JEE MainPhysicsUnits and Dimensions
The work done W by a variable force on a particle is represented by the integral W = ₀^ x₀ ( ) (- v^2 h ) dx , where v is the instantaneous velocity, h is Planck's constant, and x is the position. If and are dimensional constants, the dimensions of will be
Options
- A[ M ^2 L T ⁻¹ ]
- B[ M ^2 L ^2 T ⁻¹ ]
- C[ M ^2 L T ⁻² ]
- D[ M ^2 L ^3 T ⁻¹ ]
Correct answer
A. [ M ^2 L T ⁻¹ ]
Step-by-step solution
Given W = ₀^ x₀ ( ) (- v^2 h ) dx The argument of an exponential function must be dimensionless. Therefore, [ v^2 h ] = [ M ^0 L ^0 T ^0 ] [ ] = [ h v^2 ] The dimensional formula for Planck's constant h is [ M L ^2 T ⁻¹ ] and for velocity v is [ L T ⁻¹ ] . [ ] = [ M L ^2 T ⁻¹ ] [ L ^2 T ⁻² ] = [ M T ] Now, considering the integral equation, integrating a function with respect to dx multiplies its dimensional formula by [ L ] . [W] = [ ] [x] [ ] = [W] [ ] [x] The dimensional formula for work W is [ M L ^2 T ⁻² ] . [