Answer

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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,

$V = \dfrac{4}{3}\pi {r^3}$

Here,

$V$, will be the volume

$r$ , will be the radius

Bulk modulus,

$B = \dfrac{{ - P}}{{\dfrac{{\vartriangle V}}{V}}}$

Here,

$B$, 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$\vartriangle r$.

Therefore,

$ \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.

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