Question

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. $2 \frac{\Delta r}{r} - \frac{\Delta v_0}{v_0}$

Hint:

Errors are always added for estimating the "maximum possible" error, even if the variable is in the denominator (negative exponent).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Terminal velocity is given by Stokes' Law. We derive the expression for viscosity $\eta$ and use the propagation of errors for products and powers to find the maximum relative error.

Step 2: Key Formula or Approach:
Terminal velocity $v_0 = \frac{2 r^2 (\sigma - \rho) g}{9 \eta}$.
Rearranging for $\eta$:
\[ \eta = \frac{2 r^2 (\sigma - \rho) g}{9 v_0} \]

Step 3: Detailed Explanation:
The constants (2, 9, $\sigma$, $\rho$, $g$) have zero error. The variables with errors are $r$ and $v_0$.
The functional dependence is $\eta \propto r^2 v_0^{-1}$.
Applying the rule for maximum relative error in $Z = A^n B^m$:
\[ \frac{\Delta Z}{Z} = |n| \frac{\Delta A}{A} + |m| \frac{\Delta B}{B} \]
In our case:
\[ \frac{\Delta \eta}{\eta} = 2 \frac{\Delta r}{r} + 1 \frac{\Delta v_0}{v_0} \]
This expression gives the estimated maximum relative error in $\eta$.

Step 4: Final Answer:
The relative error is $\frac{2 \Delta r}{r} + \frac{\Delta v_0}{v_0}$.