NCERT Solutions for Physics Class 11 Mechanical Properties of Fluids Ch 9 - Download Free PDF

Radius of the mercury droplet, r = 3.00 mm = 3 $\times {10}^{-3}$ m

Surface tension, S = 4.65 $\times {10}^{-1}$ N/m

Atmospheric pressure,  $P_o$ = 1.01 $\times {10}^5$ Pa

Total pressure inside the mercury drop = Excess pressure inside mercury + Atmospheric pressure

= $\frac{2S}{r}$ + $P_o$ = $\frac{2\times 4.65\mathrm{}\times {10}^{-1}}{3\times {10}^{-3}}$ + 1.01 $\times {10}^5$ = 1.01310 $\times {10}^5$ Pa

Excess pressure inside mercury = $\frac{2S}{r}$ = $\frac{2\times 4.65\mathrm{}\times {10}^{-1}}{3\times {10}^{-3}}$ = 310 Pa

The pressure of a liquid is given by the following relation:

P = h $\rho$ g, where P = Pressure, h = height of the liquid column,  $\rho$ = is the density of the liquid and g= acceleration due to gravity

From the above relation, because of the h factor (height of the human body), the pressure is more at the feet and less at the brain

The said phenomenon is due to the $\rho$ factor. Density of air is maximum at the sea level. At height, density decreases and pressure also decreases. At 6 km height, the density of air is nearly half of that of a t sea level

When pressure is applied on the liquid, the pressure is transmitted in all directions in the liquid, Hence in absence of fixed direction, it is a scalar quantity

The angle between the tangent to the liquid surface at the point of contact and the surface inside the liquid is called the angle of contact ( $\theta )$ , as shown in the diagram

$S_{la}$ = Interfacial tension between liquid-air interface

$S_{sl}$ = Interfacial tension between solid -liquid interface

$S_{sa}$ = Interfacial tension between solid-air interface

At the line of contact of contact, the surface forces between the three media must be in equilibrium. Hence

$\cos ?\theta$ = $\frac{S_{sa}-S_{la}}{S_{la}}$

The angle of contact $\theta$ is obtuse, if $S_{sa}<S_{la}$ , as in the case of mercury on glass

This angle is acute if $S_{sl}<S_{la}$ , as in the case of water on glass

Mercury molecules (which makes an obtuse angle with glass) have a strong force of attraction between themselves and a weak force of attraction towards solids. Hence, they tend to form drops

Water molecules make acute angles with glass. They have a weak force of attraction between themselves and a strong force of attraction towards solids. Hence, they tend to spread out

Surface tension is the force acting per unit length at the interface between the plane of a liquid and any other surface. This force is independent of the area of the liquid surface. Hence, surface tension is independent of the area of the liquid surface.

Water with detergent dissolved in it has a small angle of contact( $\theta ).$ This is because for a small $\theta ,$ there is a fast capillary rise of the detergent in the cloth. The capillary rise of a liquid is directly proportion to the cosine of the angle $\theta .$ If $\theta$ is small, then cos $\theta$ will be large and then the rise of detergent in the cloth will be faster.

A liquid tend to acquire the minimum surface area because of the presence of surface tension. The surface area of a sphere is the minimum for a given volume. Hence, under no external forces, liquid drops always take spherical shape.

Height of the spirit column, $h_1$ = 12.5 cm = 0.125 m

Height of the water column, $h_2$ = 10 cm = 0.1m

$P_0$ = atmospheric pressure

${\rho }_1$ = density of spirit, ${\rho }_2$ = density of water

Pressure at point B = $P_0$ + $h_1\mathrm{}{\rho }_1\mathrm{}\mathrm{g}$

Pressure at point D = $P_0$ + $h_2\mathrm{}{\rho }_2\mathrm{}\mathrm{g}$

But pressure at point B and D are same. Hence

$P_0$ + $h_1\mathrm{}{\rho }_1\mathrm{}\mathrm{g}$ = $P_0$ + $h_2\mathrm{}{\rho }_2\mathrm{}\mathrm{g}$ or $h_1\mathrm{}{\rho }_1$ = $h_2\mathrm{}{\rho }_2$

$\frac{\mathrm{}{\rho }_1}{{\rho }_2}$ = $\frac{\mathrm{}{h}_2}{h_1}$ = $\frac{10}{12.5}$ = 0.8

So the specific gravity of spirit is 0.8

The surface tension of a liquid is inversely proportional to temperature. Decreases

Most fluids offer resistance to their motion. It is like internal mechanical friction, known as viscosity. Gas viscosity increases with temperature, whereas liquid viscosity decreases with temperature. Because, intermolecular forces weaken with temperature increase, viscosity decreases.

With reference to the elastic modulus of rigidity for solids, the shearing force is proportional to the shear strain. With reference to elastic modulus of rigidity for fluids, the shearing force is proportional to the rate of shear strain.

For a steady-flowing fluid, an increase in its flow speed at a constriction follows the conservation of mass (Bernouli' principle).

For the model of a plane in wind tunnel, turbulence occurs at a greater speed then it does for an actual plane. This follows from Bernoulli's principle and different Reynolds' numbers associated with the motions of the two planes.

When air is blown under a paper, the velocity of air is more than the upper portion of the paper. As per Bernoulli's principle, atmospheric pressure reduces under the paper and makes it fall. To keep the paper horizontal, the air needs to be blown on the upper surface of the paper.

For a smaller opening, the flow of fluid is more than when it is bigger. When we try to close the tap with our fingers, water gushes through the small openings. Area and velocity are inversely proportional to each other.

Small opening of a syringe needle controls the velocity of the blood oozing out. At the constriction point of the syringe system, the flow rate suddenly increases to a high value for a constant thumb pressure.

When a fluid flows out from a small hole in a vessel, the vessel receives backward thrust, when fluid flows out from a small hole, it attains high velocity. Due to the law of conservation of momentum, the vessel attains backward velocity.

A spinning cricket ball in air has two simultaneous motions – rotary and linear. These two types of motion, opposes each other. The velocity below the ball decreases, the pressure on upper side becomes lesser than the lower side of the ball. An upward force acts on the ball and it takes a curved path, does not follow parabolic path.

Mass of the girl, m = 50 kg

Diameter of the heel, d = 1.0 cm = 0.01 m, Radius of the heel, r = d/2 = 0.005m

Area of the heel,  $\pi r^2$ = 7.85 $\times {10}^{-5}$ $m^2$

Force exerted by heel on the floor, F = mg = 50 $\times 9.8$ N = 490 N

Pressure exerted by heel on the ground, p = F/A = 490/ (7.85 $\times {10}^{-5})$ N/ $m^2$

= 6.24 $\times {10}^6$ N/ $m^2$

Density of mercury,  ${\rho }_1$ = 13.6 $\times {10}^3$ kg/ $m^3$

Density of wine,  ${\rho }_2$ = 9.84 $\times {10}^2$ kg/ $m^3$

Height of the mercury column for atmospheric pressure,  $h_1$ = 760 mm = 0.76 m

Height of the mercury column for atmospheric pressure = $h_2$

From the relation, P = $\rho gh$ , since the pressure on both the system are equal

${\rho }_{1}g{h}_1$ = ${\rho }_{2}g{h}_2$ , we get $h_2$ = $\frac{{\rho }_{1}g{h}_1}{{\rho }_{2}g}$ = $\frac{13.6\times {10}^3\times 0.76}{9.84\times {10}^2}$ = 10.5 m

The maximum allowable stress for the structure, P = ${10}^9$ Pa

Depth of the ocean, d = 3 km = 3 $\times {10}^3$ m

Density of water $\rho$ = ${10}^3$ kg/ $m^3$

Acceleration due to gravity, g = 9.8 m/s

The pressure exerted because of sea water at the depth d = $\rho dg$ = ${10}^3\times$ 3 $\times {10}^3\times 9.8$ Pa

= 2.94 $\times {10}^7$ Pa

The maximum allowable stress is more than the pressure, hence the structure is suitable.

The maximum mass of the car can be lifted, m = 3000 kg

Area of cross-section of the load carrying piston, A = 425 ${cm}^2$ = 425 $*{10}^{-4}{m}^2$

Maximum force exerted by the load, F = mg = 3000 $*9.8$ N = 29400 N

Maximum pressure exerted, P = F/A = (29400 / 425 $*{10}^{-4}$ ) Pa = 6.917 $*{10}^5$ Pa

Height of the spirit column,  $h_1$ = 12.5 + 15 cm = 27.5 cm

Height of the water column,  $h_2$ = 10 + 15 cm = 25 cm

Density of spirit,  ${\rho }_1$ = 0.8 gm/ ${cm}^3$

Density of water,  ${\rho }_2$ = 1 gm/ ${cm}^3$

Density of mercury,  $\rho$ = 13.6 gm/ ${cm}^3$

Let h be the difference between the levels of mercury in two limbs

Pressure exerted by mercury column of h height= h $\rho g$ = 13.6hg …. (i)

Difference between pressure exerted by water and spirit columns:

= $h_2{\rho }_{2}g-h_1{\rho }_{1}g=g\left(25\times 1-27.5\times 0.8\right)=$ 3g ……. (ii)

Equating equations (i) and (ii), we get

13.6hg = 3g

h = 3/13.6 = 0.221 cm

No, Bernoulli's equation cannot be used to describe the flow of water through a rapid in a river. In rapid, the flow is turbulent whereas Bernoulli's equation is applicable for laminar flow only.

It does not matter if one uses gauge pressure, instead of absolute pressure while applying Bernoulli's equation. There should be significantly different atmospheric pressures, where Bernoulli's equation is applied.

Length of the horizontal tube, l = 1.5 m

Radius of the tube, r = 1 cm = 0.01 m

Diameter of the tube, d = 2r = 0.02m

Glycerine mass flow rate, M = 4 $\times {10}^{-3}$ kg/s

Density of glycerine, $\rho$ = 1.3 $\times {10}^3$ kg

Viscosity of glycerine, $\eta$ = 0.83 Pa-s

Now, volume of glycerine flowing per sec V = $\frac{M}{\rho }$ = $\frac{4\times {10}^{-3}}{1.3\mathrm{}\times {10}^3}$ $m^3$ /s = 3.08 $\times {10}^{-6}$ $m^3$ /s

According to Poiseville’s formula, we know the flow rate

V = $\frac{\pi \times p\times r^4}{8\times \eta \times l}$ where p is the pressure difference between two ends of the tube

p = $\frac{V\times 8\times \eta \times l}{\pi \times r^4}$ = $\frac{3.08\mathrm{}\times {10}^{-6}\times 8\times 0.83\times 1.5}{\pi \times {\left(0.01\right)}^4}$ = 976.47 Pa

Reynolds’s number is given by the relation

Re = $\frac{4\rho V}{\pi d\eta }$ = $\frac{4\times 1.3\mathrm{}\times {10}^3\times 3.08\mathrm{}\times {10}^{-6}}{3.1416\times 0.02\times 0.83}$ = 0.3

Since the Reynolds’s number is 0.3, the flow is laminar

Speed of the wind on the upper surface of the wing, $V_1$ = 70 m/s

Speed of the wind on the lower surface of the wing, $V_2$ = 63 m/s

Area of the wing, A = 2.5 $m^2$

Density of air, $\rho$ = 1.3 kg/ $m^3$

According to Bernoulli’s theorem, we have the relation:

$P_1$ + $\frac{1}{2}\rho {V_1}^2$ = $P_2$ + $\frac{1}{2}\rho {V_2}^2$

( $P_2$ - $P_1$ )= $\frac{1}{2}\rho ({V_1}^2$ - ${V_2}^2)$ , where $P_1$ = pressure on the upper surface of the wing and $P_2$ - pressure on the lower surface of the wing

The pressure difference provides lift to the aeroplane

Lift on the wing = ( $P_2$ - $P_1$ )A = $\frac{1}{2}\rho ({V_1}^2$ - ${V_2}^2)$ A = $\frac{1}{2}\times 1.3\times ({70}^2$ - ${63}^2)\times 2.5$ N = 1512.8 N = 1.5 $\times {10}^3$ N

Take the case given in figure (b)

$A_1$ = Area of pipe 1,  $A_2$ = Area of pipe 2,  $V_1$ = Speed of fluid in pipe 1,  $V_2$ = Speed of fluid in pipe 2

From the law of continuity, we have

$A_1{V}_1$ = $A_2{V}_2$

When the area of cross-section in the middle of the venturimeter is small, the speed of the flow of liquid through this part is more. According to Bernoulli's principle, if speed is more, the pressure is less. Pressure is directly proportional to height, hence the level of water in pipe 2 is less. Therefore, figure (a) is not possible.

Area of cross-section of the spray pump, $A_1$ = 8 ${cm}^2$ = 8 $\times {10}^{-4}$ $m^2$

Number of holes, n = 40

Diameter of each hole, d = 1 mm = 1 $\times {10}^{-3}$ m

Radius of each hole, r = d/2 = 0.5 $\times {10}^{-3}$ m

Area of cross-section of each hole, a = $\pi r^2$ = $\pi ({0.5\mathrm{}\times {10}^{-3})}^2$

Total area of 40 holes, $A_2$ = 40 $\pi \times ({0.5\mathrm{}\times {10}^{-3})}^2$ = 3.14 $\times {10}^{-5}$ $m^2$

Speed of liquid inside the tube, $V_1$ = 1.5 m/min = 0.025 m/s

Speed of ejection of liquid = $V_2$

According to law of continuity, we have $A_1{V}_1$ = $A_2{V}_2$ or $V_2=\frac{A_1{V}_1}{A_2}$ = $\frac{8\times {10}^{-4}\times 0.025}{3.14\times {10}^{-5}}$

= 0.637 m/s

The weight that the soap film supports, W = 1.5 $*{10}^{-2}$ N

Length of the slider, l = 30 cm = 0.3 m

A soap film has two free surfaces, hence total length = 2l = 0.6 m

Surface tension, S= $\frac{Forces/Weight}{2l}$ = $\frac{1.5\mathrm{}*{10}^{-2}}{2*0.3}$ = 2.5 $*{10}^{-2}$ N/m

The length of the liquid film supported by the weight, l = 40 cm = 0.4 m

The weight supported by the film, W = 4.5 $*{10}^{-2}$ N

Since a liquid film has two free surfaces,

Surface tension = $\frac{W}{2l}$ = $\frac{4.5\mathrm{}*{10}^{-2}}{2*0.4}$ = 5.625 $*{10}^{-2}$ N/m

In all 3 figures, the liquid is the same, temperature is also the same. Hence the surface tension in (b) and (c) are also going to be the same, with the value of 5.625 $*{10}^{-2}$ N/m and weight supported in each case is also going to be the same, since length of the film is same.

Soap bubble radius, r = 5.0 mm = 5 $*{10}^{-3}$ m

Surface tension of the soap bubble, S = 2.50 $*{10}^{-2}$ N/m

Relative density of soap solution = 1.20, hence density of soap solution,  $?$ = 1.2 $*{10}^3$ Kg/ $m^3$

Air bubble formed at a depth, h = 40 cm = 0.4 m

1 atmospheric pressure = 1.01 $*{10}^5$ Pa

Acceleration due to gravity, g = 9.8 m/ $s^2$

We know, the excess pressure inside the soap bubble is given by the relation:

P = $\frac{4S}{r}$ = $\frac{4*2.50*{10}^{-2}}{5*{10}^{-3}}$ Pa = 20 Pa

Hence, the excess pressure inside the air bubble is given by the relation, P' = $\frac{2S}{r}$ = 10 Pa

At a depth of h, the total pressure inside the air bubble = Atmospheric pressure + h $?$ g +P'

= 1.01 $*{10}^5$ + (0.4 $*1.2*{10}^3*9.8)$ + 10 Pa = 105714 Pa = 1.06 $*{10}^5$ Pa

Base area of the given tank, A = 1.0 $m^2$

Area of the hinged door, a = 20 ${cm}^2$ = 20 $\times {10}^{-4}$ $m^2$

Density of water, ${\rho }_1$ = ${10}^3$ kg/ $m^3$

Density of acid, ${\rho }_2$ = 1.7 $\times {10}^3$ kg/ $m^3$

Height of the water column, $h_1$ = 4 m

Height of the acid column, $h_2$ = 4 m

Acceleration due to gravity, g = 9.8 m/ $s^2$

The pressure due to Water column, $P_1$ = ${\rho }_{1}g{h}_1$ = ${10}^3\times 9.8\times 4$ = 3.92 $\times {10}^4$ Pa

The pressure due to Acid column, $P_2$ = ${\rho }_{2}g{h}_2$ = 1.7 $\times {10}^3\times 9.8\times 4$ = 6.64 $\times {10}^4$ Pa

Pressure difference between two columns ΔP = $P_2$ - $P_1$ = 2.774 $\times {10}^4$ Pa

Force exerted on the small hinged door = ΔP $\times a$ = 2.774 $\times {10}^4\times$ 20 $\times {10}^{-4}$ N = 55.48 N

Atmospheric pressure,  $P_0$ = 76 cm of Hg

For figure (a)

Difference between the levels of mercury in the two limbs gives gauge pressure

Hence, gauge pressure = 20 cm of Hg

Absolute pressure = Atmospheric pressure + Gauge pressure = 76 + 20 = 96 cm of Hg

For figure (b),

Difference between the levels of mercury in the two limbs gives gauge pressure

Hence, gauge pressure = - 18 cm of Hg

Absolute pressure = Atmospheric pressure + Gauge pressure = 76 - 18 = 58 cm of Hg

When 13.6 cm of water is poured into the right limb of figure (b)

Relative density of mercury = 13.6

Hence, a column of 13.6 cm of water is equivalent to 1 cm of Mercury.

Let h be the difference between the levels of mercury in the two limbs

The pressure in the right limb is given as:

$P_g$ = Atmospheric pressure + 1 cm of Hg = 76 + 1 = 77 cm of Hg

The mercury column will rise in the left limb. Hence, pressure in the left limb

$P_l$ = 58 + h

Since $P_g$ = $P_l$ , we get h = 77 – 58 = 19 cm

Two vessels, having the same base area, will have identical force and equal pressure acting on their common base area. Since the shapes of the two vessels are different, the force exerted on the sides of the vessels has non-zero vertical components.

When these vertical components are added, the total force on one vessel comes out to be more than on the other vessel. Hence, when these vessels are filled with water to the same height, they give different readings on the weighing scale.

Gauge pressure, P = 2000 Pa

Density of whole blood,  $\rho$ = 1.06 $\times {10}^3$ kg/ $m^3$

Acceleration due to gravity, g = 9.8 m/s

Let the height of the blood container = h

Pressure on the blood container P = $\rho gh$

By equating, we get h = (2000)/ ( 1.06 $\times {10}^3\times 9.8)$ = 0.1925m

Diameter of the artery, d = 2 $\times {10}^{-3}$ m

Viscosity of the blood,  $\eta$ = 2.084 $\times {10}^{-3}$ Pa-s

Density of the blood,  $\rho$ = 1.06 $\times {10}^3$ kg/ $m^3$

Reynolds’s number for laminar flow,  $R_e$ = 2000

$V_{avg}$ = $\frac{R_e\eta }{\rho d}$ = $\frac{2000\times 2.084\mathrm{}\times {10}^{-3}}{1.06\mathrm{}\times {10}^3\times 2\mathrm{}\times {10}^{-3}}$ = 1.966 m/s

As the fluid velocity increases, the dissipative forces become more important. This is because of the rise of turbulence. Turbulent flow causes dissipative loss in a fluid.

Radius of the artery, r = 2 $\times {10}^{-3}$ m

Diameter of the artery, d = 4 $\times {10}^{-3}$ m

Viscosity of blood, $\eta$ = 2.084 $\times {10}^{-3}$ Pa-s

Density of the blood, $\rho$ = 1.06 $\times {10}^3$ kg/ $m^3$

Reynolds’s number for laminar flow, $R_e$ = 2000

The largest velocity is given by the relation: $V_{avg}$ = $\frac{R_e\eta }{\rho d}$

= $\frac{2000\times 2.084\mathrm{}\times {10}^{-3}}{1.06\mathrm{}\times {10}^3\times 4\mathrm{}\times {10}^{-3}}$ = 0.983 m/s

Flow rate is given by the relation:

R = $\pi r^2{V}_{avg}$ = 3.1416 $\times ({2\mathrm{}\times {10}^{-3})}^2\times 0.983\mathrm{}$ = 1.235 $\times {10}^{-5}$ $m^3$ /s

The area of the wing of the plane, A = 2 $\times 25m^2$ = 50 $m^2$

Speed of air over the lower wing, $V_1$ = 180 km/h = 50 m/s

Speed of air over the upper wing, $V_2$ = 234 km/h = 65 m/s

Density of air, $\rho$ = 1 kg/ $m^3$

Let us assume, pressure over the lower wing = $P_1$ and pressure over the upper wing = $P_2$

The upward force on the plane can be obtained using Bernoulli’s equation:

On lower wing= $P_1$ + $\frac{1}{2}\rho V_1^2$ , On upper wing = $P_2$ + $\frac{1}{2}\rho V_2^2$

From equilibrium of momentum

$P_1$ + $\frac{1}{2}\rho V_1^2$ = $P_2$ + $\frac{1}{2}\rho V_2^2$

( $P_1$ - $P_2)=\frac{1}{2}\rho (V_2^2-V_1^2$ )

The net upward force = ( $P_1$ - $P_2)A$ = $\frac{1}{2}\rho (V_2^2-V_1^2$ ) $\times A$

= $\frac{1}{2}1({65}^2-{50}^2$ ) $\times 50$ = 43125 N

Using the relation F = mg, we get m = F/g

= (43125/9.8) Kg = 4400.51 kg

Hence the mass of the plane is about 4400 kg.

Radius of the uncharged drop, r = 2.0 $\times {10}^{-5}$ m

Density of the uncharged drop, $\rho$ = 1.2 $\times {10}^3$ kg/ $m^3$

Viscosity of air, $\eta$ = 1.8 $\times {10}^{-5}$ Pas

Density of air ${\rho }_o$ , can be taken as zero in order to neglect the buoyancy of air

Acceleration due to gravity, g = 9.8 m/ $s^2$

Terminal velocity, v can be written as

v = $\frac{2\times r^2\times (\rho -{\rho }_o)g}{9\times \eta }$ = $\frac{2\times ({2.0\mathrm{}\times {10}^{-5})}^2\times (1.2\mathrm{}\times {10}^3-0)\times 9.8}{9\times 1.8\mathrm{}\times {10}^{-5}}$ = 0.05807 m/s = 5.8 cm/s

The viscous force on the drop is given by

F = 6 $\pi \eta rv$ = 6 $\times 3.1416\times$ 1.8 $\times {10}^{-5}\times$ 2.0 $\times {10}^{-5}\times$ 0.05807

= 3.9 $\times {10}^{-10}$ N

Radius of the narrow tube, r = 1 mm= 1 $\times {10}^{-3}$ m

Surface tension of mercury at the given temperature, s= 0.465 N/m

Density of mercury $\rho =13.6\times {10}^3$ kg/ $m^3$

Dipping depth = h

Acceleration due to gravity, g = 9.8 m/ $s^2$

Surface tension related with angle of contact and dipping depth is given by:

s = $\frac{h\rho gr}{2\cos ?\theta }$ or h = $\frac{2\mathrm{scos}?\theta }{\rho gr}$ = $\frac{2\times 0.465\times \cos ?140°}{13.6\times {10}^3\times 9.8\times 1\times {10}^{-3}}$ = -5.345 $\times {10}^{-3}$ m= -5.345 mm

Diameter of the 1st bore, $d_1$ = 3 mm = 3 $\times {10}^{-3}$ m

Radius of the first bore, $r_1$ = 1.5 mm = 1.5 $\times {10}^{-3}$ m

Diameter of the 2nd bore, $d_2$ = 6 mm = 6 $\times {10}^{-3}$ m

Radius of the 2nd bore, $r_2$ = 3.0 mm = 3 $\times {10}^{-3}$ m

Surface tension of water, s= 7.3 $\times {10}^{-2}$ N/m

Angle of contact between the bore surface and water, $\theta =0$

Density of water, $\rho =1.0\times {10}^3$ kg/ $m^3$

Acceleration due to gravity. g = 9.8 m/ $s^2$

Let $h_1$ and $h_2$ be the heights to which water rises in 1st and 2nd tubes respectively. These heights are given by the relations:

$h_1={(2scos?\theta )/\rho gr}_1$ …(i)

$h_2={(2scos?\theta )/\rho gr}_2$ …(ii)

The difference in level of water in the 2 limbs

$h_1-h_2$ = $\frac{2scos?\theta }{\rho g}$ ( $\frac{1}{r_1}$ - $\frac{1}{r_2})$

= $\frac{2\times 7.3\mathrm{}\times {10}^{-2}\times cos?0}{1.0\times {10}^3\times 9.8}$ ( $\frac{1}{1.5\mathrm{}\times {10}^{-3}}$ - $\frac{1}{3\mathrm{}\times {10}^{-3}})$ = 4.965 $\times {10}^{-3}$ m = 4.965 mm

Volume of the balloon, V = 1425 ${\mathit{\rho }}_{\mathit{H}\mathit{e}}$

Mass of the payload, m = 400 kg

Acceleration due to gravity, g = 9.8 m/ $m^3$

$s^2$ = 8000 m

$y_o$ = 0.18 kg m^–3

${\rho }_{He}$ = 1.25 kg m^–3

Density of the balloon = ${\rho }_o$

Height to which the balloon will rise = y

Density of air decreases with height and the relationship is given by:

$\rho$ = $\rho ={\rho }_o{e}^{\frac{-y}{y_o}}$ ……(i)

Differentiating equation (i), we get

$\frac{\rho }{{\rho }_o}$ $e^{\frac{-y}{y_o}}$

$-\frac{d\rho }{dy}$ , where k is the constant of proportionality

$\alpha \rho$ , height changes from 0 to y, while density changes from $\frac{d\rho }{dy}=-k\rho$ to $\frac{d\rho }{\rho }=-kdy$ . Integrating both sides between the limits, we get:

${\rho }_o$

$\rho$ = -ky

${\int }_{{\rho }_o}^{\rho }\frac{d\rho }{\rho }=-{\int }_0^{y}kdy$ = ${\left[{\log }_e?\rho \right]}_{{\rho }_o}^{\rho }$ ….(ii)

From equation (i) and (ii), we get

$\frac{\rho }{{\rho }_o}$ = $e^{-ky}$ , k = $y_0$

Substituting the value of k in equation (ii), we get

$\frac{1}{k}$

Density $\frac{1}{y_o}$ = $\rho ={\rho }_o{e}^{\frac{-y}{y_o}}\dots \dots \left(iii\right)$

= $\rho =\frac{Mass}{Volume}=\frac{Massofthepayload+Massofhelium}{Volume}$ = 0.461 kg/ $\frac{m+V{\rho }_{He}}{V}$

From equation (iii), taking log on both sides

$\frac{400+1425\times 0.18}{1425}$ = - $m^3$

y = -8000 ${\log }_e?\frac{\rho }{{\rho }_o}$ = 8000 m = 8 km

The balloon will rise to 8 km