Question

Net torque applied by the tension in strings can be related as

• A. $\frac{3BV{r}^2}{R}$
• B. $\frac{BV{r}^2}{R}$
• C. $\frac{BV{r}^2}{3R}$
• D. $\frac{BV{r}^2}{2R}$

Answer 1

Answer: (B)

$\frac{BV{r}^2}{R}$

$PC$ and $CQ$ are in parallel.

So, equivalent resistance of the circuit is $\frac{R}{2}.$

In static condition, current through the battery is $I=\frac{2V}{R}$.

The current through $CP$ and $CQ$ is $i=\frac{1}{2}=\frac{V}{R}$

Hence, $F_1=F_2=Bir$

Consider the $FBD$ of ring and the blocks
$FBD$ of ring: Consider torque about '$C$',

$T_{1}r+\frac{F_{1}r}{2}+\frac{F_{2}r}{2}=T_{2}r$ (i)

FBD of blocks: $T_1=mg$ and $T_2=2mg$
So, $mgr+\frac{Bi{r}^2}{2}+\frac{Bi{r}^2}{2}=2mg$
or $i=\frac{mg}{Br}$ or $\frac{V}{R}=\frac{mg}{Br}$ or $V=\frac{mgR}{Br}$

Net torque of tension

$=(T_2-T_1)r=mgr=\frac{BV{r}^2}{R}$