Question

An external pressure $P$ is applied on a cube ${0^ \circ }c$ so that it is equally compressed from all sides. $K$ is the bulk modulus of the material of the cube and $\alpha$ is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by:
$\left( a \right)$ $3PK\alpha$
$\left( b \right)$ $\dfrac{P}{{3\alpha K}}$
$\left( c \right)$ $\dfrac{P}{{\alpha K}}$
$\left( d \right)$ $\dfrac{{3\alpha }}{{PK}}$

Answer 1

Hint First of all we will see the bulk modulus of materials and it is equal to the $\dfrac{{\vartriangle V}}{{{V_0}}} = \dfrac{{\vartriangle P}}{K}$. And by using the thermal expansion we will get the value of $\dfrac{{\vartriangle V}}{{{V_0}}}$and the result we will get from there we will find out the $\vartriangle T$ and after that, we will change it in pressure and get the required result.
Formula used
The bulk modulus of the material,
$K = \dfrac{{\vartriangle P}}{{\left( { - \vartriangle V/{V_0}} \right)}}$
Here,
$K$, will be the bulk modulus
$\vartriangle V$, will be the change in volume
$\vartriangle P$, will be the change in pressure
${V_0}$, will be the original volume

Complete Step By Step Solution
As we know the bulk modulus will be equal to
$K = \dfrac{{\vartriangle P}}{{\left( { - \vartriangle V/{V_0}} \right)}}$
And it can also be written as
$\Rightarrow \dfrac{{\vartriangle V}}{{{V_0}}} = \dfrac{{\vartriangle P}}{K}$
Since there is a rise in temperature, so due to thermal expansion
$V = {V_0}\left( {1 + \gamma \vartriangle t} \right)$
And also it can be written as
$\Rightarrow V = {V_0} + {V_0}\gamma \vartriangle t$
Now taking $\gamma \vartriangle t$on one side and rest at one side, we get
$\Rightarrow \dfrac{{V - {V_0}}}{{{V_0}}} = \gamma \vartriangle t$
And since $\left[ {\gamma = 3\alpha } \right]$
Therefore,
$\Rightarrow \dfrac{{V - {V_0}}}{{{V_0}}} = \gamma \vartriangle t = 3\alpha \vartriangle t$
So, the upper equation can be written as,
$\Rightarrow \dfrac{{\vartriangle V}}{{{V_0}}} = 3\alpha \vartriangle t$
And therefore it can be written as
$\Rightarrow \dfrac{{\vartriangle P}}{K} = 3\alpha \vartriangle t$
And from here, we get
$\Rightarrow \vartriangle t = \dfrac{{\vartriangle P}}{{3\alpha K}}$
Or we can write it as
Since $\vartriangle P = P$given in the question.
Therefore,
$\Rightarrow \vartriangle t = \dfrac{P}{{3\alpha K}}$

Hence, Option $B$ is the correct choice.

Note Bulk means wholesome, in a sense it refers to the whole of the material, i.e. Volume. Simply put, any modulus is the ratio between stresses to strain. Any Modulus is a measure of the resistance to deformation. Typically brought up as in understandability, the bulk modulus may be a measure of the flexibility of a substance to face up to changes in volume once below compression on all sides.