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 $
