The normal density of gold is $ρ$ and its bulk modules is K. The increase in density of a piece of gold when a pressure P is applied uniformly from all sides is

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$\frac{KP}{2 ρ}$

$\frac{Pρ}{2 K}$

$\frac{Kρ}{K - ρ}$

$\frac{ρP}{K - P}$

answer is D.

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Detailed Solution

$ΔV = - \frac{V . ΔP}{K} \dots ... (i); ρ = \frac{m}{V} \Rightarrow \frac{ΔV}{V} = \frac{- Δ ρ}{ρ}$

$K = \frac{- VΔP}{ΔV} = \frac{ΔP}{- (\frac{ΔV}{V})} = \frac{ΔP}{\frac{Δ ρ}{ρ}} \Rightarrow Δ ρ = \frac{P ρ}{K}$

$(Here ΔP = P)$

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