Hi all, it was a question asked in our test:

If Earth's radius is 6400Km , the height from earth's surface of a point from where 1/4 of earth's surface is visible is:

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## Comments

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TopNewestHi, it should be \(6400Km\)

Here is the solution,

Cutting into rings , the area of a small ring at an angle \(\theta\) would be \(2 \pi R^2 sin\theta d\theta\).

Hence area of visible surface for an angle \(\phi\)(Shown as \(\frac{\pi}{4}\) in Pranav A.'s image) = \(\displaystyle \int_{0}^{\phi}2 \pi R^2 sin\theta d\theta\)

= \(2 \pi R^2(1 - cos\phi) = \frac{1}{4}4 \pi R^2\)(given)

Hence, \(\phi = \frac{\pi}{3}\).

Hence, height from surface = \(\frac{R}{cos\phi} - R = R = 6400Km\)

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Great!, first, when i posted the question resonance was showing answer to be \(3200Km\), but now they have changed their answer.

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Image

Since \(\cos(\pi/4)=R/(R+h) \), h can be calculated. R is the radius of earth.

Is it possible for you to share the answer?

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I would like to request the members to stop voting up my reply. Jatin's answer is correct and nicely explained.

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Why people voted down?,they should not have the right to vote down something other than an abuse .

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Can you tell me what would be the expected cutoff.

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I don't have any idea , previous year it was 56

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Hello Jatin, I am just curious to know that you're in which class?

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Its ok.

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