A small sphere carrying a charge ‘q’ is hanging in between two parallel plates by a string of length L. Time period of pendulum is $T_{0} .$ When parallel plates are charged, the time period changes to T. The ratio $T / T_{0}$ is equal to

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None of these

${(\frac{g}{g + \frac{q E}{m}})}^{3 / 2}$

${(\frac{g}{g + \frac{q E}{m}})}^{1 / 2}$

${(\frac{g + \frac{q E}{m}}{g})}^{1 / 2}$

answer is C.

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Detailed Solution

$T_{0} = 2 π \sqrt{\frac{L}{g}} \to (1)$

If plates charged

$g^{1} = g + a [∵ a = \frac{q E}{m}]$

$T = 2 π \sqrt{\frac{L}{g + (\frac{q E}{m})}}$

$\frac{T}{T_{0}} = {[\frac{g}{(g + \frac{q E}{m})}]}^{^{1 / 2}}$

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