# A mass of $10 \mathrm{~kg}$ is suspended by a rope of length $4 \mathrm{~m}$, from the ceiling. A force $\mathrm{F}$ is applied horizontally at the midpoint of the rope such that the top half of the rope makes an angle of $45^{\circ}$ with the vertical. Then $F$ equals: (Take $g=10 \mathrm{~ms}^{-2}$ and the rope to be massless)

Question:

A mass of $10 \mathrm{~kg}$ is suspended by a rope of length $4 \mathrm{~m}$, from the ceiling. A force $\mathrm{F}$ is applied horizontally at the midpoint of the rope such that the top half of the rope makes an angle of $45^{\circ}$ with the vertical. Then $F$ equals:

(Take $g=10 \mathrm{~ms}^{-2}$ and the rope to be massless)

1. $100 \mathrm{~N}$

2. $90 \mathrm{~N}$

3. $70 \mathrm{~N}$

4. $75 \mathrm{~N}$

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

JEE Main Previous Year 7 Jan. 2020

Correct Option: 1

Solution:

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