
A particle comes round a circle of radius 1 m once. The time taken by it is 10 sec. The average velocity of motion is
A. \[0.2\pi \,m/s\]
B. \[2\pi \,m/s\]
C. \[2\,m/s\]
D. Zero
Answer
216.3k+ views
Hint: The average velocity of motion is the ratio of net displacement and the total time of journey. The net displacement is the change in position of the body. When the body returns to the same position after the journey then there is no change in the position of the body.
Formula used:
\[\overrightarrow {{v_{av}}} = \dfrac{{\overrightarrow {\Delta x} }}{{\Delta t}}\]
Here \[\overrightarrow {\Delta {v_{av}}} \] is the average velocity, \[\overrightarrow {\Delta x} \] is the net displacement and \[\Delta t\] is the total time of journey.
Complete step by step solution:
It is given that the particle comes around a circle once in 10 sec. When the particle comes to the same position after completing one revolution then the final position of the particle coincides with the initial position. The radius of the circle around which the particle is moving is given as 1 metre.
As we know that linear displacement is the change in position, so the net linear displacement of the particle is zero because the initial position is the same as the final position.
\[\overrightarrow {\Delta x} = 0m\]
The time taken to complete one revolution around a circle is 10 sec.
\[\Delta = 10\sec \]
So, using the average velocity formula, the average velocity of the particle during the journey will be,
\[\overrightarrow {{v_{av}}} = \dfrac{{0m}}{{10\sec }}\]
\[\therefore \overrightarrow {{v_{av}}} = 0\,m/s\]
Hence, the average velocity of the particle during this 10 sec of the journey is zero.
Therefore, the correct option is D.
Note: During the complete revolution the net distance is non-zero and it is equal to the circumference of the circular path. So, when it is asked to find the average speed then we calculate it as the ratio of the net distance travelled during the journey to the time taken to complete the journey. As the net distance covered is non-zero, so the average speed is not zero.
Formula used:
\[\overrightarrow {{v_{av}}} = \dfrac{{\overrightarrow {\Delta x} }}{{\Delta t}}\]
Here \[\overrightarrow {\Delta {v_{av}}} \] is the average velocity, \[\overrightarrow {\Delta x} \] is the net displacement and \[\Delta t\] is the total time of journey.
Complete step by step solution:
It is given that the particle comes around a circle once in 10 sec. When the particle comes to the same position after completing one revolution then the final position of the particle coincides with the initial position. The radius of the circle around which the particle is moving is given as 1 metre.
As we know that linear displacement is the change in position, so the net linear displacement of the particle is zero because the initial position is the same as the final position.
\[\overrightarrow {\Delta x} = 0m\]
The time taken to complete one revolution around a circle is 10 sec.
\[\Delta = 10\sec \]
So, using the average velocity formula, the average velocity of the particle during the journey will be,
\[\overrightarrow {{v_{av}}} = \dfrac{{0m}}{{10\sec }}\]
\[\therefore \overrightarrow {{v_{av}}} = 0\,m/s\]
Hence, the average velocity of the particle during this 10 sec of the journey is zero.
Therefore, the correct option is D.
Note: During the complete revolution the net distance is non-zero and it is equal to the circumference of the circular path. So, when it is asked to find the average speed then we calculate it as the ratio of the net distance travelled during the journey to the time taken to complete the journey. As the net distance covered is non-zero, so the average speed is not zero.
Recently Updated Pages
Wheatstone Bridge Explained: Working, Formula & Uses

Young’s Double Slit Experiment Derivation Explained

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Electricity and Magnetism Explained: Key Concepts & Applications

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Electromagnetic Waves and Their Importance

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Average and RMS Value in Electrical Circuits

