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 \over \pi \times 10^2 Nm-2. What is the minimum radius of the wire ? (Take g = 10 ms-2)

Properties of Matter PYQ — JEE Main 2021 · 26 Aug · Shift 2 · Q62 with Solution

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Correct answer: C

Two masses $m_1=3\,\text{kg}$ and $m_2=5\,\text{kg}$ are connected over a smooth pulley, so this is an Atwood machine.

The acceleration is

a=\frac{(m_2-m_1)g}{m_1+m_2} =\frac{(5-3)\cdot 10}{3+5} =\frac{20}{8}=2.5\,\text{m/s}^2

Now tension in the wire is

T=m_1(g+a)=3(10+2.5)=37.5\,\text{N}

(Equivalently, $T=m_2(g-a)=5(10-2.5)=37.5\,\text{N}$ .)

Breaking stress is given by

\sigma=\frac{F}{A}

At the limiting condition for minimum radius,

\sigma_{\text{break}}=\frac{T}{\pi r^2}

Given

\sigma_{\text{break}}=\frac{24}{\pi}\times 10^2\,\text{N/m}^2

So,

\frac{T}{\pi r^2}=\frac{24}{\pi}\times 10^2

Substitute $T=37.5$ N:

\frac{37.5}{\pi r^2}=\frac{24}{\pi}\times 10^2

Cancel $\pi$ :

\frac{37.5}{r^2}=24\times 10^2=2400

Hence,

r^2=\frac{37.5}{2400}=0.015625

r=\sqrt{0.015625}=0.125\,\text{m}

Convert to cm:

0.125\,\text{m}=12.5\,\text{cm}

So the correct option is

\boxed{\text{C: }12.5\,\text{cm}}