The fractional change in radius of a solid sphere is 1 0 − 6 when a gauge pressure of 2 × 1 0 7 Pa is applied, its bulk modulus is

B = 3∣Δ R / R ∣ ∣Δ P ∣ = 3 × 1 0 − 6 2 × 1 0 7 = 3 2 × 1 0 13 P a Answer (B)

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PhysicsElasticityMiscellaneousEasy2 minQB

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The fractional change in radius of a solid sphere is $1 0^{- 6}$ when a gauge pressure of $2 \times 1 0^{7} Pa$ is applied, its bulk modulus is

Answer:

Solution:

$B = \frac{∣Δ P ∣}{3∣Δ R / R ∣} = \frac{2 \times 1 0 ^{7}}{3 \times 1 0 ^{- 6}} = \frac{2}{3} \times 1 0^{13} P a$

Answer (B)

Stream:JEESubject:PhysicsTopic:ElasticitySubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

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