A simple pendulum of length $ℓ$ is set in motion such that the bob, of mass m, moves along a horizontal circular path, and string makes a constant angle $θ$ with the vertical. The time period of rotation of the bob is t and tension in the thread is T.

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

$t = 2 π \sqrt{ℓ / g}$

$t = 2 π \sqrt{ℓ \cos θ / g}$

$T = \frac{4 π^{2} m ℓ}{t^{2}}$

The bob is in equilibrium

answer is B, C.

(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

$T \sin θ = m ℓ ω^{2} o r T = m ω^{2} ℓ [∵ r = ℓ \sin θ]$

$T \cos θ = m g o r m ω^{2} / ℓ \cos θ = m g$

$o r ω^{2} = \frac{g}{ℓ \cos θ}$

$t = \frac{2 π}{ω} = 2 π \sqrt{\frac{ℓ \cos θ}{g}}$

$T = m [\frac{4 π^{2}}{t^{2}}] ℓ$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE