JEE Physics Mechanical Properties Of Fluids MCQs Set 12

Answer: (b) \(\frac{13}{5}P\)

Explanation:
1. The water contributes \(3P-P=2P\) at the start
2. After one fifth is drawn out the depth is \(\frac45\) of the original, so the water now contributes \(\frac85P\)
3. Total pressure = \(P+\frac85P=\frac{13}{5}P\)

Teacher's Note:
1. The pressure at the bottom is atmospheric pressure plus \(\rho gh\); only the second part changes with depth.

Answer: (d) \(450\ \text{g}(\text{cm}^{-2})\)

Explanation:
1. Water column 2.5 m = 250 cm, oil column 2.5 m = 250 cm
2. Pressure from water = \(250\times1=250\) g-weight per cm\(^2\)
3. Pressure from oil = \(250\times0.8=200\) g-weight per cm\(^2\)
4. Total = 450 g-weight per cm\(^2\)

Teacher's Note:
1. Add \(h\rho\) for each layer; the units in CGS gram-weight per cm\(^2\) avoid multiplying by \(g\).

Answer: (c) 2.7

Explanation:
1. Volumes: \(\frac12=0.5\) and \(\frac43=1.33\) (in cc)
2. Total mass = 5 g, total volume = 1.83 cc
3. Density = \(\frac{5}{1.83}=2.7\)

Teacher's Note:
1. Mixture density = total mass / total volume, not the average of the densities.

Answer: (c) 1 : 1

Explanation:
1. Liquid thrust on the base is \(F=\rho g h\times A\)
2. Height and base area are the same for both
3. The thrust is the same: 1 : 1 (the hydrostatic paradox)

Teacher's Note:
1. Thrust on the base depends on height, density and base area, not on the shape or the amount of liquid.

Answer: (b) 12 hrs 30 min

Explanation:
1. \(v=\frac{2r^2(\rho-\sigma)g}{9\eta}=\frac{2\times10^{-12}\times100\times10}{9\times10^{-3}}=2.22\times10^{-7}\) m/s
2. \(t=\frac{10^{-2}}{2.22\times10^{-7}}=4.5\times10^4\) s
3. \(=12.5\) hours = 12 hours 30 minutes

Teacher's Note:
1. Terminal speed in a viscous liquid follows Stokes' law. Very small particles settle extremely slowly, taking hours for a centimetre.

Answer: (a) \(\frac{m-m'}{r}\)

Explanation:
1. At terminal speed: weight = buoyancy + viscous force
2. \((m-m')g=6\pi\eta rv\)
3. \(v\propto\frac{m-m'}{r}\)

Teacher's Note:
1. Net weight after buoyancy balances Stokes' drag. Remember that \(m\) and \(m'\) are masses, so they already include the radius dependence.

Answer: (b) 2 : 1

Explanation:
1. Equal masses in a tube of uniform cross-section means equal weights, so the pressures at the bottom of the two arms are equal
2. \(\rho_wh_w=\rho_oh_o\)
3. With \(\rho_w=2\rho_o\): \(h_o=2h_w\), so the ratio is 2 : 1

Teacher's Note:
1. Liquids in contact at a vertical interface must have equal pressure. The lighter liquid therefore stands higher.

Answer: (a) (i) B (ii) D

Explanation:
1. In an accelerated tank the pressure rises opposite to the acceleration: \(\frac{dP}{dx}=-\rho a\)
2. It also increases with depth because of gravity
3. Maximum at the bottom-left corner B (back and bottom) and minimum at the top-right corner D (front and top)

Teacher's Note:
1. Pressure increases both downwards and towards the rear of an accelerating liquid.

Answer: (d) 0.130 N

Explanation:
1. A thin ring touches the film on both its inner and outer sides, so the film pulls along twice the circumference
2. Surface tension force \(=2\times75\times20=3000\) dyne = 0.03 N
3. Force needed = weight + surface tension force = 0.1 + 0.03 = 0.13 N

Teacher's Note:
1. A ring lifted from a liquid surface is held by two surfaces (inside and outside), so the force is \(2T\times\text{circumference}\).

Answer: (a) \(P\) is directly proportional to \(r\) (a straight line through the origin)

Explanation:
1. Surface energy of a bubble (two surfaces): \(E=8\pi r^2T\)
2. \(P=\frac{dE}{dt}=16\pi rT\frac{dr}{dt}\)
3. \(dr/dt\) is constant, so \(P\propto r\)

Teacher's Note:
1. Differentiate the surface energy with respect to time using the chain rule.

Answer: (b) water will stay at a height of 12 cm in the capillary tube but the radius of curvature at the top will increase

Explanation:
1. The height of rise \(h=\frac{2T\cos\theta}{\rho gR_c}\) where \(R_c\) is the radius of curvature of the meniscus
2. At 12 cm the height is smaller, so \(hR_c\) being constant means the meniscus radius \(R_c\) increases
3. The water stays at 12 cm; it does not overflow

Teacher's Note:
1. A capillary cannot create a fountain. It only adjusts the meniscus curvature to balance the pressure.

Answer: (c) \(125\times10^{-2}\) N/m

Explanation:
1. A soap bubble has two surfaces: \(\Delta P=\frac{4T}{r}\)
2. \(T=\frac{\Delta P\,r}{4}=\frac{2\times10^5\times25\times10^{-3}}{4}=1.25\) N/m = \(125\times10^{-2}\) N/m

Teacher's Note:
1. Use 4T/r for a soap bubble and 2T/r for a drop. The radius is half the 50 mm diameter.

Answer: (d) 27 : 8

Explanation:
1. Excess pressures are 0.02 atm and 0.03 atm
2. \(\Delta P=\frac{4T}{r}\) so \(r\propto\frac1{\Delta P}\): \(r_1:r_2=3:2\)
3. \(V\propto r^3\): \(V_1:V_2=27:8\)

Teacher's Note:
1. Use the excess pressure (inside minus outside), not the absolute pressure. Volume goes as the cube of the radius.

Answer: (b) \(1.6\times10^{-4}\) watt

Explanation:
1. \(E=8\pi r^2T\) so \(P=16\pi rT\frac{dr}{dt}\)
2. \(=16\pi\times0.05\times0.04\times\frac{0.01}{2\pi}\)
3. \(=1.6\times10^{-4}\) W

Teacher's Note:
1. Convert cm to m before substituting. A soap film has two surfaces, hence the factor 8 in \(8\pi r^2T\).

Answer: (d) 5 cm

Explanation:
1. In steady state, inflow = outflow
2. \(a\sqrt{2gh}=100\)
3. \(1\times\sqrt{2\times1000\,h}=100\), so \(2000h=10^4\) and \(h=5\) cm

Teacher's Note:
1. In steady state the water level stays constant so the outflow must equal the inflow.

Answer: (d) 35 cc

Explanation:
1. All 60 cc of water must lie in the other vertical arm, giving a column of 60 cm
2. The heavy liquid fills the horizontal arm (20 cc) and rises a height \(h\) in its own vertical arm
3. Pressure balance at the corner: \(4h=1\times60\), so \(h=15\) cm
4. Volume of heavy liquid = 20 + 15 = 35 cc

Teacher's Note:
1. Balance the pressure at the level of the horizontal arm. Count the horizontal arm's volume in the poured liquid.

Answer: (b) 0.45 m/s

Explanation:
1. Neglecting buoyancy: \(6\pi\eta rv=\tfrac43\pi r^3\rho g\), so \(v\propto r^2\)
2. \(v'=9\times5=45\) cm/s = 0.45 m/s

Teacher's Note:
1. Terminal speed grows as the square of the radius. Convert cm/s to m/s with care.

Answer: (d) Air resistance balances the gravitational force on a drop.

Explanation:
1. Air drag increases with speed
2. At terminal speed the drag (with buoyancy) equals the weight
3. The net force is then zero, so the speed stays constant

Teacher's Note:
1. Constant speed means zero net force. In a fluid this is called terminal velocity.

Answer: (d) \(2.5\times10^{-3}\)

Explanation:
1. The plate is midway, so each side has an oil layer of thickness 1.25 mm and both faces of the plate experience viscous drag
2. \(F=2\eta A\frac vd\)
3. \(\eta=\frac{Fd}{2Av}=\frac{1\times1.25\times10^{-3}}{2\times0.5\times0.5}=2.5\times10^{-3}\) kg/m s

Teacher's Note:
1. A thin plate between two fixed surfaces drags oil on both of its faces, so the force is doubled.