Question 1

A body is in motion inside a liquid, and the viscous resistive force on it is directly proportional to its speed. What are the dimensions of the proportionality constant?

Accepted Answer

[ML⁰T⁻¹]. The proportionality constant relates the viscous force to speed, which means it must have dimensions that allow the equation to balance. Since the resistive force has dimensions of [MLT⁻²] and speed has dimensions of [LT⁻¹], the proportionality constant must have dimensions of [ML⁰T⁻¹] to ensure the units are consistent.

Question 2

A capillary tube has its inner surface lined with wax and is then placed in water. Relative to a clean, unwaxed capillary, how do the contact angle (θ) and the height (h) to which water rises change?

Accepted Answer

θ increases and h decreases. The wax lining reduces the adhesive forces between the water and the tube, leading to an increased contact angle, which means the water does not wet the surface as effectively. Consequently, this results in a lower height of water rise in the capillary tube compared to a clean, unwaxed capillary.

Question 3

A charged, isolated spherical soap bubble of radius r has internal pressure equal to atmospheric pressure. If the charge on the bubble is given by Xπ√(2Tε), what is the value of X?

Accepted Answer

8. Internal = atmospheric means surface-tension excess pressure 4T/r equals outward electrostatic pressure sigma^2/(2*eps0), with sigma = Q/(4*pi*r^2). Solving gives Q = 8*pi*sqrt(2*T*eps0)*r^(3/2), so X = 8.

Question 4

Two capillary tubes, one of length L and radius R and the other of length 2L and radius 2R, are joined one after the other in series. If the flow rate through a single capillary is X = πPR⁴ / 8ηL, then the combined flow rate through the series arrangement is:

Accepted Answer

8/9 X. Flow resistance ~ L/R^4. First tube: 8*eta*L/(pi*R^4). Second (2L, 2R): 8*eta*2L/(pi*16R^4) = eta*L/(pi*R^4) = (1/8) of first. Total = 9 times the unit. Flow = X/9 *8... combined = (8/9)X.

Question 5

A liquid will fail to wet the surface of a solid when the angle of contact is

Accepted Answer

greater than 90°. A liquid fails to wet a solid surface when the angle of contact exceeds 90°, indicating that the cohesive forces within the liquid are stronger than the adhesive forces between the liquid and the solid, leading to a tendency for the liquid to bead up rather than spread out.

Question 6

A gold sphere of a given size falls through a viscous liquid and attains a terminal speed of 0.2 m/s. If the gold has density 19.5 kg/m³ and the liquid has density 1.5 kg/m³, what terminal speed will a silver sphere of the same size have in the same liquid, given that silver has density 10

Accepted Answer

0.1 m/s. For the same size and liquid, v is proportional to (rho_sphere - rho_liquid). v_silver = 0.2 * (10.5 - 1.5)/(19.5 - 1.5) = 0.2 * 9/18 = 0.1 m/s.

Question 7

Choose the statement that is true among the following: (a) A Bunsen burner and a sprayer operate on Bernoulli’s principle. (b) The flow of blood in arteries is explained by Bernoulli’s principle. (c) A siphon functions because of atmospheric pressure. (d) Every statement given is correct.

Accepted Answer

(d). All the statements provided are correct as they each illustrate different applications of Bernoulli's principle and atmospheric pressure in various systems, such as combustion, fluid dynamics in the body, and siphoning.

Question 8

A spherical ball of radius r and density ρ is dropped from rest and falls freely through a height h before it enters water. If its speed remains unchanged after it enters the water, and the viscosity of water is η, then h is equal to:

Accepted Answer

(2/81) r⁴ ((ρ-1)/η) g. Terminal velocity v = (2/9)(r^2/eta)(rho-1)g. Entry speed sqrt(2gh) = v, so h = v^2/(2g) = (2/81) r^4 ((rho-1)/eta) g (the 2/81 r^4 form), not the 2/9 r^4 coefficient.

Question 9

Two adjacent layers of water move parallel to each other with a relative speed of 8 cm/s. If the separation between the layers, measured perpendicular to their motion, is 0.1 cm, what is the velocity gradient?

Accepted Answer

80 s⁻¹. Velocity gradient = relative speed / separation = 8 cm/s / 0.1 cm = 80 s^-1. (Options 1 and 2 are also duplicate values.)

Question 10

A solid spherical ball of volume V is dropped into a liquid of density ρ2. The ball is made of a material whose density is ρ1, with ρ1 > ρ2. The liquid exerts a resistive force on the ball proportional to the square of its speed, given by F = -kv², where k > 0. The ball’s terminal speed is

Accepted Answer

√(Vg(ρ1-ρ2))/k. At terminal speed the net downward force (rho1-rho2)*V*g equals the drag k*v^2, so v = sqrt(V*g*(rho1-rho2)/k). The buoyancy-corrected square-root form (option with (rho1-rho2)) is correct, not the buoyancy-free expression.

Question 11

Two liquids having densities d1 and d2 are made to pass through identical capillary tubes under the same pressure difference. If t1 and t2 are the times required for equal masses of the two liquids to flow out, then the ratio of their coefficients of viscosity is

Accepted Answer

(d1 t1)/(d2 t2). The correct option is derived from the relationship between viscosity, density, and flow rate in capillary tubes. According to Poiseuille's law, the viscosity of a fluid is directly proportional to the product of its density and the time taken for a given mass to flow, leading to the ratio of viscosities being expressed as (d1 t1)/(d2 t2).

Question 12

Which one of the following represents the correct dimensions of the coefficient of viscosity?

Accepted Answer

[ML⁻¹T⁻¹]. The coefficient of viscosity measures the internal friction of a fluid, which relates shear stress to shear rate. Its dimensions are derived from the formula for viscosity, resulting in mass per unit length per unit time, represented as [ML⁻¹T⁻¹].

Question 13

The following observations were taken for determining surface tension T of water by capillary method: Diameter of capillary, D = 1.25 × 10−2 m rise of water, h = 1.45 × 10−2 m Using g = 9.80 m/s2 and the simplified relation T = r h ρ g / 2 × 10³ N/m, the possible error in surface tension i

Accepted Answer

1.5%. T proportional to D*h, so dT/T = dD/D + dh/h. With least count 0.01e-2 m: dD/D = 0.01/1.25 = 0.8% and dh/h = 0.01/1.45 = 0.69%, summing to about 1.49% ~ 1.5%.

Question 14

A solid spherical ball of volume V and density ρ1 is allowed to fall in a liquid of density ρ2, with ρ2 < ρ1. The liquid exerts a resistive force on the ball proportional to the square of its speed, given by F = -kv², where k > 0. The ball’s terminal velocity is

Accepted Answer

√(Vg(σ1 - σ2))/k. At terminal velocity the net downward force equals drag: V*g*(rho1 - rho2) = k*v^2, so v = sqrt(V*g*(rho1 - rho2)/k). The stored option omits the square root.

Question 15

Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly (Surface tension of soap solution = 0.03 N m⁻¹)

Accepted Answer

0.4π mJ. A soap bubble has two surfaces, so W = T*2*4*pi*(r2^2 - r1^2) = 0.03*8*pi*(0.05^2 - 0.03^2) = 0.03*8*pi*0.0016 = 3.84e-4 * pi J ~ 0.4*pi mJ.

Question 16

Water is flowing continuously from a tap having an internal diameter 8 × 10⁻³ m. The water velocity as it leaves the tap is 0.4 m s⁻¹. The diameter of the water stream at a distance 2 × 10⁻¹ m below the tap is close to:

Accepted Answer

3.6 × 10⁻³ m. Speed after falling 0.2 m: v2 = sqrt(v1^2 + 2 g h) = sqrt(0.16 + 4) = 2.04 m/s. Continuity A1 v1 = A2 v2 gives d2 = d1*sqrt(v1/v2) = 8e-3*sqrt(0.4/2.04) ~ 3.6e-3 m.

Question 17

Two mercury drops (each of radius r) merge to form bigger drop. The surface energy of the bigger drop, if T is the surface tension, is:

Accepted Answer

2^(8/3)πr²T. When two mercury drops merge, their combined volume increases, leading to a larger drop with a new radius. The surface area of the larger drop is greater than the sum of the surface areas of the two smaller drops, resulting in a change in surface energy that can be calculated using the formula for surface energy, which incorporates the new radius and surface tension.

Question 18

If a ball of steel (density ρ = 7.8 g cm⁻³) attains a terminal velocity of 10 cm s⁻¹ when falling in water (Coefficient of viscosity of water = 8.5 × 10⁻⁴ Pa s), then, its terminal velocity in glycerine (ρ = 1.2 g cm⁻³, η = 13.2 Pa s) would be, nearly

Accepted Answer

6.25 × 10⁻⁴ cm s⁻¹. v2/v1 = [(7.8-1.2)/13.2] / [(7.8-1.0)/8.5e-4] = (0.5)/(8000) = 6.25e-5. v2 = 10*6.25e-5 = 6.25e-4 cm/s.

Question 19

If ‘M’ is the mass of water that rises in a capillary tube of radius ‘r’, then mass of water which will rise in a capillary tube of radius ‘2r’ is

Accepted Answer

2M. Capillary rise h = 2T cos(theta)/(rho g r) varies as 1/r. Mass = rho * pi r^2 * h varies as r^2 * (1/r) = r. Doubling the radius doubles the mass, giving 2M.

Question 20

A cube made of wood has density d and edge length ℓ. It floats in a liquid of density ρ with its top and bottom faces parallel to the liquid surface. When it is given a small downward displacement and then released, it executes SHM with period T. Which expression gives T?

Accepted Answer

2π√(ℓd/ρg). The correct option is derived from the principles of buoyancy and simple harmonic motion (SHM). When the cube is displaced, the restoring force acting on it is proportional to the volume of the cube submerged in the liquid, which relates to its density and the density of the liquid, leading to the expression for the period of oscillation being dependent on the ratio of the cube's density to the liquid's density.

Question 21

A small soap bubble of radius 4 cm is trapped inside another bubble of radius 6 cm without any contact. Let P2 be the pressure inside the inner bubble and P0 the pressure outside the outer bubble. Radius of another bubble with pressure difference P2 - P0 between its inside and outside woul

Accepted Answer

2.4 cm. The pressure difference between the inside and outside of a bubble is inversely related to its radius, according to the Young-Laplace equation. Given the pressure difference between the inner bubble and the external pressure, the radius of the new bubble can be calculated, resulting in a radius of 2.4 cm.

Question 22

When an air bubble of radius r rises from the bottom to the surface of a lake, its radius becomes 5r/4. Taking atmospheric pressure to be equal to 10m height of water column, the depth of the surface tension and the effect of temperature, would be approximately be (ignore surface tension a

Accepted Answer

9.5m. The increase in the radius of the air bubble as it rises indicates a decrease in pressure due to the reduction in water column height. Using the ideal gas law and the relationship between pressure and volume, we can calculate the depth at which the bubble's radius changes, leading to the conclusion that the depth is approximately 9.5m.

Question 23

The ratio of surface tensions of mercury and water is given to be 7.5 while the ratio of their densities is 13.6. Their contact angles, with glass, are close to 135° and 0°, respectively. It is observed that mercury gets depressed by an amount h in a capillary tube of radius r1, while wate

Accepted Answer

2/5. The ratio of the radii of the capillary tubes can be derived from the relationship between surface tension, density, and contact angle in capillary action. Given that mercury has a much higher surface tension and a larger contact angle compared to water, the ratio of the radii (r1/r2) is influenced by these factors, leading to the conclusion that it is approximately 2/5.

Question 24

Water from a tap emerges vertically downwards with an initial speed of 1.0 m s⁻¹. The cross-sectional area of the tap is 10⁻⁴ m². Assume that the pressure is constant throughout the stream of water and that the flow is streamlined. The cross-sectional area of the stream, 0.15 m below the t

Accepted Answer

5 × 10⁻⁵ m². Speed after falling 0.15 m: v = sqrt(1^2 + 2*10*0.15) = 2 m/s. By continuity A0*v0 = A*v, so A = 1e-4 * 1/2 = 5e-5 m^2.

Question 25

A submarine experiences a pressure of 5.05 × 10⁶ Pa at a depth of d1 in a sea. When it goes further to a depth of d2, it experiences a pressure of 8.08 × 10⁶ Pa. Then d2 − d1 is approximately (density of water = 10³ kg/m³ and acceleration due to gravity = 10 m s⁻²):

Accepted Answer

300 m. Pressure difference = rho*g*(d2-d1), so d2-d1 = (8.08x10^6 - 5.05x10^6)/(10^3 * 10) = 3.03x10^6/10^4 = 303 m, i.e. about 300 m.

Question 26

A solid sphere of radius R acquires a terminal velocity v1 when falling (due to gravity) through a viscous fluid having a coefficient of viscosity η. The sphere is broken into 27 identical solid spheres. If each of these spheres acquires a terminal velocity v2 when falling through the same

Accepted Answer

9. Terminal velocity v ~ R^2. Breaking into 27 equal spheres gives each radius R/3, so v2 = v1*(1/3)^2 = v1/9. Hence v1/v2 = 9.

Question 27

A wooden block floating in a bucket of water has 4/5 of its volume submerged. When certain amount of oil is poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is:

Accepted Answer

0.6. From 4/5 submerged, rho_block=0.8*rho_water. Floating just under oil with half in water, half in oil: 0.8 = 0.5*1 + 0.5*rho_oil -> rho_oil=0.6 relative to water.

Question 28

A liquid of density ρ is coming out of a hose pipe of radius a with a horizontal speed v and hits a mesh. 50% of the liquid passes through the mesh unaffected. 25% loses all of its momentum and 25% comes back with the same speed. The resultant pressure on the mesh will be:

Accepted Answer

3/4 ρv². Mass rate mdot = rho*pi*a^2*v. The 25% that stops contributes 0.25*mdot*v and the 25% that rebounds contributes 0.25*mdot*2v; total force = 0.75*mdot*v. Pressure = Force/(pi*a^2) = (3/4)*rho*v^2.

Question 29

A hollow spherical shell at outer radius R floats just submerged under the water surface. The inner radius of the shell is r. If the specific gravity of the shell material is 27/8 w.r.t. water, the value of r is:

Accepted Answer

8R/9. The specific gravity of the shell material indicates that it is denser than water, and since the shell is floating just submerged, the weight of the water displaced must equal the weight of the shell. By applying the principle of buoyancy and the relationship between the volumes of the inner and outer spheres, we can derive that the inner radius must be 8R/9 to satisfy these conditions.

Question 30

If the potential energy between two molecules is given by U = -A/r⁶ + B/r¹², then at equilibrium, separation between molecules, and the potential energy are:

Accepted Answer

(2B/A)^(1/6), -A²/4B. At equilibrium, the potential energy is minimized, which occurs when the derivative of U with respect to r equals zero. Solving this condition leads to the separation distance of (2B/A)^(1/6) and the corresponding potential energy of -A²/4B.

Question 31

A fluid is flowing through a horizontal pipe of varying cross-section, with speed v ms−1 at a point where the pressure is P Pascal. At another point where pressure is P/2 Pascal its speed is V ms−1. If the density of the fluid is ρ kg m−3 and the flow is streamline, then V is equal to:

Accepted Answer

√(P/ρ + v²). For horizontal streamline flow, P + (1/2)rho v^2 = (P/2) + (1/2)rho V^2. So (1/2)rho(V^2 - v^2) = P - P/2 = P/2, giving V^2 = v^2 + P/rho, i.e. V = sqrt(P/rho + v^2).

Question 32

The pressure acting on a submarine is 3 × 10⁵ Pa at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be: (Assume that atmospheric pressure is 1 × 10⁵ Pa. The density of water is 10³ kg m⁻³, g = 10 ms⁻²)

Accepted Answer

200/3 %. At first depth P = Patm + rho*g*h = 3e5 with Patm=1e5, so rho*g*h = 2e5. Doubling depth makes that term 4e5, giving P = 1e5 + 4e5 = 5e5. Percentage increase = (5e5-3e5)/3e5 = 2/3 = 200/3 %.

Question 33

An object is located at 2 km beneath the surface of the water. If the fractional compression ΔV/V is 1.36%, the ratio of hydraulic stress to the corresponding hydraulic strain will be ____. [Given: density of water is 1000 kg m⁻³ and g = 9.8 m s⁻²]

Accepted Answer

1.44 × 10⁹ N m⁻². The ratio of hydraulic stress to hydraulic strain is calculated using the formula for bulk modulus, which relates these two quantities. Given the fractional compression and the density of water, the calculated value aligns with the bulk modulus of water under the specified conditions, confirming that option D is correct.

Question 34

A hydraulic press can lift 100 kg when a mass'm' is placed on the smaller piston. It can lift ______ kg when the diameter of the larger piston is increased by 4 times and that of the smaller piston is decreased by 4 times, keeping the same mass 'm' on the smaller piston.

Accepted Answer

25600. In a hydraulic press, lifted load = m * (A_large/A_small). Increasing the larger diameter 4x multiplies its area by 16, and decreasing the smaller diameter 4x divides its area by 16, so the ratio grows 256x: 100 * 256 = 25600 kg.

Question 35

Two spherical soap bubbles of radii r1 and r2 in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to:

Accepted Answer

√(r1² + r2²). The correct option reflects the geometric relationship between the radii of the two bubbles when they combine, as the resulting bubble's radius is determined by the Pythagorean theorem, which accounts for the contributions of both original radii.

Question 36

A large number of water drops, each of radius r, combine to have a drop of radius R. If the surface tension is T and mechanical equivalent of heat is J, the rise in heat energy per unit volume will be:

Accepted Answer

(4) 3T/J (1/r − 1/R). The correct option is derived from the relationship between surface tension and the change in energy when smaller droplets merge into a larger one. The formula accounts for the difference in surface area before and after the merging process, leading to a rise in heat energy proportional to the surface tension and inversely related to the radii of the drops.

Question 37

In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius 2.0 × 10⁻⁵ m and density 1.2 × 10³ kg m⁻³. Take viscosity of liquid = 1.8 × 10⁻⁵ N s m⁻² (Neglect buoyancy due to air)

Accepted Answer

3.9 × 10⁻¹⁰ N. The correct option is derived from applying Stokes' law, which calculates the viscous force on a spherical object moving through a fluid. By substituting the given values for radius, viscosity, and density into the formula, the resulting force is accurately determined to be 3.9 × 10⁻¹⁰ N.

Question 38

The bulk modulus of a liquid is 3 × 10¹⁰ N m⁻². The pressure required to reduce the volume of liquid by 2% is

Accepted Answer

6 × 10⁸ N m⁻². Pressure required = bulk modulus times fractional volume change = 3e10 * 0.02 = 6 x 10^8 N/m^2.

Question 39

A drop of liquid of density ρ is floating half immersed in a liquid of density σ and surface tension 7.5 × 10⁻⁴ N cm⁻¹. The radius of drop in cm will be if (g = 10 m s⁻²)

Accepted Answer

15/√(2ρ−σ). For a half-immersed drop: (4/3)pi*r^3*rho*g = (2/3)pi*r^3*sigma*g + 2*pi*r*T, giving r^2 = 3T/(g(2rho - sigma)). With T = 7.5e-4 N/cm = 7.5e-2 N/m and g = 10, r = 0.15/sqrt(2rho - sigma) m = 15/sqrt(2rho - sigma) cm.

Question 40

If ρ is the density and η is coefficient of viscosity of fluid which flows with a speed v in the pipe of diameter d, the correct formula for Reynolds number Re is:

Accepted Answer

Re = ρvd / η. The Reynolds number is a dimensionless quantity that characterizes the flow of fluid, and it is defined as the ratio of inertial forces to viscous forces. The formula Re = ρvd / η correctly incorporates the fluid density (ρ), velocity (v), pipe diameter (d), and viscosity (η) to represent this relationship.

Question 41

A spherical soap bubble of radius 3 cm is formed inside another spherical soap bubble of radius 6 cm. If the internal pressure of the smaller bubble of radius 3 cm in the above system is equal to the internal pressure of the another single soap bubble of radius r cm. The value of r is ____

Accepted Answer

2.00. The internal pressure of a soap bubble is inversely proportional to its radius, according to the formula P = 4T/r, where T is the surface tension. Since the internal pressure of the smaller bubble (radius 3 cm) equals that of another bubble with radius r, we can set up the equation 4T/3 = 4T/r, leading to r = 2 cm.

Question 42

The velocity of a small ball of mass 'm' and density d1, when dropped in a container filled with glycerine, becomes constant after some time. If the density of glycerine is d2, then the viscous force acting on the ball will be

Accepted Answer

mg(1 - d2/d1). The correct option reflects the balance of forces acting on the ball when it reaches terminal velocity. At this point, the upward viscous force equals the downward gravitational force, leading to the relationship where the viscous force is proportional to the difference in densities, specifically mg(1 - d2/d1), indicating how the buoyant force affects the ball's motion.

Question 43

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Product of Pressure (P) and time (t) has the same dimension as that of coefficient of viscosity.
Reason R: Coefficient of viscosity = Force / Velocity gradient
Choose the corr

Accepted Answer

A is true but R is false.. The assertion is correct because the product of pressure and time does indeed have the same dimensions as the coefficient of viscosity, which is derived from the relationship between force and velocity gradient. However, the reason provided is incorrect because the definition of the coefficient of viscosity does not directly relate to the dimensional analysis of pressure and time.

Question 44

The velocity of a small ball of mass 0.3 g and density 8 g/cc when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is 1.3 g/cc, then the value of viscous force acting on the ball will be x × 10⁻⁴ N, the value of x is _____. [use g

Accepted Answer

25.125. Ball volume V = m/ρ = 0.3/8 cc = 3.75×10^-8 m³. Buoyant force = Vρ_gly g = 3.75×10^-8×1300×10 = 4.875×10^-4 N; weight = 0.3×10^-3×10 = 30×10^-4 N. Viscous force = weight - buoyancy = 25.125×10^-4 N, so x = 25.125.

Question 45

A small spherical ball of radius 0.1 mm and density 10⁴ kg m⁻³ falls freely under gravity through a distance h before entering a tank of water. If, after entering the water the velocity of ball does not change and it continue to fall with same constant velocity inside water, then the value

Accepted Answer

20.00. The ball reaches a terminal velocity when the gravitational force is balanced by the drag force due to viscosity, which occurs after falling a certain distance h. Given the parameters, the calculations show that this distance h is 20.00 m, where the ball maintains a constant velocity in water.

Question 46

Given below are two statements:
Statement I: Pressure in a reservoir of water is same at all points at the same level of water.
Statement II: The pressure applied to enclosed water is transmitted in all directions equally.
In the light of the above statements, choose the correct answer fro

Accepted Answer

Both Statement I and Statement II are true. Both statements accurately describe fundamental principles of fluid mechanics: pressure at a given depth in a fluid is uniform in all directions (Statement I), and any pressure applied to a confined fluid is transmitted equally throughout the fluid (Statement II).

Question 47

Glycerin of density 1.25 × 10³ kg m⁻³ is flowing through the conical section of pipe. The area of cross-section of the pipe at its ends are 10 cm² and 5 cm² and pressure drop across its length is 3 N m⁻². The rate of flow of glycerin through the pipe is x × 10⁻⁵ m² s⁻¹. The value of x is _

Accepted Answer

4. The correct option is 4 because applying the principle of conservation of mass (continuity equation) and Bernoulli's equation allows us to calculate the flow rate. Given the pressure drop and the cross-sectional areas, the flow rate can be derived, leading to a value of 4 × 10⁻⁵ m² s⁻¹.

Question 48

Match List I with List II

List-I
(A) Surface tension
(B) Pressure
(C) Viscosity
(D) Impulse

List-II
(I) kg m⁻¹ s⁻¹
(II) kg m s⁻¹
(III) kg m⁻¹ s⁻²
(IV) kg s⁻²

Choose the correct answer from the options given below:

Accepted Answer

A-IV, B-III, C-I, D-II. The correct option matches each physical quantity with its appropriate unit: Surface tension is measured in kg s⁻² (force per unit length), Pressure in kg m⁻¹ s⁻² (force per unit area), Viscosity in kg m⁻¹ s⁻¹ (force per unit velocity), and Impulse in kg m s⁻¹ (change in momentum).

Question 49

Surface tension of a soap bubble is 2.0 × 10⁻² Nm⁻¹. Work done to increase the radius of soap bubble from 3.5 cm to 7 cm will be:
Take π = 22/7

Accepted Answer

18.48 × 10⁻⁴ J. The work done to increase the radius of a soap bubble is calculated using the formula for work done against surface tension, which is proportional to the change in surface area. In this case, the increase in radius leads to a significant increase in surface area, resulting in the calculated work of 18.48 × 10⁻⁴ J.

Question 50

A fully loaded boeing aircraft has a mass of 5.4 × 10⁵ kg. Its total wing area is 500 m². It is in level flight with a speed of 1080 km/h. If the density of air is 1.2 kg m⁻³, the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface in

Accepted Answer

10. The correct option is 10% because the lift generated by the wings is proportional to the difference in airspeed above and below the wings, which can be calculated using Bernoulli's principle. The calculations show that the increase in speed on the upper surface compared to the lower surface results in a fractional increase of approximately 0.1, translating to a 10% increase.

JEE Main Physics: Mechanical Properties of Fluids questions with solutions

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