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Question

Three strips of 10 m width each are placed along the equator (A1 ), the Tropic of Cancer (A2 ), and the Arctic Circle (A3 ), respectively. The relationship amongst the areas of the strips is

The correct answer is A 1 >  A 2  > A 3  

Understanding Areas of Strips at Different Latitudes

The question asks about the relationship between the areas of three strips of equal width placed at the Equator, the Tropic of Cancer, and the Arctic Circle.

The Earth is approximately a sphere. Lines of latitude are circles parallel to the Equator. The circumference of these circles varies depending on the latitude. The formula for the circumference of a circle of latitude at an angle \(\phi\) (latitude) on a sphere of radius \(R\) is given by:

\(C(\phi) = 2\pi R \cos(\phi)\)

From this formula, we can see that the circumference \(C(\phi)\) depends on the cosine of the latitude angle \(\phi\). The cosine function \(\cos(\phi)\) has its maximum value (1) at \(\phi = 0^\circ\) (the Equator) and decreases as \(\phi\) increases towards the poles (where \(\cos(90^\circ) = 0\)).

The locations of the strips are:

  • Strip A1: Along the Equator (\(\phi = 0^\circ\))
  • Strip A2: Along the Tropic of Cancer (\(\phi \approx 23.5^\circ\))
  • Strip A3: Along the Arctic Circle (\(\phi \approx 66.5^\circ\))

All three strips have the same width, given as 10 m.

The area of each strip can be approximated as the circumference of the circle of latitude multiplied by the constant width. Since the width is the same for all strips, the area of each strip is directly proportional to the circumference of the circle of latitude at that location.

Comparing the latitudes:

  • Equator: \(\phi_1 = 0^\circ\)
  • Tropic of Cancer: \(\phi_2 \approx 23.5^\circ\)
  • Arctic Circle: \(\phi_3 \approx 66.5^\circ\)

Comparing the cosine values for these latitudes:

  • \(\cos(0^\circ) = 1\)
  • \(\cos(23.5^\circ) \approx 0.917\)
  • \(\cos(66.5^\circ) \approx 0.399\)

Since \(0^\circ < 23.5^\circ < 66.5^\circ\) and the cosine function decreases as the angle increases from \(0^\circ\) to \(90^\circ\), we have:

\(\cos(0^\circ) > \cos(23.5^\circ) > \cos(66.5^\circ)\)

This means the circumference of the circle of latitude is largest at the Equator, smaller at the Tropic of Cancer, and smallest at the Arctic Circle among these three locations.

Let \(C_1\), \(C_2\), and \(C_3\) be the circumferences at the Equator, Tropic of Cancer, and Arctic Circle, respectively. Then:

\(C_1 > C_2 > C_3\)

Since the area of each strip is proportional to its circumference (Area \(\approx\) Circumference \(\times\) Width), the relationship between the areas \(A_1\), \(A_2\), and \(A_3\) will be the same as the relationship between the circumferences.

\(A_1 > A_2 > A_3\)

Therefore, the area of the strip along the Equator (A1) is greater than the area of the strip along the Tropic of Cancer (A2), which is greater than the area of the strip along the Arctic Circle (A3).

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Important Questions from Mensuration

  1. The volume of a wall that is 5 times as high as it is broad and 8 times as long as it is high is 12.8 cubic meters. Find the breadth of the wall.
  2. If the radius of a sphere is increased by 2 cm, then its surface area increases by 704 cm\(^2\). Using \(\pi = \frac{22}{7}\), find the radius of the sphere after the increase.
  3. What is the perimeter of an equilateral triangle with side 13 cm?
  4. The perimeter of a square is 76 m. Its area (in $m^2$) is:
  5. A cylindrical rod has an outer curved surface area of 1800 cm$^2$. If the length of the rod is 52 cm, then the outer radius (in cm) of the rod, correct to two places of decimal, is:
    Take $\pi = \frac{22}{7}$
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