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The radius of Jupiter is $11$ times the radius of the earth. So, which of the following is the approximate number of earth that Jupiter can accommodate?

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2000

1331

5523

1254

answer is B.

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Detailed Solution

Approximately

1331

piles of earth can be accommodated in Jupiter.
It is given that the radius of Jupiter is

11

times the radius of the earth. Thus we can compare the volume of the Earth with the volume of Jupiter. Considering both planets are in sphere shape.
We know that volume of the sphere of radius,

r

\frac{4}{3} {πr}^{3} . . . . . . . . . . . . . . . . . . . . (a)

From (a), the volume of the Earth is

V_{E} = \frac{4}{3} {π {(R}_{E})}^{3} . . . . . . . . . . . . . . . . . (i)

{R_{E} =}^{}

Radius of the Earth.
From (a), the volume of Jupiter is

V_{J} = \frac{4}{3} {π {(R}_{J})}^{3} . . . . . . . . . . . . . (ii)

{R_{J} =}^{}

Radius of Jupiter.
Radius of Jupiter is

11

times the radius of the earth
So

R_{J} = 11 R_{E} . . . . . . . . . . . . . (iii)

Substituting the

R_{J}

from (iii) in (ii) we get

V_{J} = \frac{4}{3} {π {(11 R}_{E})}^{3}

{\Rightarrow V}_{J} = \frac{4}{3} {π {\times 1331 \times (R}_{E})}^{3}

{\Rightarrow V}_{J} = 1331 (\frac{4}{3} {π {(R}_{E})}^{3}) . . . . . . . . . . . . . . (iv)

Now dividing (iv) with (i)

\frac{V_{J}}{V_{E}} = \frac{1331 (\frac{4}{3} {π {(R}_{E})}^{3})}{\frac{4}{3} {π {(R}_{E})}^{3}}

\Rightarrow \frac{V_{J}}{V_{E}} = \frac{1331}{1}

\Rightarrow V_{J} : V_{E} = 1331 : 1

Hence the

1331

piles of Earth can be accommodated in Jupiter if the radius of Jupiter is

11

times the radius of the Earth.

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