question_answer

A solid cyclinder 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 \[2\,\,rev/{{s}^{2}}\] is [NEET 2014]

A)       25 N    

B) 50 N                

C) 78.5 N

D) 157 N

Correct Answer: D

Solution :

\[\Rightarrow \]   \[m=50kg\Rightarrow r=0.5\Rightarrow \alpha =2rev/{{s}^{2}}\]

\[\Rightarrow \]   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),

\[=\left( \frac{M{{R}^{2}}}{2} \right)\times (2\times 2\pi )\frac{rad}{{{s}^{2}}}\]

\[\therefore \]      \[{{l}_{\text{solid cylinder}}}\,=\frac{M{{R}^{2}}}{2}\]

\[\frac{T}{2}=\frac{50\times {{(0.5)}^{2}}}{2}\times 4\pi \]

\[T=50\times \frac{1}{4}\times 4\pi \]

\[=50\pi =157N\]