Sol:

$$[v] = [LT^{-1}]$$ $$\eta = \frac{F}{A\frac{dv}{dx}}$$ $$\Rightarrow [\eta] = \frac{[MLT^{-2}]}{[L^{2}][T^{-1}]}$$ $$\Rightarrow [M^{1}L^{-1}T^{1}]$$ $$[\rho ] = [ML^{-3}]$$ $$\Rightarrow [LT^{-1}] = [M^{1}L^{-1}T^{-1}]^{x} [ML^{-3}]^{y} [L]^{z}$$

Now by equating the exponents of M, L and T on both LHS and RHS, we get

$$\Rightarrow M^{0} = M^{(x+y)}$$ $$\Rightarrow y = – x$$

For T

= -1 = -x

x= 1

y = -x = -1

For L

1 = -x – 3y + z

1 = -1 + 3 + z

1 = 2 + z

$$\Rightarrow z = -1$$

Therefore, the values of x, y and z are

x = 1, y = -1 and z = -1

Explore more such questions and answers at BYJU’S.