GATE 2024 EC Q.36: solution and answer

GATE 2024 · ECQ.36 · 2 marks · MCQ

Consider the Earth to be a perfect sphere of radius $R$ . Then the surface area of the region, enclosed by the $6 0^{\circ}$ N latitude circle, that contains the north pole in its interior is ________.

(A)
$(2 - 3) π R^{2}$
(B)
$\frac{( 2 - 1 ) π R ^{2}}{2}$
(C)
$\frac{2 π R ^{2}}{3}$
(D)
$\frac{( 2 + 3 ) π R ^{2}}{8 2}$

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Answer and solution

ANSWER

The surface area of the polar cap bounded by the 60°N latitude circle is $(2 - 3) π R^{2}$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2024 EC (IISc Bengaluru)

Use spherical coordinates with the polar angle $θ$ measured from the north pole. Latitude is measured from the equator, so latitude $6 0^{\circ}$ N lies at polar angle $θ = 9 0^{\circ} - 6 0^{\circ} = 3 0^{\circ} = π /6$ . The region containing the north pole is therefore the cap $0 \leq θ \leq π /6$ , $0 \leq ϕ \leq 2 π$ .
The surface area element on a sphere of radius $R$ is $d A = R^{2} sin θ d θ d ϕ$ .
Integrate over the cap: $A = \int_{0}^{2 π} \int_{0}^{π /6} R^{2} sin θ d θ d ϕ = 2 π R^{2} [- cos θ]_{0}^{π /6} = 2 π R^{2} (1 - cos \frac{π}{6})$
Substitute $cos (π /6) = 3 /2$ : $A = 2 π R^{2} (1 - \frac{3}{2}) = (2 - 3) π R^{2}$
Check with Archimedes' spherical-cap formula $A = 2 π R h$ . The plane of the $6 0^{\circ}$ N circle sits at height $R sin 6 0^{\circ}$ above the equatorial plane, so the cap height is $h = R - R sin 6 0^{\circ} = R (1 - \frac{3}{2})$ . This gives the same $(2 - 3) π R^{2} \approx 0.268 π R^{2}$ , about $6.7%$ of the whole sphere's $4 π R^{2}$ . That is a sensible size for a small polar cap.
This matches option (A).

KEY IDEA

A latitude circle cuts a spherical cap off the sphere. Convert latitude to polar angle ( $θ_{0} = 9 0^{\circ} - latitude$ ) and integrate $d A = R^{2} sin θ d θ d ϕ$ to get $A = 2 π R^{2} (1 - cos θ_{0})$ . This is the same as Archimedes' $A = 2 π R h$ , where $h$ is the cap height.

WATCH OUT FOR

Using $6 0^{\circ}$ as the polar angle, i.e. mixing up latitude (measured from the equator) with the angle from the pole. That gives $2 π R^{2} (1 - cos 6 0^{\circ}) = π R^{2}$ , which is not among the options. A second slip is taking the cap height as $R cos 6 0^{\circ}$ or $R - R cos 6 0^{\circ}$ instead of $R - R sin 6 0^{\circ}$ .

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GATE 2024 EC Q.36: solution and answer

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