Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Solve! 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 pulley to be a perfect uniform circular disc, the acceleration of the mass m , if the string does not slip

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 pulley to be a perfect uniform circular disc, the acceleration of the mass m , if the string does not slip on the pulley, is

Option 1)

\frac{3}{2}g

 Option 2)

g
Option 3)

\frac{2}{3}g

 Option 4)

\frac{g}{3}

Answers (1)

best_answer

As we learnt in

Rolling of a body on an inclined plane -

a= \frac{g\sin \Theta }{1+\frac{K^{2}}{R^{2}}}

f= \frac{mg\sin \Theta }{1+\frac{R^{2}}{K^{2}}}

- wherein

K=Radius of gyration

\Theta = Angle of inclination

 

 

 

a=\frac{Mg}{m+\frac{I}{R^{2}}}=\frac{Mg}{m+\frac{1}{2}\frac{mR^{2}}{R^{2}}}

a=\frac{2Mg}{3m}

a=\frac{2g}{3}


Option 1)

\frac{3}{2}g

This is an incorrect option.

Option 2)

g

This is an incorrect option.

Option 3)

\frac{2}{3}g

This is the correct option.

Option 4)

\frac{g}{3}

This is an incorrect option.

Posted by

Aadil

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024: NTA debars 39 candidates for 3 years on account of using unfair means
anam-khan

JEE Main 2024 session 2 paper 2 answer key objection window closes today; fees
anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’

Latest Question