1. Given relation

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

   We are given percentage errors in:

• $y = 18\%$
• $m = 1\%$
• $r = 0.5\%$
• $l = 4\%$
• $g = p\%$

We need to find $x$ and $p$.

2. Use error propagation for powers

For a quantity $Q = a^\alpha b^\beta c^\gamma,$ maximum percentage error is $\frac{\Delta Q}{Q}\times 100 = |\alpha|\frac{\Delta a}{a}\times 100 + |\beta|\frac{\Delta b}{b}\times 100 + |\gamma|\frac{\Delta c}{c}\times 100.$

So for $y = m^2 r^{-4} g^x l^{-3/2},$ we get

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

Now simplify each term:

$2(1) = 2$ $4(0.5) = 2$ $\frac{3}{2}(4) = 6$

Hence,

$18 = 2 + 2 + 6 + |x|p$ $18 = 10 + |x|p$ $|x|p = 8$

3. Check the options

We test each option for the condition $|x|p = 8.$

• Option A: $x=5,\; p=\pm 2$

  $|x|p = 5\times 2 = 10 \neq 8$ Not correct.

• Option B: $x=4,\; p=\pm 3$

  $|x|p = 4\times 3 = 12 \neq 8$ Not correct.

• Option C: $x=\frac{16}{3},\; p=\pm \frac{3}{2}$

  $|x|p = \frac{16}{3}\times \frac{3}{2} = 8$ Correct.

• Option D: $x=8,\; p=\pm 2$

  $|x|p = 8\times 2 = 16 \neq 8$ Not correct.

4. Final answer

Thus,

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

So the correct option is C.