JEE Mains · Physics · STD 11- Mechanical properties of solids
- JEE Mains 2024
- Hard
- Numerical
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 \(\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right]\)
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Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
\(\Delta \mathrm{L}=\frac{\mathrm{FL}}{\mathrm{AY}}\)
\(\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{\mathrm{F}}{\mathrm{AY}}\)
\(\frac{\frac{\Delta \mathrm{L}_1}{\mathrm{~L}_1}}{\frac{\Delta \mathrm{L}_2}{\mathrm{~L}_2}}=\frac{\mathrm{F}_1}{\mathrm{~F}_2}=\frac{30}{10}=3\)
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