# 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
4 years, 11 months ago

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## Comments

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

- 4 years, 11 months ago

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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.

- 4 years, 11 months ago

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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?

- 4 years, 11 months ago

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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.

- 4 years, 11 months ago

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

- 4 years, 11 months ago

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Can you tell me what would be the expected cutoff.

- 4 years, 11 months ago

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

- 4 years, 11 months ago

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

- 4 years, 11 months ago

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Its ok.

- 4 years, 10 months ago

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