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
5 years, 11 months ago

No vote yet
5 votes

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Hi, it should be 6400Km6400Km

Here is the solution,

Cutting into rings , the area of a small ring at an angle θ\theta would be 2πR2sinθdθ2 \pi R^2 sin\theta d\theta.

Hence area of visible surface for an angle ϕ\phi(Shown as π4\frac{\pi}{4} in Pranav A.'s image) = 0ϕ2πR2sinθdθ\displaystyle \int_{0}^{\phi}2 \pi R^2 sin\theta d\theta

= 2πR2(1cosϕ)=144πR22 \pi R^2(1 - cos\phi) = \frac{1}{4}4 \pi R^2(given)

Hence, ϕ=π3\phi = \frac{\pi}{3}.

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

jatin yadav - 5 years, 11 months ago

Log in to reply

Great!, first, when i posted the question resonance was showing answer to be 3200Km3200Km, but now they have changed their answer.

kushagraa aggarwal - 5 years, 11 months ago

Log in to reply

Image

Since cos(π/4)=R/(R+h)\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 - 5 years, 11 months ago

Log in to reply

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

Pranav Arora - 5 years, 11 months ago

Log in to reply

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

jatin yadav - 5 years, 11 months ago

Log in to reply

Can you tell me what would be the expected cutoff.

Divyansh Singhal - 5 years, 11 months ago

Log in to reply

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

jatin yadav - 5 years, 11 months ago

Log in to reply

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

Bhargav Das - 5 years, 11 months ago

Log in to reply

Its ok.

Divyansh Singhal - 5 years, 11 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...