Questions: Mechanical Properties of Fluids

Q1. Why is pressure a scalar even though force is a vector?

Pressure at a point in a fluid is defined as the normal force per unit area on a small surface through that point. If the surface is rotated, the normal force changes direction but the value of the force divided by area stays the same. Since the magnitude does not depend on a direction, pressure is a scalar. The force exerted by the fluid on a wall is a vector, and its direction is normal to the wall, but that force comes from the scalar pressure multiplied by the area vector of the wall.

Hinglish: Pressure ka value har direction me same rehta hai, sirf force ka direction change hota hai. Isliye pressure scalar hai aur force vector. Wall par lagne wala force pressure times area hota hai.

Q2. State Pascal's law and explain how a hydraulic lift multiplies force.

Pascal's law states that when the pressure at any point of a confined liquid is changed, the same change appears at every other point of the liquid and on the walls of the vessel. In a hydraulic lift, a small piston of area a is pushed with force f, creating pressure P = f/a. This pressure acts on a large piston of area A connected by the same fluid, so the upward force is P A = f A/a. The force is multiplied by A/a, while the small piston must move down a distance larger by the same ratio, so the work done is the same at both ends. A hydraulic brake uses the same multiplication to press the brake shoes.

Hinglish: Pascal's law kehta hai ki band liquid me pressure har jagah same transmit hota hai. Hydraulic lift me chhote piston ka pressure bade piston par lagega, isliye force A/a guna badh jaata hai, lekin displacement ulta proportion me kam hota hai, so work same rehta hai.

Q3. A body of density 800 kg m^-3 floats in water. What fraction of its volume is submerged?

For a floating body the buoyant force equals its weight, so rho_liquid times the submerged volume times g equals rho_body times the total volume times g. The fraction submerged is rho_body/rho_liquid = 800/1000 = 0.8, so 80 percent of the volume stays under water and 20 percent stays above. If the same body is placed in a liquid of density 1600 kg m^-3, the fraction becomes 800/1600 = 0.5, so it floats half submerged. The lighter the body compared with the liquid, the smaller the submerged fraction.

Hinglish: Floating body ke liye submerged fraction density ka ratio hota hai. Water me 800/1000 = 0.8, yaani 80 percent volume andar. Zyada dense liquid me fraction kam ho jaata hai.

Q4. Water flows through a pipe of radius 2 cm at 1.0 m s^-1 and then through a narrow section of radius 1 cm. Find the speed in the narrow section and the volume flow rate.

The equation of continuity gives A1v1 = A2v2. The areas are proportional to the squares of the radii, so v2 = v1 (r1/r2)^2 = 1.0 x (2/1)^2 = 4.0 m s^-1. The volume flow rate is A1v1 = pi (0.02)^2 x 1.0 = 1.26 x 10^-3 m^3 s^-1, about 1.26 litres per second. The same flow rate crosses the narrow section, where the speed is four times larger because the area is four times smaller.

Hinglish: Continuity equation se speed area ke inversely badalti hai. Area four times chhota hai to speed four times badh jaati hai. Volume flow rate A1v1 se nikalta hai.

Q5. State Bernoulli's principle and apply it to find the efflux speed from a small hole at depth h.

Bernoulli's principle says that for steady, non-viscous, incompressible flow along a streamline, P + (1/2) rho v^2 + rho g h is constant. Take point 1 at the free surface and point 2 at the small hole. Both are open to the atmosphere, so P1 = P2 = P0. The surface moves negligibly slowly compared with the efflux, so v1 is nearly zero. The height difference is h, so the equation reduces to rho g h = (1/2) rho v^2, giving v = sqrt(2gh). This is Torricelli's law, and it shows that the hole's depth alone decides the efflux speed, not the size of the tank.

Hinglish: Bernoulli equation me dono points ka pressure atmosphere ke barabar hai aur surface ki speed zero maan lete hain. Phir rho g h = half rho v squared se v = sqrt(2gh) milta hai, yeh Torricelli's law hai.

Q6. Why does a small bubble have a greater excess pressure than a large bubble?

The excess pressure inside a soap bubble is 4T/r and inside a liquid drop is 2T/r, so the pressure difference grows as the radius falls. A bubble of radius 1 mm therefore has a much larger internal pressure than a bubble of radius 1 cm made from the same soap solution. When two bubbles of different sizes are joined by a tube, air flows from the smaller bubble into the larger one, because the pressure is higher on the smaller side. The small bubble shrinks and the large one grows, which is the opposite of what many people expect.

Hinglish: Excess pressure radius ke inversely hota hai, isliye chhota bubble zyada pressure par hota hai. Do bubbles tube se jodo to hawa chhote se bade ki taraf jaati hai, small bubble sikudta hai aur bada badhta hai.

Q7. Define viscosity and explain why liquids become less viscous on heating while gases become more viscous.

Viscosity is the internal friction between layers of a fluid in relative motion. In liquids, molecules are close together and held by intermolecular forces, so the main resistance comes from these cohesive forces. Heating gives the molecules more energy, weaker effective cohesion, and easier sliding, so the viscosity of a liquid falls. In gases the molecules are far apart and cohesion is negligible; momentum transfer between molecules is the main mechanism. Heating speeds the molecules up, so they exchange momenta more frequently and the viscosity rises. This is why a cold day makes engine oil thick but makes the air's viscosity slightly lower.

Hinglish: Liquid me cohesion strong hoti hai, garam karne se molecules ko energy milti hai aur viscosity ghatti hai. Gas me momentum transfer main kaam karta hai, garam karne se molecules tez hote hain aur viscosity badhti hai.

Q8. Derive the terminal velocity of a small sphere falling through a viscous liquid.

Three forces act on the sphere: its weight (4/3) pi r^3 rho g downwards, the buoyant force (4/3) pi r^3 sigma g upwards, and the viscous drag 6 pi eta r v upwards. At terminal speed the net force is zero, so 6 pi eta r v_t = (4/3) pi r^3 (rho - sigma) g. Rearranging gives v_t = 2r^2 (rho - sigma) g/(9 eta). The terminal speed rises with the square of the radius, so bigger drops fall faster in the same liquid, and it falls as the viscosity rises. For a bubble rising through a liquid, the same formula is used with the density difference reversed.

Hinglish: Weight niche, buoyancy aur drag upar. Terminal speed par net force zero hota hai. Formula se v = 2r^2(rho minus sigma) g/(9 eta) milta hai. Bada radius matlab zyada terminal speed.

Q9. Explain surface tension as surface energy and give two everyday examples.

Molecules at the surface of a liquid have fewer neighbours above them than molecules inside, so they have higher potential energy. Creating new surface therefore costs energy, and surface tension is that energy per unit area. It is also the force per unit length along the surface, and the two views give the same number in N/m. Everyday examples: a small insect can stand on water because its weight is supported by the deformed surface, and rain falls as spherical drops because a sphere has the least surface area for a given volume. Adding soap lowers the surface tension, so soap bubbles last longer and a detergent spreads water over a greasy plate.

Hinglish: Surface ke molecules ki energy zyada hoti hai, isliye naya surface banane me energy lagti hai. Yeh energy per area hi surface tension hai. Insect paani par khada rehta hai, aur drop gol kyunki sphere ka surface area minimum hota hai.

Q10. Water rises to 10 cm in a capillary tube of radius 0.2 mm. Find the surface tension of water if the contact angle is zero and g = 9.8 m s^-2.

Capillary rise is h = 2T cos(theta)/(rho g r), with theta = 0 so cos(theta) = 1. Rearranging gives T = h rho g r/2 = 0.10 x 1000 x 9.8 x 0.2 x 10^-3/2 = 0.098 N m^-1. The answer is close to the standard value of about 0.072 N m^-1 for water at room temperature, with the difference due to the approximate height and radius. If the tube radius is halved, the rise doubles, because h is inversely proportional to r.

Hinglish: Capillary rise ka formula h = 2T cos theta/(rho g r). Yahan cos theta = 1, isliye T = h rho g r/2. Radius aadha karne par height double ho jaati hai.

Quiz below: five questions on pressure at depth, continuity, Bernoulli, viscosity and surface tension. Attempt them without looking at the formula sheet first.