In a Conical pendulum, a string of length $120cm$ is fixed at rigid support and carries a mass $150g$at the end. If the mass is revolved in a horizontal circle of radius $0.2m$ around a vertical axis, Calculate the tension in the string. $(g = 9.8\dfrac{m}{{{s^2}}})$
Answer
301.8k+ views
Hint: In order to solve this equation, we first have to draw a free body diagram of the body. After that, we have to balance the Net force that is acting on the body. All the forces should be in accordance with Newton's third law. We can easily find the solution after balancing all the forces in all the given directions.
Formula Used:
$Centripetal Force$ = $\dfrac{{m{V^2}}}{r}$
$m$ is the mass of body
$V$ is the velocity of body
$r$ is the radius of circle & Secondly, Net force on a body is equal to $0$ i. e. ${F_{net}} = 0$
Complete step by step answer:
Here, a conical pendulum, a string of length $120cm$ & mass of $150g$ & given we have to calculate the Tension $T$ in the String.
In order to understand the Question, Let’s draw a free body diagram of a pendulum.
Free Body Diagram:

Here the mass $m$ is moving in a circular motion of radius $0.2m$ with velocity $V.$ The Force acting on a mass $m$ will be
${F_{net}} = \dfrac{{m{V^2}}}{r}$
Here, the Tension $T$ will be resolved into two components $T\sin \theta $ (in vertical direction) and $ \uparrow \cos \theta $ (in horizontal direction). Now balancing all forces we get
In Vertical Direction:
$T\sin \theta = mg - (i)$
Now,
In Horizontal plane,
$T\cos \theta = $ Horizontal force on a mass $m$ (i.e. Centripetal force)
So,
$T\cos \theta = \dfrac{{m{V^2}}}{r} - (ii)$
So, from eq. we get
$T = \dfrac{{mg}}{{\sin \theta }}$
Now, In $\Delta ABC$, we get $\sin \theta = \dfrac{P}{H}$
So, $\sin \theta = \dfrac{{\sqrt {{l^2} - {R^2}} }}{l}$
Putting the value of $\sin \theta $in eq. $(1)$ we get
$T = \dfrac{{mgl}}{{\sqrt {{l^2} - {R^2}} }}$
Further,
$T = \left( {\dfrac{{150}}{{1000}}} \right)\dfrac{{(9.8) \times 120}}{{\sqrt {{{(120)}^2} - {{(20)}^2}} }}$ $[1kg = 1000g,1m = 100cm]$
$T = 1.52$ Newton.
Hence, the tension on the string will be $1.52$N.
Note: While solving this question, we have to be very careful with directions. Only forces acting in the same direction will balance each other. Also according to Newton's third law, action and reaction occur on different bodies. So we have to be careful while applying it to anybody. the units for every force should be the same.
Formula Used:
$Centripetal Force$ = $\dfrac{{m{V^2}}}{r}$
$m$ is the mass of body
$V$ is the velocity of body
$r$ is the radius of circle & Secondly, Net force on a body is equal to $0$ i. e. ${F_{net}} = 0$
Complete step by step answer:
Here, a conical pendulum, a string of length $120cm$ & mass of $150g$ & given we have to calculate the Tension $T$ in the String.
In order to understand the Question, Let’s draw a free body diagram of a pendulum.
Free Body Diagram:

Here the mass $m$ is moving in a circular motion of radius $0.2m$ with velocity $V.$ The Force acting on a mass $m$ will be
${F_{net}} = \dfrac{{m{V^2}}}{r}$
Here, the Tension $T$ will be resolved into two components $T\sin \theta $ (in vertical direction) and $ \uparrow \cos \theta $ (in horizontal direction). Now balancing all forces we get
In Vertical Direction:
$T\sin \theta = mg - (i)$
Now,
In Horizontal plane,
$T\cos \theta = $ Horizontal force on a mass $m$ (i.e. Centripetal force)
So,
$T\cos \theta = \dfrac{{m{V^2}}}{r} - (ii)$
So, from eq. we get
$T = \dfrac{{mg}}{{\sin \theta }}$
Now, In $\Delta ABC$, we get $\sin \theta = \dfrac{P}{H}$
So, $\sin \theta = \dfrac{{\sqrt {{l^2} - {R^2}} }}{l}$
Putting the value of $\sin \theta $in eq. $(1)$ we get
$T = \dfrac{{mgl}}{{\sqrt {{l^2} - {R^2}} }}$
Further,
$T = \left( {\dfrac{{150}}{{1000}}} \right)\dfrac{{(9.8) \times 120}}{{\sqrt {{{(120)}^2} - {{(20)}^2}} }}$ $[1kg = 1000g,1m = 100cm]$
$T = 1.52$ Newton.
Hence, the tension on the string will be $1.52$N.
Note: While solving this question, we have to be very careful with directions. Only forces acting in the same direction will balance each other. Also according to Newton's third law, action and reaction occur on different bodies. So we have to be careful while applying it to anybody. the units for every force should be the same.
Recently Updated Pages
Four persons A B C and D initially at the corners of class 11 physics JEE_Main

What is the difference between Conduction and conv class 11 physics JEE_Main

Moment of inertia of solid sphere about its diameter class 11 physics JEE_Main

If a piece of ice floating on the surface of water class 11 physics JEE_Main

At what temperature speed of sound in air will be doubled class 11 physics JEE_Main

A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main

Trending doubts
Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Understanding Uniform Acceleration in Physics

What Are Current and Potential Difference in Electricity?

Understanding Collisions: Types and Examples for Students

Understanding Average and RMS Value in Electrical Circuits

Other Pages
JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Understanding Inertial and Non-Inertial Frames of Reference

CBSE Notes Class 11 Physics Chapter 9 - Mechanical Properties of Fluids - 2026-27 Free PDF Download (Sign-in Required)

CBSE Notes Class 11 Physics Chapter 14 - Waves - 2026-27 Free PDF Download (Sign-in Required)

