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 rod of mass M and length L is placed in a horizontal plane with one end hinged about the vertical axis. A vertical force $F = \frac{M g}{2}$ is applied at a distance $\frac{5 L}{6}$ from the hinged end. The angular acceleration of the rod will be

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$\frac{3 g}{4 L}$

$\frac{4 g}{5 L}$

$\frac{5 g}{4 L}$

$\frac{4 g}{3 L}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

We know that $τ = I α \dots (i)$

$I = \frac{{ML}^{2}}{3}$ (moment of inertia of rod wrt axis passing from one end and perpendicular to its length)

Now $τ =$ (force) (perpendicular distance)

$τ = (\frac{Mg}{2}) (\frac{5 L}{6})$

from eq. (i)

$(\frac{Mg}{2}) (\frac{5 L}{6}) = (\frac{{ML}^{2}}{3}) (α)$

solving this

$\Rightarrow α = \frac{5 g}{4 L}$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

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

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!