Two blocks of masses 3 kg and 5 kg are connected by a metal wire going over a smooth pulley. The breaking stress of the metal is $(24/\mathrm{\pi })\times {10}^2{\mathrm{Nm}}^{-2}$. What is the minimum radius of the wire? (Take, g = 10 ms^-2)

Given, breaking stress of wire, $\mathrm{\sigma }=\frac{24}{\mathrm{\pi }}\times {10}^2 {\mathrm{Nm}}^{-2}$

Free body diagram of 5kg block is given as

where, a is common acceleration.

Value of acceleration due to gravity, $\mathrm{g}=10{ \mathrm{ms}}^{-2}$

From free body diagram of block of mass 5 kg

      $\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} 5\mathrm{g}-\mathrm{T}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=5\mathrm{a}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}5\times 10-\mathrm{T}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=5\mathrm{a}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}50-\mathrm{T}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=5\mathrm{a} ...\left(\mathrm{i}\right)\end{array}$

Free body diagram of 3 kg block is given as

From free body diagram of block of mass 3 kg,

       $\begin{array}{l} \hspace{0.25em}\hspace{0.25em} \mathrm{T}-3\mathrm{g}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=3\mathrm{a}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{T}-3\times 10\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=3\mathrm{a}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \mathrm{T}-30\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=3\mathrm{a} ...\left(\mathrm{ii}\right)\end{array}$

Add Eqs. (i) and (ii), we get

         $\begin{array}{l} 50-\mathrm{T}+\mathrm{T}-30=5\mathrm{a}+3\mathrm{a}\\ \Rightarrow 20=8\mathrm{a}\\ \Rightarrow \mathrm{a}=2.5{\mathrm{ms}}^{-2}\end{array}$

Substituting the value of a in Eq. (i), we get

         $\begin{array}{l} 50-\mathrm{T}=5\times 2.5\\ \Rightarrow \mathrm{T}=37.5\mathrm{N}\end{array}$

Let us assume the minimum radius of wire is r.

The breaking stress is expressed as

$\begin{array}{l} \mathrm{\sigma }=\frac{\mathrm{T}}{{\mathrm{\pi r}}^2}\\ \frac{24}{\mathrm{\pi }}\times {10}^2=\frac{37.5}{{\mathrm{\pi r}}^2}\\ \Rightarrow {\mathrm{r}}^2=\frac{37.5}{24\times {10}^2}=\frac{1}{64}\\ \Rightarrow \mathrm{r}=\frac{1}{8}\mathrm{m}=\frac{100}{8}\mathrm{cm}=12.5\mathrm{cm}\end{array}$

Thus, the minimum radius of wire should be 12.5 cm.