JEE MainPhysicsUnits and Dimensions
The force F acting on a particle is given by the equation F = v + t^2 , where v is the velocity and t is the time. If Energy ( E ), Momentum ( p ) and Force ( F ) are chosen as the fundamental quantities, then the dimensional formula of the ratio is
Options
- A[E^ -1/2 p^ -3/2 F^4 ]
- B[E^ 1/2 p^ -5/2 F^2 ]
- C[E⁻¹ p^3 F⁻² ]
- D[E^ -1/2 p^ 5/2 F⁻² ]
Correct answer
D. [E^ -1/2 p^ 5/2 F⁻² ]
Step-by-step solution
By the principle of dimensional homogeneity, each term in the equation must have the dimensions of force. [F] = [ M ^1 L ^1 T ⁻²] [ v ] = [F] [ ] = [ M ^1 L ^1 T ⁻²] [ L ^1 T ⁻¹]^ 1/2 = [ M ^1 L ^ 1/2 T ^ -3/2 ] [ t^2] = [F] [ ] = [ M ^1 L ^1 T ⁻²] [ T ^2] = [ M ^1 L ^1 T ⁻⁴] The dimensional formula for the ratio is: [ ] = [ M ^1 L ^ 1/2 T ^ -3/2 ] [ M ^1 L ^1 T ⁻⁴] = [ M ^0 L ^ -1/2 T ^ 5/2 ] Let [ ] = E^x p^y F^z . Writing their standard dimensional formulas: [ M ^0 L ^ -1/2 T ^ 5/2 ] = [ M ^1 L ^2 T ⁻²]^x [ M ^1