MHT CET · Physics · Gravitation

Three particles each of mass ' \(\mathrm{m}_{1}\) ' are placed at the corners of an equilateral triangle of side \(\frac{\mathrm{L}}{3}\) '. A particle of mass 'm \(_{2}\) ' is placed at the mid point of any one side of triangle. Due to the system of particles the force acting on ' \(\mathrm{m}_{2}\) ' is (G = Universal constant of gravitation)

A \(\frac{12 \mathrm{G} \mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~L}^{2}}\)
B \(\frac{2 \mathrm{G} \mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~L}^{2}}\)
C \(\frac{4 \mathrm{G} \mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~L}^{2}}\)
D \(\frac{8 \mathrm{G} \mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~L}^{2}}\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{12 \mathrm{G} \mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~L}^{2}}\)

Step-by-step Solution

Detailed explanation

Forces on mass \(\mathrm{m}_{2}\) due to masses at \(\mathrm{B}\) and \(\mathrm{C}\) will be equal and opposite and cancel each other.
\(\mathrm{h}=\frac{\mathrm{L}}{3} \cos 30^{\circ}=\frac{\mathrm{L}}{3} \frac{\sqrt{3}}{2}=\frac{\mathrm{L}}{2 \sqrt{3}}\)
Force on \(m_{2}\) due to mass \(m_{1}\) at \(A\) is given by
\(F=G \frac{m_{1} m_{2}}{\left(\frac{L}{2 \sqrt{3}}\right)^{2}}=\frac{12 G m_{1} m_{2}}{L^{2}}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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