A thin uniform annular disc (see figure) of mass M has outer radius 4R and the inner radius 3R. The work required to take a unit mass from point P on it’s axis at a distance of 4R from centre of disc to infinity is $\frac{2 GM}{xR} (4 \sqrt{2} - 5)$ Find the value of $x$ .

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Detailed Solution

$σ = \frac{M}{A}; σ = \frac{M}{16 {πR}^{2} - 9 {πR}^{2}} : σ = \frac{M}{7 {πR}^{2}}$
Potential due to small ring of mass dm at P is $dV = \frac{- Gdm}{r} = \frac{- G 2 πdx \cdot σ}{r}$
$dV = - 2 πGσ \frac{xdx}{\sqrt{16 R^{2} + x^{2}}}$
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Total potential at P,
$\begin{array}{l} V_{p} = \int_{3 R}^{4 R} dV; V_{p} = - 2 πGσ \int_{3 R}^{4 R} \frac{xdx}{\sqrt{16 R^{2} + x^{2}}} \\ V_{p} = - 2 π \frac{GM}{7 {πR}^{2}} [\sqrt{16 R^{2} + 16 R^{2}} - \sqrt{16 R^{2} + 9 R^{2}}] \\ V_{p} = \frac{- 2 GM}{7 R} [4 \sqrt{2} - 5]; V_{\infty} = 0, ω = ΔVm \\ ω = V_{\infty} - V_{p} \\ ω = \frac{2 GM}{7 R} [4 \sqrt{2} - 5] \end{array}$