### NEET Solved Paper 2014 Question 9

##### Question: A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of $ 2rev/s^{2} $ is [AIPMT 2014]

#### Options:

A) 25N

B) 50N

C) 78.5N

D) 157N

## Show Answer

#### Answer:

Correct Answer: D

#### Solution:

$ \Rightarrow \ m=50kg\Rightarrow r=0.5\Rightarrow \alpha =2rev/s^{2} $ Torque produce by the tension in the string $ =T\times R=T\times 0.5=\frac{T}{2}Nm……(i) $

We know $ \tau =/\alpha …….(ii) $ From Eqs. (i) and (ii), $ =( \frac{MR^{2}}{2} )\times (2\times 2\pi )\frac{rad}{s^{2}} $

$ \therefore $ $ l _{solid cylinder}=\frac{MR^{2}}{2} $ $ \frac{T}{2}=\frac{50\times {{(0.5)}^{2}}}{2}\times 4\pi $ $ T=50\times \frac{1}{4}\times 4\pi $ $ =50\pi =157N $