Question

Two blocks of masses $1\,kg$ and $2\,kg$ are connected by a metal wire going over a smooth pulley as shown in the figure. The breaking stress of the metal is $2\times 10^{9}Nm^{- 2}$ . If the minimum radius of the wire used is $\frac{r}{10}\times 10^{- 5}m$ so that it is not to break, find the value of $r$ . Take, $g=10\,ms^{- 2}$ .

Answer 1

First, recall the formula of surface tension in terms of mass,
$T=\left(\frac{2 m_{1} m_{2}}{m_{1} + m_{2}}\right)g=\left[\frac{2 \times 2 \times 1}{2 + 1}\right]\times 10=\frac{40}{3}N$ .
Now, recall the formula of stress,
$\text{stress}=\frac{T}{\pi r^{2}}$ .
It means,
$r=\sqrt{\frac{T}{\pi \times \text{stress}}}=\sqrt{\frac{40/3}{3 . 14 \times 2 \times 10^{9}}}=4.60\times 10^{- 5}=\frac{46}{10}\times 10^{- 5}m$ .