If speed V, area A and force F are chosen and fundamental units, then the dimension of Young,s modulus will be :

Options

A. `V^(-4)A^(2)F`
B. `V^(-2)A^(2)F^(-2)`
C. `V^(-2)A^(2)F^(2)`
D. `V^(-4)A^(-2)F`

Answer

A

Explanation

`[y]=[V_(X)A^(Y)F^(Z)]`
or `[mL^(-1)T^(-2)]=(LT^(-1))^(X)(LT^(-2))^(Y)(mLT^(-2))^(z)]or(mL^(-1)T^(-2)]=(m^(Z)L^(X+Y+Z)T^(-X-2Y-2Z))`
Comparing powers of m, L and T on both sides.
`z=1`
`X+Y+Z=-1`
`-x-2Y-2z=-2rArrx=-4`
`y=2`
`z=1therefore` Required dimensional formula is `V^(-4)A^(2)F`.