An infinite nonconducting sheet has surface charge density $σ$ . There is a small hole in the sheet as shown in the figure. A uniform rod of length l having linear charge density $λ$ is hinged in the hole as shown. If the mass of the rod is m, then the time period of oscillation for small angular displacement is

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$π \sqrt{\frac{{mε}_{0}}{3 σλ}}$

$\frac{π}{2} \sqrt{\frac{{mε}_{0}}{3 σλ}}$

$4 π \sqrt{\frac{{mε}_{0}}{3 σλ}}$

$2 π \sqrt{\frac{2 {mε}_{0}}{3 σλ}}$

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

$τ = 2 \int_{0}^{ℓ / 2} (\frac{σ}{2 ε_{0}} λdx) xsin θ = \frac{σ}{ε_{0}} λ \int_{0}^{ℓ / 2} x \cdot dxsin θ = \frac{σ}{2 ε_{0}} λ \frac{ℓ^{2}}{4} \sin θ$

$τ = - Iα or α = - τ / I = \frac{σ}{2 ε_{0}} λ \frac{ℓ^{2}}{4} \frac{12}{{mℓ}^{2}} \sin θ$

$α = - (\frac{3 σλ}{2 {mε}_{0}} θ) (for small angle)$

$T = \frac{2 π}{ω} = 2 π \sqrt{\frac{2 {mε}_{0}}{3 σλ}}$

Watch 3-min video & get full concept clarity

hear from our champions

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET