Question

One end of a metal wire is fixed to a ceiling and a load of 2 kg hangs from the other end. A similar wire is attached to the bottom of the load and another load of 1 kg hangs from this lower wire. Then the ratio of longitudinal strain of the upper wire to that of the lower wire will be:$\text{[Area of cross section of wire = 0.005 cm}^2, \, Y = 2 \times 10^{11} \, \text{N/m}^2, \, g = 10 \, \text{m/s}^2\text{]}$

Answer 1

Given: - Mass of the upper load: $2 \, \text{kg}$ - Mass of the lower load: $1 \, \text{kg}$

Step 1: Calculating the Tension in Each Wire

The tension in the upper wire ($T_{\text{upper}}$) is due to the combined weight of both loads:

$T_{\text{upper}} = (2 \, \text{kg} + 1 \, \text{kg}) \times g = 3 \times 10 = 30 \, \text{N}$

The tension in the lower wire ($T_{\text{lower}}$) is due to the weight of the lower load:

$T_{\text{lower}} = 1 \times 10 = 10 \, \text{N}$

Step 2: Calculating the Longitudinal Strain in Each Wire

The longitudinal strain ($\epsilon$) is given by:

$\epsilon = \frac{\text{Stress}}{Y} = \frac{T}{A \times Y}$

Calculating the strain in the upper wire:

$\epsilon_{\text{upper}} = \frac{T_{\text{upper}}}{A \times Y} = \frac{30}{5 \times 10^{-7} \times 2 \times 10^{11}}$ $\epsilon_{\text{upper}} = \frac{30}{2 \times 10^{6}} = 1.5 \times 10^{-5}$

Calculating the strain in the lower wire:

$\epsilon_{\text{lower}} = \frac{T_{\text{lower}}}{A \times Y} = \frac{10}{5 \times 10^{-7} \times 2 \times 10^{11}}$ $\epsilon_{\text{lower}} = \frac{10}{2 \times 10^{6}} = 0.5 \times 10^{-5}$

Step 3: Calculating the Ratio of Longitudinal Strains

The ratio of the longitudinal strain of the upper wire to that of the lower wire is given by:

$\text{Ratio} = \frac{\epsilon_{\text{upper}}}{\epsilon_{\text{lower}}} = \frac{1.5 \times 10^{-5}}{0.5 \times 10^{-5}} = 3$

Conclusion: The ratio of longitudinal strain of the upper wire to that of the lower wire is $3$.