Mechanical Properties Of FluidsClass 11 Physics NCERT Solutions

(a) The pressure in a fluid varies with depth according to the equation $P = P_0 + \rho gh$, where $P_0$ is the pressure at the top surface, $\rho$ is the density of the fluid, $g$ is the acceleration due to gravity, and $h$ is the depth. The human body contains blood, which is a fluid. The height of the blood column is greater at the feet than at the brain. Since pressure increases with the height of the fluid column (depth), the blood pressure at the feet is greater than at the brain.

(b) The density of air is not constant; it decreases with altitude. The formula for pressure variation, $P = P_a + \rho gh$, assumes a constant density, which is not true for the atmosphere. The air at higher altitudes is less dense because there is less air above it to compress it. Due to this exponential decrease in air density with height, the atmospheric pressure decreases significantly in the first few kilometers. At about 6 km, the density of air is much lower, and the cumulative weight of the air column above it reduces to nearly half the value at sea level. The upper layers of the atmosphere are very rare and contribute very little to the total pressure.

(c) Pressure is defined as the normal force acting per unit area. The force component considered is always perpendicular (normal) to the surface area, regardless of the orientation of the area. Since the direction of the force is specified by the orientation of the area itself, pressure does not have a specific direction of its own. At any point inside a fluid, the pressure is exerted equally in all directions. A quantity that has magnitude but no specific direction is a scalar. Therefore, hydrostatic pressure is a scalar quantity.

(a) The angle of contact depends on the relative strength of cohesive forces (between molecules of the liquid) and adhesive forces (between molecules of the liquid and the solid). In the case of mercury and glass, the cohesive force between mercury molecules is much stronger than the adhesive force between mercury and glass molecules. This causes mercury to pull itself inwards, forming an obtuse angle of contact. For water and glass, the adhesive force between water and glass molecules is stronger than the cohesive force between water molecules. This causes water to be pulled towards the glass surface, resulting in an acute angle of contact.

(b) This phenomenon is directly related to the angle of contact. Since the adhesive forces between water and glass are stronger than the cohesive forces of water, water tends to maximize its contact with the glass surface, causing it to spread out and wet the glass. Conversely, since the cohesive forces of mercury are much stronger than the adhesive forces with glass, mercury molecules are strongly attracted to each other and tend to minimize their contact with the glass surface. This causes mercury to pull into a bead or drop.

(c) Surface tension is defined as the force per unit length acting on an imaginary line drawn on the liquid's surface. It is an intrinsic property of the liquid at a given temperature, determined by the intermolecular cohesive forces. It depends on the nature of the liquid and the surrounding medium, not on the macroscopic dimensions like the area of the surface. While the total surface energy (Surface Tension × Area) depends on the area, the surface tension itself is a characteristic constant for the liquid.

(d) Detergents are wetting agents. When dissolved in water, they reduce the surface tension of water. According to the relation $S_{sa} = S_{sl} + S_{la} \cos \theta$, a decrease in the liquid-air surface tension ($S_{la}$) leads to an increase in $\cos \theta$, which means the angle of contact $\theta$ becomes smaller (more acute). A small angle of contact allows the detergent solution to spread over a larger area and penetrate into the pores of the fabric, enhancing its cleaning action.

(e) A liquid surface tends to acquire the minimum possible surface area due to surface tension. For a given volume, a sphere is the geometric shape that has the minimum surface area. When a liquid drop is under no external forces (like gravity), the forces of surface tension are dominant. These forces act to minimize the surface energy by minimizing the surface area, thus pulling the drop into a spherical shape.

(a) Surface tension of liquids generally decreases with temperatures. (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 (as expressed by the equation of continuity). (e) For the model of a plane in a wind tunnel, turbulence occurs at a greater speed for turbulence for an actual plane. (This is related to the concept of Reynolds number, which depends on velocity and length. To have the same Reynolds number for a smaller model, the speed must be higher).

(a) This is an application of Bernoulli's principle. When you blow over the top surface of the paper, the speed of air above the paper increases. According to Bernoulli's principle, where the speed of a fluid is high, the pressure is low. This creates a lower pressure on the upper surface of the paper compared to the atmospheric pressure on the lower surface. The pressure difference results in a net upward force (dynamic lift) that keeps the paper horizontal.

(b) This is explained by the equation of continuity ($A_1v_1 = A_2v_2$). When you place your fingers over the tap opening, you reduce the effective cross-sectional area ($A_2$) for the water to flow out. To maintain a constant volume flow rate (conservation of mass), the velocity ($v_2$) of the water must increase significantly. This is why fast jets of water gush through the small openings.

(c) The flow rate of a fluid through a narrow tube (like a needle) is described by Poiseuille's formula: $Q = \frac{\pi \Delta P r^4}{8 \eta L}$. The flow rate ($Q$) is proportional to the fourth power of the radius ($r^4$) of the needle but only directly proportional to the pressure difference ($\Delta P$) applied by the thumb. A small change in the radius of the needle has a much more significant impact on the flow rate than a change in thumb pressure. Therefore, the needle size is a more effective controller of the flow rate.

(d) This is a consequence of Newton's third law of motion and the principle of conservation of momentum. As the fluid flows out of the hole with a certain velocity, it carries momentum. To conserve the total momentum of the system (vessel + fluid), the vessel must gain an equal and opposite momentum. This results in a backward force, or thrust, on the vessel.

(e) This phenomenon is known as the Magnus effect, which is an application of Bernoulli's principle. A spinning ball drags a layer of air with it. If the ball is spinning and moving forward, the velocity of air on one side of the ball (where the spin direction is the same as the air flow) is higher than on the other side. This difference in air speeds creates a pressure difference, resulting in a sideways force that deflects the ball from its normal parabolic path.

Given: Mass of the girl, $m = 50 \text{ kg}$ Diameter of the heel, $d = 1.0 \text{ cm} = 1.0 \times 10^{-2} \text{ m}$ Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$

To Find: The pressure exerted by the heel, $P$

Formula: Pressure is defined as force per unit area: $P = \frac{F}{A}$ Force exerted by the girl is her weight, $F = mg$. The area of the circular heel is $A = \pi r^2 = \pi (\frac{d}{2})^2$.

Calculation: First, calculate the force (weight): $F = 50 \text{ kg} \times 9.8 \text{ m/s}^2 = 490 \text{ N}$

Next, calculate the area of the heel: Radius, $r = \frac{d}{2} = \frac{1.0 \times 10^{-2} \text{ m}}{2} = 0.5 \times 10^{-2} \text{ m}$ Area, $A = \pi r^2 = \pi (0.5 \times 10^{-2} \text{ m})^2 = \pi \times 0.25 \times 10^{-4} \text{ m}^2 = 7.854 \times 10^{-5} \text{ m}^2$

Now, calculate the pressure: $P = \frac{490 \text{ N}}{7.854 \times 10^{-5} \text{ m}^2} = 6.24 \times 10^6 \text{ N/m}^2 = 6.24 \times 10^6 \text{ Pa}$

Final Answer: The pressure exerted by the heel on the horizontal floor is $6.24 \times 10^6 \text{ Pa}$.

Given: Density of French wine, $\rho = 984 \text{ kg m}^{-3}$ Normal atmospheric pressure, $P_{atm} = 1.013 \times 10^5 \text{ Pa}$ Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$

To Find: The height of the wine column, $h$

Formula: The pressure exerted by a fluid column is given by: $P = \rho g h$ For the barometer to measure atmospheric pressure, the pressure of the wine column must be equal to the atmospheric pressure: $P_{atm} = \rho g h$

Calculation: Rearranging the formula to solve for $h$: $h = \frac{P_{atm}}{\rho g}$ Substituting the given values: $h = \frac{1.013 \times 10^5 \text{ Pa}}{(984 \text{ kg m}^{-3}) \times (9.8 \text{ m/s}^2)}$ $h = \frac{1.013 \times 10^5}{9643.2} \text{ m}$ $h \approx 10.5 \text{ m}$

Final Answer: The height of the wine column for normal atmospheric pressure would be approximately $10.5 \text{ m}$.

Given: Maximum stress the structure can withstand, $P_{max} = 10^9 \text{ Pa}$ Depth of the ocean, $h = 3 \text{ km} = 3 \times 10^3 \text{ m}$ Density of sea water, $\rho \approx 1.03 \times 10^3 \text{ kg m}^{-3}$ Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$

To Find: To determine if the structure is suitable by comparing the pressure at the given depth with the maximum withstandable stress.

Formula: The hydrostatic pressure exerted by the sea water at depth $h$ is given by: $P = \rho g h$

Calculation: Calculate the pressure at a depth of 3 km: $P = (1.03 \times 10^3 \text{ kg m}^{-3}) \times (9.8 \text{ m/s}^2) \times (3 \times 10^3 \text{ m})$ $P = 1.03 \times 9.8 \times 3 \times 10^6 \text{ Pa}$ $P = 30.282 \times 10^6 \text{ Pa} \approx 3.03 \times 10^7 \text{ Pa}$

Comparison: Compare the calculated pressure with the maximum stress the structure can withstand: Pressure at 3 km depth, $P = 3.03 \times 10^7 \text{ Pa}$ Maximum stress, $P_{max} = 10^9 \text{ Pa}$

Clearly, $P < P_{max}$ ($3.03 \times 10^7 \text{ Pa} < 10^9 \text{ Pa}$).

Final Answer: The pressure exerted by the ocean at a depth of 3 km is approximately $3.03 \times 10^7 \text{ Pa}$, which is significantly less than the maximum stress of $10^9 \text{ Pa}$ that the structure can withstand. Therefore, the structure is suitable for this purpose.

Given: Maximum mass of the car, $m = 3000 \text{ kg}$ Area of cross-section of the larger piston, $A_2 = 425 \text{ cm}^2 = 425 \times 10^{-4} \text{ m}^2$ Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$

To Find: The maximum pressure the smaller piston would have to bear, $P_1$.

Formula: According to Pascal's law, the pressure applied to the smaller piston is transmitted undiminished to the larger piston. Therefore, the pressure on the smaller piston ($P_1$) is equal to the pressure on the larger piston ($P_2$). $P_1 = P_2$ The pressure on the larger piston is due to the weight of the car: $P_2 = \frac{F_2}{A_2} = \frac{mg}{A_2}$

Calculation: First, calculate the force (weight of the car) on the larger piston: $F_2 = m \times g = 3000 \text{ kg} \times 9.8 \text{ m/s}^2 = 29400 \text{ N}$

Now, calculate the pressure on the larger piston: $P_2 = \frac{29400 \text{ N}}{425 \times 10^{-4} \text{ m}^2} = \frac{29400}{0.0425} \text{ Pa}$ $P_2 \approx 6.917 \times 10^5 \text{ Pa}$

Since $P_1 = P_2$, the pressure the smaller piston must bear is the same.

Final Answer: The maximum pressure the smaller piston would have to bear is approximately $6.92 \times 10^5 \text{ Pa}$.

Given: Height of the water column, $h_w = 10.0 \text{ cm} = 0.10 \text{ m}$ Height of the spirit column, $h_s = 12.5 \text{ cm} = 0.125 \text{ m}$ Density of water, $\rho_w = 1000 \text{ kg m}^{-3}$ The mercury columns in the two arms are at the same level.

To Find: The specific gravity of spirit.

Formula: Since the mercury levels are the same in both arms, the pressure exerted by the water column in one arm must be equal to the pressure exerted by the spirit column in the other arm at the level of the mercury surface. Pressure exerted by water column: $P_w = h_w \rho_w g$ Pressure exerted by spirit column: $P_s = h_s \rho_s g$ Equating the pressures: $h_w \rho_w g = h_s \rho_s g$ $h_w \rho_w = h_s \rho_s$ Specific gravity of spirit is the ratio of its density to the density of water: $\text{Specific Gravity} = \frac{\rho_s}{\rho_w}$

Calculation: From the pressure balance equation, we can find the ratio $\frac{\rho_s}{\rho_w}$: $\frac{\rho_s}{\rho_w} = \frac{h_w}{h_s}$ Substituting the given values: $\text{Specific Gravity} = \frac{10.0 \text{ cm}}{12.5 \text{ cm}} = \frac{10}{12.5} = 0.8$

Final Answer: The specific gravity of spirit is 0.8.

Given: Initial height of water, $h_{w1} = 10.0 \text{ cm}$ Initial height of spirit, $h_{s1} = 12.5 \text{ cm}$ Additional height of water and spirit poured = $15.0 \text{ cm}$ New height of water, $h_{w2} = 10.0 + 15.0 = 25.0 \text{ cm}$ New height of spirit, $h_{s2} = 12.5 + 15.0 = 27.5 \text{ cm}$ Specific gravity of spirit, $SG_s = 0.8 \implies \rho_s = 0.8 \rho_w$ Specific gravity of mercury, $SG_m = 13.6 \implies \rho_m = 13.6 \rho_w$ Density of water, $\rho_w = 1 \text{ g/cm}^3$

To Find: The difference in the levels of mercury in the two arms, $h$.

Formula: Let the mercury level in the spirit arm be higher than in the water arm by a height $h$. We can balance the pressure at the level of the lower mercury surface (which is in the water arm). The total pressure in the water arm at this level is $P_{water} = h_{w2} \rho_w g$. The total pressure in the spirit arm at this level is $P_{spirit} = h_{s2} \rho_s g + h \rho_m g$. Equating the pressures: $h_{w2} \rho_w g = h_{s2} \rho_s g + h \rho_m g$ $h_{w2} \rho_w = h_{s2} \rho_s + h \rho_m$ Divide by $\rho_w$: $h_{w2} = h_{s2} (\frac{\rho_s}{\rho_w}) + h (\frac{\rho_m}{\rho_w})$ $h_{w2} = h_{s2} (SG_s) + h (SG_m)$

Calculation: Substitute the known values into the equation: $25.0 \text{ cm} = (27.5 \text{ cm} \times 0.8) + (h \times 13.6)$ $25.0 = 22.0 + 13.6 h$ $13.6 h = 25.0 - 22.0$ $13.6 h = 3.0$ $h = \frac{3.0}{13.6} \approx 0.2205 \text{ cm}$

Final Answer: The difference in the levels of mercury in the two arms is approximately $0.22 \text{ cm}$.

No, Bernoulli's equation cannot be used to describe the flow of water through a rapid in a river. Bernoulli's equation is based on several key assumptions about the fluid flow:

The flow of water in a rapid is turbulent, not steady. In turbulent flow, the velocity at any point fluctuates erratically, and whirlpools or eddies are formed. This violates the primary condition of steady, streamline flow required for Bernoulli's principle. Furthermore, in turbulent flow, a significant amount of energy is dissipated as heat due to viscous forces, which contradicts the non-viscous assumption and the principle of energy conservation upon which Bernoulli's equation is based. Therefore, Bernoulli's equation is not applicable to such chaotic and dissipative flow.

It depends on the specific application of Bernoulli's equation. Bernoulli's equation is: $P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant}$ When applied between two points (1 and 2), it is written as: $P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2$ This can be rearranged to focus on the pressure difference: $P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2) + \rho g (h_2 - h_1)$ Absolute pressure ($P_{abs}$) is related to gauge pressure ($P_{gauge}$) and atmospheric pressure ($P_{atm}$) by $P_{abs} = P_{gauge} + P_{atm}$. If we substitute this into the pressure difference term: $P_1 - P_2 = (P_{1,gauge} + P_{atm}) - (P_{2,gauge} + P_{atm}) = P_{1,gauge} - P_{2,gauge}$ As shown, the atmospheric pressure term cancels out. Therefore, if the equation is used to relate the properties of the fluid at two different points within the flow where both points are subject to the same external atmospheric pressure, it does not matter whether gauge or absolute pressures are used, as only the difference in pressure is relevant.

However, if one of the pressures in the equation is atmospheric pressure itself (for example, at a free surface open to the atmosphere, where $P = P_{atm}$), then one must be consistent. If absolute pressure is used for one point, it must be used for the other. Using gauge pressure at one point (where $P_{gauge} = 0$ for an open surface) and absolute pressure at another would lead to an incorrect result.

Given: Length of the tube, $L = 1.5 \text{ m}$ Radius of the tube, $r = 1.0 \text{ cm} = 1.0 \times 10^{-2} \text{ m}$ Mass flow rate, $M = 4.0 \times 10^{-3} \text{ kg s}^{-1}$ Density of glycerine, $\rho = 1.3 \times 10^3 \text{ kg m}^{-3}$ Viscosity of glycerine, $\eta = 0.83 \text{ Pa s}$

To Find: The pressure difference between the two ends, $\Delta P$.

Formula: The flow of a viscous fluid through a pipe is described by Poiseuille's formula for the volume flow rate, $Q$: $Q = \frac{\pi \Delta P r^4}{8 \eta L}$ The volume flow rate $Q$ is related to the mass flow rate $M$ and density $\rho$ by: $Q = \frac{M}{\rho}$

Calculation: First, calculate the volume flow rate $Q$: $Q = \frac{4.0 \times 10^{-3} \text{ kg s}^{-1}}{1.3 \times 10^3 \text{ kg m}^{-3}} = 3.077 \times 10^{-6} \text{ m}^3 \text{s}^{-1}$

Now, rearrange Poiseuille's formula to solve for the pressure difference $\Delta P$: $\Delta P = \frac{8 \eta L Q}{\pi r^4}$ Substitute the known values: $\Delta P = \frac{8 \times (0.83 \text{ Pa s}) \times (1.5 \text{ m}) \times (3.077 \times 10^{-6} \text{ m}^3 \text{s}^{-1})}{\pi \times (1.0 \times 10^{-2} \text{ m})^4}$ $\Delta P = \frac{30.64 \times 10^{-6}}{\pi \times 10^{-8}} = \frac{30.64}{\pi} \times 10^2 \text{ Pa}$ $\Delta P = 9.754 \times 10^2 \text{ Pa} = 975.4 \text{ Pa}$

Check for Laminar Flow: To check if the flow is laminar, we calculate the Reynolds number, $R_e = \frac{\rho v d}{\eta}$, where $v$ is the average velocity and $d$ is the diameter. Average velocity $v = \frac{Q}{A} = \frac{Q}{\pi r^2} = \frac{3.077 \times 10^{-6}}{\pi (10^{-2})^2} = 9.79 \times 10^{-3} \text{ m/s}$. Diameter $d = 2r = 0.02 \text{ m}$. $R_e = \frac{(1.3 \times 10^3) \times (9.79 \times 10^{-3}) \times (0.02)}{0.83} \approx 0.307$ Since $R_e \approx 0.31$ is much less than the critical value of 2000 for the onset of turbulence, the flow is indeed laminar.

Final Answer: The pressure difference between the two ends of the tube is $9.75 \times 10^2 \text{ Pa}$.

Given: Flow speed on the upper surface, $v_2 = 70 \text{ m s}^{-1}$ Flow speed on the lower surface, $v_1 = 63 \text{ m s}^{-1}$ Area of the wing, $A = 2.5 \text{ m}^2$ Density of air, $\rho = 1.3 \text{ kg m}^{-3}$

To Find: The lift on the wing, $F_{lift}$.

Formula: The lift on the wing is due to the pressure difference between the lower and upper surfaces. According to Bernoulli's principle (ignoring the small difference in height between the surfaces): $P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2)$ The lift force is this pressure difference multiplied by the area of the wing: $F_{lift} = (P_1 - P_2) \times A = \frac{1}{2} \rho (v_2^2 - v_1^2) A$

Calculation: Substitute the given values into the formula: $F_{lift} = \frac{1}{2} \times (1.3 \text{ kg m}^{-3}) \times [(70 \text{ m s}^{-1})^2 - (63 \text{ m s}^{-1})^2] \times (2.5 \text{ m}^2)$ $F_{lift} = 0.65 \times (4900 - 3969) \times 2.5 \text{ N}$ $F_{lift} = 0.65 \times 931 \times 2.5 \text{ N}$ $F_{lift} = 1512.9 \text{ N} \approx 1.51 \times 10^3 \text{ N}$

Final Answer: The lift on the wing is approximately $1.51 \times 10^3 \text{ N}$.

Figure 9.20 (a) is incorrect.

Reason: This problem involves the application of both the equation of continuity and Bernoulli's principle.

In Figure 9.20(a), the pipe has a constriction in the middle. According to the equation of continuity, the fluid speed is highest at this constriction. According to Bernoulli's principle, the pressure should therefore be lowest at this point. The height of the liquid in the vertical tube indicates the pressure at that point in the pipe. In Figure (a), the liquid level in the central tube is the highest, indicating the highest pressure. This contradicts Bernoulli's principle.

Figure 9.20 (b) correctly depicts the situation. The liquid level in the vertical tube at the constriction is the lowest, indicating the lowest pressure, which corresponds to the highest fluid speed at that point. Therefore, Figure (a) is incorrect.

Given: Cross-sectional area of the tube, $A_1 = 8.0 \text{ cm}^2 = 8.0 \times 10^{-4} \text{ m}^2$ Number of holes, $n = 40$ Diameter of each hole, $d = 1.0 \text{ mm} = 1.0 \times 10^{-3} \text{ m}$ Speed of liquid inside the tube, $v_1 = 1.5 \text{ m min}^{-1} = \frac{1.5}{60} \text{ m s}^{-1} = 0.025 \text{ m s}^{-1}$

To Find: The speed of ejection of the liquid through the holes, $v_2$.

Formula: According to the equation of continuity, the volume flow rate is constant: $A_1 v_1 = A_2 v_2$ where $A_2$ is the total cross-sectional area of all the holes. The area of one hole is $\pi r^2 = \pi (\frac{d}{2})^2$. The total area of 40 holes is $A_2 = n \times \pi (\frac{d}{2})^2$.

Calculation: First, calculate the total area of the holes, $A_2$: Radius of one hole, $r = \frac{d}{2} = \frac{1.0 \times 10^{-3}}{2} = 0.5 \times 10^{-3} \text{ m}$ Area of one hole = $\pi (0.5 \times 10^{-3})^2 = \pi \times 0.25 \times 10^{-6} \text{ m}^2$ Total area, $A_2 = 40 \times (\pi \times 0.25 \times 10^{-6}) = 10 \pi \times 10^{-6} \text{ m}^2 = 3.142 \times 10^{-5} \text{ m}^2$

Now, use the equation of continuity to find $v_2$: $v_2 = \frac{A_1 v_1}{A_2}$ $v_2 = \frac{(8.0 \times 10^{-4} \text{ m}^2) \times (0.025 \text{ m s}^{-1})}{3.142 \times 10^{-5} \text{ m}^2}$ $v_2 = \frac{2.0 \times 10^{-5}}{3.142 \times 10^{-5}} \text{ m s}^{-1}$ $v_2 \approx 0.6365 \text{ m s}^{-1}$

Final Answer: The speed of ejection of the liquid through the holes is approximately $0.64 \text{ m s}^{-1}$.

Given: Weight supported by the film, $F = 1.5 \times 10^{-2} \text{ N}$ Length of the slider, $L = 30 \text{ cm} = 0.30 \text{ m}$

To Find: The surface tension of the soap film, $S$.

Formula: The force due to surface tension supports the weight. A soap film has two surfaces (front and back) that are in contact with the slider. Therefore, the total length over which the surface tension acts is $2L$. The force exerted by the surface tension is given by: $F = S \times (2L)$

Calculation: Rearrange the formula to solve for $S$: $S = \frac{F}{2L}$ Substitute the given values: $S = \frac{1.5 \times 10^{-2} \text{ N}}{2 \times 0.30 \text{ m}}$ $S = \frac{1.5 \times 10^{-2}}{0.60} \text{ N m}^{-1}$ $S = 2.5 \times 10^{-2} \text{ N m}^{-1}$

Final Answer: The surface tension of the film is $2.5 \times 10^{-2} \text{ N m}^{-1}$.

The weight supported by the film will be the same in all three cases: $4.5 \times 10^{-2} \text{ N}$.

Explanation: The force due to surface tension, which balances the weight, is given by the formula $F = S \times (2L)$, where $S$ is the surface tension of the liquid and $L$ is the length of the slider on which the film acts. A soap film has two surfaces, hence the factor of 2.

Since both the surface tension ($S$) and the length of the slider ($L$) are the same in all three configurations, the total upward force exerted by the film ($F = S \times 2L$) is also the same. This force balances the weight supported. Therefore, the weight supported by the film in Fig. (b) and Fig. (c) is the same as in Fig. (a), which is $4.5 \times 10^{-2} \text{ N}$.

Given: Radius of the mercury drop, $r = 3.00 \text{ mm} = 3.00 \times 10^{-3} \text{ m}$ Surface tension of mercury, $S = 4.65 \times 10^{-1} \text{ N m}^{-1}$ Atmospheric pressure (outside pressure), $P_o = 1.01 \times 10^5 \text{ Pa}$

Calculation: First, calculate the excess pressure: $P_{excess} = \frac{2 \times (4.65 \times 10^{-1} \text{ N m}^{-1})}{3.00 \times 10^{-3} \text{ m}}$ $P_{excess} = \frac{0.93}{3.00 \times 10^{-3}} \text{ Pa} = 0.31 \times 10^3 \text{ Pa} = 310 \text{ Pa}$

Next, calculate the total pressure inside the drop: $P_i = (1.01 \times 10^5 \text{ Pa}) + (310 \text{ Pa})$ $P_i = 101000 \text{ Pa} + 310 \text{ Pa} = 101310 \text{ Pa}$ $P_i = 1.0131 \times 10^5 \text{ Pa}$

Final Answer: The excess pressure inside the drop is $310 \text{ Pa}$. The total pressure inside the drop of mercury is $1.0131 \times 10^5 \text{ Pa}$.

Given: Radius of the soap bubble, $r = 5.00 \text{ mm} = 5.00 \times 10^{-3} \text{ m}$ Surface tension of soap solution, $S = 2.50 \times 10^{-2} \text{ N m}^{-1}$

To Find: Excess pressure, $P_{excess}$.

Formula: A soap bubble has two liquid-air interfaces (inner and outer). The excess pressure is given by: $P_{excess} = \frac{4S}{r}$

Calculation: $P_{excess} = \frac{4 \times (2.50 \times 10^{-2} \text{ N m}^{-1})}{5.00 \times 10^{-3} \text{ m}}$ $P_{excess} = \frac{10.0 \times 10^{-2}}{5.00 \times 10^{-3}} \text{ Pa} = 2.0 \times 10^1 \text{ Pa} = 20 \text{ Pa}$

Answer Part 1: The excess pressure inside the soap bubble is $20 \text{ Pa}$.

Given: Radius of the air bubble, $r = 5.00 \times 10^{-3} \text{ m}$ Depth, $h = 40.0 \text{ cm} = 0.40 \text{ m}$ Relative density of soap solution = 1.20 Density of soap solution, $\rho = 1.20 \times 1000 \text{ kg m}^{-3} = 1.2 \times 10^3 \text{ kg m}^{-3}$ Atmospheric pressure, $P_{atm} = 1.01 \times 10^5 \text{ Pa}$ Acceleration due to gravity, $g = 9.8 \text{ m s}^{-2}$ Surface tension, $S = 2.50 \times 10^{-2} \text{ N m}^{-1}$

To Find: The pressure inside the bubble, $P_{in}$.

Formula: An air bubble inside a liquid has only one liquid-air interface. The excess pressure is $P_{excess} = \frac{2S}{r}$. The pressure just outside the bubble ($P_{out}$) is the sum of atmospheric pressure and the hydrostatic pressure due to the liquid column: $P_{out} = P_{atm} + h \rho g$ The pressure inside the bubble is the sum of the outside pressure and the excess pressure: $P_{in} = P_{out} + P_{excess} = (P_{atm} + h \rho g) + \frac{2S}{r}$

Answer Part 2: The pressure inside the bubble would be approximately $1.06 \times 10^5 \text{ Pa}$.