Question

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

Answer 1

Hint As we know the volume of the sphere. $V={\displaystyle \frac{4}{3}}\pi r^3$. 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,
$V={\displaystyle \frac{4}{3}}\pi r^3$
Here,
$V$, will be the volume
$r$ , will be the radius
Bulk modulus,
$B={\displaystyle \frac{-P}{{\displaystyle \frac{△V}{V}}}}$
Here,
$B$, will be the bulk modulus
$P$, will be the pressure
$△V$, change in the volume

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

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.