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A charged sphere is suspended by a light string. When the sphere is allowed to oscillate, its time period of oscillation is found to be 2 sec. When the sphere is not oscillating and a vertically upward uniform electric field is switch on, the string becomes just taut and tension in it becomes zero. Now the vertical field is switched off and a horizontal electric field of same intensity is switched on. if the pendulum is now allowed to oscillate, its period of oscillation will be

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$\frac{T}{{(2)}^{0.5}}$

$\frac{T}{2}$

$\frac{T}{{(2)}^{0.25}}$

$\sqrt{2} T$

answer is C.

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Detailed Solution

Initially, T = $2 π \sqrt{\frac{l}{g}}$

When the vertical field is switched on, mg = EQ.

When the horizontal field is switched on, effective acceleration due to gravity ( $g_{e}$ ) is gives by

$g_{e} = \frac{\sqrt{{(mg)}^{2} + {(EQ)}^{2}}}{m} = \frac{\sqrt{{(mg)}^{2} + {(mg)}^{2}}}{m} = \sqrt{2} g ∴ New time period T' = 2 π \sqrt{\frac{l}{g_{e}}} = 2 π \sqrt{\frac{l}{\sqrt{2}} g} = \frac{T}{2^{0.25}}$

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