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.

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$\sqrt{\frac{g}{ℓ}}$

$\sqrt{\frac{2 g}{ℓ}}$

$\sqrt{\frac{3 g}{ℓ}}$

$\sqrt{\frac{5 g}{ℓ}}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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 $ω = \sqrt{\frac{2 g}{l}}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE