JEE PYQ: Units & Measurements - Question ID da2c301c32ea (JEE Main 2026)

Core Formula & Concept:

When a spherical body falls freely through a viscous liquid, it experiences three primary forces:

1. Gravitational Force ($F_g$): Downward force due to gravity, given by $F_g = \frac{4}{3} \pi r^3 \sigma g$, where $r$ is the radius, $\sigma$ is the density of the sphere, and $g$ is gravitational acceleration.
2. Buoyant Force ($F_b$): Upward force due to the displaced liquid, given by $F_b = \frac{4}{3} \pi r^3 \rho g$, where $\rho$ is the density of the liquid.
3. Viscous Drag Force ($F_d$): Opposing force due to viscosity, given by Stokes' Law: $F_d = 6 \pi \eta r v$, where $\eta$ is the viscosity of the liquid, and $v$ is the velocity of the sphere.

At terminal velocity ($v_0$), the net force on the sphere is zero, meaning the downward forces balance the upward forces:

$F_g - F_b - F_d = 0$ Substituting the expressions for $F_g$, $F_b$, and $F_d$: $\frac{4}{3} \pi r^3 \sigma g - \frac{4}{3} \pi r^3 \rho g - 6 \pi \eta r v_0 = 0$ Simplifying, we solve for $\eta$: $6 \pi \eta r v_0 = \frac{4}{3} \pi r^3 (\sigma - \rho) g$ $\eta = \frac{2}{9} \frac{r^2 (\sigma - \rho) g}{v_0}$

Since errors in $\sigma$, $\rho$, and $g$ are ignored, the maximum error in $\eta$ depends only on the errors in $r$ and $v_0$.

Step-by-Step Derivation:

To find the maximum error in $\eta$, we use the formula for relative error in a product or quotient. For a function of the form:

$\eta = k \cdot r^2 \cdot v_0^{-1}$ where $k = \frac{2}{9} (\sigma - \rho) g$ is a constant (since errors in $\sigma$, $\rho$, and $g$ are ignored), the relative error in $\eta$ is given by: $\frac{\Delta \eta}{\eta} = \left| \frac{\partial \eta}{\partial r} \frac{\Delta r}{\eta} \right| + \left| \frac{\partial \eta}{\partial v_0} \frac{\Delta v_0}{\eta} \right|$ Calculating the partial derivatives: 1. Partial derivative with respect to $r$: $\frac{\partial \eta}{\partial r} = 2 k r v_0^{-1} = \frac{2 \eta}{r}$ Thus, $\left| \frac{\partial \eta}{\partial r} \frac{\Delta r}{\eta} \right| = \left| \frac{2 \eta}{r} \cdot \frac{\Delta r}{\eta} \right| = 2 \frac{\Delta r}{r}$ 2. Partial derivative with respect to $v_0$: $\frac{\partial \eta}{\partial v_0} = -k r^2 v_0^{-2} = -\frac{\eta}{v_0}$ Thus, $\left| \frac{\partial \eta}{\partial v_0} \frac{\Delta v_0}{\eta} \right| = \left| -\frac{\eta}{v_0} \cdot \frac{\Delta v_0}{\eta} \right| = \frac{\Delta v_0}{v_0}$ Adding these contributions, the maximum relative error in $\eta$ is: $\frac{\Delta \eta}{\eta} = 2 \frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$

Therefore, the maximum error in $\eta$ is:

$\Delta \eta = \eta \left( 2 \frac{\Delta r}{r} + \frac{\Delta v_0}{v_0} \right)$ However, the question asks for the estimated maximum error in the quantity $\eta$, which is typically expressed as the relative error $\frac{\Delta \eta}{\eta}$. The options are given in terms of this relative error, so the correct choice is: $\frac{\Delta \eta}{\eta} = 2 \frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$ Common Traps & Exam Tip:

Students often make the following mistakes in this question:

1. Incorrect Sign Handling: Some students forget that errors are always added in quadrature or as absolute values, leading to incorrect signs (e.g., subtracting $\frac{\Delta v_0}{v_0}$ instead of adding).
2. Misapplying the Error Formula: The error in $r^2$ is $2 \frac{\Delta r}{r}$, not $\frac{\Delta r}{r}$. Students may overlook the exponent and treat $r^2$ as $r$.
3. Ignoring the Negative Exponent: The term $v_0^{-1}$ contributes $\frac{\Delta v_0}{v_0}$ to the error, but students may mistakenly treat it as $2 \frac{\Delta v_0}{v_0}$ due to confusion with the $r^2$ term.
4. Confusing Absolute and Relative Errors: The question asks for the relative error in $\eta$, not the absolute error. Students may incorrectly compute $\Delta \eta$ instead of $\frac{\Delta \eta}{\eta}$.

Exam Tip: Always write down the formula for $\eta$ explicitly and identify which variables contribute to the error. Use the rule that for a function $f = k x^a y^b$, the relative error is $\frac{\Delta f}{f} = |a| \frac{\Delta x}{x} + |b| \frac{\Delta y}{y}$. This will help avoid mistakes with exponents and signs.