JEE Mains · Physics · STD 11- 8. mechanical properties of solids
Two metallic wires \(P\) and \(Q\) have same volume and are made up of same material. If their area of cross sections are in the ratio \(4: 1\) and force \(F_1\) is applied to \(\mathrm{P}\), an extension of \(\Delta l\) is produced. The force which is required to produce same extension in \(Q\) is \(\mathrm{F}_2\).The value of \(\frac{\mathrm{F}_1}{\mathrm{~F}_2}\) is _______.
- A \(16\)
- B \(14\)
- C \(20\)
- D \(50\)
Answer & Solution
Correct Answer
(A) \(16\)
Step-by-step Solution
Detailed explanation
\( \mathrm{Y}=\frac{\text { Stress }}{\text { Strain }}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell}=\frac{\mathrm{F} \ell}{\mathrm{A} \Delta \ell} \) \( \Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{AY}} \)…
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