MHT CET202020 Oct 2020Evening ShiftPhysicsUnits and DimensionsActual
In the expression A=B+ C D+E , the dimensions of physical quantities B and C are [ L ¹ M ⁰ ~T ⁻¹ ] and [ L ¹ M ⁰ ~T ⁰ ] respectively. The dimensions of quantities A , D and E are
Options
- A[ A ]= [ L ¹ M ⁰ ~T ⁻¹ ], [ D ]= [ T ¹ ],[ E ]= [ T ¹ ]
- B[ A ]= [ L ⁰ M ⁰ ~T ⁻¹ ], [ D ]= [ T ¹ ],[ E ]= [ L ¹ ~T ¹ ]
- C[ A ]= [ L ¹ M ¹ ~T ⁰ ], [ D ]= [ T ² ],[ E ]= [ L ¹ ~T ² ]
- D[ A ]= [ L ¹ M ⁰ ~T ⁻¹ ], [ D ]= [ M ¹ ~T ¹ ],[ E ]= [ M ¹ ~T ¹ ]
Correct answer
A. [ A ]= [ L ¹ M ⁰ ~T ⁻¹ ], [ D ]= [ T ¹ ],[ E ]= [ T ¹ ]
Step-by-step solution
Given, Dimension of B is [M⁰ L T⁻¹ ] & C is [M⁰ L T⁰ ] Now we know that, same dimensions can be added and the results we get are in same dimensions. So, dimension of A=B= [M⁰ L T⁻¹ ] Let, the dimension of D & E is [M^ X L^ Y T^ Z ] M⁰ L T⁻¹= M⁰ L T⁰ M^ X L^ Y T^ Z M^ X L^ Y T^ Z =M⁰ L⁰ T¹X=0, Y=0 & Z=1 So, the dimension of D & E is [T]