Question

Two blocks of masses 1 kg and 2 kg are connected by a metal wire going over a smooth pulley as shown in figure. The breaking stress of the metal is $\left(\frac{40}{3\pi }\right)\times {10}^{6}N/m^2$. If $g=10m{s}^{-2}$, then what should be the minimum radius of the wire used if it is not to break?

• A. 0.5 mm
• B. 1 mm
• C. 1.5 mm
• D. 2 mm

Answer 1

The correct option is B

1 mm

$T=\frac{2m_1{m}_2}{m_1+m_2}g=\frac{2\times 1\times 2}{1+2}\times 10N=\frac{40}{3}N$

If r is the minimum radius, then

Breaking stress $=\frac{\frac{40}{3}}{\pi r^2}$ or $\frac{40}{3\pi }\times {10}^6=\frac{40}{3\pi r^2}$

or $r^2=\frac{1}{{10}^6}$ or $r=\frac{1}{{10}^3}m$

or $r=\frac{1}{{10}^3}\times {10}^{3}mm=1mm$