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 sress 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. $3.8\times {10}^{-6}m$
• B. $4.6\times {10}^{-5}m$
• C. $3.4\times {10}^{-8}m$
• D. $2.4\times {10}^{-4}m$

Answer 1

The correct option is B $4.6\times {10}^{-5}m$

The stress in the wire = $\frac{\text{Tension}}{\text{Area of cross section}}$.
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,
and T = 10 N = (1 kg) a
20 N - T = (2 kg) a.
Eliminating a from these equations,
T = (40/3) N.
(40/3) N
The stress = $\pi r^2$

If the minimum radius needed to avoid breaking is r,
$2\times {10}^9\frac{N}{m^2}=\frac{\left(40/3\right)N}{\pi r^2}$

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