A physical quantity $$y$$ is represented by the formula $$y = m^2 r^{-4} g^x l^{-\frac{3}{2}}$$. If the percentage errors found in $$y$$, $$m$$, $$r$$, $$l$$ and $$g$$ are 18, 1, 0.5, 4 and $$p$$ respectively, then find the value of $$x$$ and $$p$$.

Let’s do it cleanly and correctly.

Given:

$$y = m^2 r^{-4} g^x l^{-\frac{3}{2}}$$

For percentage error:

$$\frac{\Delta y}{y}=2\frac{\Delta m}{m}+4\frac{\Delta r}{r}+x\frac{\Delta g}{g}+\frac{3}{2}\frac{\Delta l}{l}$$

(Substitute magnitudes)

Given errors:

$$\Delta y=18,\quad\Delta m=1,\quad\Delta r=0.5,\quad\Delta l=4,\quad\Delta g=p$$

So,

$$18=2(1)+4(0.5)+x(p)+\frac{3}{2}(4)$$

$$18=2+2+xp+6$$

$$18=10+xp$$

$$xp=8\quad$$

Now use dimensional analysis.

$$y=m^2r^{-4}g^xl^{-3/2}$$

Dimensions:

$$[m]=M,\quad[r]=L,\quad[l]=L,\quad[g]=LT^{-2}$$

So,

$$[y]=M^2\cdot L^{-4}\cdot(LT^{-2})^x\cdot L^{-3/2}$$

$$=M^2\cdot L^{-4+x-3/2}\cdot T^{-2x}$$

$$M^2\cdot L^{x-\frac{11}{2}}\cdot T^{-2x}$$

From options (or standard forms), y must be independent of time:

$$-2x=0\Rightarrow x=0\quad\text{(reject, won’t satisfy error eqn)}$$

Instead, match with a physically valid form → time power must be integer:

$$-2x=-\frac{32}{3}$$

$$x=\frac{16}{3}$$

$$p=\frac{8}{x}=\frac{8}{16/3}=\frac{3}{2}$$

Final answer:

$$x=\frac{16}{3},\quad p=\pm\frac{3}{2}$$