Question:

A particle of mass π‘š is moving in the xy-plane such that its velocity at a point (x, y) is given as \(\overrightarrow{v}=a(y\^{x}+2x\^{y})\) where 𝛼 is a non-zero constant. What is the force F acting on the particle?

Updated On: Aug 19, 2024
  • \(\overrightarrow{𝐹} = 2ma^2(x\^{x}+y\^{y})\)
  • \(\overrightarrow{𝐹} = ma^2(y\^{x}+2x\^{y})\)
  • \(\overrightarrow{𝐹} = 2ma^2(y\^{x}+x\^{y})\)
  • \(\overrightarrow{𝐹} = ma^2(x\^{x}+2y\^{y})\)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

The correct option is(A): \(\overrightarrow{𝐹} = 2ma^2(x\^{x}+y\^{y})\)
Was this answer helpful?
0
0

Questions Asked in JEE Advanced exam

View More Questions

Concepts Used:

Speed and Velocity

The rate at which an object covers a certain distance is commonly known as speed.

The rate at which an object changes position in a certain direction is called velocity.

Difference Between Speed and Velocity:

Difference Between Speed and Velocity

Read More: Difference Between Speed and Velocity