Answer: (C) \(\frac{16}{3}\) and \(\pm \frac{3}{2}\)
Given, \(y = m^2r^{-4}g^xl^{-3/2}\)
As we know that, relative errors can be written as
\(\frac{\Delta y}{y}=\frac{2\Delta m}{m}+\frac{4\Delta r}{r}+\frac{x\Delta g}{g}+\frac{3\Delta l}{2l}\)
On multiplying each term by 100, each term will represent percentage error.
So, putting the given values in the above equation we get,
\(18=2(1)+4(0.5)+xp+\frac{3}{2}(4)\)
\(\Rightarrow 18=10+xp\)
By trial and error, only option (C) is valid for this scenario. Thus,
x= \(\frac{16}{3}, p = \pm \frac{3}{2}\)
