TS EAMCET · Physics · Electrostatics

Four identical masses of $m$ are kept at corners of a square. If the gravitational force exerted on one of masses by the other masses is $(\frac{2 \sqrt{2} + 1}{32}) \frac{G m^{2}}{L^{2}},$ then the length of the side of the square is

A $L$
B $2 L$
C $4 L$
D $L / 2$

Verified Solution

Answer & Solution

Correct Answer

(C) $4 L$

Step-by-step Solution

Detailed explanation

Let suppose that the length of the square is,a, then the net force acted on any one mass will be, the vector sum of the forces,F→24+F→34+F→14=F→net, it means,F→net=F→24+F→34+F→14⇒F→net=F→342+F→142+2F→34F→14cos90°+F→14⇒F→net=2F→34+F→14⇒F→net=2Gm2a2+Gm22a2=Gm2a22+12 Which should…

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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Four identical masses of m are kept at corners of a square. If the gravitational force exerted on one of masses by the other masses is 22+132Gm2 L2, then the length of the side of the square is

Step-by-step Solution

Four identical masses of $m$ are kept at corners of a square. If the gravitational force exerted on one of masses by the other masses is $(\frac{2 \sqrt{2} + 1}{32}) \frac{G m^{2}}{L^{2}},$ then the length of the side of the square is