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NVAMediumJEE 2025Newton's Law of Gravitation

JEE Physics 2025 Question with Solution

Three identical spheres of mass mm, are placed at the vertices of an equilateral triangle of length aa. When released, they interact only through gravitational force and collide after a time T=4 sT = 4 \, \text{s}. If the sides of the triangle are increased to length 2a2a and also the masses of the spheres are made 2m2m, then they will collide after _____ seconds.

Answer

Correct answer:8

Step-by-step solution

Standard Method

Given: Three identical spheres of mass mm are at the vertices of an equilateral triangle of side aa and collide after T=4 sT = 4 \, \text{s}.

Find: The new collision time when mass becomes 2m2m and side becomes 2a2a.

From dimensional analysis, let

T∝mxGyazT \propto m^x G^y a^z

Using dimensions,

[T]=[M]x[M−1L3T−2]y[L]z[T] = [M]^x [M^{-1}L^3T^{-2}]^y [L]^z

So,

[T]=Mx−yL3y+zT−2y[T] = M^{x-y} L^{3y+z} T^{-2y}

Comparing powers of dimensions:

x−y=0x-y=0 −2y=1-2y=1 3y+z=03y+z=0

Hence,

y=−12,x=−12,z=32y=-\frac{1}{2}, \quad x=-\frac{1}{2}, \quad z=\frac{3}{2}

Therefore,

T∝m−1/2G−1/2a3/2T \propto m^{-1/2} G^{-1/2} a^{3/2}

Since GG is constant,

T∝a3/2mT \propto \frac{a^{3/2}}{\sqrt{m}}

For the new situation,

T′T=(2a)3/2/2ma3/2/m=23/22=2\frac{T'}{T} = \frac{(2a)^{3/2}/\sqrt{2m}}{a^{3/2}/\sqrt{m}} = \frac{2^{3/2}}{\sqrt{2}} = 2

Thus,

T′=2T=2×4=8 sT' = 2T = 2 \times 4 = 8 \, \text{s}

Therefore, the spheres will collide after 8 s8 \, \text{s}.

Force and acceleration scaling

Given: Initially each sphere has mass mm and side length aa. In the modified case, mass becomes 2m2m and side becomes 2a2a.

Find: The new collision time.

The gravitational force between two spheres is

F=Gm2a2F = \frac{Gm^2}{a^2}

So the effective acceleration scale is proportional to force divided by mass:

aeff∝Gm2/a2m=Gma2a_{\text{eff}} \propto \frac{Gm^2/a^2}{m} = \frac{Gm}{a^2}

In the new configuration,

aeff′∝G(2m)(2a)2=2Gm4a2=12Gma2a'_{\text{eff}} \propto \frac{G(2m)}{(2a)^2} = \frac{2Gm}{4a^2} = \frac{1}{2}\frac{Gm}{a^2}

Thus,

aeff′=12aeffa'_{\text{eff}} = \frac{1}{2} a_{\text{eff}}

For motion starting from rest over a similar geometric path, time scales as

T∝lengthaccelerationT \propto \sqrt{\frac{\text{length}}{\text{acceleration}}}

Equivalently here, using the derived proportionality,

T∝a3/2mT \propto \frac{a^{3/2}}{\sqrt{m}}

Hence doubling the scale gives

T′=2T=8 sT' = 2T = 8 \, \text{s}

The correct numerical answer is 88.

Common mistakes

  • Assuming the force remains the same, so the collision time also remains the same. This is wrong because even if pairwise force magnitude is unchanged, the acceleration changes due to the increased mass and larger distance scale. Use the time scaling with aa and mm instead.

  • Using T∝1/aT \propto 1/a or T∝1/aT \propto 1/\sqrt{a}. This is incorrect because dimensional analysis gives T∝a3/2/mT \propto a^{3/2}/\sqrt{m}. Always match dimensions carefully before concluding the dependence.

  • Applying only acceleration scaling and forgetting that the travel distance also doubles. This gives an incomplete time estimate. Since both the size of the triangle and the acceleration scale change, both effects must be included.

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