
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
232.8k+ 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
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

