Question

A uniform follow sphere has internal radius a and external radius b. The ratio of gravitational potential at a point on its outer surface to that of a point on its inner surface is

• A. $\frac{2(b^3-a^3)}{3b(b^2-a^2)}$
• B. $\frac{2(b^3+a^3)}{3b(b^2+a^2)}$
• C. $\frac{a(a^2-b^2)}{3b(a^2+b^2)}$
• D. $1$

Answer 1

The correct option is A $\frac{2(b^3-a^3)}{3b(b^2-a^2)}$

The inner radius of sphere is $a$ and outer radius is $b$. Therefore the mass of the sphere is whose density is $p$ is

$M=\frac{4}{3}\pi p(b^3-a^3)$

Gravitational potential at a point on outer surface is

$V_b=\frac{-GM}{b}=\frac{-G\frac{4}{3}\pi p(b^3-a^3)}{b}$

Now to find gravitational potential at point on inner surface consider a point at a distance $r$ from $a$ and $b$

Mass of sphere $=\frac{4}{3}\pi p(r^3-a^3)$

Therefore Gravitational field strength is $M=\frac{G\frac{4}{3}\pi p(r^3-a^3)}{r^2}$

Now the relation is $g=\frac{-dV}{dr}$

$\therefore$ $-{\int }_b^{a}g={\int }_b^a\frac{dV}{dr}=V_a-V_b$

$\therefore$ $V_a-V_b=\frac{4\pi pa}{3}{\int }_b^a\frac{r^3-a^3}{r^2}dr$

$\therefore$ $V_a-V_b=\frac{4\pi pa}{3}{[\frac{r^2}{2}+\frac{a^3}{r}]}_b^a$

$=\frac{4}{3}\pi pa[\frac{a^2}{2}-\frac{b^2}{2}+\frac{a^3}{a}-\frac{a^3}{b}]$

$\therefore$ $V_a-V_b=\frac{4\pi pa}{3}[\frac{{3a}^2}{2}-\frac{b^2}{2}-\frac{a^3}{b}]$

Therefore we get

$V_a=V_b+\frac{4\pi pa}{3}[\frac{{3a}^2}{2}-\frac{b^2}{2}-\frac{a^3}{b}]$

$\therefore$ $V_a=-\frac{G\frac{4}{3}\pi p(b^3-a^3)}{b}+\frac{4}{3}\pi ph[\frac{{3a}^2}{2}-\frac{b^2}{2}-\frac{a^3}{b}]$

$\therefore$ $V_a=\frac{4\pi ph}{3}[\frac{{3a}^2}{2}-\frac{b^2}{2}-\frac{a^3}{b}-b^2+\frac{a^3}{b}]$

$\therefore$ $V_a=\frac{4\pi pG}{3}[\frac{{3a}^2}{2}-\frac{{3b}^2}{2}]$

$\therefore$ Ratio is

$\frac{V_b}{V_a}=\frac{\frac{-G\frac{4}{3}\pi p(b^3-a^3)}{b}}{\frac{4\pi ph}{3}[\frac{{3a}^2}{2}-\frac{{3b}^2}{2}]}$

$=-\frac{\frac{b^3-a^3}{b}}{\frac{{3a}^2-{3b}^2}{2}}$

$=\frac{-2(b^3-a^3)}{3b(a^2-b^2)}=\frac{2(b^3-a^3)}{3b(b^2-a^2)}$