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4-1.Newton's Laws of Motion
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A mass of $10 \,kg$ is suspended vertically by a rope of length $5 \,m$ from the roof. A force of $30 \,N$ is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is $\theta=\tan ^{-1}\left(x \times 10^{-1}\right)$. The value of $x$ is ................
$\text { (Given } g =10 \,m / s ^{2} \text { ) }$
A
$2$
B
$5$
C
$4$
D
$3$
(JEE MAIN-2022)
Solution
$ T \sin \theta=30 $
$ T \cos \theta=100 $
$ \Rightarrow \quad \tan \theta=0.3 $
Standard 11
Physics
