ACB is a smooth quater circular path of radius R. Four forces are acting at a particle placed at A. F1 is always horizontal, F2 is always vertical, F3 is always tangential to the path, F4 is always directed from particle 's position to point B. Magnitude of all forces are equal to F.

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A→R, B→P, C→R, D→Q

A→P, B→P, C→R, D→Q

A→Q, B→R, C→P, D→P

A→S, B→P, C→Q, D→R

answer is B.

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Detailed Solution

Work done by F1= F1 x displacement in the direction of force F1 = F1 RWork done by F2 = F2 x displacement in the direction of force F2 = F2 R Work done by F3=∫0π/2 F3⋅R⋅dθ=F3×πR2vmax2=16×13k−39k−(13)22k=84.5k=84.5845=110(m/s)2=1000(cm/s)2.⇒vmax=1010cm/s

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