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Physics - Elasticity Question with Solution | TestHub

PhysicsElasticityMiscellaneousMedium2 minPYQ_2023

PhysicsMediumsingle choice

A force is applied to a steel wire $A$ , rigidly clamped at one end. As a result elongation in the wire is $0.2 mm$ . If same force is applied to another steel wire $B$ of double the length and a diameter $2.4$ times that of the wire $A$ , the elongation in the wire $B$ will be (wires having uniform circular cross sections)

Options:

Answer:

Solution:

As we know, $Y = \frac{F}{\frac{A}{\frac{∆ l}{l}}}$

$\Rightarrow F = \frac{Y A}{l} ∆ l$

For both cases force applied is the same, therefore

${(\frac{Y A ∆ l}{l})}_{1} = {(\frac{Y A ∆ l}{l})}_{2}$

$\Rightarrow \frac{∆ l_{2}}{∆ l_{1}} = \frac{A_{1}}{A_{2}} \times \frac{l_{2}}{l_{1}}$

$\Rightarrow \frac{∆ l_{2}}{0.2} = \frac{π R^{2}}{π {(2.4 R)}^{2}} \times \frac{2}{1}$

$\Rightarrow ∆ l_{2} = 6.9 \times 10^{- 2} mm$

Stream:JEESubject:PhysicsTopic:ElasticitySubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2023

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