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Question: The normal density of gold is \(\rho \) and its bulk modulus is \(K\) .The Increase in the density o...

The normal density of gold is ρ\rho and its bulk modulus is KK .The Increase in the density of a lump of gold, when pressure PP is applied uniformly on all sides is
A. ρPK\dfrac{{\rho P}}{K}
B. ρKP\dfrac{{\rho K}}{P}
C. PρK\dfrac{P}{{\rho K}}
D. KρP\dfrac{K}{{\rho P}}

Explanation

Solution

We first write the bulk modulus formula, then we know in bulk modules there is no change in mass, applying this statement in the problem we get a relation between volume and density of the gold. Use this relation in the bulk modulates formula.Now after rearrangement we will get the increase in the density of the lump gold.

Formula used:
K=−dPdVVK = - \dfrac{{dP}}{{\dfrac{{dV}}{V}}}
Where, Bulk modulus = KK, Pressure applied = dPdP and Change in its volume =dVV\dfrac{{dV}}{V}.
ρ=mV\rho = \dfrac{m}{V}
Where, Density = ρ\rho , Mass = mm and Volume = VV.

Complete step by step answer:
As given in the problem, there is a lump of gold whose normal density is ρ\rho , bulk modulus is KK and the pressure applied is PP.Due to this applied pressure there is an increase in the density of the lump of gold.Bulk modulus is defined as the pressure that we applied divided by the fractional change in its volume due to application of this pressure.Mathematically we can write,
K=−dPdVVK = - \dfrac{{dP}}{{\dfrac{{dV}}{V}}}

Now by applying this bulk formula we will get,
K=−dPdVVK = - \dfrac{{dP}}{{\dfrac{{dV}}{V}}}
⇒K=−VdPdV……(1)\Rightarrow K = - V\dfrac{{dP}}{{dV}} \ldots \ldots \left( 1 \right)
We can define density as mass divided by volume of the lump of gold.
Mathematically,
ρ=mV\rho = \dfrac{m}{V}
⇒ρ=mV−1\Rightarrow \rho = m{V^{ - 1}}
Now differentiating the above expression we will get,
 Δρ=V−1Δm−mV−2 ΔV\,\Delta \rho = {V^{ - 1}}\Delta m - m{V^{ - 2}}\,\Delta V
Divind both side by ρ\rho we will get,
 Δρρ=V−1Δmρ−m V−2ΔVρ\,\dfrac{{\Delta \rho }}{\rho } = \dfrac{{{V^{ - 1}}\Delta m}}{\rho } - \dfrac{{m\,{V^{ - 2}}\Delta V}}{\rho }

Putting ρ\rho value in RHS we will get,
 Δρρ=V−1ΔmmV−1−mV−2 ΔVmV−1\,\dfrac{{\Delta \rho }}{\rho } = \dfrac{{{V^{ - 1}}\Delta m}}{{m{V^{ - 1}}}} - \dfrac{{m{V^{ - 2}}\,\Delta V}}{{m{V^{ - 1}}}}
Cancelling the common terms we will get,
 Δρρ=Δmm− ΔVV\,\dfrac{{\Delta \rho }}{\rho } = \dfrac{{\Delta m}}{m} - \dfrac{{\,\Delta V}}{V}
We know that in bulk modulus change in mass is equal to zero.
Hence the above equation will become,
 Δρρ=− ΔVV\,\dfrac{{\Delta \rho }}{\rho } = - \dfrac{{\,\Delta V}}{V}
Small change in the term,
 dρρ=− dVV……(2)\,\dfrac{{d\rho }}{\rho } = - \dfrac{{\,dV}}{V} \ldots \ldots \left( 2 \right)

Replacing equation (1)\left( 1 \right) with equation (2)\left( 2 \right) we will get,
K=−VdPdVK = - V\dfrac{{dP}}{{dV}}
⇒K=−dPdVV\Rightarrow K = - \dfrac{{dP}}{{\dfrac{{dV}}{V}}}
By putting,
K=dPdρρK = \dfrac{{dP}}{{\dfrac{{d\rho }}{\rho }}}
⇒K=ρdPdρ\Rightarrow K = \rho \dfrac{{dP}}{{d\rho }}
As given in the problem here dP=PdP = P
Now replacing the value in the above equation we will get,
K=ρPdρK = \rho \dfrac{P}{{d\rho }}
Rearranging the above equation we will get,
dρ=ρPKd\rho = \rho \dfrac{P}{K}
Where, change in density = dρd\rho

Therefore the correct option is (A)\left( A \right).

Note: Be careful while differentiating the density term. Differentiate the RHS term into two different parts: first differentiate with respect to mass by keeping volume constant and then again differentiate with respect to volume by keeping mass constant. This eases the calcul;ations in the problem.