JEE Main201911 Jan 2019Evening ShiftPhysicsUnits and DimensionsActual
If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young's modulus will be :
Options
- AV ⁻² ~A ² ~F ⁻²
- BV ⁻² ~A ² ~F ²
- CV ⁻⁴ ~A ⁻² ~F
- DV ⁻⁴ ~A ² ~F
Correct answer
D. V ⁻⁴ ~A ² ~F
Step-by-step solution
Let [Y] = [V] ^ a [ ~F ]^ b [ A ]^ c [ ML ⁻¹ ~T ⁻² ]= [ LT ⁻¹ ]^ a [ MLT ⁻² ]^ b [ LT ⁻² ]^ c [ ML ⁻¹ ~T ⁻² ]= [ M ^ b ~L ^ a +b+ c T ^ - a -2 ~b -2 c ] Comparing power both side of similar terms we get, b=1, a+b+c=-1,-a-2 b-2 c=-2 solving above equations we get a=-4, b=1, c=2 so [ Y ]= [ V ⁻⁴ FA ² ]= [ V ⁻⁴ ~A ² ~F ]