JEE Physics Mechanical Properties Of Fluids MCQs Set 22

Answer: (c) constant acceleration

Explanation:
1. The upward buoyant force \( V\rho_wg \) and the weight of the block are both constant while the block is fully submerged.
2. So the net force and the acceleration are constant (neglecting viscous drag).

Teacher's Note:
1. While fully submerged, the buoyant force does not depend on the depth, so the acceleration is constant.

Answer: (b) \( \dfrac{\rho_1-\rho}{\rho_1-\rho_2} \)

Explanation:
1. Let a fraction x be in the lower liquid. Equilibrium: \( \rho V = x\rho_2V + (1-x)\rho_1V \).
2. \( x(\rho_2-\rho_1) = \rho-\rho_1 \), so \( x = \dfrac{\rho-\rho_1}{\rho_2-\rho_1} = \dfrac{\rho_1-\rho}{\rho_1-\rho_2} \).

Teacher's Note:
1. The weight equals the sum of the buoyant forces from the two liquids.

Answer: (d) 90 N

Explanation:
1. By Pascal's law the pressure is the same: \( \dfrac{F_1}{A_1} = \dfrac{F_2}{A_2} \).
2. \( F_2 = 10\times\left(\dfrac{3}{1}\right)^2 = 90 \) N.

Teacher's Note:
1. Area goes as the square of the diameter. A diameter ratio of 3 gives a force ratio of 9.

Answer: (a) \( \dfrac vn\left(\dfrac Rr\right)^{2} \)

Explanation:
1. The equation of continuity: \( \pi R^2v = n\pi r^2v' \).
2. \( v' = \dfrac{vR^2}{nr^2} = \dfrac vn\left(\dfrac Rr\right)^2 \).

Teacher's Note:
1. The volume flow rate in the tube equals the total volume flow rate through all the holes.

Answer: (c) \( \dfrac{VR^2}{nr^2} \)

Explanation:
1. Continuity: \( \pi R^2V = n\pi r^2v \).
2. \( v = \dfrac{VR^2}{nr^2} \).

Teacher's Note:
1. Total area of the holes is nπr², so the speed increases in the ratio of the tube area to the total hole area.

Answer: (b) 50 m/s

Explanation:
1. The water must rise to 125 m, where its speed becomes zero.
2. \( v^2 = 2gh = 2\times10\times125 = 2500 \Rightarrow v = 50 \) m/s.

Teacher's Note:
1. Minimum speed means the water just reaches the fire at the top of its path.

Answer: (d) \( 5.0\times10^{-5} \) m²

Explanation:
1. Speed after falling 3.75 m: \( v^2 = 5^2+2\times10\times3.75 = 100 \), so \( v = 10 \) m/s (g = 10 m/s²).
2. Continuity: \( A_2 = A_1\dfrac{v_1}{v_2} = 10^{-4}\times\dfrac{5}{10} = 5\times10^{-5} \) m².

Teacher's Note:
1. The water speeds up as it falls, so the stream narrows to keep the flow rate constant.

Answer: (d) \( 2.4\times10^5 \) N, upwards

Explanation:
1. By Bernoulli's theorem, the pressure drop over the roof is \( \Delta P = \tfrac12\rho v^2 = \tfrac12\times1.2\times1600 = 960 \) Pa.
2. Force \( = \Delta P\times A = 960\times250 = 2.4\times10^5 \) N.
3. The pressure inside the house is higher than outside above the roof, so the force is upwards.

Teacher's Note:
1. Fast air has lower pressure, so the roof is pushed outwards, which is why roofs lift off in storms.

Answer: (a) \( 2\rho v^2A \)

Explanation:
1. Pressure difference \( = \tfrac12\rho(v_{upper}^2-v_{lower}^2) = \tfrac12\rho(5v^2-v^2) = 2\rho v^2 \).
2. Lift \( = 2\rho v^2A \).

Teacher's Note:
1. Dynamic lift comes from the pressure difference between the lower and upper surfaces, given by Bernoulli's theorem.

Answer: (d) \( y = h \)

Explanation:
1. The speed from the hole is \( \sqrt{2gy} \). The hole is at height \( (h-y)+h = 2h-y \) above the ground, so the fall time is \( \sqrt{\dfrac{2(2h-y)}{g}} \).
2. Range \( x = 2\sqrt{y(2h-y)} \), which is maximum when \( y = 2h-y \), i.e. \( y = h \).
3. Then \( x_m = 2h \).

Teacher's Note:
1. The product y(2h − y) is maximum when the two factors are equal, so the hole must be at the middle of the total height.

Answer: (b) \( 10^{-3} \) m³/s

Explanation:
1. Speed of efflux \( v = \sqrt{2gh} = \sqrt{2\times10\times5} = 10 \) m/s.
2. Volume rate \( = Av = 10^{-4}\times10 = 10^{-3} \) m³/s.

Teacher's Note:
1. Torricelli's theorem gives the efflux speed. Convert 1 cm² to 10⁻⁴ m².

Answer: (a) Bernoulli's theorem

Explanation:
1. Air blown quickly across the top of a tube dipped in perfume creates a low pressure there.
2. The higher atmospheric pressure on the liquid surface pushes the liquid up the tube, which is Bernoulli's principle.

Teacher's Note:
1. Fast-moving fluid has lower pressure. The same idea explains the Bunsen burner and the paint sprayer.

Answer: (c) \( \dfrac{L}{\sqrt{2\pi}} \)

Explanation:
1. Flow rate = area × speed: \( L^2\sqrt{2gy} = \pi R^2\sqrt{2g\cdot4y} = 2\pi R^2\sqrt{2gy} \).
2. \( L^2 = 2\pi R^2 \Rightarrow R = \dfrac{L}{\sqrt{2\pi}} \).

Teacher's Note:
1. The deeper hole has double the speed (√4 = 2), so it needs half the area to give the same flow rate.

Answer: (b) \( 3.45\times10^{-3} \) Pa s

Explanation:
1. At constant speed the viscous force equals the pull of the hanging mass: \( F = mg = 0.010\times9.8 = 0.098 \) N.
2. \( F = \eta A\dfrac vd \Rightarrow \eta = \dfrac{Fd}{Av} = \dfrac{0.098\times0.30\times10^{-3}}{0.10\times0.085} \approx 3.45\times10^{-3} \) Pa s.

Teacher's Note:
1. Constant speed means zero net force, so the viscous drag balances the weight of the hanging mass.

Answer: (a) 2 mm

Explanation:
1. 0.01 poise = 0.001 Pa s.
2. \( F = \eta A\dfrac vd \Rightarrow d = \dfrac{\eta Av}{F} = \dfrac{0.001\times1\times4}{2} = 2\times10^{-3} \) m = 2 mm.

Teacher's Note:
1. 1 poise = 0.1 Pa s. Convert the viscosity to SI units before substituting.

Answer: (d) 1.52 poise

Explanation:
1. The terminal speed of a rising bubble is \( v = \dfrac{2r^2\rho g}{9\eta} \) with r = 0.1 cm.
2. \( \eta = \dfrac{2\times(0.1)^2\times1.47\times980}{9\times2.1} \approx 1.52 \) poise.

Teacher's Note:
1. For a bubble (negligible density) the net upward force is the buoyancy, which is balanced by the viscous drag at terminal speed.

Answer: (a) almost independent of h

Explanation:
1. The drop quickly reaches terminal velocity, where the weight is balanced by the viscous drag and buoyancy of the air.
2. After that the speed does not change, so the final velocity is almost independent of h.

Teacher's Note:
1. For a large fall height, the final speed is the terminal speed and does not depend on h.

Answer: (c) 0.1 m/s

Explanation:
1. \( v\propto(\rho-\sigma) \) for the same size and liquid.
2. \( \dfrac{v_{Ag}}{v_{Au}} = \dfrac{10.5-1.5}{19.5-1.5} = \dfrac{9}{18} = \dfrac12 \), so \( v_{Ag} = 0.1 \) m/s.

Teacher's Note:
1. Terminal speed is proportional to the density of the sphere minus the density of the liquid.

Answer: (d) \( 6.25\times10^{-4} \) cm s⁻¹

Explanation:
1. \( v\propto\dfrac{\rho-\sigma}{\eta} \) for the same ball.
2. \( \dfrac{v_g}{v_w} = \dfrac{(7.8-1.2)/13.2}{(7.8-1)/8.5\times10^{-4}} = \dfrac{0.5}{8000} = 6.25\times10^{-5} \).
3. \( v_g = 10\times6.25\times10^{-5} = 6.25\times10^{-4} \) cm/s.

Teacher's Note:
1. The very high viscosity of glycerine makes the terminal speed far smaller than in water.

Answer: (d) 6 m/s

Explanation:
1. By the equation of continuity, \( A_Pv_P = A_Qv_Q \).
2. \( v_Q = \dfrac{80\times3}{40} = 6 \) m/s.

Teacher's Note:
1. The heights are not needed for continuity. They matter only when you apply Bernoulli's equation to find pressure.

Answer: (c) 0.0125 N/m

Explanation:
1. The film has two surfaces, each pulling along the slider, so \( F = 2T\ell \).
2. \( T = \dfrac{1.5\times10^{-2}}{2\times0.6} = 0.0125 \) N/m.

Teacher's Note:
1. A film has two free surfaces, so the total force is 2Tℓ.

Answer: (b) an elastic membrane is formed on water due to the property of surface tension

Explanation:
1. Surface tension makes the surface of water behave like a stretched elastic membrane.
2. The small weight of the insects is supported by this membrane, which is slightly depressed.

Teacher's Note:
1. Surface tension supports small objects like needles and insects only if their weight is small enough.

Answer: (b) (b) and (d)

Explanation:
1. Detergent molecules lower the surface tension at the oil-water interface, so water spreads and wets the greasy surface.
2. They also surround the dirt particles, holding them suspended so they can be washed away.

Teacher's Note:
1. Detergents are surface-active agents. Lower surface tension means better wetting and cleaning.

Answer: (c) \( 2\times10^{-3} \) m

Explanation:
1. The surface tension force supports the weight: \( W = 2\pi rT\cos\theta \) with \( \theta = 0 \).
2. \( r = \dfrac{6.28\times10^{-4}}{2\pi\times5\times10^{-2}} = 2\times10^{-3} \) m.

Teacher's Note:
1. The vertical component of the surface tension force along the circumference holds up the column.