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Q. 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 θ=tan1(x×101). The value of x is _____.
(Given g=10m/s2)

JEE MainJEE Main 2022Laws of Motion

Solution:

image
Tsinθ=30
Tcosθ=100
tanθ=0.3