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

Q.

A uniform disc of radius R and mass M is free to rotate only about its axis. A string is wrapped over its rim and a body of mass m is tied to the free end of the string as shown in the figure. The body is released from rest. Then the acceleration of the body is :

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

a

$\frac{2 mg}{2 m + M}$

b

$\frac{2 mg}{2 M + m}$

c

$\frac{2 Mg}{2 M + m}$

d

$\frac{2 Mg}{2 m + M}$

answer is B.

(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

From the figure,

$ma = mg - T______(1)$

The MI of uniforms disc

$I = \frac{1}{2} {MR}^{2}$

also, we know that acceleration of the disc,

$a = αR$

$Torque = Iα = RT$

$∴ RT = \frac{1}{2} {MR}^{2} \times \frac{a}{R} = \frac{MaR}{2}$

$T = \frac{Ma}{2}$

substituting the value in equation 1,

$ma = mg - \frac{Ma}{2}$

$a (m + \frac{Ma}{2}) = mg$

$a = \frac{2 mg}{2 m + M}$

Hence the correct answer is $\frac{2 mg}{2 m + M} .$

Watch 3-min video & get full concept clarity

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

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

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

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

Not sure what to do in the future? Don’t worry! We have a FREE career guidance session just for you!

A uniform disc of radius R and mass M is free to rotate only about its axis. A string is wrapped over its rim and a body of mass m is tied to the free end of the string as shown in the figure. The body is released from rest. Then the acceleration of the body is :