NCERT Solutions for Class 11 Physics Chapter 10

Q.1 Explain why
(a) The blood pressure in humans is greater at the feet than at the brain
(b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km
(c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.

Ans.

(a) Pressure of a liquid is given as:

P = hρg

Here, ρ = density of liquid

We can conclude that pressure is directly proportional to height. Thus, the blood pressure at the feet is more than it is at the brain.

(b) The density of air is the maximum near the sea level and it decreases rapidly with increase in height. At a height of about 6 km, density decreases to approximately half of its value at the sea level..

(c) As force is applied on a liquid, the pressure in the liquid is transmitted equally in all directions. Therefore, hydrostatic pressure does not have a fixed direction. It is a scalar quantity.

Q.2 Explain why
(a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute.
(b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.)
(c) Surface tension of a liquid is independent of the area of the surface
(d) Water with detergent dissolved in it should have small angles of contact.
(e) A drop of liquid under no external forces is always spherical in shape

Ans.

(a) The angle formed between the tangent to the liquid surface at the point of contact and the surface inside the liquid is called as the angle of contact (θ). In the given figure, S_la, S_sa, and S_sl are the surface tensions between the liquid-air, solid-air, and solid-liquid interfaces respectively. The surface forces between the three media should be in equilibrium at the line of contact, i.e.,

\begin{array}{l}\text{cos θ =}\frac{{\text{S}}_{\text{sa}}{\text{– S}}_{\text{sl}}}{{\text{S}}_{\text{la}}}\\ {\text{The angle of contact θ, will be obtuse if S}}_{\text{sa}}\text{<}{\text{S}}_{\text{la}}\text{(as in the case of mercury on glass)}\text{.}\\ \text{This angle will be}{\text{acute if S}}_{\text{sl}}{\text{< S}}_{\text{la}}\text{(as in the case of water on glass)}\text{.}\end{array}

(b) The mercury makes an obtuse angle of contact with glass. The molecules of mercury have a strong force of attraction between themselves called cohesion and a weak force of attraction toward solids called adhesion. Therefore, they have tendency to form drops. On the contrary, the molecules of water have a weak force of cohesion and a strong force of adhesion. Therefore, they make acute angles with glass and have tendency to spread out.

(c) Surface tension is defined as the force acting per unit length at the interface between the plane of a liquid and any other surface. Surface tension does not depend on the area of the surface of the liquid; therefore, it is also independent of the area of the liquid surface.

(d) The cloth has small spaces in the form of capillaries. The liquid with detergent dissolved in it has small angles of contact (θ), because for a small θ, there is a fast capillary rise of detergent in the cloth. The rise of a liquid in a capillary tube is directly proportional to the cosine of the angle of contact (θ). If angle of contact is small, then cosine of angle of contact will be large and the penetration of the detergent water in the cloth will be fast.

(e) The liquid drop has the tendency to acquire the minimum surface area due to the presence of surface tension. Since the surface area of a sphere is the minimum for a given volume, therefore, in the absence of external forces, liquid drops always take spherical shape.

Q.3 Fill in the blanks using the word(s) from the list appended with each statement:
(a) Surface tension of liquids generally . .. with temperatures (increases /decreases)
(b) Viscosity of gases . . . with temperature, whereas viscosity of liquids . . . with temperature (increases / decreases)
(c) For solids with elastic modulus of rigidity, the shearing force is proportional to . . . , while for fluids it is proportional to . . . (shear strain / rate of shear strain)
(d) For a fluid in a steady flow, the increase in flow speed at a constriction follows
(conservation of mass / Bernoulli’s principle)
(e) For the model of a plane in a wind tunnel, turbulence occurs at a . . . speed for turbulence for an actual plane (greater / smaller)

Ans.

(a) decreases

(b) Increases; decreases

(c) Shear strain; Rate of shear strain

(d) Conservation of mass; Bernoulli’s principle

(e) Greater

Q.4 Explain why
(a) To keep a piece of paper horizontal, you should blow over, not under, it
(b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers
(c) The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection
(d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel
(e) A spinning cricket ball in air does not follow a parabolic trajectory

Ans.

(a) When we blow under a paper, the velocity of air blow increases under the paper. According to Bernoulli’s principle, atmospheric pressure decreases under the paper and the paper falls. On blowing over the paper the velocity increases above the paper. According to Bernoulli’s principle, atmospheric pressure decreases above the paper and the paper remains horizontal.

(b) From the equation of continuity:

Area × Velocity = Constant

If we try to close a tap of water with our fingers, the area of the outlet of the water jet is reduced, so velocity of water increases. Therefore, area is inversely proportional to velocity.

(c) The size of the needle of the syringe controls the velocity of the blood flowing outside

\begin{array}{l}\text{From}\text{Bernoulli’s}\text{theorem,}\\ \text{P + ρgh +}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\text{= C}\text{onstant}\end{array}

From the above equation, it can be observed that the velocity have more influence on the flow of a liquid as it occurs with power two. Therefore, the needle has a better control over the flow.

(d) From the Torricelli’s theorem, the velocity of a fluid flowing outside from a small hole is given as:

\text{v}\text{=}\sqrt{\text{2gH}}

Here, H = Height of fluid above the hole

When a fluid flows out from a small hole in a vessel, it acquires a large velocity. Therefore, its momentum increases to a large amount in the outward direction.

Since no external forces are acting on the system, therefore, according to the law of conservation of momentum, the vessel experiences a recoil momentum in the backward direction. As a result, the vessel experiences a backward thrust.

(e) The spinning cricket ball has two types of simultaneous motions. These are rotatory motion and linear motion. These motions oppose the effect of each other. As a result, the velocity of air flowing below the ball decreases. Therefore, the pressure on the lower side of the ball becomes more than that on the upper side. Because of it, the spinning ball experiences an upward force. Therefore, the ball takes a curved path and it deviates from its parabolic path.

Q.5 A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?

Ans.

\begin{array}{l}\text{Here,}\text{mass of the girl, m = 50 kg}\\ \text{Diameter of the}\text{circular}\text{heel, d}\text{=}\text{1 cm = 0}\text{.01 m}\\ \therefore \text{Radius of the}\text{circular}\text{heel,r}\text{=}\frac{\text{d}}{\text{2}}\text{= 0}\text{.005}\text{m}\\ \text{Area of the}\text{circular}\text{heel}\text{=}\pi {\text{r}}^{\text{2}}\text{=}\pi {\left(\text{0}\text{.005}\right)}^{\text{2}}\\ \text{a =7}\text{.85}\text{×}{\text{10}}^{\text{–5}}{\text{m}}^{\text{2}}\\ \text{Force exerted by the heel on the horizontal}\text{floor}\text{is}\text{given}\text{as,}\\ \text{F = mg}\\ \therefore \text{F = 50}\text{×}\text{9}\text{.8}\text{=}\text{490 N}\\ \text{Pressure exerted by}\text{circular heel on}\text{the}\text{horizontal}\text{floor,}\\ \text{P}\text{=}\frac{\text{Force}}{\text{Area}}\\ \text{P =}\frac{\text{490}}{\text{7}\text{.85}\text{×}{\text{10}}^{\text{–5}}}\text{=}\text{6}\text{.24}\text{×}{\text{10}}^{\text{6}}{\text{N m}}^{\text{–2}}\\ \therefore \text{Pressure exerted by}\text{heel on the horizontal floor =}\text{6}\text{.24}\text{×}{\text{10}}^{\text{6}}{\text{Nm}}^{\text{–2}}\end{array}

Q.6 Torricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m^–3. Determine the height of the wine column for normal atmospheric pressure.

Ans.

\begin{array}{l}\text{Here,}{\text{density of mercury,ρ}}_{\text{m}}\text{=}\text{13}\text{.6}\text{×}{\text{10}}^{\text{3}}{\text{kgm}}^{\text{-3}}\\ {\text{Height of mercury column,h}}_{\text{m}}\text{=}\text{0}\text{.76 m}\\ \text{Density of French wine,}{\text{ρ}}_{\text{w}}{\text{= 984 kgm}}^{\text{-3}}\\ \text{Let}\text{height of French wine column}\text{=}{\text{h}}_{\text{w}}\\ \text{Acceleration due to gravity, g}\text{=}\text{9}{\text{.8 ms}}^{\text{-2}}\\ \text{As}\text{the pressure in the two}\text{columns is equal,}\\ \text{Pressure in mercury column}\text{=}\text{Pressure in French wine column}\\ {\text{ρ}}_{\text{m}}{\text{h}}_{\text{m}}\text{g}\text{=}{\text{ρ}}_{\text{w}}{\text{h}}_{\text{w}}\text{g}\\ \therefore {\text{h}}_{\text{w}}\text{=}\frac{{\text{ρ}}_{\text{m}}{\text{h}}_{\text{m}}}{{\text{ρ}}_{\text{w}}}\\ {\text{h}}_{\text{w}}\text{=}\frac{\text{13}{\text{.6 × 10}}^{\text{3}}{\text{kgm}}^{\text{-3}}\text{× 0}\text{.76 m}}{{\text{984 kgm}}^{\text{-3}}}\text{= 10}\text{.5 m}\\ \therefore \text{Height of the French wine column for normal}\\ \text{atmospheric pressure}\text{=}\text{10}\text{.5 m}\end{array}

Q.7 A vertical off-shore structure is built to withstand a maximum stress of 10⁹ Pa. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly 3 km, and ignore ocean currents.

Ans.

Here, Maximum stress for the off-shore structure, P = 10⁹ Pa

Depth of ocean, d = 3 km = 3 × 10³ m

Density of the sea water, ρ = 10³ kgm^-3

Acceleration due to gravity, g = 9.8 ms^-2

Hydrostatic pressure exerted due to sea water at the bottom of sea,

d = ρdg = 3 × 10³ m × 10³ kgm^-3 × 9.8ms^-2

d = 2.94 × 10⁷ Pa

The maximum stress which the off-shore structure can withstand is more than the pressure of the sea water (2.94 × 10⁷ Pa). Hence, the pressure exerted by the ocean is less than the pressure that the off shore structure can withstand. Thus, the off shore structure will be suitable for putting up on top of the oil well in the ocean.

Q.8 A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm². What maximum pressure would the smaller piston have to bear?

Ans.

\begin{array}{l}\text{Here,}\text{maximum mass of the car ,m}\text{=}\text{3000 kg}\\ \text{Cross-sectional}\text{area of the load-carrying piston,}\\ {\text{A = 425 cm}}^{\text{2}}{\text{= 425 × 10}}^{\text{–4}}{\text{m}}^{\text{2}}\\ \text{Maximum force exerted by the load}\text{is}\text{given}\text{as,}\\ \text{F = mg}\\ \therefore \text{F = 3000 kg × 9}{\text{.8 ms}}^{\text{-2}}\text{= 29400 N}\\ \text{Maximum pressure exerted on the piston}\text{carrying}\text{load,}\\ \text{P =}\frac{\text{F}}{\text{A}}\text{=}\frac{\text{29400 N}}{{\text{425 × 10}}^{\text{–4}}{\text{m}}^{\text{2}}}\text{= 6}{\text{.917 × 10}}^{\text{5}}\text{Pa}\\ \text{Pressure is transmitted in all directions equally}\text{in the liquids}\text{.}\\ \therefore \text{Maximum pressure that the smaller piston can bear = 6}{\text{.917×10}}^{\text{5}}\text{Pa}\end{array}

Q.9 A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit?

Ans.

\begin{array}{l}\text{Here,}\text{height of the}\text{water}\text{column}\text{in}\text{one}\text{arm}\text{of}\text{U}\text{tube,}\\ {\text{h}}_{\text{1}}\text{=}\text{10 cm}\text{=}\text{0}\text{.1 m}\\ \text{Height of the}\text{water}\text{column}\text{in}\text{other}\text{arm}\text{of}\text{U}\text{tube,}\\ {\text{h}}_{\text{2}}\text{=}\text{12}\text{.5 cm}\text{=}\text{0}\text{.125 m}\\ {\text{Density of water,ρ}}_{\text{1}}\text{= 1}\text{g}{\text{cm}}^{\text{-3}}\\ {\text{Let density of spirit = ρ}}_{\text{2}}\\ \text{Since}\text{the}\text{mercury}\text{columns}\text{in}\text{the}\text{two}\text{arms}\text{of}\text{U}\text{tube are}\text{in}\text{level,}\\ \text{therefore,}\text{the}\text{pressure}\text{exerted}\text{by}\text{both of}\text{them}\text{is}\text{equal}\text{.}\\ \therefore {\text{h}}_{\text{1}}{\text{ρ}}_{\text{1}}{\text{g = h}}_{\text{2}}{\text{ρ}}_{\text{2}}\text{g}\\ \therefore \frac{{\text{ρ}}_{\text{1}}}{{\text{ρ}}_{\text{2}}}\text{=}\frac{{\text{h}}_{\text{2}}}{{\text{h}}_{\text{1}}}\text{=}\frac{\text{10}}{\text{12}\text{.5}}\text{= 0}\text{.8}\\ \therefore \text{Relative}\text{density}\text{of}\text{Spirit}\text{=}\text{0}\text{.8}\end{array}

Q.10 In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Specific gravity of mercury = 13.6)

Ans.

\begin{array}{l}{\text{Here,height of water column, h}}_{\text{w}}\text{= 10 cm + 15 cm = 25 cm}\\ {\text{Height of spirit column, h}}_{\text{s}}\text{= 12}\text{.5 cm + 15 cm = 27}\text{.5 cm}\\ {\text{Density of water, ρ}}_{\text{w}}{\text{= 1 gcm}}^{\text{–3}}\\ {\text{Density of spirit, ρ}}_{\text{s}}\text{= 0}{\text{.8 gcm}}^{\text{–3}}\\ \text{Density of mercury,}\text{ρ = 13}{\text{.6 g cm}}^{\text{–3}}\\ \text{Let difference between the levels of mercury}\text{in}\text{the}\text{two}\text{arms}\text{=}\text{h}\\ \text{Pressure}\text{exerted}\text{by}\text{h}\text{cm}\text{of}\text{mercury}\text{column}\text{=}\text{The}\text{difference}\text{in}\text{pressure}\\ \text{exerted}\text{by}\text{water}\text{and}\text{spirit}\\ \therefore {\text{hρg = h}}_{\text{w}}{\text{ρ}}_{\text{w}}{\text{g – h}}_{\text{s}}{\text{ρ}}_{\text{s}}\text{g}\\ \text{h}\text{×}\text{13}\text{.6 g}\text{=}\text{g}\left({\text{h}}_{\text{w}}{\text{ρ}}_{\text{w}}{\text{– h}}_{\text{s}}{\text{ρ}}_{\text{s}}\right)\\ \text{h}\text{×}\text{13}\text{.6 g = g}\left(\text{25 × 1 – 27}\text{.5 × 0}\text{.8}\right)\approx \text{= 3g}\\ \therefore \text{h =}\frac{\text{3}}{\text{13}\text{.6}}\text{= 0}\text{.220588}\approx \text{0}\text{.221}\text{cm}\\ \therefore \text{Difference between the levels of mercury in}\text{the two}\text{arms = 0}\text{.221 cm}\text{.}\end{array}

Q.11 Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.

Ans.

No, Bernoulli’s equation can only be applied to streamline flow, Rapid in a river is a turbulent flow.

Q.12 Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation? Explain.

Ans.

It does not matter if one uses gauge instead of absolute pressure while using Bernoulli’s equation. The two points where bernoulli’s equation will be applied will have different atmospheric pressures.

Q.13 Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10^–3 kg s^–1, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 × 10³ kg m^–3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of lamina r flow in the tube is correct].

Ans.

\begin{array}{l}\text{Here, length of tube, l = 1}\text{.5 m}\\ \text{Radius of tube, r}\text{=}\text{1 cm}\text{=}\text{0}\text{.01 m}\\ \text{Diameter of tube,}\text{d}\text{=}\text{2r}\text{=}\text{0}\text{.02 m}\\ \text{Mass}\text{of}\text{glycerine flowing}\text{per}\text{second,}\text{M}\text{=}\text{4}\text{.0}\text{×}{\text{10}}^{\text{–3}}{\text{kgs}}^{\text{–1}}\\ \text{Density of glycerine, ρ}\text{=}\text{1}\text{.3}\text{×}{\text{10}}^{\text{3}}{\text{kgm}}^{\text{–3}}\\ \text{Viscosity of glycerine,}\text{η}\text{=}\text{0}\text{.83 Pa s}\\ \text{Volume of glycerine flowing per sec,}\text{V}\text{=}\frac{\text{M}}{\text{ρ}}\text{=}\frac{\text{4}\text{.0}\text{×}{\text{10}}^{\text{–3}}\text{kg}}{\text{1}\text{.3}\text{×}{\text{10}}^{\text{3}}{\text{kgm}}^{\text{–3}}}\\ \text{V}\text{=}\text{3}\text{.08}\text{×}{\text{10}}^{\text{–6}}{\text{m}}^{\text{3}}\\ \text{Let pressure difference between the two ends of the}\text{tube}\text{=}\text{p}\\ \text{Using}\text{Poiseuille’s formula, we have}\\ \text{V =}\frac{\pi {\text{pr}}^{\text{4}}}{\text{8ηl}}\\ \therefore \text{p =}\frac{\text{V8ηl}}{\pi {\text{r}}^{\text{4}}}\text{=}\frac{\text{3}{\text{.08 ×10}}^{\text{–6}}{\text{m}}^3\text{× 8 × 0}\text{.83 Pa s × 1}\text{.5 m}}{\pi \text{×}{\left(\text{0}\text{.01 m}\right)}^{\text{4}}}\\ \text{p = 9}{\text{.8 × 10}}^{\text{2}}\text{Pa}\\ \therefore \text{Pressure difference between the two ends of the}\\ \text{tube =}\text{9}{\text{.8 × 10}}^{\text{2}}\text{Pa}\\ \text{Reynold’s number is given by the relation, R =}\frac{\text{4ρV}}{\text{πdη}}\\ \text{R =}\frac{\text{4×1}\text{.3}\text{×}{\text{10}}^{\text{3}}{\text{kgm}}^{\text{–3}}\text{× 3}\text{.08}\text{×}{\text{10}}^{\text{–6}}{\text{m}}^{\text{3}}}{\text{3}\text{.14 × 0}\text{.02 m × 0}\text{.83 Pa s}}\text{=0}\text{.3}\\ \text{Thus, the flow is laminar}\text{.}\end{array}

Q.14 In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s^–1 and 63 m s^-1 respectively. What is the lift on the wing if its area is 2.5 m²? Take the density of air to be 1.3 kg m^–3.

Ans.

\begin{array}{l}\text{Here,}{\text{speed of wind on upper surface of the wing, v}}_{\text{1}}=\text{7}0{\text{ms}}^{\text{-1}}\\ \text{Speed of wind on lower surface of the wing},{\text{v}}_{\text{2}}={\text{63 ms}}^{\text{-1}}\\ \text{Area of wing},\text{A}=\text{2}.{\text{5 m}}^{\text{2}}\\ \text{Density of air},\text{ρ}=\text{1}.{\text{3 kg m}}^{–\text{3}}\\ \text{Let}{\text{pressure on the upper surface of the wing = P}}_{\text{1}}\\ \text{Let}{\text{pressure on the lower surface of the wing = P}}_2\\ \text{From Bernoulli’s theorem,}\text{we}\text{have,}\\ {\text{P}}_{\text{1}}+\frac{1}{2}{\text{ρv}}_{\text{1}}{}^2={\text{P}}_2+\frac{1}{2}{\text{ρv}}_2{}^2\\ \therefore {\text{P}}_2-{\text{P}}_{\text{1}}=\frac{1}{2}\text{ρ}\left({\text{v}}_{\text{1}}{}^2-{\text{v}}_2{}^2\right)\\ \text{The difference}\text{of pressure}\text{between the}\text{upper and lower}\\ \text{surfaces of the wing gives the}\text{lift to the plane}\text{.}\\ \text{Lift on the aeroplane}\text{=}P\text{ressure}\text{difference}\text{×}\text{area}\text{of}\\ \text{wings}\text{=}\left({\text{P}}_2-{\text{P}}_{\text{1}}\right)A\\ =\frac{1}{2}\text{ρ}\left({\text{v}}_{\text{1}}{}^2-{\text{v}}_2{}^2\right)A\\ =\frac{1}{2}\times 1.3{\text{kg m}}^{–\text{3}}\left[{\left(70{\text{ms}}^{\text{-1}}\right)}^2-{\left(63{\text{ms}}^{\text{-1}}\right)}^2\right]\times 2.5{\text{m}}^{\text{2}}\\ \text{= 1512}.\text{87}=\text{1}.\text{51}\times \text{1}{0}^{\text{3}}\text{N}\\ \therefore \text{Lift on the wing of the aeroplane}=\text{1}.\text{51}\times \text{1}{0}^{\text{3}}\text{N}\end{array}

Q.15 Figures 10.23 (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?

Ans.

Let us take the case given in figure (b).

From the equation of continuity, we have

av = constant

When the cross-sectional area in the middle portion of the venturimeter is small, the speed of the flow of liquid is more through this part.

From Bernoulli’s theorem, if speed is more, then pressure will be less. Since pressure is directly proportional to the height; therefore the water level in pipe 2 is less.

Hence, figure (a) is incorrect.

Q.16 The cylindrical tube of a spray pump has a cross-section of 8.0 cm² one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min^–1, what is the speed of ejection of the liquid through the holes?

Ans.

\begin{array}{l}{\text{Here, cross-sectional area of the spray pump, a}}_{\text{1}}{\text{= 8 cm}}^{\text{2}}{\text{= 8 × 10}}^{\text{–4}}{\text{m}}^{\text{2}}\\ \text{Number of holes}\text{in}\text{the}\text{cylinderical}\text{tube, N = 40}\\ {\text{Diameter of hole, d = 1 mm = 1 × 10}}^{\text{–3}}\text{m}\\ \text{Radius of hole, r =}\frac{\text{d}}{\text{2}}\text{= 0}{\text{.5 × 10}}^{\text{–3}}\text{m}\\ \text{Cross-sectional area of each hole, a =}\pi {\text{r}}^{\text{2}}\\ \text{a =}\pi {\left(\text{0}{\text{.5× 10}}^{\text{–3}}\right)}^{\text{2}}{\text{m}}^{\text{2}}\\ \text{Total area of}\text{cross-section}\text{of 40 holes,}\\ {\text{a}}_{\text{2}}\text{= N × a = 40 ×}\pi {\left(\text{0}{\text{.5× 10}}^{\text{–3}}\right)}^{\text{2}}{\text{m}}^{\text{2}}\\ {\text{a}}_{\text{2}}\text{= 31}{\text{.41 × 10}}^{\text{–6}}{\text{m}}^{\text{2}}\\ \text{Speed of the}{\text{liquid inside the tube,v}}_{\text{1}}\text{=}\text{1}\text{.5 m}{\text{min}}^{\text{-1}}\\ {\text{v}}_{\text{1}}\text{= 0}{\text{.025 ms}}^{\text{-1}}\\ \text{Let}\text{speed of discharge of the}\text{liquid through the holes}\text{=}{\text{v}}_{\text{2}}\\ \text{From the equation of continuity, we have}\\ {\text{a}}_{\text{1}}{\text{v}}_{\text{1}}{\text{= a}}_{\text{2}}{\text{v}}_{\text{2}}\\ \therefore {\text{v}}_{\text{2}}\text{= a =}\frac{{\text{8 × 10}}^{\text{–4}}{\text{m}}^{\text{2}}\text{× 0}{\text{.025 ms}}^{\text{-1}}}{\text{31}{\text{.41 × 10}}^{\text{–6}}{\text{m}}^{\text{2}}}\\ {\text{v}}_{\text{2}}\text{= 0}{\text{.6367 ms}}^{\text{-1}}\\ \therefore \text{Speed of ejection of the liquid through the}\text{holes}\text{=}\text{0}{\text{.633 ms}}^{\text{-1}}\end{array}

Q.17 A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 × 10^–2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?

Ans.

\begin{array}{l}\text{Here,weight supported}\text{by the soap film, W = 1}\text{.5}\text{×}{\text{10}}^{\text{–2}}\text{N}\\ \text{Length of slider,}\text{l =}\text{30 cm}\text{=}\text{0}\text{.3 m}\\ \text{As}\text{a}\text{soap film has two free surfaces,}\\ \therefore \text{Total length to}\text{be}\text{supported}\text{=}\text{2l}\text{=}\text{2}\text{×}\text{0}\text{.3}\text{=}\text{0}\text{.6 m}\\ \text{Surface tension}\text{is}\text{given}\text{as:}\\ \text{S =}\frac{\text{Force}}{\text{Total}\text{length}}\text{=}\frac{\text{F}}{\text{2l}}\text{=}\frac{\text{1}{\text{.5 × 10}}^{\text{–2}}\text{N}}{\text{0}\text{.6 m}}\\ \text{S = 2}\text{.5}\text{×}{\text{10}}^{\text{–2}}{\text{Nm}}^{\text{-1}}\\ \text{Thus, surface tension}\text{of}\text{the}\text{film}\text{=}\text{2}{\text{.5×10}}^{\text{–2}}{\text{Nm}}^{\text{-1}}\end{array}

Q.18 Figure 10.24 (a) shows a thin liquid film supporting a small weight = 4.5 × 10^–2 N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c)? Explain your answer physically.

Ans.

\begin{array}{l}\text{For case}\left(\text{a}\right)\text{:}\\ \text{Length of the liquid film,l = 40 cm = 0}\text{.4 cm}\\ \text{Weight supported by the liquid}\text{film,}\text{W = 4}{\text{.5 × 10}}^{\text{–2}}\text{N}\\ \text{As}\text{a liquid film has two free surfaces,}\\ \therefore \text{Surface tension}\text{of}\text{the}\text{liquid}\text{=}\frac{\text{W}}{\text{2l}}\\ \frac{\text{W}}{\text{2l}}\text{=}\frac{\text{4}{\text{.5 × 10}}^{\text{–2}}\text{N}}{\text{2 × 0}\text{.4 cm}}\text{= 5}{\text{.625 × 10}}^{\text{–2}}{\text{Nm}}^{\text{-1}}\\ \text{For all the three figures, the liquid film}\text{of}\text{same}\text{length}\text{is}\\ \text{used}\text{and}\text{the}\text{temperature is also the same for each case}\text{.}\\ \text{Therefore, the}\text{weight supported in all}\text{the}\text{three}\text{figures is}\\ \text{exactly same}\text{.}\\ \text{As,}\text{length of film in each}\text{case}\text{=}\text{40 cm}\\ \therefore \text{Weight supported in each case}\text{=}\text{4}\text{.5}\text{×}{\text{10}}^{\text{–2}}\text{N}\end{array}

Q.19 What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65 × 10^–1 N m^-1. The atmospheric pressure is 1.01 × 10⁵ Pa. Also give the excess pressure inside the drop.

Ans.

\begin{array}{l}\text{Here,}\text{radius of mercury drop, r}\text{=}\text{3}\text{.00 mm}\text{=}\text{3}\text{×}{\text{10}}^{\text{–3}}\text{m}\\ \text{Surface tension of mercury}\text{at}\text{the}\text{given}\text{temperature, S = 4}{\text{.65 × 10}}^{\text{–1}}{\text{Nm}}^{\text{–1}}\\ {\text{Atmospheric pressure,P}}_{\text{0}}\text{=}\text{1}\text{.01}\text{×}{\text{10}}^{\text{5}}\text{Pa}\\ \text{Excess pressure inside mercury}\text{=}\frac{\text{2S}}{\text{r}}\\ \frac{\text{2S}}{\text{r}}\text{=}\frac{\text{2×4}{\text{.65 × 10}}^{\text{–1}}{\text{Nm}}^{\text{–1}}}{{\text{3 × 10}}^{\text{–3}}m}\text{=310}\text{Pa}\\ \therefore \text{Excess pressure inside mercury}\text{drop}\text{=}\text{310}\text{Pa}\\ \text{Total pressure inside the mercury drop =}\text{Atmospheric pressure}\text{+}\text{Excess pressure inside}\\ \text{mercury}\text{=}{\text{P}}_{\text{0}}\text{+}\frac{\text{2S}}{\text{r}}\\ \text{= 1}{\text{.01 × 10}}^{\text{5}}\text{Pa + 310}\text{Pa}\\ \text{= 1}{\text{.0131 × 10}}^{\text{5}}\text{Pa}\\ \text{= 1}{\text{.01 ×10}}^{\text{5}}\text{Pa}\\ \text{Thus, total pressure inside the mercury drop = 1}{\text{.01×10}}^{\text{5}}\text{Pa}\end{array}

Q.20 What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20°C) is 2.50 ×10^–2 N m^–1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 × 10⁵ Pa).

Ans.

\begin{array}{l}\text{Here, radius}\text{of}\text{Soap bubble,}\text{r = 5}{\text{.00 mm = 5 × 10}}^{\text{–3}}\text{m}\\ \text{Surface tension of soap solution,S = 2}{\text{.50 × 10}}^{\text{–2}}{\text{Nm}}^{\text{-1}}\\ \text{Relative density of soap solution}\text{=}\text{1}\text{.20}\\ \therefore \text{Density of the soap solution,}\text{ρ}\text{=}\text{1}\text{.2}\text{×}{\text{10}}^{\text{3}}{\text{kgm}}^{\text{-3}}\\ \text{Air bubble formed at the depth,}\text{h}\text{=}\text{40 cm}\text{=}\text{0}\text{.4 m}\\ \text{Radius of air bubble,r}\text{=}\text{5 mm}\text{=}\text{5}\text{×}{\text{10}}^{\text{–3}}\text{m}\\ \text{Atmospheric pressure}\text{=}\text{1}\text{.01}\text{×}{\text{10}}^{\text{5}}\text{Pa}\\ \text{Acceleration due to gravity, g}\text{=}\text{9}{\text{.8 ms}}^{\text{-2}}\\ \text{The}\text{excess pressure inside the soap bubble}\text{is}\text{given}\text{as,}\text{P}\text{=}\frac{\text{4S}}{\text{r}}\\ \text{P}\text{=}\frac{\text{4}\text{×}\text{2}\text{.50}\text{×}{\text{10}}^{\text{–2}}{\text{Nm}}^{\text{-1}}}{\text{5}\text{×}{\text{10}}^{\text{–3}}\text{m}}\text{=}\text{20}\text{Pa}\\ \therefore \text{Excess pressure inside the soap bubble}\text{is 20 Pa}\text{.}\\ \text{The}\text{excess pressure inside the air bubble}\text{is}\text{given}\text{as,}\\ \text{P’ =}\frac{\text{2S}}{\text{r}}\text{=}\frac{\text{2}\text{×}\text{2}\text{.50}\text{×}{\text{10}}^{\text{–2}}}{\text{5}\text{×}{\text{10}}^{\text{–3}}}\text{=}\text{10}\text{Pa}\\ \therefore \text{Excess pressure inside the air bubble is}\text{10 Pa}\text{.}\\ \text{Total pressure inside the air bubble}\text{at the depth 0}\text{.4 m}\\ \text{=}\text{Atmospheric pressure}\text{+}\text{hρg}\text{+}\text{P’}\\ \text{= 1}{\text{.01 × 10}}^{\text{5}}\text{Pa + 0}\text{.4 m × 1}{\text{.2 × 10}}^{\text{3}}{\text{kgm}}^{\text{-3}}\text{× 9}{\text{.8 ms}}^{\text{-2}}\text{+ 10 Pa}\\ \text{= 1}{\text{.057 ×10}}^{\text{5}}\text{Pa}\\ \text{Thus, the pressure inside the air bubble is}\text{1}{\text{.057×10}}^{\text{5}}\text{Pa}\\ \approx \text{1}{\text{.06×10}}^{\text{5}}\text{Pa}\end{array}

Q.21 A tank with a square base of area 1.0 m² is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm². The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. compute the force necessary to keep the door close.

Ans.

\begin{array}{l}\text{Here, base area of given tank, A}\text{=}\text{1}{\text{.0 m}}^{\text{2}}\\ \text{Area of hinged door, a}\text{=}{\text{20 cm}}^{\text{2}}\text{=}\text{20}\text{×}{\text{10}}^{\text{-4}}{\text{m}}^{\text{2}}\\ \text{Density of water,}{\text{ρ}}_{\text{w}}\text{=}{\text{10}}^{\text{3}}{\text{kgm}}^{\text{-3}}\\ \text{Density of acid,}{\text{ρ}}_{\text{a}}\text{=}\text{1}\text{.7}\text{×}{\text{10}}^{\text{3}}{\text{kgm}}^{\text{-3}}\\ \text{Height of water column,}{\text{h}}_{\text{w}}\text{=}\text{4 m}\\ \text{Height of acid column,}{\text{h}}_{\text{a}}\text{=}\text{4 m}\\ \text{Acceleration due to gravity, g}\text{=}\text{9}\text{.8}{\text{ms}}^{\text{-2}}\\ \text{Pressure due to water is}\text{given}\text{by,}{\text{P}}_{\text{w}}\text{=}{\text{h}}_{\text{w}}{\text{ρ}}_{\text{w}}\text{g}\\ {\text{P}}_{\text{w}}{\text{= 4 m × 10}}^{\text{3}}{\text{kgm}}^{\text{-3}}\text{× 9}{\text{.8 ms}}^{\text{-2}}\text{= 3}{\text{.92 × 10}}^{\text{4}}\text{Pa}\\ \text{Pressure due to acid is given by,}{\text{P}}_{\text{a}}{\text{= h}}_{\text{a}}{\text{ρ}}_{\text{a}}\text{g}\\ {\text{P}}_{\text{a}}\text{= 4 m × 1}{\text{.7 × 10}}^{\text{3}}{\text{kgm}}^{\text{-3}}\text{× 9}{\text{.8ms}}^{\text{-2}}\\ {\text{P}}_{\text{a}}\text{= 6}{\text{.664×10}}^{\text{4}}\text{Pa}\\ \text{The}\text{pressure difference between water and acid}\\ \text{columns}\text{is}\text{given}\text{as,}\\ {\text{ΔP = P}}_{\text{a}}{\text{– P}}_{\text{w}}\\ \text{ΔP = 6}{\text{.664 × 10}}^{\text{4}}\text{Pa – 3}{\text{.92×10}}^{\text{4}}\text{Pa}\\ \text{= 2}{\text{.744 ×10}}^{\text{4}}\text{Pa}\\ \therefore \text{Force exerted on the door}\text{=}\text{ΔP × a}\\ \text{F = 2}{\text{.744 × 10}}^{\text{4}}{\text{Pa × 20 × 10}}^{\text{–4}}{\text{m}}^{\text{2}}\text{=}\text{54}\text{.88 N}\\ \therefore \text{Force necessary to keep the door closed = 54}\text{.88 N}\end{array}

Q.22 A manometer reads the pressure of a gas in an enclosure as shown in Fig. 10.25 (a) When a pump removes some of the gas, the manometer reads as in Fig. 10.25 (b)

The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury.
(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm of mercury.
(b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury) are poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).

Ans.

\begin{array}{l}{\text{Here, atmospheric pressure, P}}_{\text{0}}\text{= 76 cm of Hg}\\ \left(\text{a}\right)\text{For figure}\left(\text{a}\right)\\ \text{The difference between the levels of mercury in the two}\\ \text{Limbs give the gauge pressure}\text{.}\\ \therefore \text{Gauge pressure = 20 cm of Hg}\\ \text{Absolute pressure}\text{=}\text{Atmospheric pressure}\text{+}\text{Gauge pressure}\\ \text{=}\text{76}\text{+}\text{20}\text{=}\text{96 cm of Hg}\\ \text{For figure}\left(\text{b}\right)\\ \text{The}\text{difference between the levels of mercury in the}\\ \text{two limbs = –18 cm}\\ \therefore \text{Gauge pressure = –18 cm of Hg}\\ \text{Absolute pressure}\text{=}\text{Atmospheric pressure}\text{+}\text{Gauge}\text{pressure}\\ \text{= 76 cm – 18 cm = 58 cm}\\ \left(\text{b}\right)\text{Here,}\text{13}\text{.6 cm of water is poured into the right limb}\text{of figure}\left(\text{b}\right)\\ \text{Relative density of mercury}\text{=}\text{13}\text{.6}\\ \therefore \text{A column of 13}\text{.6 cm of water is equivalent to}\frac{\text{13}\text{.6}}{\text{13}\text{.6}}\text{=}\text{1}\text{cm of mercury}\\ \text{Let difference between the levels of mercury in the two}\text{limbs}\text{=}\text{h}\\ \text{Pressure in the right limb,}{\text{P}}_{\text{R}}\text{= Atmospheric pressure + 1 cm of Hg = 76 + 1 = 77 cm of Hg}\left(\text{i}\right)\\ \text{Mercury}\text{column}\text{will}\text{rise}\text{in}\text{the}\text{left}\text{limb}\text{.}\\ \therefore {\text{Pressure in the left limb, P}}_{\text{L}}\text{= 58 + h}\text{(ii)}\\ \text{Equating equations}\left(\text{i}\right)\text{and}\left(\text{ii}\right)\text{, we get,}\\ \text{77 = 58 + h}\\ \therefore \text{h = 19 cm}\\ \text{Thus, difference between the levels of mercury in}\text{the two}\text{limbs}\text{= 19 cm}\end{array}

Q.23 Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill up to a particular common height. Is the force exerted by the water on the base of the vessel the same in the two cases? If so, why do the vessels filled with water to that same height give different readings on a weighing scale?

Ans.

Two vessels with same base area will have same force and equal pressure acting on their common base area. The force exerted on the sides of the vessels has non-zero vertical components as their shapes are different. On adding these vertical components the total force on one vessel comes out to be more than that on the other vessel. Thus, the two vessels filled with water to the same height, give different readings on a weighing scale.

Q.24 During blood transfusion the needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein? [Use the density of whole blood from Table 10.1].

Ans.

\begin{array}{l}\text{Here,}\text{gauge pressure,}\text{P}\text{=}\text{2000 Pa}\\ \text{Density of whole blood, ρ}\text{=}\text{1}\text{.06}\text{×}{\text{10}}^{\text{3}}{\text{kgm}}^{\text{–3}}\\ \text{Acceleration due to gravity, g}\text{=}\text{9}{\text{.8 ms}}^{\text{-2}}\\ \text{Let}\text{height of the blood container = h}\\ \text{Pressure of the blood container}\text{is}\text{given}\text{as:}\\ \text{P}\text{=}\text{hρg}\\ \therefore \text{h}\text{=}\frac{\text{P}}{\text{ρg}}\text{=}\frac{\text{2000}}{\text{1}{\text{.06 × 10}}^{\text{3}}{\text{kgm}}^{\text{–3}}\text{× 9}{\text{.8 ms}}^{\text{-2}}}\text{=}\text{0}\text{.1925}\text{m}\\ \text{Blood may enter the vein if the height}\text{of}\text{the}\text{blood}\\ \text{container is slightly}\text{> 0}\text{.1925 m, i}\text{.e}\text{. 0}\text{.2 m}\text{.}\end{array}

Q.25 In deriving Bernoulli’s equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy. (a) What is the largest average velocity of blood flow in an artery of diameter 2 × 10^–3 m if the flow must remain laminar? (b) Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively.

Ans.

\begin{array}{l}\left(\text{a}\right)\text{Here,}\text{diameter of the}\text{artery, d}\text{=}\text{2}\text{×}{\text{10}}^{\text{–3}}\text{m}\\ \text{Viscosity of the}\text{blood,}\text{η}\text{=}\text{2}\text{.084}\text{×}{\text{10}}^{\text{–3}}\text{Pa s}\\ \text{Density of the}\text{blood,ρ}\text{=}\text{1}\text{.06}\text{×}{\text{10}}^{\text{3}}{\text{kgm}}^{\text{-3}}\\ {\text{Reynolds’ number for laminar flow,N}}_{\text{R}}\text{=}\text{2000}\\ \text{The largest average velocity of blood}\text{is}\text{given}\text{by}\text{the}\text{relation:}\\ {\text{V}}_{\text{avg}}\text{=}\frac{{\text{N}}_{\text{R}}\text{η}}{\text{ρd}}\\ {\text{V}}_{\text{avg}}\text{=}\frac{\text{2000 × 2}{\text{.084 × 10}}^{\text{–3}}\text{Pa s}}{\text{1}{\text{.06 × 10}}^{\text{3}}{\text{kgm}}^{\text{-3}}\text{×}{\text{2 × 10}}^{\text{–3}}\text{Pa s}}\text{=}\text{1}\text{.966}{\text{ms}}^{\text{-1}}\\ \therefore \text{The largest average velocity of blood = 1}{\text{.966 ms}}^{\text{-1}}\\ \\ \text{(b) When the velocity of the liquid increases, the dissipative forces}\\ \text{become more important}\text{. This is due to the rise of turbulence}\text{. The}\\ \text{dissipative loss in the fluid is due to turbulent flow}\text{.}\end{array}

Q.26 (a) What is the largest average velocity of blood flow in an artery of radius 2 × 10^–3 m if the flow must remain laminar? (b) What is the corresponding flow rate? (Take viscosity of blood to be 2.084 × 10^–3 Pa s).

Ans.

\begin{array}{l}\left(\text{a}\right)\text{Here,radius of the artery, r}\text{=}\text{2}\text{×}{\text{10}}^{\text{–3}}\text{m}\\ \text{Diameter of the artery, d}\text{=}\text{2r}\text{=}\text{2}\text{×}\text{2}\text{×}{\text{10}}^{\text{–3}}\text{m}\\ \text{d =}\text{4}\text{×}{\text{10}}^{\text{–3}}\text{m}\\ \text{Viscosity of the}\text{blood,}\text{η}\text{=}\text{2}{\text{.084 × 10}}^{\text{-3}}\text{Pa}\text{s}\\ \text{Density of the}\text{blood,ρ = 1}{\text{.06 × 10}}^{\text{3}}{\text{kgm}}^{\text{-3}}\\ \text{Reynolds’ number for the}{\text{laminar flow,N}}_{\text{R}}\text{= 2000}\\ \text{The largest average velocity of blood is given}\text{as:}\\ {\text{V}}_{\text{avg}}\text{=}\frac{{\text{N}}_{\text{R}}\text{η}}{\text{ρd}}\text{=}\frac{\text{2000 × 2}{\text{.084 × 10}}^{\text{-3}}\text{Pa}\text{s}}{\text{1}{\text{.06 × 10}}^{\text{3}}{\text{kgm}}^{\text{-3}}{\text{× 4 × 10}}^{\text{–3}}m}\text{= 0}\text{.983}{\text{ms}}^{\text{-1}}\\ \therefore \text{The largest average velocity of blood}\text{=}\text{0}{\text{.983 ms}}^{\text{-1}}\\ \left(\text{b}\right)\text{The}\text{corresponding}\text{flow rate is given as:}\\ \text{R =}\pi {\text{r}}^{\text{2}}{\text{V}}_{\text{avg}}\\ \text{R = 3}\text{.14 ×}{\left({\text{2 × 10}}^{\text{-3}}m\right)}^{\text{2}}\text{×0}\text{.983}{\text{ms}}^{\text{-1}}\\ \text{= 1}{\text{.235 × 10}}^{\text{-5}}{\text{m}}^{\text{3}}{\text{s}}^{\text{-1}}\\ \therefore \text{The corresponding flow rate}\text{=}\text{1}{\text{.235 × 10}}^{\text{-5}}{\text{m}}^{\text{3}}{\text{s}}^{\text{-1}}\end{array}

Q.27 A plane is in level flight at constant speed and each of its two wings has an area of 25 m². If the speed of the air is 180 km h^-1 over the lower wing and 234 kmh^-1 over the upper wing surface, determine the plane’s mass. (Take air density to be 1 kg m^–3).

Ans.

\begin{array}{l}\text{Here, area of the wings of the plane, A}\text{=}\text{2}\text{×}\text{25}{\text{m}}^{\text{2}}\text{=}{\text{50 m}}^{\text{2}}\\ \text{Speed of air over the lower wing of plane,}\\ {\text{v}}_{\text{1}}\text{=}{\text{180 kmh}}^{\text{-1}}\text{=}{\text{50 ms}}^{\text{-1}}\\ \text{Speed of the}\text{air over upper wing}\text{of}\text{plane,}\\ {\text{v}}_{\text{2}}\text{=}{\text{234 kmh}}^{\text{-1}}\text{=}{\text{65 ms}}^{\text{-1}}\\ \text{Density of air,ρ}\text{=}{\text{1 kg m}}^{\text{–3}}\\ \text{Let}\text{pressure of air over the lower wing}\text{of}\text{plane}\text{=}{\text{P}}_{\text{1}}\\ \text{Let}\text{pressure of air over the upper wing}\text{of}\text{plane}\text{=}{\text{P}}_{\text{2}}\\ \text{From}\text{Bernoulli’s equation, we}\text{have:}\\ {\text{P}}_{\text{1}}\text{+}\frac{\text{1}}{\text{2}}{\text{ρv}}_{\text{1}}{}^{\text{2}}{\text{= P}}_{\text{2}}\text{+}\frac{\text{1}}{\text{2}}{\text{ρv}}_{\text{2}}{}^{\text{2}}\\ \therefore {\text{P}}_{\text{1}}{\text{– P}}_{\text{2}}\text{=}\frac{\text{1}}{\text{2}}\text{ρ}\left({\text{v}}_{\text{2}}{}^{\text{2}}{\text{– v}}_{\text{1}}{}^{\text{2}}\right)\text{(i)}\\ \text{The upward force}\left(\text{F}\right)\text{on the plane}\text{=}\left({\text{P}}_{\text{1}}{\text{– P}}_{\text{2}}\right)\text{A}\\ \text{From}\text{equation}\text{(i),}\text{we}\text{obtain:}\\ \text{F =}\frac{\text{1}}{\text{2}}\text{ρ}\left({\text{v}}_{\text{2}}{}^{\text{2}}{\text{– v}}_{\text{1}}{}^{\text{2}}\right)\text{A}\\ \text{=}\frac{\text{1}}{\text{2}}{\text{× 1 kg m}}^{\text{–3}}\text{×}\left[{\left({\text{65 ms}}^{\text{-1}}\right)}^{\text{2}}\text{–}{\left({\text{50 ms}}^{\text{-1}}\right)}^{\text{2}}\right]{\text{×50 m}}^{\text{2}}\text{= 43125}\text{N}\\ \text{According}\text{to Newton’s force equation, the mass}\left(\text{m}\right)\text{of the}\\ \text{plane can}\text{be}\text{obtained}\text{as:}\\ \text{F}\text{=}\text{mg}\\ \therefore \text{m}\text{=}\frac{\text{43125}}{\text{9}\text{.8}}\text{=}\text{4400}\text{.51}\text{kg~4400 kg}\\ \therefore \text{Mass of the plane is approximately}\text{4400 kg}\text{.}\end{array}

Q.28 In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 × 10^–5 m and density 1.2 × 10³ kg m^–3? Take the viscosity of air at the temperature of the experiment to be 1.8 × 10^–5 Pa s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.

Ans.

\begin{array}{l}\text{Here,}\text{radius}\text{of}\text{drop = 2}{\text{.0 × 10}}^{\text{-5}}\text{m}\\ \text{Density}\text{of}\text{drop,}\text{ρ = 1}{\text{.2 × 10}}^{\text{3}}{\text{kgm}}^{\text{-3}}\\ \text{Viscosity}\text{of}\text{air,}\text{η = 1}{\text{.8 × 10}}^{\text{-5}}\text{Pa}\text{s}\\ \text{Acceleration}\text{due}\text{to}\text{gravity,}\text{g}\text{=}\text{9}\text{.8}{\text{ms}}^{\text{-2}}\\ \text{In}\text{order}\text{to}\text{neglect}\text{buoyancy}\text{of}\text{air,}\text{take}\text{density}\text{of}\text{air,}\\ {\text{ρ}}_{\text{o}}\text{=}\text{0}\\ \text{The}\text{relation}\text{for}\text{terminal}\text{velocity}\text{is}\text{given}\text{as,}\\ \text{v}\text{=}\frac{{\text{2r}}^{\text{2}}\text{×}\left({\text{ρ – ρ}}_{\text{o}}\right)\text{g}}{\text{9η}}\\ \text{=}\frac{\text{2 ×}{\left(\text{2}{\text{.0 × 10}}^{\text{-5}}\text{m}\right)}^{\text{2}}\left(\text{1}{\text{.2 × 10}}^{\text{3}}\text{– 0}\right){\text{kgm}}^{\text{-3}}\text{× 9}{\text{.8 ms}}^{\text{-2}}}{\text{9 × 1}{\text{.8 × 10}}^{\text{-5}}\text{Pa}\text{s}}\\ \text{= 5}{\text{.807 × 10}}^{\text{-2}}{\text{ms}}^{\text{-1}}\text{= 5}\text{.8}{\text{cms}}^{\text{-1}}\\ \therefore \text{Terminal}\text{speed}\text{of}\text{the}\text{drop = 5}\text{.8}{\text{cms}}^{\text{-1}}\\ \text{Viscous}\text{force}\text{on}\text{the}\text{drop}\text{is}\text{given}\text{by}\text{the}\text{relation,}\\ \text{F}\text{=}\text{6πηrv}\\ \therefore \text{F = 6 × 3}\text{.14 × 1}{\text{.8 × 10}}^{\text{-5}}\text{Pa}\text{s × 2}{\text{.0 × 10}}^{\text{-5}}\text{m × 5}\text{.8}{\text{×10}}^{\text{-2}}\text{m}\\ \text{F = 3}{\text{.9 × 10}}^{\text{-10}}\text{N}\\ \therefore \text{The}\text{viscous}\text{force}\text{on}\text{the}\text{drop}\text{=}\text{3}{\text{.9 × 10}}^{\text{-10}}\text{N}\end{array}

Q.29 Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? Surface tension of mercury at the temperature of the experiment is 0.465 Nm^–1. Density of mercury = 13.6 × 10³ kgm^–3.

Ans.

\begin{array}{l}\text{Here,}\text{radius of the tube, r}\text{=}\text{1 mm}\text{=}\text{1}\text{×}{\text{10}}^{\text{–3}}\text{m}\\ \text{Angle of contact between mercury and soda lime}\\ \text{glass,}\text{θ}\text{=}\text{140°}\\ \text{Surface tension of mercury at the temperature}\text{of}\text{the}\\ \text{experiment,}\text{s}\text{=}\text{0}{\text{.465 Nm}}^{\text{–1}}\\ \text{Density of mercury, ρ}\text{=}\text{13}\text{.6}\text{×}{\text{10}}^{\text{3}}{\text{kgm}}^{\text{-3}}\\ \text{Let}\text{dip in the height of mercury}\text{= h}\\ \text{Acceleration due to gravity, g}\text{=}\text{9}{\text{.8 ms}}^{\text{-2}}\\ \text{Relation}\text{connecting}\text{surface tension, angle of contact}\\ \text{and}\text{the dip in the height is}\text{given}\text{as,}\\ \text{s =}\frac{\text{hρgr}}{\text{2cosθ}}\\ \therefore \text{h =}\frac{\text{2scosθ}}{\text{rρg}}\text{=}\frac{\text{2 × 0}{\text{.465 Nm}}^{\text{–1}}\text{× cos140°}}{{\text{1 × 10}}^{\text{–3}}\text{m ×13}{\text{.6 × 10}}^{\text{3}}{\text{kgm}}^{\text{-3}}\text{×9}{\text{.8 ms}}^{-2}}\\ \text{h = -0}\text{.00534}\text{m = -5}\text{.34}\text{mm}\\ \text{Here,}\text{the}\text{negative}\text{sign}\text{indicates}\text{the}\text{the}\text{level}\text{of}\text{mercury}\\ \text{is}\text{decreasing}\text{.}\\ \therefore \text{Dip in the height of mercury}\text{=}\text{-5}\text{.34}\text{mm}\end{array}

Q.30 Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7.3 × 10^–2 N m^–1. Take the angle of contact to be zero and density of water to be 1.0 × 10³ kg m^–3 (g = 9.8 m s^–2).

Ans.

\begin{array}{l}\text{Here,}{\text{diameter of first bore, d}}_{\text{1}}\text{=}\text{3}\text{.0 mm =}\text{3}\text{×}{\text{10}}^{\text{–3}}\text{m}\\ \therefore \text{Radius}\text{of}\text{first}\text{bore,}{\text{r}}_{\text{1}}\text{=}\frac{{\text{d}}_{\text{1}}}{\text{2}}\text{=}\text{1}\text{.5}\text{×}{\text{10}}^{\text{–3}}\text{m}\\ \text{Diameter of second bore,}{\text{d}}_{\text{2}}\text{=}\text{6}\text{.0 mm}\\ \therefore \text{Radius}\text{of}\text{second}\text{bore,}{\text{r}}_{\text{2}}\text{=}\frac{{\text{d}}_{\text{2}}}{\text{2}}\text{=}\text{3}\text{×}{\text{10}}^{\text{–3}}\text{m}\\ \text{Surface tension of the}\text{water, s}\text{=}\text{7}\text{.3}\text{×}{\text{10}}^{\text{–2}}{\text{Nm}}^{\text{–1}}\\ \text{The}\text{angle of contact between the bore surface and}\\ \text{water,}\text{θ}\text{=}\text{0}°\\ \text{Density of water, ρ}\text{=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{3}}{\text{kgm}}^{\text{–3}}\\ \text{Acceleration due to gravity, g}\text{=}\text{9}{\text{.8 ms}}^{\text{-2}}\\ \text{Let the heights to which the}\text{water rises in}\text{first tube}\text{and}\\ \text{second tube be}{\text{h}}_{\text{1}}{\text{and h}}_{\text{2}}\text{respectively}\text{.}\\ {\text{Then, h}}_{\text{1}}\text{=}\frac{\text{2scosθ}}{{\text{r}}_{\text{1}}\text{ρg}}\text{(i)}\\ {\text{h}}_{\text{2}}\text{=}\frac{\text{2scosθ}}{{\text{r}}_{\text{2}}\text{ρg}}\text{(i)}\\ \text{Difference between the water levels}\text{in the two limbs of}\\ \text{the tube}\text{=}\frac{\text{2s}\text{cos}\text{θ}}{{\text{r}}_{\text{1}}\text{ρg}}\text{–}\frac{\text{2s}\text{cos}\text{θ}}{{\text{r}}_{\text{2}}\text{ρg}}\\ \text{=}\frac{\text{2scosθ}}{\text{ρg}}\left[\frac{\text{1}}{{\text{r}}_{\text{1}}}\text{–}\frac{\text{1}}{{\text{r}}_{\text{2}}}\right]\\ \text{=}\frac{\text{2}\text{×}\text{7}\text{.3}\text{×}{\text{10}}^{\text{–2}}{\text{Nm}}^{\text{–1}}\text{×}\text{1}}{\text{1}\text{.0}\text{×}{\text{10}}^{\text{3}}{\text{kgm}}^{\text{–3}}\text{× 9}{\text{.8 ms}}^{\text{-2}}}\left[\frac{\text{1}}{\text{1}\text{.5}\text{×}{\text{10}}^{\text{–3}}\text{m}}\text{–}\frac{\text{1}}{\text{3}\text{×}{\text{10}}^{\text{–3}}\text{m}}\right]\\ \text{=}\text{4}\text{.966}\text{×}{\text{10}}^{\text{–3}}\text{m =}\text{4}\text{.97}\text{mm}\\ \therefore \text{The}\text{difference between the}\text{levels of water in the two bores}\\ \text{is 4}\text{.97 mm}\text{.}\end{array}

Q.31 (a) It is known that density ρ of air decreases with height y as
ρ = ρ₀e^-y/y₀

Where ρ₀ =1.25 kg m^–3 is the density at sea level, and y₀ is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant (isothermal conditions). Also assume that the value of g remains constant.
(b) A large He balloon of volume 1425 m³ is used to lift a payload of 400 kg. Assume that the balloon maintains constant radius as it rises. How high does it rise?
[Take y₀ = 8000 m and ρ_He= 0.18 kgm^–3].

Ans.

\begin{array}{l}\left(\text{a}\right)\text{Acceleration due to gravity, g}\text{=}\text{9}{\text{.8 ms}}^{\text{-2}}\\ \text{Let}\text{density of}\text{balloon}\text{=}\text{ρ}\\ \text{Let}\text{height attained}\text{by the balloon}\text{=}\text{y}\\ \text{We}\text{know}\text{that}\text{density}\left(\text{ρ}\right)\text{of air decreases with height}\left(\text{y}\right)\text{as:}\\ \text{ρ}\text{=}{\text{ρ}}_{\text{0}}{\text{e}}^{\frac{\text{-y}}{{\text{y}}_{\text{0}}}}\\ \therefore \frac{\text{ρ}}{{\text{ρ}}_{\text{0}}}{\text{= e}}^{\frac{\text{-y}}{{\text{y}}_{\text{0}}}}\text{(i)}\\ \text{From equation}\left(\text{i}\right)\text{it}\text{is}\text{clear}\text{that the rate of}\text{decrease}\\ \text{of density with height is directly proportional to}\text{ρ}\text{.}\\ \text{–}\frac{\text{dρ}}{\text{dy}}\propto \text{ρ}\\ \frac{\text{dρ}}{\text{dy}}\text{= -kρ}\\ \frac{\text{dρ}}{\text{ρ}}\text{= -kdy}\\ \text{Here, k is the constant of proportionality}\text{.}\\ \text{Height changes from 0 to y , as density changes from}{\text{ρ}}_{\text{0}}\text{to ρ}\text{.}\\ \text{Integrating both}\text{sides}\text{within these limits, we get}\\ {\displaystyle \underset{{\text{ρ}}_{\text{0}}}{\overset{\text{ρ}}{\int }}\frac{\text{dρ}}{\text{ρ}}}\text{=}{\displaystyle \underset{\text{0}}{\overset{\text{y}}{\int }}\text{kdy}}\\ {\left[{\text{log}}_{\text{e}}\text{ρ}\right]}_{{\text{ρ}}_{\text{0}}}^{\text{ρ}}\text{= -ky}\\ {\text{log}}_{\text{e}}\frac{\text{ρ}}{{\text{ρ}}_{\text{0}}}\text{=-ky}\\ \frac{\text{ρ}}{{\text{ρ}}_{\text{0}}}{\text{= e}}^{\text{-ky}}\text{(ii)}\\ \text{From}\text{equation}\text{(i)}\text{and}\text{(ii),}\text{we}\text{get}\\ {\text{y}}_{\text{0}}\text{=}\frac{\text{1}}{\text{k}}\\ \text{k =}\frac{\text{1}}{{\text{y}}_{\text{0}}}\text{(iii)}\\ \text{From}\text{equation}\text{(i)}\text{and}\text{(iii),}\text{we}\text{obtain}\\ {\text{ρ = ρ}}_{\text{0}}{\text{e}}^{\frac{\text{-y}}{{\text{y}}_{\text{0}}}}\\ \left(\text{b}\right)\text{Here,}\text{volume of the balloon,}\text{V}\text{=}{\text{1425 m}}^{\text{3}}\\ \text{Mass of payload,}\text{m}\text{=}\text{400 kg}\\ \text{Density}\text{is}\text{given}\text{by}\text{the}\text{relation:}\\ \text{ρ =}\frac{\text{Mass}}{\text{Volume}}\\ \text{ρ =}\frac{\text{Mass}\text{of}\text{payload}\text{+}\text{Mass}\text{of}\text{helium}}{\text{Volume}}\\ \rho \text{=}\frac{{\text{m + Vρ}}_{\text{He}}}{\text{V}}\\ \text{=}\frac{{\text{400kg +1425 m}}^{\text{3}}\text{×0}{\text{.18 kgm}}^{\text{-3}}}{{\text{1425 m}}^{\text{3}}}\\ \text{= 0}\text{.46}{\text{kgm}}^{\text{-3}}\\ \text{Using}\text{equation}\text{(ii)}\text{and}\text{(iii),}\text{we}\text{get}\\ \text{ρ}\text{=}{\text{ρ}}_{\text{0}}{\text{e}}^{\frac{\text{-y}}{{\text{y}}_{\text{0}}}}\\ {\text{log}}_{\text{e}}\frac{\text{ρ}}{{\text{ρ}}_{\text{0}}}\text{=}\text{–}\frac{\text{y}}{{\text{y}}_{\text{0}}}\\ \therefore \text{y = -8000}\text{×}{\text{log}}_{\text{e}}\frac{\text{0}\text{.46}}{\text{1}\text{.25}}\text{= 8 km}\\ \therefore \text{The height}\text{attained}\text{by}\text{the}\text{balloon is 8 km}\text{.}\end{array}