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Question

A block of mass 5kg is suspended by a massless rope of length 2m from the ceiling. A force of 50N is applied in the horizontal direction at the midpoint P of the rope, as shown in the figure.
The angle made by the rope with the vertical in equilibrium is (Take g=10 ms2)
939613_a940fbb3fd424fd3a93959cf7dfd1285.png

A
300
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B
400
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C
600
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D
450
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Solution

The correct option is D 450
Let θ be the angle made by the rope with the vertical in equilibrium. Let T1 and T2 be the tension in the rope as shown in figure.
The free body diagram of 5kg block is as shown in fig (b).
In equilibrium, T2=5g=5×10=50N

The free body diagram of the point P is as shown in fig (c).
In equilibrium, T1sin(θ)=50N ..............1
T1cos(θ)=T2=50N ..............2

Dividing equation 1 by 2, we get
tan(θ)=5050=1
θ=tan1=450

942206_939613_ans_9e953ce4ad8b480e8913bec6e813a352.png

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