
In a conical pendulum, the axial height of a right circular cone described by the string is $\sqrt 3$ times the radius of the circular path of its bob. The tension in the string is:
A) \[2{\text{ }}mg\]
B) $\sqrt {3mg} $
C) $\dfrac{2}{{\sqrt 3 }}mg$
D) $\dfrac{1}{{\sqrt 3 }}mg$
Answer
147k+ views
Hint: A conical pendulum is a string with a mass attached at the end. The mass moves in a horizontal circular path and forms a right circular cone.A right circular cone is a cone where it’s axis is the line meeting the vertex to the midpoint of the radius of the circle. The center point of the base of the circle is joined with the apex of the cone and it forms a right angle.

Formula used:
\[T\cos \theta = mg\]
Here, $T$ = tension created in the string
$r$ =radius of the circular path
$h$ = Axial height of the right circular cone
$l$ = height of the right circular cone that is slightly slant.
$mg$ = weight of the bob
$\theta $= $h$ and the string angle in between them measured in degrees
Complete step by step answer:
Let us consider the given tension diagram.

Here we divide the tension $T$ into two components that is a vertical and a horizontal.
The horizontal component is \[ = Tsin\theta \]
The vertical component \[ = Tcos\theta \].
So, according to the diagram,
\[T\cos \theta = mg\], where \[\cos \theta = \dfrac{h}{L}\]
Given, \[h = \sqrt 3 r\]
\[\therefore L = \sqrt {{r^2} + {{(\sqrt 3 r)}^2}} \] this can also be written as:
\[ \Rightarrow L = \sqrt {{r^2} + 3{r^2}} \]
Or,
\[ \Rightarrow L = \sqrt {4{r^2}} \]
Or,
\[ \Rightarrow L = 2r\]
So from,
\[T\cos \theta = mg\]
\[ \Rightarrow T = \dfrac{{mg}}{{\cos \theta }}\]
\[ \Rightarrow T = \dfrac{{mg}}{{\dfrac{h}{l}}}\]
\[ \Rightarrow T = \dfrac{{mgl}}{h}\]
\[ \Rightarrow T = mg \times \dfrac{{2r}}{{\sqrt 3 r}}\]
\[\therefore T = \dfrac{2}{{\sqrt 3 }}mg\]
Therefore the tension of the string \[T = \dfrac{2}{{\sqrt 3 }}mg\].
Hence the answer is the option (C).
Note: When two or more objects that are in contact, they exert force on each other and this particular force is called a Tension. Tension is the force that acts along a medium of the rope especially where the force is carried. Since the tension does not cause any displacement, the work done in the tension is zero and it acts in the opposite direction of the gravity force.

Formula used:
\[T\cos \theta = mg\]
Here, $T$ = tension created in the string
$r$ =radius of the circular path
$h$ = Axial height of the right circular cone
$l$ = height of the right circular cone that is slightly slant.
$mg$ = weight of the bob
$\theta $= $h$ and the string angle in between them measured in degrees
Complete step by step answer:
Let us consider the given tension diagram.

Here we divide the tension $T$ into two components that is a vertical and a horizontal.
The horizontal component is \[ = Tsin\theta \]
The vertical component \[ = Tcos\theta \].
So, according to the diagram,
\[T\cos \theta = mg\], where \[\cos \theta = \dfrac{h}{L}\]
Given, \[h = \sqrt 3 r\]
\[\therefore L = \sqrt {{r^2} + {{(\sqrt 3 r)}^2}} \] this can also be written as:
\[ \Rightarrow L = \sqrt {{r^2} + 3{r^2}} \]
Or,
\[ \Rightarrow L = \sqrt {4{r^2}} \]
Or,
\[ \Rightarrow L = 2r\]
So from,
\[T\cos \theta = mg\]
\[ \Rightarrow T = \dfrac{{mg}}{{\cos \theta }}\]
\[ \Rightarrow T = \dfrac{{mg}}{{\dfrac{h}{l}}}\]
\[ \Rightarrow T = \dfrac{{mgl}}{h}\]
\[ \Rightarrow T = mg \times \dfrac{{2r}}{{\sqrt 3 r}}\]
\[\therefore T = \dfrac{2}{{\sqrt 3 }}mg\]
Therefore the tension of the string \[T = \dfrac{2}{{\sqrt 3 }}mg\].
Hence the answer is the option (C).
Note: When two or more objects that are in contact, they exert force on each other and this particular force is called a Tension. Tension is the force that acts along a medium of the rope especially where the force is carried. Since the tension does not cause any displacement, the work done in the tension is zero and it acts in the opposite direction of the gravity force.
Recently Updated Pages
How to find Oxidation Number - Important Concepts for JEE

How Electromagnetic Waves are Formed - Important Concepts for JEE

Electrical Resistance - Important Concepts and Tips for JEE

Average Atomic Mass - Important Concepts and Tips for JEE

Chemical Equation - Important Concepts and Tips for JEE

Concept of CP and CV of Gas - Important Concepts and Tips for JEE

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

Degree of Dissociation and Its Formula With Solved Example for JEE

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry
