A spherical body of radius r and density falls freely through a viscous liquid having density and viscosity and attains a terminal velocity v0. Estimated maximum error in the quantity is: (Ignore…

Given: A spherical body has radius r, density, and terminal velocity v0 in a liquid of density and viscosity. Find: The estimated maximum relative error in. Using Stokes' law for terminal velocity, v0 = 2 r^2 ( -) g9 Rearranging for viscosity, = 2 r^2 ( -) g9 v0 So the functional dependence is r^2 v0^-1 For maximum possible error, errors are added with the magnitudes of the powers: = 2 rr + v0v0 Therefore, the estimated maximum relative error in is 2 rr + v0v0. The correct option is C.

A spherical body of radius r and density falls freely through a v… — Solution | JEE Physics

MCQEasyJEE 2026Significant Figures & Error Analysis

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)

A
$2 \left[ \frac{\Delta r}{r} - \frac{\Delta v_0}{v_0} \right]$
B
$2 \left[ \frac{\Delta r}{r} + \frac{\Delta v_0}{v_0} \right]$
C
$\frac{2 \Delta r}{r} + \frac{\Delta v_0}{v_0}$
D
$2 \frac{\Delta r}{r} - \frac{\Delta v_0}{v_0}$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: A spherical body has radius $r$ , density $\sigma$ , and terminal velocity $v_0$ in a liquid of density $\rho$ and viscosity $\eta$ .

Find: The estimated maximum relative error in $\eta$ .

Using Stokes' law for terminal velocity,

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

Rearranging for viscosity,

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

So the functional dependence is

\eta \propto r^2 v_0^{-1}

For maximum possible error, errors are added with the magnitudes of the powers:

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

Therefore, the estimated maximum relative error in $\eta$ is $\frac{2 \Delta r}{r} + \frac{\Delta v_0}{v_0}$ . The correct option is C.

Common mistakes

Using a minus sign for the error contribution of $v_0$ because it appears in the denominator is incorrect. For maximum error, absolute values of the powers are used and all contributions are added.
Not rearranging the terminal velocity formula for $\eta$ before applying error propagation can lead to using the wrong powers of variables. First write $\eta \propto r^2 v_0^{-1}$ , then apply the error rule.
Including errors in $\sigma$ , $\rho$ , or $g$ is incorrect here because the question explicitly says to ignore errors associated with these quantities. Only $r$ and $v_0$ contribute.

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Answer

Step-by-step solution

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