Class 11th

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.

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

Consider the diagram,

The scale is adjusted to zero, therefore, when the block suspended to a spring is immersed

In water, then the reading of the scale will be equal to the thrust on the block due to water.

Thrust= weight of water displaced

=V $\rho$ _wg (where V is volume of the block and $\rho$ _w is density of water)

= $\frac{m}{\rho }{\rho }_{w}g$ = $\frac{{\rho }_w}{\rho }mg$

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

Given density of ice $\mathit{\rho }$ _i = 0.917 g cm^-3

Density of water ${\rho }_w=1g/cm$ ³

Let V be the total volume of iceberg and V' of its volume be submerged in water.

In floating condition

Weight of the iceberg= weight of the water displaced by the submerged part by ice

V $\rho$ _ig = V' ${\rho }_w$ g

$\frac{V\text{'}}{V}=\frac{{\rho }_i}{{\rho }_w}$ = $\frac{0.917}{1}=0.917$

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

No surface tension is a scalar quantity.

Surface tension = work done/ surface area, where work done and surface area both are scalar quantities.

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

Viscosity is a property of liquid it does not have any direction hence it is scalar quantity.