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:

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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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 = \(\frac{R}{cos\phi} - R = R = 6400Km\) – Jatin Yadav · 3 years, 9 months ago

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– Kushagraa Aggarwal · 3 years, 9 months ago

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

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? – Pranav Arora · 3 years, 9 months ago

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– Pranav Arora · 3 years, 9 months ago

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

– Jatin Yadav · 3 years, 9 months ago

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

Can you tell me what would be the expected cutoff. – Divyansh Singhal · 3 years, 9 months ago

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– Jatin Yadav · 3 years, 9 months ago

I don't have any idea , previous year it was 56Log in to reply

– Bhargav Das · 3 years, 9 months ago

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

– Divyansh Singhal · 3 years, 9 months ago

Its ok.Log in to reply