Two resistances are given as $R_1=(10\pm 0.5)\mathrm{\Omega }\text{ and }{R}_2=(15\pm 0.5)\mathrm{\Omega }\text{. }$The percentage error in the measurement of equivalent resistance when they are connected in parallel is

For resistors in parallel effective resistance $\Rightarrow \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R}=\frac{1}{10}+\frac{1}{15}$

R = 6

$\frac{dR}{R^2}=\frac{d{R}_1}{R_1^2}+\frac{d{R}_2}{R_2^2}$

$\Rightarrow \frac{dR}{R}\times 100=R\left(\frac{d{R}_1}{R_1^2}+\frac{d{R}_2}{R_2^2}\right)\times 100$

$\begin{array}{l}\mathrm{\%}\text{ error }=\left(\frac{0.5}{100}+\frac{0.5}{225}\right)6\times 100\\ =300\left(\frac{325}{100\times 225}\right)\\ =\frac{13}{3}=4.33\%\end{array}$