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Assume for a moment that the earth is a perfectly uniform
sphere of radius 6400 km. Suppose a thread equal to the
length of the circumference of the earth was placed along
the equator, and drawn to a tight fit.
Now suppose that the length of the thread is increased by 12
cm, and that it is pulled away uniformly in all directions.
By how many cm. will the thread be separated from the
earth's surface?
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Answer / guest
The cicumference of the earth is
= 2 * PI * r
= 2 * PI * 6400 km
= 2 * PI * 6400 * 1000 m
= 2 * PI * 6400 * 1000 * 100 cm
= 1280000000 * PI cm
where r = radius of the earth, PI = 3.141592654
Hence, the length of the thread is = 1280000000 * PI cm
Now length of the thread is increasd by 12 cm. So the new
length is = (1280000000 * PI) + 12 cm
This thread will make one concentric circle with the earth
which is slightly away from the earth. The circumfernce of
that circle is nothing but (1280000000 * PI) + 12 cm
Assume that radius of the outer circle is R cm
Therefore,
2 * PI * R = (1280000000 * PI) + 12 cm
Solving above equation, R = 640000001.908 cm
Radius of the earth is r = 640000000 cm
Hence, the thread will be separatedfrom the earth by
= R - r cm
= 640000001.908 - 640000000
= 1.908 cm
| Is This Answer Correct ? | 6 Yes | 2 No |
Answer / superpogi
R = old radius
R' = new radius
x = 12 cm
old length = 2piR
new length = 2piR' = 2piR + x
R' = R + x/(2pi)
separation = x/(2pi) = 12/(2pi) = 1.91 cm
| Is This Answer Correct ? | 5 Yes | 2 No |
Answer / roop anand
radius=6400000cm
circumference=2*pi*r
=2*3.14*6400000
=40192000cm
length is elongated by 12cm..
so circumference is increased br 12cm..
2*pi*r=40192000+12
=40192012cm
the new radius = 2*pi*r/2*pi
=40192000/2*(3.14)
=6400001.911
therefore the length of the distance its seperated from the
surface is = 6400001.911-640000
=1.911cm
| Is This Answer Correct ? | 3 Yes | 1 No |
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