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 $\mathrm{T}=4$ seconds. If the sides of the triangle are increased to length 2a and also the masses of the spheres are made 2 m , then they will collide after ______ seconds.

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Gravitation

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 2a and also the masses of the spheres are made 2 m , then they will collide after ______ seconds.

Answer: 8

Solution

The collision time is fixed by mutual gravitation, so it can depend only on the mass

m

, the side

a

and the gravitational constant

G

. Balancing dimensions in

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

gives

x = - \frac{1}{2}

y = - \frac{1}{2}

z = \frac{3}{2}

, hence

T \propto \frac{a ^{3}}{G m} .

Doubling both the side and the mass,

\frac{T ^{'}}{T} = \frac{( 2 a ) ^{3}}{2 m} \cdot \frac{m}{a ^{3}} = \frac{8}{2} = 2.

Therefore

T^{'} = 2 \times 4 = 8

seconds.

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Integer answer

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JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Gravitation

Solution

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