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Step-by-step solutions with formulas and diagrams for all questions from NCERT Class 11 Physics Chapter 9.
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Q1
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 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.
In a fluid at rest, pressure arises due to the weight of the fluid column above a given point. For a fluid of density ρ, the pressure difference between two points separated by vertical height h is
ΔP = ρgh
This relation explains how pressure varies with depth in liquids and with altitude in the atmosphere. Another important property of fluids at rest is that pressure acts equally in all directions (Pascal’s law).
(a) Blood pressure is greater at the feet than at the brain
Blood behaves like a fluid inside the circulatory system. According to hydrostatic pressure relation
P = ρgh
pressure increases with the vertical depth of the fluid column.
Because the feet lie lower, the vertical height of the blood column above them is greater. Therefore the hydrostatic pressure at the feet becomes larger than at the brain.
Hence, blood pressure is greater at the feet due to the larger hydrostatic column of blood.
(b) Atmospheric pressure decreases rapidly with height
Atmospheric pressure at any point is due to the weight of the air column above it.
Because density decreases with height, the upper layers of the atmosphere contain much less air. Hence the weight of the air column above a height of about 6 km becomes roughly half of that at sea level.
Therefore atmospheric pressure becomes nearly half at about 6 km altitude even though the atmosphere extends beyond 100 km.
(c) Hydrostatic pressure is a scalar quantity
Pressure is defined as
P = F / A
Although force is a vector quantity, pressure does not have a specific direction because:
Thus pressure itself is completely described by its magnitude only and has no fixed direction.
Hence hydrostatic pressure is a scalar physical quantity.
Q2
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.
(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.
When a liquid comes in contact with a solid surface, the shape of the liquid surface near the solid depends on two types of intermolecular forces:
The angle between the tangent to the liquid surface and the solid surface inside the liquid is called the angle of contact (θ).
(a) Angle of contact: Mercury obtuse, Water acute with glass
Hence, water forms an acute angle with glass while mercury forms an obtuse angle.
(b) Water spreads on glass while mercury forms drops
Thus water wets glass while mercury does not.
(c) Surface tension is independent of surface area
Surface tension is defined as
T = F / L
Hence surface tension is a characteristic property of the liquid and is independent of surface area.
(d) Water with detergent has smaller angle of contact
Therefore the angle of contact decreases when detergent is added to water.
(e) A liquid drop is spherical in the absence of external forces
Surface tension tries to minimize the surface area of a liquid for a given volume.
Hence a liquid drop takes a spherical shape in the absence of external forces due to surface tension.
Q3 Fill in the blanks using the word(s) from the list appended with each statement:
Several properties of fluids change with temperature and flow conditions.
(a) Surface tension of liquids generally decreases with temperature.
When temperature increases, the kinetic energy of molecules increases and intermolecular attraction becomes weaker. Hence the surface tension decreases.
(b) Viscosity of gases increases with temperature, whereas viscosity of liquids decreases with temperature.
(c) For solids with elastic modulus of rigidity, the shearing force is proportional to shear strain, while for fluids it is proportional to rate of shear strain.
(d) For a fluid in a steady flow, the increase in flow speed at a constriction follows conservation of mass.
For incompressible flow,
A₁v₁ = A₂v₂
If the cross-sectional area decreases, velocity must increase to maintain constant mass flow rate.
(e) For the model of a plane in a wind tunnel, turbulence occurs at a greater speed than for an actual plane.
Turbulence depends on the Reynolds number
Re = ρvL / η
Hence turbulence occurs at a greater speed in the wind-tunnel model.
Q4
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
Several principles of fluid dynamics explain these phenomena:
(a) To keep a piece of paper horizontal, we blow over it.
Hence the paper rises when we blow over it.
(b) Fast jets emerge through gaps between fingers.
Therefore fast jets gush through the gaps between fingers.
(c) Needle size controls syringe flow rate better than thumb pressure.
Hence the needle size controls the flow rate more effectively.
(d) Fluid leaving a hole produces backward thrust.
This is an example of Newton’s third law.
(e) A spinning cricket ball does not follow a parabolic path.
Hence the trajectory of a spinning cricket ball deviates from a parabola.
Q5 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 ?
Pressure is defined as the force acting per unit area.
P = F / A
If the same force acts on a smaller area, the pressure becomes very large. This is why pointed objects such as needles or high heels exert very large pressures.
Step 1: Write the given data
Step 2: Force exerted on the floor
The force exerted is equal to the weight of the girl.
$$ \begin{aligned} F &= mg \\ &= 50 \times 9.8 \\ &= 490 \, N \end{aligned} $$Step 3: Area of contact of the heel
The heel is circular, so
$$ \begin{aligned} A &= \pi r^2 \\ &= \frac{22}{7} \times (0.005)^2 \\ &= 7.85 \times 10^{-5}\, m^2 \end{aligned} $$Step 4: Calculate pressure
$$ \begin{aligned} P &= \frac{F}{A} \\ &= \frac{490}{7.85 \times 10^{-5}} \\ &\approx 6.24 \times 10^6 \, Pa \end{aligned} $$Pressure exerted on the floor ≈ \(6.2 \times 10^6\) Pa
Physical Interpretation
Because the contact area of the heel is extremely small, the pressure becomes very large. This explains why high heels can easily damage wooden floors or soft ground.
Q6 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.
A barometer measures atmospheric pressure using the pressure exerted by a vertical column of liquid. At equilibrium,
P = ρgh
where \(ρ\) is the density of the liquid, \(g\) is gravitational acceleration, and \(h\) is the height of the liquid column. Liquids with smaller density require a larger column height to balance atmospheric pressure.
Step 1: Write the given data
Step 2: Use hydrostatic pressure relation
For a barometer,
P = ρgh
Therefore,
$$ \begin{aligned} h &= \frac{P}{ρg} \end{aligned} $$Step 3: Substitute values
$$ \begin{aligned} h &= \frac{1.01 \times 10^5}{984 \times 9.8} \end{aligned} $$ $$ \begin{aligned} 984 \times 9.8 &= 9643.2 \end{aligned} $$ $$ \begin{aligned} h &= \frac{1.01 \times 10^5}{9643.2} \\ h &\approx 10.47\,m \end{aligned} $$Height of wine column ≈ \(10.5\;m\)
Physical Interpretation
Mercury barometers require only about 0.76 m of mercury because mercury has a very high density. Wine has a much lower density, so a much taller column (about 10.5 m) is needed to balance atmospheric pressure.
Q7 A vertical off-shore structure is built to withstand a maximum stress of \(10^9\) 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.
The pressure at a depth inside a liquid is given by the hydrostatic pressure relation
P = P_0 + ρgh
where \(P_0\) is atmospheric pressure, \(ρ\) is the density of the liquid, \(g\) is acceleration due to gravity, and \(h\) is the depth. As depth increases, pressure increases due to the weight of the water column above.
Step 1: Given data
Step 2: Calculate pressure at the ocean bottom
Using the hydrostatic pressure formula:
$$ P = P_0 + ρgh $$ Substituting values, $$ \begin{aligned} P &= 1.01 \times 10^5 + (1.03 \times 10^3)(9.8)(3 \times 10^3) \end{aligned} $$ $$ \begin{aligned} ρgh &= 1030 \times 9.8 \times 3000 \\ &\approx 3.03 \times 10^7 \, Pa \end{aligned} $$ Therefore, $$ \begin{aligned} P &= 1.01 \times 10^5 + 3.03 \times 10^7 \\ &\approx 3.04 \times 10^7 \, Pa \end{aligned} $$Step 3: Compare with maximum allowable stress
\(3.0 \times 10^7\; Pa \ll 10^9\; Pa\)
the pressure at the ocean bottom is much smaller than the maximum stress the structure can withstand.Therefore, the structure is safe and suitable for installation on the oil well.
Conceptual Illustration
Q8 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 \(425\ cm^2\). What maximum pressure would the smaller piston have to bear?
Hydraulic lifts work on Pascal’s law, which states that pressure applied to a confined fluid is transmitted equally in all directions.
P = F / A
Thus the same pressure acts on both pistons of the hydraulic system.
Step 1: Given data
Step 2: Calculate force exerted by the car
\[ \begin{aligned} F &= mg \\ &= 3000 \times 9.8 \\ &= 2.94 \times 10^4\ N \end{aligned} \]Step 3: Calculate pressure on the large piston
\[ \begin{aligned} P &= \frac{F}{A} \\ &= \frac{2.94 \times 10^4}{0.0425} \\ &= 6.92 \times 10^5\ Pa \end{aligned} \]Step 4: Apply Pascal’s law
According to Pascal’s law, the pressure is transmitted equally throughout the fluid. Therefore the same pressure acts on the smaller piston.
Maximum pressure on the smaller piston ≈ \(6.9 \times 10^5\ Pa\)
Conceptual Illustration
Q9 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?
In a liquid at rest, pressure at the same horizontal level must be equal. Therefore, at the same level in the mercury columns of a U-tube, the pressure exerted by the liquid columns above must be equal.
ρ₁ g h₁ = ρ₂ g h₂
Here \(ρ_1\) and \(ρ_2\) are densities of the two liquids and \(h_1\) and \(h_2\) are the heights of their columns.
Step 1: Given data
Step 2: Apply pressure equality
At the same horizontal level in mercury, \[ ρ_1 g h_1 = ρ_2 g h_2 \] Substituting \(ρ_2 = S\), \[ 1 \times g \times 10 = S \times g \times 12.5 \] Cancel \(g\), \[ 10 = 12.5 S \]Step 3: Solve for specific gravity
\[ S = \frac{10}{12.5} \] \[ S = 0.8 \]Specific gravity of methylated spirit = 0.8
Conceptual Illustration
Q10 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)
In a connected liquid system, the pressure at the same horizontal level must be equal. If two different liquids produce unequal pressure, the denser liquid (mercury here) shifts until hydrostatic equilibrium is restored.
ρw g hw = ρs g hs + ρm g hm
where \(h_m\) is the difference in mercury levels between the two arms.
Step 1: Known values
Step 2: Apply pressure balance
At the same horizontal level inside mercury, \[ ρ_w g h_w = ρ_s g h_s + ρ_m g h_m \] Substitute relative densities: \[ 1 \times 25 = 0.8 \times 27.5 + 13.6\,h_m \]Step 3: Solve the equation
\[ 25 = 22 + 13.6h_m \] \[ 13.6h_m = 3 \] \[ h_m = \frac{3}{13.6} \] \[ h_m \approx 0.22\ \text{cm} \]Difference in mercury levels ≈ 0.22 cm
Conceptual Illustration
Q11 Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river ? Explain.
Bernoulli’s equation describes the conservation of mechanical energy in a moving fluid.
P + ρgh + \(\frac{1}{2}\)ρv² = constant
However, it is valid only when the following conditions are satisfied:
In a river rapid, water flows very irregularly and violently.
Because of these effects, the assumptions required for Bernoulli’s equation are not satisfied.
Therefore, Bernoulli’s equation cannot accurately describe the flow of water through a rapid.
Conceptual Illustration
Q12 Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation ? Explain.
Pressure in a fluid can be expressed in two ways:
These are related by
Pabsolute = Pgauge + Patm
Bernoulli’s equation between two points in a fluid flow is
\[ P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2 \]If absolute pressures are used,
\[ P_1 = P_{g1} + P_{atm}, \quad P_2 = P_{g2} + P_{atm} \] Substituting in Bernoulli’s equation, \[ (P_{g1}+P_{atm}) + \frac{1}{2}\rho v_1^2 + \rho g h_1 = (P_{g2}+P_{atm}) + \frac{1}{2}\rho v_2^2 + \rho g h_2 \] Since atmospheric pressure \(P_{atm}\) appears on both sides, it cancels out. Therefore, \[ P_{g1} + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_{g2} + \frac{1}{2}\rho v_2^2 + \rho g h_2 \]Hence Bernoulli’s equation gives the same result whether we use gauge pressure or absolute pressure.
This happens because Bernoulli’s equation depends only on pressure differences, not on the reference level of pressure.
Conceptual Illustration
Q13 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 \(4.0 \times 10^{-3}\ kg\ s^{-1}\), what is the pressure difference between the two ends of the tube ? (Density of glycerine = \(1.3 \times 10^3\ kg\ m^{-3}\), viscosity of glycerine = \(0.83\ Pa\,s\)). [Also check whether the assumption of laminar flow is valid.]
For viscous flow of a liquid through a cylindrical tube, Poiseuille’s law gives the volume flow rate:
\[ Q = \frac{\pi r^{4}\Delta P}{8\eta l} \]
where \(r\) is the tube radius, \(l\) its length, \(η\) the viscosity of the liquid, and \(\Delta P\) the pressure difference between the ends.
Step 1: Given data
Step 2: Convert mass flow rate to volume flow rate
\[ Q = \frac{\dot{m}}{\rho} \] \[ Q = \frac{4.0 \times 10^{-3}}{1.3 \times 10^{3}} \] \[ Q \approx 3.08 \times 10^{-6}\ m^3 s^{-1} \]Step 3: Use Poiseuille’s equation
\[ \Delta P = \frac{8\eta l Q}{\pi r^4} \] Substituting values: \[ \Delta P = \frac{8 \times 0.83 \times 1.5 \times 3.08 \times 10^{-6}} {\pi \times (0.01)^4} \] \[ \Delta P \approx 9.8 \times 10^{2}\ Pa \]Pressure difference ≈ \(1.0 \times 10^{3}\ Pa\)
Step 4: Check whether flow is laminar
Average velocity: \[ v = \frac{Q}{\pi r^2} \] \[ v = \frac{3.08 \times 10^{-6}}{\pi (0.01)^2} \] \[ v \approx 9.8 \times 10^{-3}\ m\,s^{-1} \] Reynolds number: \[ Re = \frac{\rho v D}{\eta} \] \[ Re = \frac{(1.3 \times 10^3)(9.8 \times 10^{-3})(0.02)}{0.83} \] \[ Re \approx 0.3 \]Since \(Re \ll 2000\), the flow is laminar. Therefore the use of Poiseuille’s law is justified.
Conceptual Illustration
Q14 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.
Lift on an aeroplane wing arises due to the pressure difference between the upper and lower surfaces of the wing. According to Bernoulli’s principle, higher speed of airflow corresponds to lower pressure.
\[ P + \frac{1}{2}\rho v^2 = \text{constant} \]
If air moves faster over the upper surface than the lower surface, the pressure above the wing becomes smaller, producing an upward lift force.
Step 1: Given data
Step 2: Calculate pressure difference
Using Bernoulli’s equation at the same height: \[ \Delta P = \frac{1}{2}\rho (v_1^2 - v_2^2) \] \[ v_1^2 - v_2^2 = 70^2 - 63^2 \] \[ = 4900 - 3969 \] \[ = 931 \] \[ \Delta P = \frac{1}{2} \times 1.3 \times 931 \] \[ \Delta P \approx 605\,Pa \]Step 3: Calculate lift force
\[ L = \Delta P \times A \] \[ L = 605 \times 2.5 \] \[ L \approx 1.51 \times 10^3\,N \]Lift on the wing ≈ \(1.5 \times 10^3\,N\)
Conceptual Illustration
Q15 Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?
For steady incompressible flow of a liquid through a pipe, the continuity equation must be satisfied.
\[ A_1 v_1 = A_2 v_2 \]
This means that the volume flow rate remains constant. If the cross-sectional area of the pipe decreases, the velocity of the fluid must increase.
In a narrowing section of a pipe, the cross-sectional area decreases. According to the continuity equation, the velocity of the fluid must therefore increase.
In Figure (a), the fluid speed increases as the pipe becomes narrower. This is consistent with the continuity equation and represents correct steady flow.
However, in Figure (b), the pipe narrows but the velocity is shown to decrease. This violates the continuity condition \(A_1 v_1 = A_2 v_2\).
Therefore, Figure (b) is incorrect.
Q16 The cylindrical tube of a spray pump has a cross-section of \(8.0\ cm^2\). One end 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?
For steady incompressible flow, the continuity equation states that the volume flow rate remains constant.
\[ A_1 v_1 = A_2 v_2 \]
If the total exit area becomes smaller, the velocity of the fluid must increase.
Step 1: Given data
Step 2: Area of one hole
\[ a = \pi r^2 \] \[ a = \pi (0.5 \times 10^{-3})^2 \] \[ a = 7.85 \times 10^{-7}\ m^2 \]Step 3: Total area of 40 holes
\[ A_2 = 40a \] \[ A_2 = 40 \times 7.85 \times 10^{-7} \] \[ A_2 = 3.14 \times 10^{-5}\ m^2 \]Step 4: Apply continuity equation
\[ A_1 v_1 = A_2 v_2 \] \[ v_2 = \frac{A_1 v_1}{A_2} \] \[ v_2 = \frac{8.0 \times 10^{-4} \times 0.025}{3.14 \times 10^{-5}} \] \[ v_2 \approx 0.64\ m\,s^{-1} \]Speed of liquid through the holes ≈ \(0.64\ m\,s^{-1}\)
Conceptual Illustration
Q17 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 \(1.5 \times 10^{-2}\ N\). The length of the slider is 30 cm. What is the surface tension of the film?
Surface tension is defined as the force acting per unit length along the surface of a liquid.
\[ T = \frac{F}{L} \]
In a soap film there are two surfaces (front and back). Therefore the effective length along which surface tension acts becomes \(2L\).
Step 1: Given data
Step 2: Effective length of soap film
Since the film has two surfaces, \[ l = 2L \] \[ l = 2 \times 0.30 \] \[ l = 0.60\ m \]Step 3: Calculate surface tension
\[ T = \frac{F}{l} \] \[ T = \frac{1.5 \times 10^{-2}}{0.60} \] \[ T = 2.5 \times 10^{-2}\ N\,m^{-1} \]Surface tension of the soap film ≈ \(2.5 \times 10^{-2}\ N\,m^{-1}\)
Conceptual Illustration
Q18 Figure 9.21 (a) shows a thin liquid film supporting a small weight \(4.5 \times 10^{-2}\,N\). 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.
A soap film has two surfaces. If a slider of length \(L\) is in contact with the film, the upward force due to surface tension is
\[ F = 2TL \]
where \(T\) is the surface tension of the liquid. Thus the supported weight depends only on surface tension and the length of the slider, not on the shape of the film.
Step 1: Determine surface tension from Fig. (a)
Given weight supported: \[ W_a = 4.5 \times 10^{-2}\,N \] Length of slider: \[ L = 40\,cm = 0.40\,m \] Using \(W = 2TL\), \[ 4.5 \times 10^{-2} = 2T(0.40) \] \[ T = \frac{4.5 \times 10^{-2}}{0.80} \] \[ T = 5.63 \times 10^{-2}\,N\,m^{-1} \]Step 2: Weight supported in Fig. (b)
The length of the slider is still \(0.40\,m\). \[ W_b = 2TL \] \[ W_b = 2 \times 5.63 \times 10^{-2} \times 0.40 \] \[ W_b = 4.5 \times 10^{-2}\,N \]Step 3: Weight supported in Fig. (c)
Again the slider length remains \(0.40\,m\). \[ W_c = 2TL \] \[ W_c = 2 \times 5.63 \times 10^{-2} \times 0.40 \] \[ W_c = 4.5 \times 10^{-2}\,N \]Therefore, the weight supported in Fig. (b) and Fig. (c) is also \(4.5 \times 10^{-2}\,N\).
Physical Explanation
The force due to surface tension depends only on the length of the slider and the surface tension of the liquid. Since both are the same in figures (a), (b), and (c), the supported weight remains the same. The shape of the soap film does not affect the force.
Q19 What is the pressure inside a drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20 °C) is \(4.65 \times 10^{-1}\ N\,m^{-1}\). The atmospheric pressure is \(1.01 \times 10^5\ Pa\). Also give the excess pressure inside the drop.
Surface tension causes the pressure inside a liquid drop to be higher than the outside pressure. For a spherical liquid drop, the excess pressure inside is
\[ \Delta P = \frac{2T}{r} \]
where \(T\) is the surface tension and \(r\) is the radius of the drop.
Step 1: Given data
Step 2: Calculate excess pressure
\[ \Delta P = \frac{2T}{r} \] \[ \Delta P = \frac{2 \times 4.65 \times 10^{-1}}{3.0 \times 10^{-3}} \] \[ \Delta P = \frac{0.93}{3.0 \times 10^{-3}} \] \[ \Delta P \approx 3.1 \times 10^{2}\,Pa \]Step 3: Pressure inside the drop
\[ P = P_0 + \Delta P \] \[ P = 1.01 \times 10^{5} + 3.1 \times 10^{2} \] \[ P \approx 1.0131 \times 10^{5}\,Pa \]Excess pressure inside the drop ≈ \(3.1 \times 10^{2}\,Pa\)
Pressure inside the drop ≈ \(1.0131 \times 10^{5}\,Pa\)
Conceptual Illustration
Q20 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 20 °C is \(2.50 \times 10^{-2}\,N\,m^{-1}\)? If an air bubble of the same dimension were formed at a depth of 40.0 cm inside a container containing the soap solution (relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure = \(1.01 \times 10^5\,Pa\))
Surface tension produces extra pressure inside curved liquid surfaces.
Step 1: Given data
Step 2: Excess pressure inside soap bubble
\[ \Delta P = \frac{4T}{r} \] \[ \Delta P = \frac{4 \times 2.50 \times 10^{-2}}{5.0 \times 10^{-3}} \] \[ \Delta P = \frac{0.10}{5.0 \times 10^{-3}} \] \[ \Delta P = 20\,Pa \]Excess pressure in soap bubble = \(20\,Pa\)
Step 3: Pressure outside air bubble at depth 40 cm
Relative density = 1.20 \[ \rho = 1.20 \times 10^3\ kg\,m^{-3} \] Depth: \[ h = 40\,cm = 0.40\,m \] Pressure at that depth: \[ P_{out} = P_0 + \rho g h \] \[ P_{out} = 1.01 \times 10^{5} + (1.20 \times 10^3)(9.8)(0.40) \] \[ P_{out} = 1.01 \times 10^{5} + 4.70 \times 10^{3} \] \[ P_{out} \approx 1.057 \times 10^{5}\,Pa \]Step 4: Excess pressure inside air bubble
For an air bubble in liquid: \[ \Delta P' = \frac{2T}{r} \] \[ \Delta P' = \frac{2 \times 2.50 \times 10^{-2}}{5.0 \times 10^{-3}} \] \[ \Delta P' = 10\,Pa \]Step 5: Pressure inside the air bubble
\[ P_{in} = P_{out} + \Delta P' \] \[ P_{in} = 1.057 \times 10^{5} + 10 \] \[ P_{in} \approx 1.058 \times 10^{5}\,Pa \]Pressure inside air bubble ≈ \(1.058 \times 10^{5}\,Pa\)
Conceptual Illustration
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