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Question

A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about 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
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B
50 N
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C
78.5 N
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D
157 N
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Solution

The correct option is D 157 N
$$I_{Cylinder} = \dfrac{MR^2}{2} $$ 
$$ = \dfrac{50 \times \left  (\dfrac{1}{2}  \right  )^{2}}{2}  $$ 
$$ = \dfrac{50 \times 1}{2 \times 4} = \dfrac{50}{8}\ kgm^{2} $$ 
Torque due to tension $$ = TR$$
$$ z = I \alpha $$ 
$$TR = \left  (\dfrac{50}{8}  \right  ) \times 2 $$ 
$$T \times \dfrac{1}{2} = \dfrac{50}{4} $$
$$ \boxed{T = 25\ N} $$ 
 $$ \alpha = \dfrac{2 rev}{5^2} = \dfrac{2 \times (2 \pi)}{5^2} $$ rad 
$$1$$ revolution $$ = 2 \pi $$ rad.
$$ \alpha = 4 \pi\ rad/s^{2} $$ 
$$ z = I \alpha $$ 
$$ T \times \left  (\dfrac{1}{2}  \right  ) = \left  (\dfrac{50}{5}  \right  ) \times 4 \pi = 50 \times \pi = 157.07\ N $$ 
$$ \boxed{T = 157\ N} $$
 

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Physics

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