Physics - Elasticity Question with Solution | TestHub

PhysicsElasticityMiscellaneousHard2 minQB

PhysicsHardmultiple choice

Passage / Comprehension

A 4.0 kg mass is hung from a vertical light steel wire 2.0 m long and $5.0 \times 1 0^{- 3} cm^{2}$ in cross-sectional area. The wire is securely fastened to the ceiling. (Young's modulus $= 2 \times 1 0^{11} N / m^{2}$ )

Now assume that the mass is very slowly pulled downward 0.06 cm from its equilibrium position by an external force $F$ .

Options:(select one or more)

Answer:

A, C, D

Solution:

Work by gravity: $W_{1} = M g s = 4 \times 10 \times 0.06 \times 1 0^{- 2} = 0.024 J$

Change in elastic PE: $W_{2} = \int_{0}^{6 \times 1 0^{- 4}} k (x + Δ ℓ) d x = 0.033 J$ with $k = \frac{Y A}{ℓ}$

Work by external force: $W_{3} = W_{2} - W_{1} = 0.009 J$

Answer (A, C, D)

Stream:JEESubject:PhysicsTopic:ElasticitySubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

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