
A uniform meter stick of mass 200 g is suspended from the ceiling through two vertical strings of equal lengths fixed to the ends. A small object of a mass of 20 g is placed on the stick at a distance of 70 cm from the left end. Find the tension in two strings.
Answer
233.1k+ views
Hint: Tension is defined as the pulling force transmitted axially using a string or a chain or a cable. Tension force is the force transmitted through a string. Torque is defined as a measure of force that causes an object to rotate about an axis. The direction depends on the direction of the force on the axis.

Complete step by step solution:
Mass of the stick, M = 200 gm = 0.2 kg
Mass of the object, m = 20 gm = 0.02 kg
The weight of the stick is $0.2 \times 10 = 2N$
The weight of the object is $0.02 \times 10 = 2N$
Let the tension in two strings be $T_1$ and $T_2$,
The system is in equilibrium and thus
\[{T_1} + {T_2} = 2 + 0.2 = 2.2N\_\_\_\_\_\_\_\_\_\_\_\_(1)\]
As the stick does not rotate, the net torque acting on it is zero.
But the torque about the midpoint is given by,
\[\left( {0.2 \times 0.2} \right) + \left( {{T_1} \times 0.5} \right) = \left( {{T_2} \times 0.5} \right)\]
\[ \Rightarrow 0.04 + 0.5{T_1} = 0.5{T_2}\]
\[ \Rightarrow \left( {{T_2} - {T_1}} \right) = 0.08\_\_\_\_\_\_\_\_\_\_(2)\]
Thus from equations 1 and 2, we get
\[2{T_2} = 2.28N\]
\[ \Rightarrow {T_2} = 1.14N\]
\[ \Rightarrow {T_1} = \left( {1.14 - 0.08} \right)N\]
\[ \Rightarrow {T_1} = 1.06N\]
Thus the tension in two strings, $T_1$ and $T_2$ are $1.06 N$ and $1.14 N$.
Note: 1. An object is said to be in a state of equilibrium when all the forces that are acting upon the object are balanced and thus the net force will be zero.
2. For the object to be in equilibrium, the object must have no acceleration which means that the net force and the net torque acting on the object must be zero. For the object to be in equilibrium, the sum of the torques about any point must also be equal to zero.
3. Tension can also be defined as the action-reaction pair of forces acting on it.

Complete step by step solution:
Mass of the stick, M = 200 gm = 0.2 kg
Mass of the object, m = 20 gm = 0.02 kg
The weight of the stick is $0.2 \times 10 = 2N$
The weight of the object is $0.02 \times 10 = 2N$
Let the tension in two strings be $T_1$ and $T_2$,
The system is in equilibrium and thus
\[{T_1} + {T_2} = 2 + 0.2 = 2.2N\_\_\_\_\_\_\_\_\_\_\_\_(1)\]
As the stick does not rotate, the net torque acting on it is zero.
But the torque about the midpoint is given by,
\[\left( {0.2 \times 0.2} \right) + \left( {{T_1} \times 0.5} \right) = \left( {{T_2} \times 0.5} \right)\]
\[ \Rightarrow 0.04 + 0.5{T_1} = 0.5{T_2}\]
\[ \Rightarrow \left( {{T_2} - {T_1}} \right) = 0.08\_\_\_\_\_\_\_\_\_\_(2)\]
Thus from equations 1 and 2, we get
\[2{T_2} = 2.28N\]
\[ \Rightarrow {T_2} = 1.14N\]
\[ \Rightarrow {T_1} = \left( {1.14 - 0.08} \right)N\]
\[ \Rightarrow {T_1} = 1.06N\]
Thus the tension in two strings, $T_1$ and $T_2$ are $1.06 N$ and $1.14 N$.
Note: 1. An object is said to be in a state of equilibrium when all the forces that are acting upon the object are balanced and thus the net force will be zero.
2. For the object to be in equilibrium, the object must have no acceleration which means that the net force and the net torque acting on the object must be zero. For the object to be in equilibrium, the sum of the torques about any point must also be equal to zero.
3. Tension can also be defined as the action-reaction pair of forces acting on it.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

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

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

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

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

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

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

