System of Particles and Rotational Motion - Result Question 28
28. 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 revolutions $s^{-2}$ is :
(a) $25 N$
(b) $50 N$
(c) $78.5 N$
(d) $157 N$
[2014]
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Answer:
Correct Answer: 28. (d)
Solution:
- (d) Here $\alpha=2$ revolutions $/ s^{2}=4 \pi rad / s^{2}$ (given)
$I _{\text{cylinder }}=\frac{1}{2} MR^{2}=\frac{1}{2}(50)(0.5)^{2}=\frac{25}{4} Kg-m^{2}$
As $\tau=I \alpha \quad$ so $TR=I \alpha$
$\Rightarrow T=\frac{I \alpha}{R}=\frac{(\frac{25}{4})(4 \pi)}{(0.5)} N=50 \pi N=157 N$
