JEE Physics Mechanical Properties Of Fluids MCQs Set 18

Answer: (a) 6 × 1000 × 10 × 80 N

Explanation:
1. Pressure at the bottom (gauge) is \( \rho g h=1000\times10\times6 \) Pa.
2. Base area = 10 × 8 = 80 m².
3. Thrust = pressure × area = 6 × 1000 × 10 × 80 N.

Teacher's Note:
1. The thrust on the base depends only on the depth, density and base area, not on the shape of the sides.
2. Here the sides are vertical, so it also equals the weight of the water.

Answer: (a) 3.75 kg

Explanation:
1. Pressure is transmitted equally, so \( \frac{F_1}{A_1}=\frac{F_2}{A_2} \).
2. Area ratio = (20)² : 1 = 400 : 1.
3. Mass on small piston = 1500 / 400 = 3.75 kg.

Teacher's Note:
1. Pascal's law: pressure change in an enclosed liquid is passed on undiminished.
2. Force ratio equals area ratio, which is the square of the radius ratio.

Answer: (c) 3 : 4

Explanation:
1. For a floating body, density = (immersed fraction) × density of water.
2. \( \rho_A=\tfrac12\rho_w,\ \rho_B=\tfrac23\rho_w \)
3. \( \rho_A:\rho_B=\tfrac12:\tfrac23=3:4 \).

Teacher's Note:
1. Weight equals buoyant force, so the immersed fraction directly gives the density ratio with the liquid.
2. The more a body sinks, the denser it is.

Answer: (c) \( \frac{d-d_0}{d} \)

Explanation:
1. Floating: \( V_{in}\,d=V_0 d_0 \), so \( V_{in}/V_0=d_0/d \).
2. Fraction above the surface = \( 1-\frac{d_0}{d}=\frac{d-d_0}{d} \).

Teacher's Note:
1. First find the submerged fraction from weight = buoyancy, then subtract it from 1.
2. The result must be positive, so d must be greater than d₀.

Answer: (d) \( m\left(\frac1y-\frac1x\right) \)

Explanation:
1. Volume of ice = m/x; volume of water formed = m/y.
2. Change = final − initial = \( \frac my-\frac mx=m\left(\frac1y-\frac1x\right) \).
3. Since x < y this is negative, i.e. the volume decreases.

Teacher's Note:
1. Mass stays constant on melting, so volume change is mass times the difference of reciprocal densities.
2. Write final minus initial to get the sign right.

Answer: (b) decreases

Explanation:
1. In an elevator accelerating down with a, the effective gravity is g' = g − a.
2. Spring force = buoyancy − weight = \( (\rho_w-\rho_c)V g' \), which decreases as g' decreases.
3. A smaller spring force means a smaller extension, so the stretched spring becomes shorter.

Teacher's Note:
1. Both weight and buoyant force scale with effective gravity, so their difference shrinks.
2. In free fall the spring would be at its natural length.

Answer: (a) V/2

Explanation:
1. Initially weight = buoyancy = \( \rho_w\frac V2 g \).
2. To just sink it must be fully immersed: buoyancy = \( \rho_w V g \).
3. Extra weight needed = \( \rho_w \frac V2 g \), i.e. water of volume V/2.

Teacher's Note:
1. Just sinking means total weight equals the buoyancy of the fully immersed sphere.
2. The extra weight needed is the difference of the two buoyant forces.

Answer: (c) energy

Explanation:
1. Bernoulli's equation \( P+\tfrac12\rho v^2+\rho g h=\text{const} \) is the work-energy theorem applied to a flowing fluid.
2. So it expresses conservation of energy.

Teacher's Note:
1. Continuity equation is conservation of mass; Bernoulli's equation is conservation of energy.
2. Each term is energy per unit volume.

Answer: (d) for an ideal liquid in streamline flow, on the same streamline

Explanation:
1. The equation holds for an ideal (non-viscous, incompressible) liquid in steady, streamline flow.
2. It relates the pressure, speed and height of points that lie on the same streamline.

Teacher's Note:
1. Different streamlines may have different Bernoulli constants if the flow is rotational.
2. All the conditions, ideal fluid, steady flow and the same streamline, must be met.

Answer: (c) H/2

Explanation:
1. Speed of efflux \( v=\sqrt{2gh} \); fall time from the hole to the floor \( t=\sqrt{2(H-h)/g} \).
2. Range \( x=vt=2\sqrt{h(H-h)} \).
3. This is maximum when h(H − h) is maximum, i.e. h = H/2.

Teacher's Note:
1. The product h(H − h) of two quantities with a fixed sum is greatest when they are equal.
2. Maximum range is then equal to H.

Answer: (c) \( (\sigma_b-\sigma_l)Vgh \)

Explanation:
1. The block gains energy \( \sigma_b Vgh \).
2. An equal volume V of liquid moves down through h, losing \( \sigma_l Vgh \).
3. Net change = \( (\sigma_b-\sigma_l)Vgh \) (negative here, so the energy actually decreases).

Teacher's Note:
1. Raising a body inside a liquid is equivalent to exchanging it with an equal volume of liquid.
2. Never forget the liquid that moves the opposite way.

Answer: (c) 3.5 × 10³ kg/m³

Explanation:
1. One-fifth of the volume is in the heavy liquid and four-fifths in water.
2. Equilibrium: \( \rho V=\tfrac15(13.5)V+\tfrac45(1)V\ (\times10^3) \).
3. ρ = 2.7 + 0.8 = 3.5 × 10³ kg/m³.

Teacher's Note:
1. The total buoyant force is the sum of the buoyant forces in each liquid.
2. The density of the body is the volume-weighted average of the two liquid densities.

Answer: (a) will increase

Explanation:
1. Buoyant force = (density of water) × (volume of iron) × g.
2. Between 2 °C and 4 °C water becomes denser, so the water density rises (by about 2 × 10⁻⁵ of itself).
3. Iron expands slightly, so its volume also rises (about 3.6 × 10⁻⁵). Both factors increase, so the buoyant force increases.

Teacher's Note:
1. Water has its maximum density at 4 °C, so below 4 °C heating makes it denser.
2. Here both the density of water and the volume of iron increase.

Answer: (c) \( M\left[\frac1\rho-\frac1\sigma\right] \)

Explanation:
1. Volume of ice = M/ρ, volume of water = M/σ.
2. Decrease = \( \frac M\rho-\frac M\sigma=M\left[\frac1\rho-\frac1\sigma\right] \).

Teacher's Note:
1. The mass is the same before and after melting, only the volume changes.
2. Check units: M times reciprocal density gives volume.

Answer: (a) 3

Explanation:
1. Buoyant force = 60 − 40 = 20 N.
2. Specific gravity = weight in air / loss of weight = 60/20 = 3.

Teacher's Note:
1. Loss of weight in water equals the weight of water displaced.
2. Specific gravity is the ratio of density of the body to that of water.

Answer: (a) 6 × 5 × 5 × 5 gf

Explanation:
1. Volume = 5 × 5 × 5 cm³.
2. Weight in air = 5 × 5 × 5 × 7 gf; buoyant force = 5 × 5 × 5 × 1 gf.
3. Apparent weight = 5 × 5 × 5 × (7 − 1) = 6 × 5 × 5 × 5 gf.

Teacher's Note:
1. In CGS units, 1 cm³ of water weighs 1 gf, so relative density and gf make the working direct.
2. Apparent weight = V(ρ − ρw)g.

Answer: (c) 3

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

Teacher's Note:
1. Equilibrium on a balance in water means equal apparent weights.
2. Apparent weight = m(1 − ρw/ρ).

Answer: (b) 20

Explanation:
1. By Torricelli's theorem \( v=\sqrt{2gh} \).
2. \( v=\sqrt{2\times10\times20}=20\ \text{m/s} \).

Teacher's Note:
1. Efflux speed equals the speed of free fall through the depth of the hole below the surface.
2. The hole is near the bottom, so h is about the full height.

Answer: (c) t = 2t₀

Explanation:
1. In water the effective gravity is \( g'=g\left(1-\frac{\rho_w}{\rho_b}\right)=g\left(1-\frac34\right)=\frac g4 \).
2. Period \( T=2\pi\sqrt{L/g'} \) so \( t=t_0\sqrt{g/g'}=2t_0 \).

Teacher's Note:
1. The buoyant force reduces the net downward acceleration of the bob.
2. Smaller effective g means a longer period.

Answer: (d) 3.6 × 10⁻³ m

Explanation:
1. Speed after falling 0.2 m: \( v^2=0.4^2+2(10)(0.2)=4.16 \), v ≈ 2.04 m/s.
2. Continuity: \( d^2v=\text{const} \), so \( d=8\times10^{-3}\sqrt{0.4/2.04} \).
3. d ≈ 3.5 × 10⁻³ m, closest to 3.6 × 10⁻³ m.

Teacher's Note:
1. Falling water speeds up, so by continuity the stream must become narrower.
2. Area is proportional to d², so the diameter ratio is the square root of the inverse speed ratio.

Answer: (a) dynamic lift of an aeroplane

Explanation:
1. Faster air above a wing means lower pressure than below, which gives a net upward lift.
2. A viscosity meter uses Stokes' law, capillary rise uses surface tension and a hydraulic press uses Pascal's law.

Teacher's Note:
1. Lift on an aerofoil comes from the pressure difference caused by different air speeds.
2. Match each device with its principle.

Answer: (b) 2d

Explanation:
1. Mixture density = total mass / total volume.
2. \( \rho=\frac{(d+2d+3d)V}{3V}=2d \).

Teacher's Note:
1. For equal volumes the density is the simple average.
2. For equal masses it would be the harmonic mean.

Answer: (b) \( \frac{18d}{11} \)

Explanation:
1. Take mass m of each: total volume \( =m\left(\frac1d+\frac1{2d}+\frac1{3d}\right)=\frac{11m}{6d} \).
2. Density = 3m / (11m/6d) = 18d/11.

Teacher's Note:
1. For equal masses the density is the harmonic mean of the densities.
2. Always divide total mass by total volume.

Answer: (b) 10 m/s

Explanation:
1. Pressure drop = 0.5 × 10⁵ N/m², which becomes kinetic energy per unit volume.
2. \( \tfrac12\rho v^2=0.5\times10^5\Rightarrow v^2=\frac{10^5}{1000}=100 \), v = 10 m/s.

Teacher's Note:
1. Bernoulli's equation at the same level: the pressure drop equals ½ρv².
2. Water density is 1000 kg/m³.

Answer: (a) 46.26 s

Explanation:
1. Area ratio \( A/a=(0.08/0.005)^2=256 \).
2. Time to drain from height H: \( t=\frac Aa\sqrt{\frac{2H}{g}} \).
3. \( t=256\sqrt{\frac{2(0.16)}{9.8}}=256\times0.1807\approx46.3\ \text{s} \).

Teacher's Note:
1. From continuity and Torricelli: A dh/dt = −a√(2gh), which integrates to this result.
2. Only the water height (0.16 m) matters, not the tank height.

Answer: (b) 2 cm

Explanation:
1. ω = 2π × 2 = 4π rad/s, so ω² = 16π² = 160.
2. Height difference \( h=\frac{\omega^2r^2}{2g}=\frac{160\times0.0025}{20}=0.02\ \text{m} \) = 2 cm.

Teacher's Note:
1. The free surface of a rotating liquid is a paraboloid, with h = ω²r²/2g.
2. Convert 0.02 m into cm at the end.

Answer: (a) 4095.0 Pa

Explanation:
1. Bernoulli at nearly the same level: \( \Delta P=\tfrac12\rho(v_1^2-v_2^2) \).
2. \( =\tfrac12(1.3)(14400-8100)=0.65\times6300=4095\ \text{Pa} \).

Teacher's Note:
1. The pressure difference does not depend on the wing area, only on the air speeds and density.
2. Faster flow above means lower pressure above.

Answer: (a) 80 kg

Explanation:
1. Volume of log = 120/600 = 0.2 m³.
2. Maximum buoyant mass when fully immersed = 0.2 × 1000 = 200 kg.
3. Extra load = 200 − 120 = 80 kg.

Teacher's Note:
1. Just sinking means the log is fully immersed with total weight equal to maximum buoyancy.
2. Load = V(ρw − ρ).

Answer: (c) 8/3 g/cc

Explanation:
1. In water: \( \rho=\tfrac23\rho_w=\tfrac23 \) g/cc.
2. In the liquid only 1/4 is immersed: \( \rho=\tfrac14\rho_L \).
3. So \( \rho_L=4\times\tfrac23=\tfrac83 \) g/cc.

Teacher's Note:
1. Density of floater = immersed fraction × density of liquid.
2. Use the same floater (same density) in both cases.

Answer: (a) 4 : 9

Explanation:
1. Continuity: \( A_1v_1=A_2v_2 \), so \( v_1/v_2=(r_2/r_1)^2=4/9 \).

Teacher's Note:
1. Speed is inversely proportional to the cross-sectional area, which goes as r².
2. The wider end has the slower flow.