A thin uniform annular disc (see figure) of mass $M$ has outer radius $4 R$ and inner radius $3 R$ . The work required to take a unit mass from point P on its axis to infinity is [2010]

Question

Smart Achievers · Accepted Answer

(1)

Mass per unit area of the shaded part

$σ = \frac{mass}{area} = \frac{M}{π ({(4 R)}^{2} - {(3 R)}^{2})} = \frac{M}{7 π R^{2}}$

Let us consider a ring of radius $x$ and thickness $d x$ as shown in the figure.

$Mass of the ring, d M = σ 2 π x d x = \frac{2 π M x d x}{7 π R^{2}}$

Potential at point $P$ due to shaded part

$V_{P} = \int_{3 R}^{4 R} - \frac{G d M}{\sqrt{{(4 R)}^{2} + x^{2}}} = - \frac{G M 2 π}{7 π R^{2}} \int_{3 R}^{4 R} \frac{x d x}{\sqrt{16 R^{2} + x^{2}}}$

Solving, we get

$V_{P} = - \frac{G M 2 π}{7 π R^{2}} {[\sqrt{16 R^{2} + x^{2}}]}_{3 R}^{4 R} = - \frac{2 G M}{7 R} (4 \sqrt{2} - 5)$

Work done in moving a unit mass from $P to \infty = V_{\infty} - V_{P}$

$or W_{P \to \infty} = 0 - (- \frac{2 G M}{7 R} (4 \sqrt{2} - 5)) = \frac{2 G M}{7 R} (4 \sqrt{2} - 5)$

Solution

Q.
A thin uniform annular disc (see figure) of mass $M$ has outer radius $4 R$ and inner radius $3 R$ . The work required to take a unit mass from point P on its axis to infinity is [2010]

1 $\frac{2 G M}{7 R} (4 \sqrt{2} - 5)$

2 $- \frac{2 G M}{7 R} (4 \sqrt{2} - 5)$

3 $\frac{G M}{4 R}$

4 $\frac{2 G M}{5 R} (\sqrt{2} - 1)$

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Solution

Q. A thin uniform annular disc (see figure) of mass M has outer radius 4R and inner radius 3R. The work required to take a unit mass from point P on its axis to infinity is [2010]

1 2GM7R (42-5)

2 -2GM7R (42-5)

3 GM4R