A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R. Assuming the pulley to be a perfect uniform circular disc, the acceleration of the mass m, if the string does not slip on the pulley, is

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

$\frac{2}{3} g$

$\frac{3}{2} g$

$\frac{g}{3}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

For the mass, m:

$mg - T = ma$ .......(1)

The pulley experiences a torque. If $α$ is its angular acceleration, then

$τ = TR = Iα$

$Now, I = \frac{1}{2} {MR}^{2} and α = \frac{a}{R}, therefore,$

$TR = (\frac{1}{2} {MR}^{2}) \frac{a}{R} \Rightarrow T = \frac{1}{2} Ma$

$substituting in Eq . (1),$

$mg - \frac{1}{2} ma = ma \Rightarrow a = \frac{2 g}{3}$

Watch 3-min video & get full concept clarity

No courses found

Ready to Test Your Skills?

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

Take Free Test