# A mass of $10 \mathrm{~kg}$ is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of $45^{\circ}$ at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$

Question:

A mass of $10 \mathrm{~kg}$ is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of $45^{\circ}$ at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$

1. $200 \mathrm{~N}$

2. $140 \mathrm{~N}$

3. $70 \mathrm{~N}$

4. $100 \mathrm{~N}$

JEE Main Previous Year Single Correct Question of JEE Main from Physics Laws of Motion chapter.

JEE Main Previous Year 9 Jan. 2019

Correct Option: 4

Solution:

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