JEE MAIN - Physics (2023 - 11th April Morning Shift - No. 5)
The radii of two planets 'A' and 'B' are 'R' and '4R' and their densities are $$\rho$$ and $$\rho / 3$$ respectively. The ratio of acceleration due to gravity at their surfaces $$\left(g_{A}: g_{B}\right)$$ will be:
3 : 16
4 : 3
1 : 16
3 : 4
Explanation
The acceleration due to gravity at the surface of a planet can be expressed as:
$$g \propto \rho R$$
Now let's find the ratio of acceleration due to gravity at the surfaces of planets A and B:
$$\frac{g_A}{g_B} = \frac{\rho_A R_A}{\rho_B R_B}$$
Given that the densities are $$\rho$$ and $$\frac{\rho}{3}$$ and the radii are $$R$$ and $$4R$$ for planets A and B, respectively, we have:
$$\frac{g_A}{g_B} = \frac{\rho \cdot R}{\left(\frac{\rho}{3}\right) \cdot (4R)} = \frac{3}{4}$$
So, the ratio of acceleration due to gravity at their surfaces is 3 : 4
$$g \propto \rho R$$
Now let's find the ratio of acceleration due to gravity at the surfaces of planets A and B:
$$\frac{g_A}{g_B} = \frac{\rho_A R_A}{\rho_B R_B}$$
Given that the densities are $$\rho$$ and $$\frac{\rho}{3}$$ and the radii are $$R$$ and $$4R$$ for planets A and B, respectively, we have:
$$\frac{g_A}{g_B} = \frac{\rho \cdot R}{\left(\frac{\rho}{3}\right) \cdot (4R)} = \frac{3}{4}$$
So, the ratio of acceleration due to gravity at their surfaces is 3 : 4
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