JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2025
- Medium
- Numerical
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.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 8
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{T} \propto \mathrm{m}^{\mathrm{x}} \mathrm{G}^{\mathrm{y}} \mathrm{a}^{\mathrm{z}} \\ & \mathrm{T} \propto \mathrm{M}^{\mathrm{x}}\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^y[\mathrm{~L}]^{\mathrm{z}} \\ & \mathrm{T} \propto \mathrm{M}^{\mathrm{x}-\mathrm{y}} \mathrm{L}^{3 \mathrm{y}+\mathrm{z}} \mathrm{T}^{-2 \mathrm{y}} \\ & \mathrm{x}-\mathrm{y}=0 \Rightarrow \mathrm{x}=\mathrm{y} \\ & -2 \mathrm{y}=1 \Rightarrow \mathrm{y}=-\frac{1}{2}, \mathrm{x}=-\frac{1}{2} \\ & \Rightarrow 3 \mathrm{y}+\mathrm{z}=0 \\ & \mathrm{z}=-3 \mathrm{y}=\frac{3}{2}\end{aligned}\)
Hence
\(\begin{aligned}
& \mathrm{T} \propto \mathrm{~m}^{-1 / 2} \mathrm{G}^{-1 / 2} \mathrm{a}^{3 / 2} \\
& \mathrm{~T} \propto\left(\frac{\mathrm{a}^3}{\mathrm{~m}}\right)^{1 / 2} \\
& \mathrm{~T}=4 \times\left(\frac{2^3}{2}\right)^{1 / 2}=8 \mathrm{~s}
\end{aligned}\)
Hence
\(\begin{aligned}
& \mathrm{T} \propto \mathrm{~m}^{-1 / 2} \mathrm{G}^{-1 / 2} \mathrm{a}^{3 / 2} \\
& \mathrm{~T} \propto\left(\frac{\mathrm{a}^3}{\mathrm{~m}}\right)^{1 / 2} \\
& \mathrm{~T}=4 \times\left(\frac{2^3}{2}\right)^{1 / 2}=8 \mathrm{~s}
\end{aligned}\)
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