Question 24 of 118

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

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Answer: 3

As we know, change in length is given by $\Delta L=\dfrac{FL}{AY}$
Therefore, strain required $\dfrac{\Delta L}{L}=\dfrac{F}{AY}$
Now the ratio will be, $\dfrac{\dfrac{\Delta {L}_{1}}{{L}_{1}}}{\dfrac{\Delta {L}_{2}}{{L}_{2}}}=\dfrac{{F}_{1}}{{F}_{2}}=\dfrac{{T}_{1}}{{T}_{2}}=\dfrac{30}{10}=3$