Question

A uniform solid sphere of radius R is in equilibrium inside a liquid whose density varies with depth from free surface as $\rho ={\rho }_0(1+\frac{h}{h_0})$, where h is the depth from free surface. If the density of sphere is $\sigma ={\rho }_0(1+\frac{xd}{y{h}_0})$ then find (x + y).

Answer 1

Step 1: Draw a diagram of given problem.

Step 2: Calculate x and y respectively.

Given,
$\text{Density of liquid}=\rho ={\rho }_0(1+\frac{h}{h_0})$
$\text{Density of sphere},\sigma ={\rho }_0(1+\frac{xd}{y{h}_0})$
As we know, the height from center of axis of sphere to liquid is d, let x be the distance from center for taking differential element and their volume will be dV. Therefore, remaining distance will be d-x and d +x respectively.
Therefore, buoyant force is acting on differential element with volume dV.
Step 2: Calculate x and y respectively.
$\sigma Vg=\int \left[{\rho }_0(1+\frac{d-x}{h_0})\right(dv)g+{\rho }_0(1+\frac{d+x}{h_0})(dV\left)g\right]$
$\sigma V=2{\rho }_0(1+\frac{d}{h_0}){\int }_0^{V/2}dV$
$={\rho }_{0}V(1+\frac{d}{h_0})$
$\sigma ={\rho }_0(1+\frac{d}{h_0})$
Hence, the value of x and y will be $1$ and $1$.
Therefore, $(x+y)=2$
Final answer:$\left(2\right)$