QJEEM25Ph90ab: JEE Main 2025 Physics - Class 11 JEEM25Ph90ab Solution: Acceleration due to gravity on the surface of earth is ' '. If…

$\because$ acceleration due to gravity on surface is given by

$$\mathrm{g}=\frac{\mathrm{GM}}{{\mathrm{R}}_{\mathrm{e}}^2}$$

Now since diameter is reduced to $1 / 3^{\text {rd }}$ , radius also reduces to $1 / 3^{\text {rd }}$ , keeping mass constant

New value of acceleration due to gravity on Earth's surface is

$$g^{\prime }=\frac{\mathrm{GM}}{{\left(\frac{{\mathrm{R}}_{\mathrm{e}}}{3}\right)}^2}=9\frac{\mathrm{GMe}}{{\mathrm{R}}_{\mathrm{e}}^2}=9\mathrm{g}$$