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Question

A block of mass $$5 kg$$ is suspended by a massless rope of length $$2m$$ from the ceiling. A force of $$50 N$$ 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 $$g=10 \ ms^{-2}$$)
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 $$45^0$$
Let $$\theta$$ be the angle made by the rope with the vertical in equilibrium. Let $$T_1$$ and $$T_2$$ be the tension in the rope as shown in figure.
The free body diagram of $$5 kg$$ block is as shown in fig (b).
In equilibrium,  $$\Rightarrow T_2=5g=5\times 10 = 50N$$

The free body diagram of the point $$P$$ is as shown in fig (c).
In equilibrium, $$\Rightarrow T_1sin(\theta)=50N$$         ..............1
$$\Rightarrow T_1cos(\theta) = T_2 =50N$$  ..............2

Dividing equation 1 by 2, we get
$$tan(\theta) =\dfrac{50}{50}=1$$
$$\Rightarrow \theta =tan^{-1}=45^0$$

942206_939613_ans_9e953ce4ad8b480e8913bec6e813a352.png

Physics

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