question_answer

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 \[2\,\,rev/{{s}^{2}}\] is [AIPMT 2014]

A)  25N

B)  50N

C)  78.5N

D)  157N

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), \[=\left( \frac{M{{R}^{2}}}{2} \right)\times (2\times 2\pi )\frac{rad}{{{s}^{2}}}\] \[\therefore \] \[{{l}_{solid}}{{\,}_{cylinder}}=\frac{M{{R}^{2}}}{2}\] \[\frac{T}{2}=\frac{50\times {{(0.5)}^{2}}}{2}\times 4\pi \] \[T=50\times \frac{1}{4}\times 4\pi \] \[=50\pi =157N\]