Question

A cord is wrapped on a pulley (disk) of mass $M$ and radius $R$ as shown in figure. To one end of the cord, a block of mass $M$ is connected as shown and to other end in (a) a force of $2Mg$ and in (b) a block of mass $2M$ . Let angular acceleration of the disk in $A$ and $B$ is ${\alpha }_A$ and ${\alpha }_B$ respectively, then (cord is not slipping on the pulley)

• A. ${\alpha }_A={\alpha }_B$
• B. ${\alpha }_A>{\alpha }_B$
• C. ${\alpha }_A<{\alpha }_B$
• D. None of these

Answer 1

Answer: (B)

In figure, $\left(T_2-T_1\right)R=\frac{M{R}^2}{2}{\left(\alpha \right)}_A$
$T_1-Mg=M{a}_A$
$T_2=2Mg$
$a_A=R{\alpha }_A$
$R(2Mg-Mg-M{a}_A)=\frac{M{R}^2}{2}{\left(\alpha \right)}_A$
$g-R{\alpha }_A=\frac{R{\alpha }_A}{2}$
${\alpha }_A=\frac{2g}{3R}$

From Figure, $\left(T_2-T_1\right)\times R=\frac{M{R}^2}{2}{\left(\alpha \right)}_B$
$T_1-Mg=M{a}_B$
$2Mg-T_2=2M{a}_B$
$a_B=R{\alpha }_B$
$\Rightarrow$ ${\alpha }_B=\frac{2g}{7R}$
So, ${\alpha }_A>{\alpha }_B$