In the adjoining figure all surfaces are frictionless. The force F must be applied to M₁to keep
M₃free from rising or falling

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$\frac{M_{1}}{M_{3}} (M_{1} + M_{2} + M_{3}) g$

$\frac{M_{3}}{M_{2}} (M_{1} + M_{2} + M_{3}) g$

$\frac{M_{2}}{M_{3}} (M_{1} + M_{2} + M_{3}) g$

$\frac{M_{3}}{M_{1}} (M_{1} + M_{2} + M_{3}) g$

answer is B.

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Detailed Solution

$T o k e e p M_{3} f r e e f r o m f a l l i n g T = m_{2} a T = m_{3} g ∴ M_{2} a = M_{3} g a = \frac{M_{2}}{M_{3}} g B u t F = (M_{1} + M_{2} + M_{3}) a \Rightarrow F = \frac{M_{2}}{M_{3}} (M_{1} + M_{2} + M_{3}) g$

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