A wire of length 10 m is subjected to a force of 100 N along its length and hence the lateral strain produced is 0 .01 × 10 - 3 m . The Poisson's ratio was found to be 0 . 4 . If the area of cross-section of wire is 0.025 m 2 , its Young's modulus will be

Poisson's ratio = L a t e r a l s t r a i n L o n g i t u d i n a l s t r a i n i e , 0.4 = 0.01 × 10 - 3 l L or L l = 0.4 0.01 × 10 - 3 = 4 × 10 4 Young's modulus Y = F L A l = 100 0.025 × 4 × 10 4 = 1.6 × 10 8 N m - 2

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Physics - Elasticity Question with Solution | TestHub

PhysicsElasticityMiscellaneousEasy2 minPYQ_2020

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A wire of length $10 m$ is subjected to a force of $100 N$ along its length and hence the lateral strain produced is $0 .01 \times 10^{- 3} m$ . The Poisson's ratio was found to be $0.4$ . If the area of cross-section of wire is $0.025 m^{2}$ , its Young's modulus will be

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Solution:

Poisson's ratio $= \frac{L a t e r a l s t r a i n}{L o n g i t u d i n a l s t r a i n}$

$i e, 0.4 = \frac{0.01 \times 10^{- 3}}{\frac{l}{L}}$

or $\frac{L}{l} = \frac{0.4}{0.01 \times 10^{- 3}} = 4 {\times 10}^{4}$

Young's modulus

$Y = \frac{F L}{A l}$

$= \frac{100}{0.025} \times 4 \times 10^{4} = 1.6 \times 10^{8} {N m}^{- 2}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:ElasticitySubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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