RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Physics · 10 questions · 24 marks
Liquids and gases cannot resist shear, and that single fact produces pressure that acts equally in all directions, buoyancy, streamlines and drag. Here you will use Pascal's law and Bernoulli's principle to explain everything from a hydraulic lift to the lift on an aeroplane wing. Surface tension and viscosity then account for the behaviour of small drops and slow-moving spheres.
Water flows through a horizontal pipe whose cross-sectional area narrows to half its original value. In the narrow section the speed and the pressure respectively:
Answer
Double and decrease — by the equation of continuity A₁v₁ = A₂v₂, halving the area doubles the speed. By Bernoulli's equation for a horizontal pipe, P + ½ρv² is constant, so a rise in speed must be paid for by a fall in pressure. This is why a constriction in a pipe is a low-pressure region, the principle behind the venturimeter.
The excess pressure inside a soap bubble of radius r, where S is the surface tension, is:
Answer
4S/r — a soap bubble has two liquid surfaces, an inner one and an outer one, each contributing 2S/r, giving a total excess pressure of 4S/r. A liquid drop, which has only one surface, has an excess pressure of just 2S/r. Confusing the two is the standard mistake here.
A block floats in water with three quarters of its volume submerged. The density of the block is (density of water = 1000 kg/m³):
Answer
750 kg/m³ — for floating equilibrium, weight = upthrust: ρ_block V g = ρ_water (0.75V) g. Cancelling V and g gives ρ_block = 0.75 × 1000 = 750 kg/m³. In general the submerged fraction equals the ratio of the densities, which is why ice with density about 917 kg/m³ floats with roughly 92% of its volume under water.
The dimensional formula of the coefficient of viscosity η is:
Answer
[ML⁻¹T⁻¹] — from F = ηA(dv/dx), η = F/(A × velocity gradient). The velocity gradient has dimensions [T⁻¹], so η = [MLT⁻²]/([L²][T⁻¹]) = [ML⁻¹T⁻¹]. Its SI unit is Pa·s, and the option [ML⁻¹T⁻²] belongs to pressure, not viscosity.
Assertion (A): A raindrop falling from a great height reaches the ground with a steady speed rather than accelerating throughout its fall. Reason (R): The viscous drag on a falling drop increases with speed until it balances the effective weight of the drop.
Answer
Both A and R are true and R is the correct explanation of A — as the drop speeds up, the drag given by Stokes' law, F = 6πηrv, grows in proportion to the speed while the weight stays fixed. At the moment when drag plus buoyancy equals weight, the net force is zero, acceleration ceases and the drop continues at its terminal velocity v_t = 2r²(ρ − σ)g/9η. Without this effect raindrops falling from a kilometre up would strike the ground at over 100 m/s.
The small piston of a hydraulic lift has an area of 5×10⁻⁴ m² and the large piston has an area of 0.2 m². What force must be applied to the small piston to raise a car of mass 1200 kg (g = 10 m/s²)?
Answer
By Pascal's law the pressure is the same on both pistons: F₁/A₁ = F₂/A₂. Required output force F₂ = mg = 1200 × 10 = 12000 N. F₁ = F₂ × (A₁/A₂) = 12000 × (5×10⁻⁴/0.2) = 12000 × 2.5×10⁻³ = 30 N. The lift multiplies force by a factor of 400, though the small piston must move 400 times as far, so no energy is created.
Calculate the height to which water rises in a capillary tube of internal radius 0.25 mm. Take the surface tension of water as 0.072 N/m, the contact angle as 0°, ρ = 1000 kg/m³ and g = 10 m/s².
Answer
Capillary rise: h = 2S cos θ/(rρg). With θ = 0°, cos θ = 1. r = 0.25 mm = 2.5×10⁻⁴ m. h = (2 × 0.072)/(2.5×10⁻⁴ × 1000 × 10) h = 0.144/2.5 = 0.0576 m. So the water rises about 5.8 cm. Note that h is inversely proportional to r, so a narrower tube gives a higher rise.
State Bernoulli's principle and derive it for the steady flow of an ideal fluid using the work-energy theorem. Water flows through a horizontal pipe that narrows from an area of 4×10⁻³ m² to 1×10⁻³ m²; if the speed in the wide section is 1 m/s and the pressure there is 2.0×10⁵ Pa, find the pressure in the narrow section (ρ = 1000 kg/m³).
Answer
Statement: for the steady, streamline flow of an incompressible non-viscous fluid, the sum of the pressure energy, kinetic energy and potential energy per unit volume is constant along a streamline: P + ½ρv² + ρgh = constant. Derivation: consider a tube of flow. At the lower end the cross-sectional area is A₁, the speed v₁, the pressure P₁ and the height h₁; at the upper end the corresponding quantities are A₂, v₂, P₂ and h₂. In a small time Δt, a mass Δm = ρA₁v₁Δt enters at the lower end and an equal mass leaves at the upper end, since the fluid is incompressible and the flow is steady. Work done by the pressure pushing fluid in at the lower end: W₁ = P₁A₁(v₁Δt) = P₁ΔV, where ΔV = Δm/ρ. Work done against the pressure at the upper end: W₂ = −P₂A₂(v₂Δt) = −P₂ΔV. Net work by pressure forces: W_pressure = (P₁ − P₂)ΔV. Work done against gravity in raising the mass Δm from h₁ to h₂: W_gravity = −Δm g(h₂ − h₁). By the work-energy theorem, the total work equals the change in kinetic energy: (P₁ − P₂)ΔV − Δm g(h₂ − h₁) = ½Δm v₂² − ½Δm v₁². Substituting Δm = ρΔV and dividing throughout by ΔV: P₁ − P₂ − ρg(h₂ − h₁) = ½ρv₂² − ½ρv₁². Rearranging: P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂. This is Bernoulli's equation. It assumes the fluid is non-viscous (no energy lost to friction), incompressible, and in steady streamline flow. Numerical: by continuity, A₁v₁ = A₂v₂, so v₂ = (4×10⁻³ × 1)/(1×10⁻³) = 4 m/s. The pipe is horizontal, so the ρgh terms cancel: P₂ = P₁ + ½ρ(v₁² − v₂²) P₂ = 2.0×10⁵ + ½(1000)(1 − 16) P₂ = 2.0×10⁵ + 500(−15) = 2.0×10⁵ − 7500 P₂ = 1.925×10⁵ Pa.
(a) Define terminal velocity and derive an expression for the terminal velocity of a small sphere falling through a viscous liquid. (b) A steel ball of radius 1×10⁻³ m and density 8000 kg/m³ falls through oil of density 900 kg/m³ and viscosity 0.99 Pa·s. Find its terminal velocity (g = 9.9 m/s²).
Answer
(a) Terminal velocity is the constant maximum velocity attained by a body falling through a viscous fluid, reached when the net force on it becomes zero. Derivation: let a sphere of radius r and density ρ fall through a fluid of density σ and coefficient of viscosity η. Three forces act: 1. Weight, acting downward: W = mg = (4/3)πr³ρg. 2. Buoyant upthrust, acting upward: F_B = (4/3)πr³σg, the weight of the displaced fluid. 3. Viscous drag, acting upward and given by Stokes' law: F_v = 6πηrv. As the sphere speeds up, only the drag grows. At terminal velocity v_t the acceleration is zero, so the upward forces balance the downward one: 6πηrv_t + (4/3)πr³σg = (4/3)πr³ρg 6πηrv_t = (4/3)πr³(ρ − σ)g Cancelling π and one factor of r: 6ηv_t = (4/3)r²(ρ − σ)g v_t = 2r²(ρ − σ)g/(9η). Notice that v_t ∝ r², so a drop of twice the radius falls four times as fast, and v_t ∝ 1/η, so a thicker liquid slows the fall. (b) Substituting the given values: r² = (1×10⁻³)² = 1×10⁻⁶ m². ρ − σ = 8000 − 900 = 7100 kg/m³. v_t = 2 × 1×10⁻⁶ × 7100 × 9.9/(9 × 0.99) Numerator: 2 × 1×10⁻⁶ × 7100 = 1.42×10⁻². Times 9.9 gives 0.14058. Denominator: 9 × 0.99 = 8.91. v_t = 0.14058/8.91 ≈ 1.58×10⁻² m/s. So the ball settles at about 1.6 cm/s, slow enough to time by eye — which is exactly how this experiment is used to measure the viscosity of a liquid.
Read the following and answer the questions that follow: A rooftop tank is filled with water to a depth of 3 m. A small circular hole of area 2×10⁻⁵ m² is accidentally punched in the vertical wall of the tank, 0.8 m above its base, and water squirts out horizontally. The tank is open to the atmosphere and its cross-section is very large compared with the hole. Take ρ = 1000 kg/m³, g = 10 m/s² and atmospheric pressure as 1.0×10⁵ Pa. (a) Find the gauge pressure of the water at the level of the hole. (b) Using Bernoulli's equation, find the speed with which the water emerges. (c) Calculate the volume of water escaping per second. (d) Explain why the speed of efflux is the same as that of a body dropped freely through the height of water above the hole, and name this result.
Answer
(a) The water surface is at a height of 3 m from the base and the hole is at 0.8 m, so the depth of water above the hole is h = 3 − 0.8 = 2.2 m. Gauge pressure = ρgh = 1000 × 10 × 2.2 = 2.2×10⁴ Pa. (The absolute pressure there would be 1.0×10⁵ + 2.2×10⁴ = 1.22×10⁵ Pa.) (b) Apply Bernoulli's equation along a streamline from the top surface (point 1) to the hole (point 2). Both points are open to the atmosphere, so P₁ = P₂ = P_atm and the pressure terms cancel. Because the tank's cross-section is very large, the surface descends so slowly that v₁ ≈ 0. ρgh₁ = ½ρv₂² + ρgh₂ g(h₁ − h₂) = ½v₂² v₂ = √(2gh) = √(2 × 10 × 2.2) = √44 ≈ 6.63 m/s. (c) Volume flow rate Q = Av = 2×10⁻⁵ × 6.63 ≈ 1.33×10⁻⁴ m³/s, that is about 0.13 litres per second. (In practice the emerging jet contracts slightly just outside the hole, so the real discharge is a little less than this ideal figure.) (d) In the derivation, the pressure terms cancelled because both the free surface and the hole are exposed to the atmosphere, leaving an equation that says the gravitational potential energy per unit mass, gh, is converted entirely into kinetic energy per unit mass, ½v². That is precisely the energy statement for a body falling freely through a height h, which gives v = √(2gh) as well. The fluid emerges as though it had simply dropped from the surface to the hole. This result is known as Torricelli's law, or the law of efflux.
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