Mechanical Properties Of Solids Question 2
Question 2 - 2024 (01 Feb Shift 2)
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 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 $g=10 ms^{-2}$ ]
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Answer: (3)
Solution:
$T _1=30 N$ $2 kg$ $T _2=10 N$ $1 kg$
$\Delta L=\frac{FL}{AY}$
$\frac{\Delta L}{L}=\frac{F}{AY}$
$\frac{\frac{\Delta L _1}{L _1}}{\frac{\Delta L _2}{L _2}}=\frac{F _1}{F _2}=\frac{30}{10}=3$
