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 $2 a$ and also the masses of the spheres are made 2 m , then they will collide after__________seconds.

<p><strong>Dimensional Analysis</strong>:</p> <p>$ \mathrm{T} \propto \mathrm{m}^{\mathrm{x}} \mathrm{G}^{\mathrm{y}} \mathrm{a}^{\mathrm{z}} $</p> <p>Using dimensional analysis for gravitational interactions, we have:</p> <p>$ \mathrm{T} \propto \mathrm{M}^{\mathrm{x}} G^{\mathrm{y}} a^{\mathrm{z}} $</p> <p>Where $ G $ has dimensions $\left[\mathrm{M}^{-1} \mathrm{L}^3 \mathrm{T}^{-2}\right]$.</p></p> <p><p><strong>Solving for Exponents</strong>:</p> <p>$ \mathrm{T} \propto \mathrm{M}^{\mathrm{x-y}} \mathrm{L}^{3y+z} \mathrm{T}^{-2y} $</p> <p>Equating dimensions, we solve:</p> <p>$ \mathrm{x}-\mathrm{y}=0 \Rightarrow \mathrm{x}=\mathrm{y} $</p> <p>$ -2\mathrm{y}=1 \Rightarrow \mathrm{y}=-\frac{1}{2}, \mathrm{x}=-\frac{1}{2} $</p> <p>$ 3\mathrm{y}+\mathrm{z}=0 \implies \mathrm{z}=-3\mathrm{y}=\frac{3}{2} $</p></p> <p><p><strong>Time Proportionality</strong>:</p> <p>$ \mathrm{T} \propto \mathrm{~m}^{-1/2} \mathrm{G}^{-1/2} \mathrm{a}^{3/2} $</p> <p>Which simplifies to:</p> <p>$ \mathrm{T} \propto \left(\frac{a^3}{m}\right)^{1/2} $</p></p> <p><p><strong>Applying New Conditions</strong>:</p> <p>When the side length becomes $2a$ and mass becomes $2m$:</p> <p>$ \mathrm{T} = 4 \times \left(\frac{(2a)^3}{2m}\right)^{1/2} $</p> <p>Simplifying:</p> <p>$ \mathrm{T} = 4 \times \left(\frac{8a^3}{2m}\right)^{1/2} = 4 \times (4)^{1/2} = 8 \text{ seconds} $</p></p> <p>Thus, the spheres will collide after 8 seconds under the new conditions.</p>