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Q.
Escape velocity at surface of earth is $11.2\,km/s$. Escape velocity from a planet whose mass is the same as that of earth and radius $1/4$ that of earth, is :
BHUBHU 2004Gravitation
Solution:
At a certain velocity of projection the body will go out of the gravitational field of the earth and will never return of the earth, this velocity is known as escape velocity
$v_{e}=\sqrt{\frac{2 G M_{e}}{R_{e}}}$
Given, $M_{e}=M_{p}, R_{p}=\frac{R_{e}}{4}$
$\therefore \frac{v_{p}}{v_{e}}=\sqrt{\frac{M_{e}}{M_{e}} \times \frac{R_{e}}{R_{e / 4}}}$
$=\sqrt{4}=2$
$\Rightarrow v_{p}=2 v_{e}=2 \times 11.2$
$=22.4 \,km / s$