JEE MainPhysicsUnits and Dimensions
The work done W by a force on a particle is given by the integral: W = ₀^x ( t + y^2 + ) dy where t is time, and x and y are position coordinates. If , , and are dimensional constants, the dimensional formula for the ratio is:
Options
- A[ M ^1 L ^2 T ⁻³]
- B[ M ^1 L ^1 T ⁻²]
- C[ M ^1 L ^1 T ⁻³]
- D[ M ⁻¹ L ⁻¹ T ^3]
Correct answer
C. [ M ^1 L ^1 T ⁻³]
Step-by-step solution
The dimension of work done W is [ M L ^2 T ⁻²] . The integral is taken with respect to position y , so the differential dy has the dimension of length [ L ] . Therefore, the dimension of the integrand must be [W] [dy] = [ M L ^2 T ⁻²] [ L ] = [ M L T ⁻²] . By the principle of dimensional homogeneity, each term in the integrand must have the dimension [ M L T ⁻²] . For the second term, we directly get [ ] = [ M L T ⁻²] . For the first term t + y^2 , the denominator must have the same dimension as y^2 . Thus, [ ] = [