Fluid Mechanics

Force due to surface tension of the film on the wire,

As, there are two free surfaces,

$ F=2 \times T \times 10 \mathrm{~cm} $

∴ Work done in increasing the distance by 1 mm ,

$ \begin{aligned} W & =F \times 0.1 \\ & =2 \times T \times 10(0.1) \\ & =2 \times 72 \times 10 \times 0.1=144 \mathrm{ergs} \end{aligned} $

In equilibrium,weight $=$ upthrust

i.e.  $ V \rho_i g=V_i \rho_w g $

or  $ \frac{V_i}{V}=\frac{\rho_i}{\rho_w}=\frac{0.9}{1.0}=0.9 $

i.e. $90 \%$ or 9 m of ice block is inside the water. Therefore, minimum length $l$ required to scoop out a bucket of water is $10-9=1.0 \mathrm{~m}$

To determine the shear stress between layers of water at $y=a$, we start by using the given velocity profile and the definition of shear stress for a fluid in laminar flow.

The shear stress $\tau$ in a fluid is given by:

$ \tau = \eta \frac{du}{dy} $

where $\eta$ is the dynamic viscosity of the fluid, and $\frac{du}{dy}$ is the velocity gradient in the direction perpendicular to the flow.

Given the velocity profile:

$ v = k \left( \frac{2 y^2}{a^2} - \frac{y^3}{a^3} \right) $

First, we need to calculate the velocity gradient $\frac{dv}{dy}$:

$ \frac{dv}{dy} = \frac{d}{dy} \left( k \left( \frac{2 y^2}{a^2} - \frac{y^3}{a^3} \right) \right) $

Applying the differentiation, we get:

$ \frac{dv}{dy} = k \left( \frac{d}{dy} \left( \frac{2 y^2}{a^2} \right) - \frac{d}{dy} \left( \frac{y^3}{a^3} \right) \right) $

Differentiating term by term, we have:

$ \frac{dv}{dy} = k \left( \frac{2 \cdot 2y}{a^2} - \frac{3y^2}{a^3} \right) = k \left( \frac{4y}{a^2} - \frac{3y^2}{a^3} \right) $

We need to evaluate this gradient at $y = a$:

$ \left. \frac{dv}{dy} \right|_{y=a} = k \left( \frac{4a}{a^2} - \frac{3a^2}{a^3} \right) = k \left( \frac{4a}{a^2} - \frac{3a^2}{a^3} \right) = k \left( \frac{4}{a} - \frac{3}{a} \right) = k \left( \frac{4-3}{a} \right) = \frac{k}{a} $

Now, using the given viscosity of the water, $2 \eta$, the shear stress $\tau$ is given by:

$ \tau = 2 \eta \left( \frac{k}{a} \right) = \frac{2 \eta k}{a} $

Therefore, the shear stress between layers of water at $y=a$ is:

$ \tau = \frac{2 \eta k}{a} $

The correct option is Option A:

$ \frac{2 \eta k}{a} $

Liquid volume coming out of the tube per second

$V = {{p\pi {r^4}} \over {8\eta l}}$

$ \Rightarrow V\,\alpha {r^4}$

$ \Rightarrow {{{V_2}} \over {{V_1}}} = {\left[ {{{{r_2}} \over {{r_1}}}} \right]^4} \Rightarrow {V_2} = {V_1}{\left[ {{{125} \over {100}}} \right]^4} = {V_1}{\left( {{5 \over 4}} \right)^4}$

$ \Rightarrow {V_2} = {V_1}(244)$

So, now ${{\Delta V} \over V} = {{{V_2} - {V_1}} \over {{V_1}}} = {{(244){V_1} - {V_1}} \over {{V_1}}} = 144\% $

Volume of bigger drop = Volume of two smaller drops

$ \Rightarrow {4 \over 3}\pi R{'^3} = 2 \times {4 \over 3}\pi {R^3}$

where, R' is the radius of bigger drop.

$ \Rightarrow R{'^3} = 2{R^3}$

$ \Rightarrow R' = {2^{1/3}}\,R$

Initial surface energy of two small drops, ${U_1} = 8\pi {R^2}\sigma $ ..... (i)

Final surface energy of big drop, ${U_2} = 4\pi R{'^2}\sigma $

$ = 4\pi \times {2^{2/3}}{R^2}\sigma $ ..... (ii)

On dividing Eq. (ii) by Eq. (i), we get

${{{U_2}} \over {{U_1}}} = {{4\pi \times {2^{2/3}}\,{R^2}\sigma } \over {8\pi {R^2}\sigma }} = {{{2^{2/3}}} \over 2} = {{{2^{ - 1/3}}} \over 1}$

Ratio of surface energy of bigger drop to smaller one is 2^$-$1/3 : 1.