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 $2 \mathrm{~kg}$ hangs from the other end. A similar wire is attached to the bottom of the load and another load of $1 \mathrm{~kg}$ 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 $=0.005 \mathrm{~cm}^{2}, \mathrm{Y}=2 \times 10^{11} \mathrm{Nm}^{-2}$ and $\mathrm{g}=10 \mathrm{~ms}^{-2}$ ]
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Answer: (3)
Solution:
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$T_{1}=30 \mathrm{~N}$ $2 \mathrm{~kg}$ $T_{2}=10 \mathrm{~N}$ $1 \mathrm{~kg}$
$\Delta \mathrm{L}=\frac{\mathrm{FL}}{\mathrm{AY}}$
$\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{\mathrm{F}}{\mathrm{AY}}$
$\frac{\frac{\Delta \mathrm{L}{1}}{\mathrm{~L}{1}}}{\frac{\Delta \mathrm{L}{2}}{\mathrm{~L}{2}}}=\frac{\mathrm{F}{1}}{\mathrm{~F}{2}}=\frac{30}{10}=3$
