Question

A cord is wrapped on a pulley (disk) of mass M and radius R as shown in fig. To one end of the cord, a block of mass M is connected as shown and to other end in (a) a force of 2 Mg and in (b) a block of mass 2M. Let angular acceleration of the disk in A and B is ${\alpha }_2$ 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

The correct option is B ${\alpha }_A>{\alpha }_B$

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

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