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One end of a metal wire is fixed to a ce...

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. Find the longitudinal strain in both wires. Area of cross section of each wire is `0.005 cm^2` and Young modulus of the metal is `2.0xx10^11 Nm^-2.` Take `g=10ms^-2`

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The situtionis described in ure.As the 1 kg mass is in equilibrium, the tension in the lower wire equals the weight of the load

Thus, `T_1=10N`
Stress`=10N/0.005cm^2`
=2xx1067Nm^-2`
`LongitudiN/Al strain =stress/Y=(2xx10^7Nm^-2)/(2xx10^11m^-2)=10^-4`
Considering the equilirium of the upper block, we can write
`T_2=20N+T_1, or, T_2=30N`
`Stress=30N/0.005cm^2`
`=6xx10^7Nm^-2`
`LongitudiN/Al strain=(6xx10^7Nm^-2)/(2xx10^11Nm^i-2)=3xx10^-4`
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