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

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Solution

Q.
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 _____.

[Area of cross-section of wire = 0.005 ${cm}^{2}$ , Y = $2 \times 10^{11}$ ${Nm}^{- 2}$ and g = 10 ${ms}^{- 2}$ ] [2024]

Ans.
(3)

$\sum f_{net} = 0$

$T_{2} = T_{1} + 2 g = g + 2 g = 3 g$

$T_{1} = g$ ...(i)

${(stress)}_{L} = Y \times {(strain)}_{L}$

${(strain)}_{L} = \frac{stress}{Y} = \frac{F}{Y A}$

${(strain)}_{upper} = \frac{T_{2}}{Y A}$

${(strain)}_{lower} = \frac{T_{1}}{Y A}$

$\frac{{(strain)}_{upper}}{{(strain)}_{lower}} = \frac{T_{2}}{T_{1}} = \frac{3 g}{g} = 3$

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