Physics · Gravitation

JEE Main 2025 — 3 April, Morning Shift — Question 65

Three identical spheres of mass $m$ , are placed at the vertices of an equilateral triangle of length $a$ .

When released, they interact only through gravitational force and collide after a time $T=4$ seconds.

If the sides of the triangle are increased to length $2 a$ and also the masses of the spheres are

made $2 m$ , then they will collide after $\qquad$ seconds.

Answer: 8

Numerical answer — enter this value.

As limiting case of elliptical path is straight line therefore proportionality

$T^{2} \propto \frac{a^{3}}{M}$ will holds from $\left(T^{2}=\frac{4 \pi^{2}}{G M} r^{3}\right)$

$\frac{4^{2}}{T^{2}}=\frac{a^{3}}{m}(2 a)^{3}=\frac{2}{8}=\frac{1}{4}$

$\frac{4}{T}=\frac{1}{2}$

$T=8$ seconds

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.