A steel rod (Young's modulus = 2 × 1 0 11 N m − 2 ) has an area of cross-section 3 × 1 0 − 4 m 2 and length 1 m . A force of 6 × 1 0 4 N stretched it axially. The elongation of the rod is

Δ L = A Y F L = 3 × 1 0 − 4 × 2 × 1 0 11 6 × 1 0 4 × 1 = 6 × 1 0 7 6 × 1 0 4 = 1 0 − 3 m

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PhysicsElasticityMiscellaneousEasy2 minQB

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A steel rod (Young's modulus $= 2 \times 1 0^{11} N m^{- 2}$ ) has an area of cross-section $3 \times 1 0^{- 4} m^{2}$ and length $1 m$ . A force of $6 \times 1 0^{4} N$ stretched it axially. The elongation of the rod is

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Answer:

Solution:

$Δ L = \frac{F L}{A Y}$

$= \frac{6 \times 1 0 ^{4} \times 1}{3 \times 1 0 ^{- 4} \times 2 \times 1 0 ^{11}}$

$= \frac{6 \times 1 0 ^{4}}{6 \times 1 0 ^{7}} = 1 0^{- 3} m$

Stream:JEESubject:PhysicsTopic:ElasticitySubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

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