JEE PYQ: Units & Measurements - Question ID 172287b8399b (JEE Main 2023)

In the problem, we are given two resistances, $R_1$ and $R_2$, each with a certain measurement error, $\Delta R_1$ and $\Delta R_2$. These resistances are connected in parallel, and we are asked to find the percentage error in the equivalent resistance of this combination.

The formula for the equivalent resistance $R$ of two resistors $R_1$ and $R_2$ in parallel is:

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$

We want to find the percentage error in $R$, which is given by $(\Delta R / R) \times 100\%$.

In order to find $\Delta R / R$, we differentiate both sides of the above equation with respect to $R$, $R_1$, and $R_2$. This gives us:

$\frac{\Delta R}{R^2} = \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}$

We can then solve this equation for $\Delta R / R$:

$\frac{\Delta R}{R} = \left(\frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}\right)R$

Substituting the given values, $R_1 = 10 \, \Omega$, $R_2 = 15 \, \Omega$, $\Delta R_1 = \Delta R_2 = 0.5 \, \Omega$, and $R = R_1R_2/(R_1+R_2) = 6 \, \Omega$, we get:

$\frac{\Delta R}{R} = \left(\frac{0.5}{100} + \frac{0.5}{225}\right) \times 6 = \frac{13}{300}$

Finally, to convert this to a percentage, we multiply by 100, giving:

$\frac{\Delta R}{R} \times 100 = \frac{13}{3} = 4.33 \%$

This tells us that the percentage error in the equivalent resistance of the two resistances in parallel is $4.33\%$.