Two blocks of masses


Two blocks of masses $3 \mathrm{~kg}$ and $5 \mathrm{~kg}$ are connected by a metal wire going over a smooth pulley. The

breaking stress of the metal is $\frac{24}{\pi} \times 10^{2} \mathrm{Nm}^{-2}$.

What is the minimum radius of the wire?

$\left(\right.$ Take $\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$

  1. $125 \mathrm{~cm}$

  2. $1250 \mathrm{~cm}$

  3. $12.5 \mathrm{~cm}$

  4. $1.25 \mathrm{~cm}$

Correct Option: , 3


$\mathrm{T}=\frac{2 \mathrm{~m}_{1} \mathrm{~m}_{2} \mathrm{~g}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}=\frac{2 \times 3 \times 5 \times 10}{8}$


Stress $=\frac{\mathrm{T}}{\mathrm{A}}$

$\frac{24}{\pi} \times 10^{2}=\frac{75}{2 \times \pi R^{2}}$

$\mathrm{R}^{2}=\frac{75}{2 \times 24 \times 100}=\frac{3}{8 \times 24}$

$\Rightarrow \mathrm{R}=0.125 \mathrm{~m}$

$\mathrm{R}=12.5 \mathrm{~cm}$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now