A steel wire ( L = 4.0 m , A = 2.0 m m 2 , Y = 2.0 × 1 0 11 Pa ) is stretched 2.0 mm .

ϵ = 4 0.002 = 5 × 1 0 − 4 . u = 2 1 Y ϵ 2 = 2.5 × 1 0 4 J/ m 3 . U = u A L = 0.20 J .

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PhysicsElasticityMiscellaneousMedium2 minQB

PhysicsMediummultiple choice

A steel wire ( $L = 4.0 m$ , $A = 2.0 m m^{2}$ , $Y = 2.0 \times 1 0^{11} Pa$ ) is stretched $2.0 mm$ .

Answer:

A, B

Solution:

$ϵ = \frac{0.002}{4} = 5 \times 1 0^{- 4}$ .

$u = \frac{1}{2} Y ϵ^{2} = 2.5 \times 1 0^{4} J/ m^{3}$ .

$U = u A L = 0.20 J$ .

Stream:JEESubject:PhysicsTopic:ElasticitySubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

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