A thin uniform rod of mass M and length L is hinged at an end and released from rest in the horizontal position. The tension at a point located a distance L/3 from the hinge point, when the rod becomes vertical, will be

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

$\frac{22 Mg}{27}$

2Mg

$\frac{11 Mg}{13}$

$\frac{6 Mg}{11}$

answer is D.

(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

When the rod becomes vertical, apply conservation of energy
$\begin{array}{l} \frac{MgL}{2} = \frac{1}{2} \frac{{ML}^{2}}{3} w^{2} \\ \Rightarrow W = \sqrt{3 g / L} \end{array}$
Centre of mass of lower two third part moves in a circle of radius 2L/3.

We can apply Newton’s second law on this part
$T - \frac{2 Mg}{3} = (\frac{2 M}{3}) (\frac{2 L}{3}) w^{2} \Rightarrow T = 2 Mg$

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