JEE Physics Mechanical Properties Of Fluids MCQs Set 19

Answer: (a) \( \left(-\frac32,-\frac12,\frac12\right) \)

Explanation:
1. Dimensions: \( [\nu]=T^{-1},\ [r]=L,\ [\rho]=ML^{-3},\ [s]=MT^{-2} \).
2. Time: \( -2c=-1\Rightarrow c=\frac12 \). Mass: \( b+c=0\Rightarrow b=-\frac12 \). Length: \( a-3b=0\Rightarrow a=-\frac32 \).

Teacher's Note:
1. Equate the powers of M, L and T on both sides.
2. Surface tension has dimensions of force per length: M T⁻².

Answer: (a) 7

Explanation:
1. At constant (terminal) velocity the viscous force equals the weight minus the buoyant force: \( F=\frac43\pi r^3(\rho-\sigma)g \).
2. \( F=\frac43\times\frac{22}7\times(10^{-3})^3\times(10500-1500)\times9.8=3.696\times10^{-4}\ \text{N}=3696\times10^{-7}\ \text{N} \), so x = 7.

Teacher's Note:
1. Convert densities to kg/m³: 10.5 g/cc = 10500 kg/m³.
2. The viscosity value is not needed for the viscous force at terminal velocity.

Answer: (a) 1

Explanation:
1. Volume conservation: \( 1000\,r^3=R^3\Rightarrow r=\frac R{10} \).
2. Surface area ratio: \( \frac{1000\times4\pi r^2}{4\pi R^2}=\frac{1000}{100}=10 \).
3. \( \frac{u_f}{u_i}=10=\frac{10}{x} \), so x = 1.

Teacher's Note:
1. Surface energy is surface tension times surface area.
2. Splitting a drop into n drops multiplies the area by n^(1/3).

Answer: (c) 18.48 x 10⁻⁴ J

Explanation:
1. A soap bubble has two surfaces: \( W=S\,\Delta A=S\times2\times4\pi\left(r_2^2-r_1^2\right) \).
2. \( W=0.02\times8\times\frac{22}{7}\times(0.07^2-0.035^2)=0.02\times25.14\times0.003675=1.848\times10^{-3}\ \text{J} \).
3. So \( W=18.48\times10^{-4}\ \text{J} \).

Teacher's Note:
1. Count both the inner and outer surfaces of the film.
2. Convert cm to m before squaring.

Answer: (a) 25 x 10⁻³ m/s

Explanation:
1. At constant speed the applied force equals the viscous force: \( F=\eta A\frac vd \).
2. \( v=\frac{Fd}{\eta A}=\frac{0.1\times0.25\times10^{-3}}{5\times10^{-3}\times0.2}=2.5\times10^{-2}\ \text{m/s}=25\times10^{-3}\ \text{m/s} \).

Teacher's Note:
1. The velocity gradient across the film is v/d.
2. Convert the thickness to metres.

Answer: (a) 0.05 m

Explanation:
1. \( h=\frac{2S\cos\theta}{r\rho g}\propto\frac S\rho \).
2. Both S and \( \rho \) double, so the ratio is unchanged: \( h_B=5\ \text{cm}=0.05\ \text{m} \).

Teacher's Note:
1. The height depends on the ratio of surface tension to density.
2. Doubling both leaves the ratio unchanged.

Answer: (b) 7.92 x 10⁻⁴ J

Explanation:
1. Volume conservation: \( R=10r=10\ \text{mm} \).
2. \( E=S\left[1000\times4\pi r^2-4\pi R^2\right]=0.07\times4\times\frac{22}7\times\left(1000\times10^{-6}-10^{-4}\right) \).
3. \( E=0.07\times12.57\times9\times10^{-4}=7.92\times10^{-4}\ \text{J} \).

Teacher's Note:
1. The big drop has less surface area than the 1000 small drops, so energy is released.
2. Release = initial surface energy - final surface energy.

Answer: (a) 2.26 x 10⁻⁵ J

Explanation:
1. Volume conservation: \( 125r^3=R^3\Rightarrow r=\frac R5 \).
2. \( \Delta A=125\times4\pi\frac{R^2}{25}-4\pi R^2=16\pi R^2 \).
3. \( \Delta E=S\,\Delta A=0.45\times16\pi\times10^{-6}=2.26\times10^{-5}\ \text{J} \).

Teacher's Note:
1. Splitting increases the surface area, so energy is absorbed.
2. 125 drops of radius R/5 have 5 times the area of the original drop.

Answer: (a) 10

Explanation:
1. Lift balances weight: \( \Delta P=\frac{mg}{A}=\frac{5.4\times10^5\times10}{500}=1.08\times10^4\ \text{Pa} \).
2. Bernoulli: \( \Delta P=\frac12\rho(v_2^2-v_1^2)\approx\rho\bar v\,\Delta v \) with \( \bar v=300\ \text{m/s} \) (1080 km/h).
3. \( \Delta v=\frac{10800}{1.2\times300}=30\ \text{m/s} \), so \( \frac{\Delta v}{v}=\frac{30}{300}=10\% \).

Teacher's Note:
1. The pressure difference across the wings supplies the lift.
2. Convert 1080 km/h into 300 m/s.

Answer: (a) 2

Explanation:
1. Equation of continuity: \( A\left|\frac{dh}{dt}\right|=av \).
2. \( \left|\frac{dh}{dt}\right|=\frac{500\times10^{-6}\times0.3}{750\times10^{-4}}=2\times10^{-3}\ \text{m/s} \), so x = 2 (the level is falling).

Teacher's Note:
1. Volume flow rate out equals the rate of decrease of volume in the tank.
2. Convert both areas to m²: 750 cm² = 0.075 m² and 500 mm² = 5 x 10⁻⁴ m².