Question 94 of 210

A mass of $10kg$ is suspended vertically by a rope of length $5m$ from the roof. A force of $30N$ is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is $\alpha ={\tan }^{-1}\left(x\times {10}^{-1}\right)$. The value of $x$ is _____ .
(Given, $g=10m{s}^{-2}$)

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Answer: 3

$\Rightarrow T\sin \theta =100$ and $T\cos \theta =30$
Therefore, $\tan \theta =\dfrac{10}{3}$
Then, $\tan \alpha =\dfrac{3}{10}\Rightarrow \alpha ={\tan }^{-1}\left(3\times {10}^{-1}\right)$
Hence, value of $x=3$.