Q 11-01-143JEE MainJEE Main 2023 (6 Apr, Shift 1)Medium
Two resistances are given as $R_1=(10\pm0.5)\ \Omega$ and $R_2=(15\pm0.5)\ \Omega$. The percentage error in the measurement of equivalent resistance when they are connected in parallel is
Answer: (D) 4.33
$R_p=\dfrac{10\times15}{25}=6\ \Omega$. From $\dfrac1{R_p}=\dfrac1{R_1}+\dfrac1{R_2}$: $\dfrac{\Delta R_p}{R_p^2}=\dfrac{\Delta R_1}{R_1^2}+\dfrac{\Delta R_2}{R_2^2}$.
$$\Delta R_p=36\left(\frac{0.5}{100}+\frac{0.5}{225}\right)=0.26\ \Omega,\qquad\frac{0.26}{6}\times100\approx4.33\%$$
Solution by Sreeraj P, M.Sc Physics