Three identical spheres of mass m are placed | vedclass

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(A) The time taken for the spheres to collide is proportional to the orbital period of a body in a gravitational field,which follows Kepler's Third La

Vedclass · Accepted Answer

$8$

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(A) The time taken for the spheres to collide is proportional to the orbital period of a body in a gravitational field,which follows Kepler's Third Law: $T^2 \propto \frac{a^3}{M}$.Here,$a$ is the side length of the triangle and $M$ is the mass of the spheres.Given the initial condition: $T_1 = 4 	ext{ s}$,$a_1 = a$,and $M_1 = m$.For the second condition: $a_2 = 2a$ and $M_2 = 2m$.Using the proportionality $T \propto \sqrt{\frac{a^3}{M}}$,we have:$\frac{T_2}{T_1} = \sqrt{\frac{a_2^3}{M_2} \cdot \frac{M_1}{a_1^3}}$$\frac{T_2}{4} = \sqrt{\frac{(2a)^3}{2m} \cdot \frac{m}{a^3}}$$\frac{T_2}{4} = \sqrt{\frac{8a^3}{2m} \cdot \frac{m}{a^3}}$$\frac{T_2}{4} = \sqrt{4} = 2$$T_2 = 4 	imes 2 = 8 	ext{ s}$.

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