Laws of Motion

JEE Main level, single correct, medium

41 of 210

Question 41 of 210

A mass of 10kg10kg is suspended by a rope of length 4m4m, from the ceiling. A force FF is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of 45∘45^{\circ} with the vertical. Then FF equals: (Take g=10ms−2g=10m{s}^{-2} and the rope to be massless)

Show answer and solution

Answer: (A) 100N100N

Let the tension in the string is TT.
Applying the condition of equilibrium in the vertical direction,
Tcos⁡(45∘)=100T\cos \left(45^{\circ}\right)=100
⇒T2=100\Rightarrow \dfrac{T}{\sqrt{2}}=100 ...(1)...\left(1\right)
In the horizontal direction,
Tsin⁡(45∘)=FT\sin \left(45^{\circ}\right)=F
⇒T2=F\Rightarrow \dfrac{T}{\sqrt{2}}=F
Put the value of TT from equation (1)\left(1\right),
⇒F=100N\Rightarrow F=100N
So, the horizontal force applied on the rope will be 100N100N.

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Laws of motion, friction and pulleys simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Laws of Motion (Newton's laws of motion), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.