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)

Question

A coil has $N$ turns and current passes through it is $I$ ampere then we obtain $L$ of self inductance. Now if current is made doubled then new self inductance be ______ H.

Answer 1

To solve this problem, we need to understand the relationship between the terminal velocity v_0, the viscosity \eta, the radius r, and other parameters of the spherical body falling through a viscous fluid.

According to Stokes' Law, the terminal velocity v_0 for a sphere falling under gravity in a viscous fluid is given by:

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

Here, \sigma is the density of the sphere, \rho is the density of the fluid, and g is the acceleration due to gravity. We are to find the maximum error in the viscosity \eta.

The relative error in viscosity can be derived from the formula for v_0:

v_0 \propto \frac{r^2}{\eta}

Taking logarithms and differentiating, the expression for relative errors is:

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

Rearranging the terms to find the error in \eta:

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

Thus, the estimated maximum error in the viscosity \eta is:

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

The correct option is therefore \frac{2 \Delta r}{r} + \frac{\Delta v_0}{v_0}.

Answer 2