First slide

Stress and strain

Question

The two wires shown in figure are made of the same material which has a breaking stress of 8 x $10^{8}$ $N / m^{2}$ . The area of the cross-section of the upper wire is 0.006 ${cm}^{2}$ and that of the lower wire is 0.003 ${cm}^{2}$ . The mass $m_{1} = 10 kg, m_{2} =$ 20 kg and the hanger is light. The maximum load that can be put on the hanger without breaking a wire is

Moderate

A

14 kg
B

1.4 kg
C

1.2 kg
D

12 kg

Solution

Let load put on the hanger is F, then stress in lower wire
$S_{1} = \frac{m_{1} g + F}{0.003 \times 10^{- 4}}$
Let $S_{1} = 8 \times 108 N / m^{2}, then$
$8 \times 10^{8} = \frac{10 \times 10 + F}{3 \times 10^{- 7}} \Rightarrow F = 140 N$
Let stress developed in upper wire is $S_{2}$ , then
$S_{2} = \frac{(m_{1} + m_{2}) g + F}{0.006 \times 10^{- 4}}$
Let $S_{2} = 8 \times 10^{8} N / m^{2}$
$8 \times 10^{8} = \frac{(10 + 20) 10 + F}{6 \times 10^{- 7}} \Rightarrow F = 180 N$
We have to select the lower value of F. Hence maximum load that can be suspended is $\frac{140}{g}$ = 14 kg .

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