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(D) The relative error in $y$ is given by the formula:$\frac{\Delta y}{y} = 2 \frac{\Delta m}{m} + 4 \frac{\Delta r}{r} + x \frac{\Delta g}{g} + \frac

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$\frac{16}{3}$ and $\pm \frac{3}{2}$

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(D) The relative error in $y$ is given by the formula:$\frac{\Delta y}{y} = 2 \frac{\Delta m}{m} + 4 \frac{\Delta r}{r} + x \frac{\Delta g}{g} + \frac{3}{2} \frac{\Delta l}{l}$Given percentage errors are: $\frac{\Delta y}{y} 	imes 100 = 18$,$\frac{\Delta m}{m} 	imes 100 = 1$,$\frac{\Delta r}{r} 	imes 100 = 0.5$,$\frac{\Delta l}{l} 	imes 100 = 4$,and $\frac{\Delta g}{g} 	imes 100 = p$.Substituting these values into the error equation:$18 = 2(1) + 4(0.5) + x(p) + \frac{3}{2}(4)$$18 = 2 + 2 + xp + 6$$18 = 10 + xp$$xp = 8$Checking the options:For option $D$,$x = \frac{16}{3}$ and $p = \pm \frac{3}{2}$.$xp = \frac{16}{3} 	imes \frac{3}{2} = 8$.Thus,the correct values are $x = \frac{16}{3}$ and $p = \pm \frac{3}{2}$.

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

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