Escape Velocity

NCERT Class 11, Chapter 8 (Gravitation) gives escape velocity as $v_e=\sqrt{\frac{2GM}{R}}$, the speed at which an object's total mechanical energy just becomes zero so it can barely reach \infty. When the density $\rho$ is held constant, the mass is tied to the size through $M=\rho \cdot \frac{4}{3}\pi R^3$. Substituting, $v_e=\sqrt{\frac{2G}{R}\cdot \rho \cdot \frac{4}{3}\pi R^3}=\sqrt{\frac{8\pi G\rho }{3}}R$, which shows that at fixed density the escape velocity is directly proportional to the radius, a less obvious result than the bare square-root scaling first suggests. Doubling the radius therefore doubles the escape velocity: $v_e^{\prime }=2\times 11.2=22.4\text{km/s}$. The option $11.2\text{km/s}$ wrongly assumes the radius has no net effect once density is fixed. The option $15.8\text{km/s}$ applies a $\sqrt{2}$ scaling, which would hold only if the mass were unchanged. The option $44.8\text{km/s}$ scales by four, confusing this with an area dependence. As a plausibility check, a larger same-density planet packs far more mass that grows as the cube of the radius, so a clean doubling of the escape speed is consistent with the derived proportionality.