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IIT-JEE 2010 · Paper 1, Physics question 64

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

An annular disc (shaded ring) with inner radius 3R and outer radius 4R; point P lies on its axis at height 4R above the centre.
Choose one option
Answer and worked solution

Official answer: A

Potential on the axis of an annulus. Surface density σ=Mπ(16−9)R2=M7πR2\sigma=\dfrac M{\pi(16-9)R^2}=\dfrac M{7\pi R^2}. A ring of radius ss at axial distance zz contributes −Gσ2πs dss2+z2-\dfrac{G\sigma2\pi s\,ds}{\sqrt{s^2+z^2}}:
V=−2πGσ[s2+z2]3R4R.V=-2\pi G\sigma\left[\sqrt{s^2+z^2}\right]_{3R}^{4R}.

At z=4Rz=4R. 32R−25R=(42−5)R\sqrt{32}R-\sqrt{25}R=(4\sqrt2-5)R, so
VP=−2GM7R(42−5).V_P=-\frac{2GM}{7R}(4\sqrt2-5).

Work to take unit mass to infinity: W=0−VP=2GM7R(42−5)W=0-V_P=\dfrac{2GM}{7R}(4\sqrt2-5).

(A)

Topics:Gravitational field and potential
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