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Physics - Elasticity Question with Solution | TestHub

PhysicsElasticityMiscellaneousMedium2 minPYQ_2016

PhysicsMediumsingle choice

An iron rod of length $2 m$ and cross-sectional area of $50 mm^{2}$ stretched by $0.5 mm$ , when a mass of $250 kg$ is hung from its lower end. Young's modulus of iron rod is

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Answer:

Solution:

$Y = \frac{F / A}{Δ ℓ / ℓ} = \frac{\frac{250 \times 9.8}{50 \times 1 0 ^{- 6}}}{\frac{0.5 \times 1 0 ^{- 3}}{2}} = \frac{250 \times 9.8}{50 \times 1 0 ^{- 6}} \times \frac{2}{0.5 \times 1 0 ^{- 3}} \Rightarrow 19.6 \times 1 0^{10} N / m^{2}$

Stream:VITEEESubject:PhysicsTopic:ElasticitySubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2016

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