Question

For the arrangement shown in the figure, tension in the string is :

• A. $\frac{Mg}{2}$
• B. $\frac{2Mg}{3}$
• C. $\frac{3Mg}{3}$
• D. $\frac{Mg}{3}$

Answer 1

The correct option is D $\frac{Mg}{3}$

According to the figure

Torque on disc $\tau =Fr$

$\tau =TR=I\alpha$

$($ Therefore $\alpha =\frac{a}{R}$ assume pure rolling of disc$)$

So $T=\frac{ma}{2}$ $-(!)$

Now observe block of mass $M$

$Mg-T=Ma$ $-\left(2\right)$

put equation $\left(1\right)$ in $\left(2\right)$

$Mg-\frac{Ma}{2}=Ma$

$Mg=\frac{3Ma}{2}$

$a=\frac{2}{3}g$

so $T=\frac{ma}{2}=\frac{m}{2}.\frac{2}{3}g$

$T=\frac{mg}{3}$

Hence option $\left(D\right)$ is correct