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Q.

The escape velocity from the earth's surface is 11 km,/sec. A certain planet has a radius twice that of the earth but its mean density is the same as that of the earth. The value of the escape velocity from this planet would be

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a

22 km/sec

b

11 km/sec

c

5-5 km/sec

d

16.5 km/sec

answer is A.

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Detailed Solution

ve=2GMR where M=43πR3d∴ ve=8GπR3d3R=8GπR2d3=8πGd3R∝RNow (v)earth veplanet =(R)earth (R)planet 11ve=R2R  or  ve=22km/sec
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