A mass of 10kg is suspended vertically by a rope of length 5m from...

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, $g = 10m{s}^{- 2}$ )

Correct answer

Step-by-step solution

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

$⇒ Tsin\theta = 100$ and $Tcos\theta = 30$

Therefore, $tan\theta = \dfrac{10}{3}$

Then, $tan\alpha = \dfrac{3}{10} ⇒ \alpha = \tan^{- 1}\left(3 \times {10}^{- 1}\right)$

Hence, value of $x = 3$ .

Practice Laws of Motion on Quantrex Academy →

More from Laws of Motion

A block takes t time to slide down a plane inclined at 45° to the horizontal. If the… JEE Main 2026 Two blocks ( P and Q ) with respectively masses 2 kg and 1.5 kg are joined by a massless… JEE Main 2026 A mass of 1 kg is kept on an inclined plane with 30° inclination with respect to… JEE Main 2026 Three masses m_1 = 4 kg, m_2 = 4 kg and m_3 = 6 kg are suspended from a fixed smooth… JEE Main 2026 A wedge Y with mass of 10 kg and all frictionless surfaces and the inclined surface… JEE Main 2026 At t=0 , a body of mass 100 g starts moving under the influence of a force (5 i +10 j )… JEE Main 2026 The time taken by a block of mass m to slide down from the highest point to the lowest… JEE Main 2026 A 0.5 kg mass is in contact against the inner wall of a cylindrical drum of radius 4 m… JEE Main 2026 Full Laws of Motion list All JEE Main PYQs