Physics Q46 – JEE-MAIN 2025

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 _______ seconds.

Enter Calculated Numerical Value:

Clue 1 of 3: Gravitational Interaction

The collision time in a system of mutually interacting particles under gravity depends on the initial configuration and the masses involved.

🧠 Full Solution Path

Step 1: Identify the scaling relationship for collision time✦ Active

For a system of particles interacting gravitationally, the characteristic time $T$ for a collision or a specific evolution scales with the initial characteristic length $a$ and mass $m$ . From dimensional analysis or by considering the equation of motion $m \frac{d^{2} x}{d t^{2}} \propto \frac{G m^{2}}{x^{2}}$ , which simplifies to $\frac{d^{2} x}{d t^{2}} \propto \frac{G m}{x^{2}}$ . If $x \sim a$ and $\frac{d^{2} x}{d t^{2}} \sim \frac{a}{T^{2}}$ , then $\frac{a}{T^{2}} \propto \frac{G m}{a^{2}}$ . This implies $T^{2} \propto \frac{a^{3}}{G m}$ , so $T \propto \sqrt{\frac{a^{3}}{m}}$ .

💡 Teacher's Secret Hint

This scaling relation is crucial for problems involving gravitational collapse or orbital periods in similar configurations.

Step 2: Apply the scaling relation to the given initial and final conditions○ Expand

Initial conditions: $T_{1} = 4$ s, $a_{1} = a$ , $m_{1} = m$ . Final conditions: $a_{2} = 2 a$ , $m_{2} = 2 m$ . Using the scaling relation $\frac{T_{2}}{T_{1}} = \sqrt{\frac{a_{2}^{3} / m_{2}}{a_{1}^{3} / m_{1}}}$ :

\frac{T_{2}}{T_{1}} = \sqrt{\frac{(2 a)^{3} / (2 m)}{a^{3} / m}}

\frac{T_{2}}{T_{1}} = \sqrt{\frac{8 a^{3} / (2 m)}{a^{3} / m}}

\frac{T_{2}}{T_{1}} = \sqrt{\frac{4 a^{3} / m}{a^{3} / m}}

\frac{T_{2}}{T_{1}} = \sqrt{4}

\frac{T_{2}}{T_{1}} = 2

💡 Teacher's Secret Hint

Be careful with the exponents and cancellations when substituting the new values.

Step 3: Calculate the new collision time○ Expand

Since $\frac{T_{2}}{T_{1}} = 2$ , we have $T_{2} = 2 T_{1}$ . Given $T_{1} = 4$ seconds, the new collision time is:

T_{2} = 2 \times 4 = 8 seconds

💡 Teacher's Secret Hint

Double-check the final arithmetic.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 46 – JEE-MAIN 2025

Clue 1 of 3: Gravitational Interaction

🧠 Full Solution Path