Question

Two blocks of masses $1kg$ and $2kg$ are connected by a metal wire going over a smooth pulley as shown in the figure. The breaking stress of the metal is $2\times 10^{9}N{m}^{-2}$ . 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, $g=10m{s}^{-2}$ .

Answer 1

Answer: 46

First, recall the formula of surface tension in terms of mass,
$T=\left(\frac{2m_1{m}_2}{m_1+m_2}\right)g=\left[\frac{2\times 2\times 1}{2+1}\right]\times 10=\frac{40}{3}N$ .
Now, recall the formula of stress,
$\text{stress}=\frac{T}{\pi r^2}$ .
It means,
$r=\sqrt{\frac{T}{\pi \times \text{stress}}}=\sqrt{\frac{40/3}{3.14\times 2\times 10^9}}=4.60\times 10^{-5}=\frac{46}{10}\times 10^{-5}m$ .