physics ncert solutions class 11th

This is a multiple choice answer as classified in NCERT Exemplar

(b), (c) Pressure is defined as the ratio of magnitude of component of the force normal to the area and the area under consideration.

As magnitude of component is considered, hence, it will not have any direction. So, Pressure is a scalar quantity.

This is a multiple choice answer as classified in NCERT Exemplar

(b), (d) Consider the diagram where two molecules of a liquid are shown. One is well inside the liquid and other is on the surface. The molecule (A) which is well inside experiences equal forces from all directions, hence net force on it will be zero.

And molecules on the liquid surface have some extra energy as it surrounded by Only lower half side of liquid molecules.

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(c) According to the question, the observed meniscus is of convex figure shape. Which is only possible when angle of contact is obtuse. Hence, the combination will be of mercury-glass (140°)

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(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.

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(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.

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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]$