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Question: A uniform pressure p is exerted on all sides of a solid cube at temperature \(t^{\circ}C\). By what ...

A uniform pressure p is exerted on all sides of a solid cube at temperature t∘Ct^{\circ}C. By what amount should the temperature of the cube be raised in order to bring its volume back to the value it had before the pressure was applied? The coefficient of volume expansion of the cube y and the bulk modulus is B.
A). p2yB\dfrac{p}{\sqrt{2yB}}
B). p2yB\dfrac{p}{2yB}
C). 2pyB\dfrac{2p}{yB}
D). pyB\dfrac{p}{yB}

Explanation

Solution

Recall that the bulk modulus is the ratio of the volumetric stress to the volumetric strain. Use this to arrive at an expression for the change in volume accounting for the expansion of the cube to bring it back to its initial volume. Then, we see that the cube is heated so as to increase its volume. Therefore, use the expression for the coefficient of volumetric expansion and determine the change in volume in terms of the parameters that the relation entails. Equate the two expressions for change in volume and arrive at the appropriate expression for the change in temperature.
Formula Used: Bulk modulus: B=−ΔPΔVVB = -\dfrac{\Delta P}{\dfrac{\Delta V}{V}}
Coefficient of volume expansion y=ΔVVΔty = \dfrac{\dfrac{\Delta V}{V}}{\Delta t}

Complete step-by-step solution:
Let us begin by first understanding what the bulk modulus means.
The bulk modulus of a substance is a numerical measure of how resistant to compression the given material is. It describes the elastic properties of the material when it is subject to pressure on all surfaces. It is numerically defined as the ratio between an increase in pressure and the corresponding decrease in the material’s volume, or the proportion of volumetric stress related to the volumetric strain of the material.
The bulk modulus B=−ΔPΔVV=−VΔPΔVB = -\dfrac{\Delta P}{\dfrac{\Delta V}{V}} = -V\dfrac{\Delta P}{\Delta V}, where P is the pressure, V is the initial volume, and ΔPΔV\dfrac{\Delta P}{\Delta V} represent the change in pressure with respect to change in volume.
From this, we can get the initial volume, or the volume expansion to get the initial volume as:
ΔV=−PB.V\Delta V = -\dfrac{P}{B}.V
Now, this volume expansion is brought about by raising the temperature of the cube by Δt∘C\Delta t^{\circ}C. This is a thermal expansion process, and hence can be modelled in terms of the coefficient of volume expansion y as follows:
y=ΔVVΔty = \dfrac{\dfrac{\Delta V}{V}}{\Delta t}
Since this is an expansion process, ΔV=−ΔV\Delta V = -\Delta V
⇒−ΔV=VyΔt⇒ΔV=−VyΔt\Rightarrow -\Delta V = Vy\Delta t \Rightarrow \Delta V = -Vy\Delta t
Equating the two equation for change in volume, we get:
−PB.V=−VyΔt-\dfrac{P}{B}.V = -Vy\Delta t
⇒Δt=PBy\Rightarrow \Delta t = \dfrac{P}{By}
Therefore, the correct choice would be D. pyB\dfrac{p}{yB}

Note: Note that we have used the bulk modulus and the coefficient of volumetric expansion since the cube undergoes stress and produces corresponding compressive strains in al three dimensions contributing to volumetric changes.
In case the whole process was linear, then we would have to use Young’s modulus and the linear coefficient of expansion:
Y=FA.L0ΔLY = \dfrac{F}{A}. \dfrac{L_0}{\Delta L}, where L represents the length of the material. And,
αL=ΔLLΔT\alpha_L = \dfrac{\Delta L}{L\Delta T}
And in case the whole process was confined to the area of the material and subsequent changes in the same, then we would have the shear modulus and coefficient of area expansion
S=FA.LΔxS =\dfrac{F}{A}.\dfrac{L}{\Delta x} , where Δx\Delta x is the lateral displacement. And,
αA=ΔAAΔT\alpha_A = \dfrac{\Delta A}{A\Delta T}, where A is the area under consideration.