JEE Physics Mechanical Properties Of Fluids MCQs Set 25

Answer: (c) 2 cm

Explanation:
1. \( \omega=2\pi(2)=4\pi \) rad/s.
2. \( h=\dfrac{\omega^2R^2}{2g}=\dfrac{16\pi^2(0.0025)}{20}=\dfrac{16(10)(0.0025)}{20}=0.02 \) m.

Teacher's Note:
1. The free surface of a rotating liquid is a paraboloid.
2. Use g = 10 m/s² to match \( \pi^2=10 \).

Answer: (b) \( h=\dfrac{3H}{8},\ x_m=\dfrac{3H}{4} \)

Explanation:
1. Range \( x=v\sqrt{2h/g}=\sqrt{h(3H-4h)} \).
2. \( \dfrac{d}{dh}\bigl[h(3H-4h)\bigr]=3H-8h=0 \), so \( h=\dfrac{3H}{8} \).
3. \( x_m=\sqrt{\dfrac{3H}{8}\cdot\dfrac{3H}{2}}=\dfrac{3H}{4} \).

Teacher's Note:
1. The range is the product of the efflux speed and the fall time to the base level.
2. Use calculus to maximise the product.

Answer: (a) \( D=D_0\sqrt{v_0/v} \)

Explanation:
1. \( \rho_w\pi D_0^2v_0/4=\rho_w\pi D^2v/4 \).
2. \( D^2=D_0^2v_0/v \), so \( D=D_0\sqrt{v_0/v} \).

Teacher's Note:
1. The diameter enters through the area, which varies as \( D^2 \).
2. A faster stream is thinner.

Answer: (b) \( \dfrac{\rho gA(h_1-h_2)^2}{4} \)

Explanation:
1. Initial potential energy: \( \tfrac12\rho gA(h_1^2+h_2^2) \). Final: each level is \( (h_1+h_2)/2 \), so \( \tfrac14\rho gA(h_1+h_2)^2 \).
2. Work by gravity = loss in potential energy = \( \tfrac14\rho gA(h_1-h_2)^2 \).

Teacher's Note:
1. Gravity does positive work because the liquid centre of mass moves down overall.
2. The energy is dissipated as the liquid sloshes and settles.

Answer: (a) 10 s

Explanation:
1. Under acceleration a the surface slopes with \( \tan\theta=a/g \); the level rises at the back by \( \dfrac{aL}{2g} \).
2. The rise allowed is 2 − 1.5 = 0.5 m, so \( \dfrac{a(10)}{20}=0.5 \) and a = 1 m/s².
3. t = v/a = 10/1 = 10 s.

Teacher's Note:
1. The surface pivots about the middle, so the rise at the back is half of the total height difference.
2. Smaller acceleration means longer time without spilling.

Answer: (a) \( \sqrt{2p_{atm}/\rho} \)

Explanation:
1. Bernoulli: \( p_{atm}=p+\tfrac12\rho v^2 \) with \( p\ge0 \).
2. So \( v_{max}=\sqrt{2p_{atm}/\rho} \).

Teacher's Note:
1. Pressure cannot fall below zero in this model.
2. This gives about 400 m/s for sea level air, far above wing speeds.

Answer: (c) less in salt water

Explanation:
1. Salt water is denser, so the buoyant force is larger.
2. The reading (weight − buoyancy) is smaller in salt water.

Teacher's Note:
1. Buoyant force is proportional to the liquid density.
2. The person's volume is the same in both cases.

Answer: (b) 187.5 Pa

Explanation:
1. Speeds: \( v_y=0.5 \) m/s, \( v_x=0.25 \) m/s (the water slows as it falls).
2. \( P_x-P_y=\rho gh+\tfrac12\rho(v_y^2-v_x^2)=93.75+93.75=187.5 \) Pa.

Teacher's Note:
1. Both the height drop and the slowing add to the pressure at the lower, wider end.
2. Compare with the case where x is above y, where the two terms cancel.

Answer: (a) 6.0 N

Explanation:
1. Exit speed \( v=\sqrt{2gh} \), and the speed at the wide end is \( vs/S \).
2. Momentum balance with the pressure at the wide section (from Bernoulli) gives the force \( F=\dfrac{\rho gh(S-s)^2}{S} \).
3. \( F=\dfrac{1000(9.8)(4.6)(2\times10^{-4})^2}{3\times10^{-4}}\approx6.0 \) N.

Teacher's Note:
1. The pressure forces on the water at the wide end and the momentum flux both matter.
2. The force vanishes when S = s (a straight tube).

Answer: (b) 150 N

Explanation:
1. Mean density = \( \tfrac14(0.4)+\tfrac34\left(\tfrac43\right)=1.1 \), so \( V=165/1100=0.15 \) m³.
2. Buoyant force = \( 1000(0.15)(10)=1500 \) N and weight = 1650 N.
3. Reading = 1650 − 1500 = 150 N.

Teacher's Note:
1. Find the volume from the mass and mean density.
2. The reading is only the unbalanced part of the weight.

Answer: (d) \( 3.5\times10^3 \) kg/m³

Explanation:
1. Weight = buoyancy: \( \rho V=\tfrac45(10^3)V+\tfrac15(13.5\times10^3)V \).
2. \( \rho=800+2700=3500 \) kg/m³.

Teacher's Note:
1. Each liquid contributes buoyancy equal to its density times the volume in it.
2. The mercury-like liquid is denser than the metal, so the sphere floats.

Answer: (b) 760 g

Explanation:
1. Buoyant force = weight of oil displaced + weight of water displaced.
2. \( m=100(6)(0.6)+100(4)(1)=360+400=760 \) g.

Teacher's Note:
1. Each part of the cube displaces the liquid it is in.
2. The base area of the cube is 100 cm².

Answer: (a) 46.26 s

Explanation:
1. \( A\,\dfrac{dh}{dt}=-a\sqrt{2gh} \), which integrates to \( T=\dfrac{A}{a}\sqrt{\dfrac{2h_0}{g}} \).
2. \( \dfrac{A}{a}=\left(\dfrac{0.08}{0.005}\right)^2=256 \) and \( \sqrt{0.32/9.8}=0.1807 \) s.
3. T = 256 × 0.1807 ≈ 46.26 s.

Teacher's Note:
1. The time to drain depends on the area ratio, which is large here.
2. The result follows from integrating the changing height, not from a constant speed.

Answer: (d) \( [2g(b+x)]^{1/2} \)

Explanation:
1. Bernoulli between the tank surface and the point x below the tap end: \( \tfrac12v^2=g(b+x) \).
2. So \( v=[2g(b+x)]^{1/2} \).

Teacher's Note:
1. The point is a total vertical distance b + x below the water surface of the tank.
2. At x = 0 this reduces to \( \sqrt{2gb} \).

Answer: (d) Only (3)

Explanation:
1. Buoyant force = \( V_{immersed}\rho_{liquid}g \).
2. It involves g, so it changes on the moon; it does not depend on orientation, the body's own density or the depth of a fully immersed body.

Teacher's Note:
1. Only the volume displaced, the liquid density and g matter.
2. On the moon g is smaller, so the buoyant force is smaller.

Answer: (d) 29025 J/m³

Explanation:
1. Continuity: \( v_Q=v_P(A_P/A_Q)=0.5 \) m/s.
2. Bernoulli: \( P_P-P_Q=\rho g(h_Q-h_P)+\tfrac12\rho(v_Q^2-v_P^2) \).
3. = 1000(9.8)(3) + 500(0.25 − 1) = 29400 − 375 = 29025 J/m³.

Teacher's Note:
1. Work done per unit volume by pressure equals the pressure drop.
2. The rise in height costs more energy than is gained by the slowing.

Answer: (b) the same for \( O_1 \) and \( O_2 \)

Explanation:
1. Buoyant force = (volume immersed) × (liquid density) × effective g.
2. These quantities are the same for both observers, since a real force does not depend on the observer's frame.

Teacher's Note:
1. A pseudo force affects only the way the observer writes the equation of motion.
2. The real force is the same in all frames.

Answer: (c) 2 : 1 : 3

Explanation:
1. For A alone: \( \rho_A=\tfrac12\rho_L \).
2. For the pair, total weight = buoyancy: \( (\rho_A+\rho_B)V=\rho_L(2V) \).
3. \( \rho_B=2\rho_L-\tfrac12\rho_L=\tfrac32\rho_L \). Ratio \( \rho_L:\rho_A:\rho_B=1:\tfrac12:\tfrac32=2:1:3 \).

Teacher's Note:
1. Just floating means the pair is fully immersed with weight equal to maximum buoyancy.
2. Reduce the ratio to whole numbers.

Answer: (a) \( 2\sqrt{D(H-D)} \)

Explanation:
1. Speed of efflux \( v=\sqrt{2gD} \).
2. Time to fall a height H − D: \( t=\sqrt{2(H-D)/g} \).
3. \( x=vt=2\sqrt{D(H-D)} \).

Teacher's Note:
1. Torricelli's speed and projectile fall time combine to give the range.
2. The range is the same for depths D and H − D.

Answer: (c) Only (2) and (3)

Explanation:
1. Volume = 10/2000 = 0.005 m³; buoyant force = \( 1000(0.005)(10)=50 \) N.
2. Spring force = 100 − 50 = 50 N, so the elongation is 50/100 = 0.5 m.
3. Energy = \( \tfrac12(100)(0.5)^2=12.5 \) J. The spring force (50 N) is less than the weight (100 N).

Teacher's Note:
1. The spring and buoyancy together balance the weight.
2. Elongation is 0.5 m, not 1 cm.

Answer: (d) \( m\left(\dfrac1y-\dfrac1x\right) \)

Explanation:
1. Mass stays m. Volume of ice = m/x and volume of water = m/y.
2. ΔV = \( m/y-m/x \).

Teacher's Note:
1. The mass is conserved on melting, not the volume.
2. Since y > x the result is negative: the volume shrinks.

Answer: (d) \( P_0+\dfrac{3dgH}{2}+\dfrac{dgL}{4} \)

Explanation:
1. Pressure from liquids without the cylinder: \( P_0+dg\tfrac H2+2dg\tfrac H2=P_0+\tfrac32dgH \).
2. The cylinder weight \( \tfrac54d\,L\,\tfrac A5g=\tfrac14dLAg \) is shared over the base area A, adding \( \tfrac14dgL \).

Teacher's Note:
1. Total force on the bottom = weight of the liquids + weight of the cylinder + atmospheric force.
2. Divide by the base area to get pressure.

Answer: (c) 3

Explanation:
1. Apparent mass of the first = \( 36\left(1-\tfrac19\right)=32 \) g.
2. For the second: \( 48\left(1-\tfrac1\rho\right)=32 \), so \( \rho=3 \) g/cc.

Teacher's Note:
1. The balance is in equilibrium when the apparent weights are equal.
2. Apparent mass = m(1 − ρw/ρ).

Answer: (b) 4.1 m/s

Explanation:
1. Continuity: speed at throat = 5v.
2. Bernoulli: \( 2\times10^5+\tfrac12\rho v^2=0+\tfrac12\rho(5v)^2 \), so \( 12000\,v^2=2\times10^5 \).
3. \( v=4.08 \) m/s (throat speed 20.4 m/s).

Teacher's Note:
1. The throat has five times the speed, so the kinetic energy term is 25 times as large.
2. Zero pressure at the throat marks the onset of cavitation.

Answer: (d) 19.6 m and 4 s

Explanation:
1. Speed on reaching water: \( v=\sqrt{2gh}=19.6 \) m/s.
2. In water the net upward force is \( (\rho_w-\rho)Vg=\rho Vg \), so the retardation is g.
3. Depth = \( v^2/2g=19.6 \) m. Time down = v/g = 2 s, and the same to rise, so 4 s.

Teacher's Note:
1. Buoyancy is twice the weight, so the net force upward equals the weight.
2. The motion in water is symmetric, so rising time equals sinking time.

Answer: (b) \( \sqrt2\,\rho VL \)

Explanation:
1. Mass flow rate = \( \rho L \), and the speed is V both before and after the corner.
2. The momentum changes from \( \rho LV \) upward to \( \rho LV \) horizontal; the change has magnitude \( \sqrt2\,\rho LV \).

Teacher's Note:
1. The two momentum vectors are at right angles, so the change is \( \sqrt2 \) times each.
2. The force on the pipe is equal and opposite to the force on the water.