If the velocity `(V)` , acceleration `(A)` , and force `(F)` are taken as fundamental quantities instead of mass `(M)` , length `(L) , and time (T)` , the dimensions of young's modulus `(Y)` would be

Explanation

Let Young.s modulus is expressed as follows :
` Y propto V^(a)A^(b)F^(c )`
`[ML^(-1)T^(-2)]=[LT^(-1)]^(a)[LT^(-2)]^(b)[MLT^(-2)]^(c )`
`[ML^(-1)T^(-2)]= [M^(c )L^(a+b+c)T^(-a-2b-2c)]`
Comparing dimensions on both the sides we get the following :
`c= 1" ""............."(i)`
`a+b+c= -1`
`implies a+b+1= -1`
`implies a+b= -2" "".........."(ii)`
`-a -2b-2c= -2`
`implies -a-2b-2= -2`
`implies a+2b=0" ""..........."(iii)`
Subtracting equation (ii) from equation (iii) we get
`b=2" ""..........."(iv)`
Substituting equation (iv) in equation (ii) we get
`a = -4" ""............"(v)`
Using values of a,b and c we can write the dimensional formula for Young.s modulus as follws :
`[Y]= [V^(-4)A^(2)F^(1)]`.