Question 113 of 280

A uniform disc with mass $M=4kg$ and radius $R=10cm$ is mounted on a fixed horizontal axle as shown in figure. A block with mass $m=2kg$ hangs from a massless cord that is wrapped around the rim of the disc. During the fall of the block, the cord does not slip and there is no friction at the axle. The tension in the cord is _____ $N$.
(Take $g=10{ms}^{-2}$)

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Answer: 10

Let $a$ be the acceleration of block and $\alpha$ be the angular acceleration of disc.
Now, for unwrapping without slipping, $R\alpha =a$.
Forces acting on block are shown in FDB.
Using Newton's second law
$mg-T=ma...\left(1\right)$
Torque about centre of disc is $\tau =Ia$
$TR=\dfrac{M{R}^{2}}{2}\alpha =\dfrac{4{R}^{2}}{2}\alpha =\Rightarrow T=2R\alpha ...\left(2\right)$
Putting the value of tension in equation $\left(1\right)$, we get
$mg-2R\alpha =ma \Rightarrow mg-2R\dfrac{a}{R}=ma \Rightarrow mg-2a=ma \Rightarrow ma+2a=mg \Rightarrow a=\dfrac{mg}{m+2}=\dfrac{2\times 10}{2+2}=5m{s}^{-2}$
Now, $\alpha =\dfrac{5}{0.1}=50rad{s}^{-2}$
Tension in the cord is $T=2\times 0.1\times 50=$$10N$.