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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,303 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=303 N

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