A weight \[\omega \;\] is suspended from the midpoint of a rope, whose ends are at the same level. In order to make the rope perfectly horizontal, the force applied to each of its ends must be
(A) Less than \[\omega \;\]
(B) Equal to \[\omega \;\]
(C) Equal to \[2\omega \;\]
(D) Infinitely large
Answer
252.3k+ views
Hint: We draw the free body diagram of the following system to solve this question. We find the sum of forces acting upwards and downwards and equate them since there is horizontal force. Both the horizontal components of the two sides of the string get cancelled. So, the weight acting downwards is equal to the sum of upward forces.
The angle between the string and the x-axis should be zero for the body to be horizontal. Substituting this condition in the final equation we get the answer.
Complete step by step solution: The free body diagram of the system is

The weight is at the centre of the string so the tension is distributed equally on both sides of the string. The tension force on both the sides of the string is resolved into
$T = Tcos\theta + T\sin \theta $
As shown in the diagram
The horizontal components are equal and opposite in direction so they get cancelled.
$Tcos\theta - Tcos\theta = 0$
Since the body is at equilibrium the sum of upwards forces is equal to downwards force (weight of the body)
Forces acting upwards are the vertical components of the tension force
$T\sin \theta + T\sin \theta = 2T\sin \theta $
Weight of the body is $\omega \ $
Upward force is equal to downwards force
$\omega \ = 2T\sin \theta $
For the rope to be perfectly horizontal $\theta $ should be $0^\circ $
Substituting $\theta = 0^\circ $
$ \omega \ = 2T\sin \theta $
$ \omega \ = 2T\sin 0^\circ $
$ T = \dfrac{{\omega \ }}{{2\sin 0^\circ }} = \dfrac{{\omega \ }}{0} $
Anything by zero is infinite
Hence the force applied on each end should be infinitely large.
Option (D) infinitely large is the correct answer.
Note: The angle $\theta $ is also equal to the angle between the tension force and the Horizontal which is the reason we can resolve the tension force with angle $\theta $
Students might go wrong by taking tensions in the two sides of the string with different values. The tension remains the same on both the sides because the weight is being suspended from the center.
The angle between the string and the x-axis should be zero for the body to be horizontal. Substituting this condition in the final equation we get the answer.
Complete step by step solution: The free body diagram of the system is

The weight is at the centre of the string so the tension is distributed equally on both sides of the string. The tension force on both the sides of the string is resolved into
$T = Tcos\theta + T\sin \theta $
As shown in the diagram
The horizontal components are equal and opposite in direction so they get cancelled.
$Tcos\theta - Tcos\theta = 0$
Since the body is at equilibrium the sum of upwards forces is equal to downwards force (weight of the body)
Forces acting upwards are the vertical components of the tension force
$T\sin \theta + T\sin \theta = 2T\sin \theta $
Weight of the body is $\omega \ $
Upward force is equal to downwards force
$\omega \ = 2T\sin \theta $
For the rope to be perfectly horizontal $\theta $ should be $0^\circ $
Substituting $\theta = 0^\circ $
$ \omega \ = 2T\sin \theta $
$ \omega \ = 2T\sin 0^\circ $
$ T = \dfrac{{\omega \ }}{{2\sin 0^\circ }} = \dfrac{{\omega \ }}{0} $
Anything by zero is infinite
Hence the force applied on each end should be infinitely large.
Option (D) infinitely large is the correct answer.
Note: The angle $\theta $ is also equal to the angle between the tension force and the Horizontal which is the reason we can resolve the tension force with angle $\theta $
Students might go wrong by taking tensions in the two sides of the string with different values. The tension remains the same on both the sides because the weight is being suspended from the center.
Recently Updated Pages
States of Matter Chapter For JEE Main Chemistry

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Circuit Switching vs Packet Switching: Key Differences Explained

Mass vs Weight: Key Differences Explained for Students

[Awaiting the three content sources: Ask AI Response, Competitor 1 Content, and Competitor 2 Content. Please provide those to continue with the analysis and optimization.]

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Hybridisation in Chemistry – Concept, Types & Applications

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Other Pages
JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2025-26

CBSE Notes Class 11 Physics Chapter 4 - Laws of Motion - 2025-26

