value of g is increased, hence time period T of oscillation decreases.

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