JEE Main 2026 — Physics Mechanics

2026mcqmedium

A spherical body of radius $r$ and density $\sigma$ falls freely through a viscous liquid having density $\rho$ and viscosity $\eta$ and attains a terminal velocity $v_{0}$ . Estimated maximum error in the quantity $\eta$ is : (Ignore errors associated with $\sigma, \rho$ and $g$ , gravitational acceleration)

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

Options

A
$\frac{2 \Delta r}{r}+\frac{\Delta v_{\mathrm{o}}}{v_{\mathrm{o}}}$
B
$2 \frac{\Delta r}{r}-\frac{\Delta v_{\mathrm{o}}}{v_{\mathrm{o}}}$
C
$2\left[\frac{\Delta r}{r}-\frac{\Delta v_{\mathrm{o}}}{v_{\mathrm{o}}}\right]$
D
$2\left[\frac{\Delta r}{r}+\frac{\Delta v_{\mathrm{o}}}{v_{\mathrm{o}}}\right]$

Did you get this right?

Solution

The terminal velocity $v_0$ of a spherical body falling through a viscous liquid is given by Stokes' Law:

$v_0 = \frac{2r^2(\sigma - \rho)g}{9\eta}$

Rearranging the formula to express viscosity $\eta$ in terms of other variables:

$\eta = \frac{2r^2(\sigma - \rho)g}{9v_0}$

Since we are told to ignore errors associated with $\sigma$ , $\rho$ , and $g$ , these quantities along with the numerical factor $\frac{2}{9}$ are treated as constants.

The expression for $\eta$ can be written as $\eta = k \cdot \frac{r^2}{v_0}$ , where $k = \frac{2(\sigma - \rho)g}{9}$ is a constant.

Taking the natural logarithm on both sides:

$\ln \eta = \ln k + 2\ln r - \ln v_0$

Differentiating both sides to find the relative error:

$\frac{d\eta}{\eta} = 0 + 2\frac{dr}{r} - \frac{dv_0}{v_0}$

For the estimated maximum fractional error, we take the absolute sum of the individual relative errors:

$\frac{\Delta \eta}{\eta} = 2\frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$

This matches option (1).

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Options

$\frac{2 \Delta r}{r}+\frac{\Delta v_{\mathrm{o}}}{v_{\mathrm{o}}}$

$2 \frac{\Delta r}{r}-\frac{\Delta v_{\mathrm{o}}}{v_{\mathrm{o}}}$

$2\left[\frac{\Delta r}{r}-\frac{\Delta v_{\mathrm{o}}}{v_{\mathrm{o}}}\right]$

$2\left[\frac{\Delta r}{r}+\frac{\Delta v_{\mathrm{o}}}{v_{\mathrm{o}}}\right]$

Solution

The terminal velocity $v_0$ of a spherical body falling through a viscous liquid is given by Stokes' Law:

$v_0 = \frac{2r^2(\sigma - \rho)g}{9\eta}$

Rearranging the formula to express viscosity $\eta$ in terms of other variables:

$\eta = \frac{2r^2(\sigma - \rho)g}{9v_0}$

Since we are told to ignore errors associated with $\sigma$ , $\rho$ , and $g$ , these quantities along with the numerical factor $\frac{2}{9}$ are treated as constants.

The expression for $\eta$ can be written as $\eta = k \cdot \frac{r^2}{v_0}$ , where $k = \frac{2(\sigma - \rho)g}{9}$ is a constant.

Taking the natural logarithm on both sides:

$\ln \eta = \ln k + 2\ln r - \ln v_0$

Differentiating both sides to find the relative error:

$\frac{d\eta}{\eta} = 0 + 2\frac{dr}{r} - \frac{dv_0}{v_0}$

For the estimated maximum fractional error, we take the absolute sum of the individual relative errors:

$\frac{\Delta \eta}{\eta} = 2\frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$

This matches option (1).

JEE Main 2026 — Physics Mechanics

Options

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Solution

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JEE Main 2026 — Physics Mechanics

Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics