Hint: \(B = -\frac{ p}{\frac{\Delta V}{V}}\)
Step 1: Find the change in the volume of the sphere.
The bulk modulus of the sphere is given by;
\(B = -\frac{p}{\frac{\Delta V}{V}}\)
\(\Rightarrow \frac{\Delta V}{V} = -\frac{p}{B}~~~…(1)\)
The volume of the sphere is given by;
\(V = \frac{4}{3}\pi R^3\)
\(\Rightarrow \frac{\Delta V}{V} = 3\frac{\Delta R}{R}\)
\(\Rightarrow \frac{\Delta R}{R} = \frac{1}{3}\frac{\Delta V}{V}~~~…(2)\)
Step 2: Find the change in the cross-sectional area of the sphere.
The surface area of the sphere is given by;
\(A= 4\pi R^2\)
\(\Rightarrow \frac{\Delta A}{A} = 2\frac{\Delta R}{R}~~~…(3)\)
From the equation \((1), (2),\) and \((3)\) we get;
\(\frac{\Delta A}{A} = -\frac{2p}{3B}\)
\(\Rightarrow \Delta A = -\frac{2pA}{3B}\)
Therefore, the magnitude of the cross-sectional area of the sphere is \(\frac{2pA}{3B}\).
Hence, option (3) is the correct answer.
