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:

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Easy Math Editor

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Comments

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

jatin yadav - 7 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.

kushagraa aggarwal - 7 years, 7 months ago

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 - 7 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.

Pranav Arora - 7 years, 7 months ago

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

jatin yadav - 7 years, 7 months ago

Can you tell me what would be the expected cutoff.

Divyansh Singhal - 7 years, 7 months ago

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

jatin yadav - 7 years, 7 months ago

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

Bhargav Das - 7 years, 7 months ago

Its ok.

Divyansh Singhal - 7 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$