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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22 km/sec

11 km/sec

5-5 km/sec

16.5 km/sec

answer is A.

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Detailed Solution

$\begin{array}{l} v_{e} = \sqrt{(\frac{2 GM}{R})} where M = \frac{4}{3} {πR}^{3} d \\ ∴ v_{e} = \sqrt{(\frac{8 {GπR}^{3} d}{3 R})} = \sqrt{(\frac{8 {GπR}^{2} d}{3})} \\ = \sqrt{(\frac{8 πGd}{3})} R \propto R \end{array}$
Now $\frac{(v)_{earth}}{{(v_{e})}_{planet}} = \frac{(R)_{earth}}{(R)_{planet}}$
$\frac{11}{v_{e}} = \frac{R}{2 R} or v_{e} = 22 km / \sec$