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The bulk modulus of a spherical object is$B$ . It is subjected to uniform pressure $P$ the fractional decrease in radius is:
$\left( a \right)$ $\dfrac{P}{{3B}}$
$\left( b \right)$ $\dfrac{P}{B}$
$\left( c \right)$ $\dfrac{B}{{3P}}$
$\left( d \right)$ $\dfrac{{3P}}{B}$

Last updated date: 20th Jun 2024
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Hint As we know the volume of the sphere. $V = \dfrac{4}{3}\pi {r^3}$. We will differentiate both sides with respect to $r$ and after dividing the equation by $V$, we will get the new equation, and then by using the bulk modulus we would be able to get the fractional decrease in the radius.
Formula used:
The volume of the sphere will be given by,
$V = \dfrac{4}{3}\pi {r^3}$
$V$, will be the volume
$r$ , will be the radius
Bulk modulus,
$B = \dfrac{{ - P}}{{\dfrac{{\vartriangle V}}{V}}}$
$B$, will be the bulk modulus
$P$, will be the pressure
$\vartriangle V$, change in the volume

Complete Step By Step Solution
As we already know,
The volume of the sphere is given by
$V = \dfrac{4}{3}\pi {r^3}$
So we will now differentiate the above equation both sides with respect to $r$
We get,
$ \Rightarrow \dfrac{{dV}}{{dr}} = 3\left( {\dfrac{4}{3}\pi {r^2}} \right)$
So on simplifying we get,
$ \Rightarrow \dfrac{{dV}}{{dr}} = 4\pi {r^2}$
Here the term $dV$can be written as $\vartriangle V$and similarly $dr$as$\vartriangle r$.
$ \Rightarrow \vartriangle V = 4\pi {r^2}\vartriangle r$
Now dividing the above equation by$V$, and also putting the value of $V$on the RHS side, we get
$ \Rightarrow \dfrac{{\vartriangle v}}{v} = \dfrac{{4\pi {r^2}\vartriangle r}}{{\dfrac{4}{3}\pi {r^3}}}$
So on solving the above equation, we get
$ \Rightarrow \dfrac{{\vartriangle v}}{v} = 3\dfrac{{\vartriangle r}}{r}$
Now by using the bulk modulus, we get
$B = \dfrac{{ - P}}{{\dfrac{{\vartriangle V}}{V}}}$
Substituting the values, we get
$ \Rightarrow B = \dfrac{{ - P}}{{\dfrac{{3\vartriangle r}}{r}}}$
And it can be written as,
$ \Rightarrow \dfrac{{\vartriangle r}}{r} = \dfrac{P}{{3B}}$

Therefore, the option $A$ will be the correct one.

Note Bulk modulus, mathematical consistency that portrays the versatile properties of a strong or liquid when it is feeling the squeeze on all surfaces. The applied weight lessens the volume of a material, which re-visitations of its unique volume when the weight is taken out. At times alluded to as the inconceivability, the mass modulus is a proportion of the capacity of a substance to withstand changes in volume when under pressure on all sides. It is equivalent to the remainder of the applied weight isolated by the relative distortion.