Mechanical Properties of Fluids

This is a multiple choice answer as classified in NCERT Exemplar

(a) d₁= diameter at 1^st point is 2.5

d₂= diameter at 2^nd  point is 3.75

Applying equation of continuity for cross section A₁ and A₂

A₁V₁=A₂V₂

^{$\frac{V_1}{V_2}=\frac{A_2}{A_1}=\frac{\pi r_2^2}{\pi r_1^2}=\left(\frac{r_2}{r_1}\right)$ 2}

= ${\left(\frac{\frac{3.75}{2}}{\frac{2.5}{2}}\right)}^2$ = ${\left(\frac{3.75}{2.5}\right)}^2=\frac{9}{4}$

This is a multiple choice answer as classified in NCERT Exemplar

(b) As we know for a streamline flow of a liquid velocity of each particle at a particular cross-section is constant, because Av = constant (law of continuity) between two cross-section of a tube of flow.

This is a multiple choice answer as classified in NCERT Exemplar

(d) In a streamline flow at any given point, the velocity of each passing fluid particles remains constant. If we consider a cross-sectional area, then a point on the area cannot have different velocities at the same time, hence two streamlines of flow cannot cross each other.

This is a multiple choice answer as classified in NCERT Exemplar

(c) When the pebble is falling through the viscous oil the viscous force is

F=6 $\pi \eta rv$

where r is radius of the pebble, v is instantaneous speed, his coefficient of viscosity. As the force is variable, hence acceleration is also variable so v-t graph will not be straight line. First velocity increases and then becomes constant known as terminal velocity.

This is a short answer type question as classified in NCERT Exemplar

Surface tension of water S= 7.28*10^-2 Nm^-1

Vapour pressure p= 2.33*10³ Pa

The drop will evaporate, if the water pressure is greater than the vapour pressure.

Let a water droplet or radius R can be formed without evaporating.

Vapour pressure= Excess pressure in drop.

P=2S/R

R= 2S/P= $\frac{2*7.28*{10}^{-2}}{2.33*{10}^3}=6.25*{10}^{-5}m$

This is a short answer type question as classified in NCERT Exemplar

When a big drop of radius R, breaks into N droplets each of radius r, the volume remains constant.

Volume of big drop = N $*$ volume of each small drop

$\frac{4}{3}\pi R^3=N*\frac{4}{3}\pi r^3$

R³ =Nr³

N=R³/r³

Now change in surface area = 4 $\pi R^3$ -N4 $\pi$ r²

= 4 $\pi \left(R^2-N{r}^2\right)$

Energy released = T $*?A$  = S $*4\pi \left(R^2-N{r}^2\right)$

Due to releasing of this energy , the temperature is lowered.

If $\rho$  is the density and s is specific heat of liquid and its temperature is lowered by $?\theta$ , then energy released = ms $?\theta$

T $*4\pi$ ( $R^2-N{r}^2$ )= $(\frac{4}{3}*R^3*\rho )s?\theta$

$?\theta$ = $\frac{T*4\pi (R^2-N{r}^2)}{\frac{4}{3}\pi R^3\rho s}$

= $\frac{3T}{\rho s}\left[\frac{R^2}{R^3}-\frac{N{r}^2}{R^3}\right]$

= $\frac{3T}{\rho s}[\frac{1}{R}-\frac{\frac{R^3}{r^3}*r^2}{R^3}]$

$\frac{3T}{\rho s}$$\left[\frac{1}{R}-\frac{1}{r}\right]$

This is a short answer type question as classified in NCERT Exemplar

Consider the diagram

Radii of mercury droplets  r₁=0.1cm = 1 $*{10}^{-3}m$

r₂=0.2cm=2 $*{10}^{-3}m$

Surface tension T= 435.5 * 10^-3 Nm^-1

Let the radius of the big drop formed by collapsing be R

Therefore,

volume of big drop = volume of small droplets

4/3 $\pi$ R³=4/3 $\pi r_1^3+$ 4/3 $\pi r_2^3$

R³= $r_1^3+r_2^3$

= 0.1³+0.2³

= 0.001+0.008

=0.009

R= 0.21 cm = 2.1 $*{10}^{-3}m$

Change in surface area $?A$ = 4 $\pi R^2-(4\pi r_1^2+4\pi r_2^2)$

=4 $\pi (R^2-(r_1^2+r_2^2\left)\right)$

Energy released =T $?A$

= T $*4\pi \left[\right(R^2-(r_1^2+r_2^2)\left)\right]$

= 435.5 $*{10}^{-3}*4*3.14[2.1*{10}^{-3}]$ 2-(1 $*{10}^{-6}+4*{10}^{-6}$ )

= 435.5 $*4*3.14\left[4.41-5\right]*{10}^{-6}*{10}^{-3}$

= -32.23 $*{10}^{-7}$

Therefore -3.22 $*{10}^{-6}J$ energy will be absorbed.

This is a short answer type question as classified in NCERT Exemplar

Consider the diagram where a tanker is accelerating with acceleration a.

Consider an elementary particle of the fluid of mass dm.

The acting forces on the particle with respect to the tanker are shown above.

Now, balancing forces (as the particle is in equilibrium) along the inclined direction

component of weight= component of pseudo force dmg sin $\theta$  =dma cos $\theta$   (we have assumed that the surface is inclined at an angle q) where, dma is pseudo force

g sin $\theta$ =acos $\theta$

a=g tan $\theta$

tan $\theta$ = a/g = slope

This is a short answer type question as classified in NCERT Exemplar

Given radius r= 2.5 $*{10}^{-5}m$

Surface tension S= 7.28 $*{10}^{-2}N/m$

Angle of contact = 0⁰

The maximum height to which SAP can rise in trees through capillarity action is given by

h = $\frac{2Scos\theta }{r\rho g}$ where S = surface tension, $\rho$ = density, r= radius

h= $\frac{2*7.28*{10}^{-2}COS{O}^0}{2.5*{10}^{-5}*1*{10}^{-3}*9.8}=0.6m$

This is the maximum height to which the SAP can rise due to surface tension. Since, many trees have heights much more than this, capillary action alone cannot account for the rise of water in all trees.