# KVPY 2013 question

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: Note by Kushagraa Aggarwal
6 years, 7 months ago

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Hi, 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 = $\frac{R}{cos\phi} - R = R = 6400Km$

- 6 years, 7 months ago

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

- 6 years, 7 months ago

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?

- 6 years, 7 months ago

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

- 6 years, 7 months ago

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

- 6 years, 7 months ago

Can you tell me what would be the expected cutoff.

- 6 years, 7 months ago

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

- 6 years, 7 months ago

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

- 6 years, 7 months ago

Its ok.

- 6 years, 7 months ago