The density of a metal at normal pressure p is ρ. When it is subjected to an excess pressure, the density becomes ρ'. If K is the bulk modulus of the metal, then the ratio (ρ')/(ρ) is (a) (1)/(1-K/p) (b) 1 + K/p (c) (1)/(1-p/K) (d) 1 + p/K

Answer: (c). Mass is constant, so (ρ')/(ρ) = V/V' = (V)/(V - Δ V). (Δ V)/(V) = p/K (with p the excess pressure). (ρ')/(ρ) = (1)/(1 - p/K) So the answer is (c).

The density of a metal at normal pressure p is ρ. When it is… | MHT CET 2020 (16 Oct, Shift 1)

Q16MHT CET 2020 · 16 OCT · SHIFT 1Medium

Bulk Modulus & Compressibility

The density of a metal at normal pressure $p$ is $\rho$ . When it is subjected to an excess pressure, the density becomes $\rho'$ . If $K$ is the bulk modulus of the metal, then the ratio $\frac{\rho'}{\rho}$ is

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Answer: (c)

Mass is constant, so $\frac{\rho'}{\rho} = \frac{V}{V'} = \frac{V}{V - \Delta V}$ .
$\frac{\Delta V}{V} = \frac{p}{K}$ (with $p$ the excess pressure).
$\frac{\rho'}{\rho} = \frac{1}{1 - \frac{p}{K}}$
So the answer is (c).

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Bulk Modulus & Compressibility question from MHT CET 2020 (16 Oct, Shift 1)

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