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Question

# Three light strings are connected at the point P. A weight W is suspended from one of the strings. End A of string AP and end B of string PB are fixed as shown. In equilibrium, PB is horizontal and PA makes an angle of 60∘ with the horizontal. If the tension in PB is 30 N, then the tension in PA and weight W are respectively given by

A
60 N,30 N
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B
603 N,303 N
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C
60 N,303 N
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D
603 N,303 N
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Solution

## The correct option is C 60 N,30√3 N Here W=mg For equilibrium of block of weight W, W=T2 For horizontal equilibrium at point P, we can say T1cos60∘=30 N ⟹T1=60 N For vertical equilibrium at point P, we can say T1sin60∘=T2=W ⟹W=60×√32=30√3 N

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