Escape Velocity

NCERT Class 11, Chapter 8 (Gravitation) derives escape velocity by setting the kinetic energy supplied at launch equal to the magnitude of the gravitational potential energy binding the object to the surface, so that the total mechanical energy becomes zero. Expressed using surface gravity, this gives $v_e=\sqrt{2gR}$, a form that is convenient when $G$ and the mass are not provided but $g$ and $R$ are known. Converting the radius to $R=1.6\times 10^6\text{m}$ and using $g=1.6{\text{m/s}}^2$: $v_e=\sqrt{2\times 1.6\times 1.6\times 10^6}=\sqrt{5.12\times 10^6}\approx 2263\text{m/s}=2.26\text{km/s}$. The option $1.60\text{km/s}$ mistakes $\sqrt{gR}$, the near-surface orbital speed, for the escape speed and omits the factor of two inside the root. The option $3.20\text{km/s}$ doubles the orbital value rather than multiplying inside the square root. The option $5.06\text{km/s}$ drops the square root entirely and uses raw numbers. As a plausibility check, this airless moon's weak gravity should give an escape speed far below Earth's value of 11.2 km/s, and 2.26 km/s is sensibly small, confirming the result is physically reasonable.