JEE Main 2025 — Gravitation Question with Solution

Dimensional Analysis:

$\mathrm{T}\propto {\mathrm{m}}^{\mathrm{x}}{\mathrm{G}}^{\mathrm{y}}{\mathrm{a}}^{\mathrm{z}}$

Using dimensional analysis for gravitational interactions, we have:

$\mathrm{T}\propto {\mathrm{M}}^{\mathrm{x}}{G}^{\mathrm{y}}{a}^{\mathrm{z}}$

Where $G$ has dimensions $\left[{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}\right]$.

Solving for Exponents:

$\mathrm{T}\propto {\mathrm{M}}^{\mathrm{x}-\mathrm{y}}{\mathrm{L}}^{3y+z}{\mathrm{T}}^{-2y}$

Equating dimensions, we solve:

$\mathrm{x}-\mathrm{y}=0\Rightarrow \mathrm{x}=\mathrm{y}$

$-2\mathrm{y}=1\Rightarrow \mathrm{y}=-\frac{1}{2},\mathrm{x}=-\frac{1}{2}$

$3\mathrm{y}+\mathrm{z}=0⟹\mathrm{z}=-3\mathrm{y}=\frac{3}{2}$

Time Proportionality:

$\mathrm{T}\propto {\mathrm{m}}^{-1/2}{\mathrm{G}}^{-1/2}{\mathrm{a}}^{3/2}$

Which simplifies to:

$\mathrm{T}\propto {\left(\frac{a^3}{m}\right)}^{1/2}$

Applying New Conditions:

When the side length becomes $2a$ and mass becomes $2m$:

$\mathrm{T}=4\times {\left(\frac{(2a{)}^3}{2m}\right)}^{1/2}$

Simplifying:

$\mathrm{T}=4\times {\left(\frac{8a^3}{2m}\right)}^{1/2}=4\times (4{)}^{1/2}=8\text{ seconds}$

Thus, the spheres will collide after 8 seconds under the new conditions.