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Normal Force

In the arrang...

Question

In the arrangements shown in the figure, the pulleys are smooth and massless and all the strings are light. Let $F_{1}$ be the force exerted on the pulley in case (i) and $F_{2}$ be the force in case (ii). Then

F_{1} > F_{2}

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F_{1} < F_{2}

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F_{1} = F_{2}

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F_{1} = 2 F_{2}

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Solution

The correct option is C $F_{1} = F_{2}$

Here, the net force on the system is same in both cases, so it will have same acceleration will result similar tension in both string.
So, $F_{1} = F_{2} = 2 T$

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