A mass of 10kg is suspended vertically by a rope of length 5m 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 q = tan^-1 (x × 10^-1). The value of x is……………..

(Given, g = 10 m/s²)

F.B.D. of point ‘P’

$Tsin\theta =30$            - (1)

$Tcos\theta =100$          - (2)

$\left(1\right)÷\left(2\right)$

$tan\theta =\frac{3}{10}\Rightarrow \theta =ta{n}^{-1}\left(3\times 10^{-1}\right)$    

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