First slide

Elastic behaviour of solids

Question

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

Moderate

A

0.5 mm
B

1 mm
C

1.5 mm
D

2mm

Solution

$T = \frac{2 m_{1} m_{2}}{m_{1} + m_{2}} g = \frac{2 \times 1 \times 2}{1 + 2} \times 10 N = \frac{40}{3} N$
If r is the minimum radius, then
$Breaking stress = \frac{\frac{40}{3}}{{πr}^{2}} or \frac{40}{3 π} \times 10^{6} = \frac{40}{3 {πr}^{2}}$
$or r^{2} = \frac{1}{10^{6}} or r = \frac{1}{10^{3}} m = 1 mm$

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