Question

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 ____________.

Options

• `1/(1 - "K"/"p")`

• `1 + "K"/"p"`

• `1/(1 - "p"/"K")`

• `1 + "p"/"K"`

Answer 1

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 `underline(1/(1 - "p"/"K"))`.

Explanation:

The bulk modulus of metal is

K = `- (Δ"p")/((Δ"V")/"V")` or ΔV = `- (Δ"pV")/"K"`

New volume of metal, V' = V + ΔV

= `"V" - (Δ"pV")/"K" = "V" (1 - (Δ"p")/"K")`

Since, mass of metal remains same,

ρV = ρ'V'

`(ρ"′")/ρ = "V"/"V'" = "V"/("V"(1 - (Δ"p")/"K")) = 1/((1 - (Δ"p")/"K"))`

When Δp = p, then `(ρ"′")/ρ = 1/((1 - ("p")/"K"))`