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In space, a horizontal electric field (E=mg/q) exist as shown in figure and a charged partical of charge ‘q’ and mass m attached at the end of a light rod of light l . If mass m is released from the position shown in figure find the angular velocity of the rod when it passes through the bottom most position.

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$\sqrt{\frac{5 g}{ℓ}}$

$\sqrt{\frac{2 g}{ℓ}}$

$\sqrt{\frac{3 g}{ℓ}}$

$\sqrt{\frac{g}{ℓ}}$

answer is B.

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Detailed Solution

Charge to the point mass = q

W_{electrostatic force} + W_gravity = $∆$ KE

We have work done by electrostatic force is

$W_{e} = qE l sinθ$

work done by the gravitational force is

$W_{g} = mg (l - l \cos θ)$

$\Rightarrow qE l \sin θ + mg (l - l \cos θ) = \frac{1}{2} {mv}^{2}$

as θ=45^o ,we get

$mg l = \frac{1}{2} {mv}^{2}$

also as v=ωl

we get $ω = \sqrt{\frac{2 g}{l}}$

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