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 $2\times {10}^{9}N{m}^{-2}$. What should be the minimum radius of the wire used if it is not to break? Take $g=10m{s}^{-2}$.

• A. $2.3\times {10}^{-5}m$
• B. $4.6\times {10}^{-4}m$
• C. $4.6\times {10}^{-5}m$
• D. $2.3\times {10}^{-4}m$

Answer 1

The correct option is C

$4.6\times {10}^{-5}m$

The stress in the wire = $\frac{Tension}{Areaofcrosssection}$
To avoid breaking, this stress should not exceed the breaking stress.
Let the tension in the wire be T. The equations of motion of the two blocks are,
T - 10 N = (1 kg)a
And 20 N - T = (2kg) a.
Eliminating a from these equations,
T = (40/3) N.
The stress =$\frac{F}{A}=\frac{F}{\pi r^2}$
If the minimum radius needed to avoid breaking is r =$\sqrt{\frac{40}{3\pi \times 2\times {10}^9}}$

Solving this,

r = 4.6 $\times$ ${10}^{-5}$ m.