Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A simple pendulum of length ‘l’ has a bob of mass ‘m’. If it is released when the string makes an angle $60 °$ with the vertical, the tension in the string at the lowest position is

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

\frac{5}{2} mg

\frac{3}{2} mg

2 mg

3mg

answer is 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

v^{2} = v_{0}^{2} - 2 g l (1 - \cos θ)

$v = 0$

$v_{0}^{2} = 2 g l (1 - \cos θ)$

$v_{0} = \sqrt{2 g l (1 - \cos 60 °)}$

$θ = 60 °$

$v_{0} = \sqrt{2 g l \times \frac{1}{2}}$

$v_{0} = \sqrt{g l}$

$T = \frac{{mv}^{2}}{l} + mg \cos θ$

$θ = 0 °$

$T = \frac{{(m \sqrt{g l})}^{2}}{l} + mg \cos 0 °$

$T = \frac{mg l}{l} + mg = 2 mg$

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

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!