JEE Physics Mechanical Properties Of Fluids MCQs Set 33

Answer: (a) (ii), (iii) and (iv) only

Explanation:
1. At 20 m/s the drag equals the weight \( mg \), so with no gravity the net force is \( mg \) opposite to the motion, giving \( 9.8\ \text{m/s}^2 \) upward (against the motion).
2. As the sphere slows, the drag decreases, so the deceleration decreases.
3. The sphere keeps slowing down and eventually stops.

Teacher's Note:
1. Drag at terminal velocity equals the weight.
2. Without gravity only the drag acts.

Answer: (b) 6

Explanation:
1. Inside pressures: \( P_A=8+\dfrac{4\times0.04}{0.02}=16 \) N/m² and \( P_B=8+\dfrac{4\times0.04}{0.04}=12 \) N/m².
2. At the same temperature \( n\propto PV\propto Pr^{3} \).
3. \( \dfrac{n_B}{n_A}=\dfrac{12\times4^{3}}{16\times2^{3}}=6 \).

Teacher's Note:
1. Find the pressure inside each bubble using \( 4T/r \) first.
2. Then compare \( PV \) values.

Answer: (c) 8 : 9

Explanation:
1. \( \rho=\dfrac{PM}{RT} \) at the same temperature.
2. \( \dfrac{\rho_1}{\rho_2}=\dfrac{P_1M_1}{P_2M_2}=\dfrac43\times\dfrac23=\dfrac89 \).

Teacher's Note:
1. Density is proportional to \( PM \) at fixed temperature.

Answer: (c) \( \dfrac{2S}{b\rho g}\cos\left(\theta+\dfrac\alpha2\right) \)

Explanation:
1. The wall at the meniscus makes angle \( \alpha/2 \) with the vertical, so the effective angle between the liquid surface and the vertical axis changes to \( \theta+\alpha/2 \).
2. Vertical force balance: \( 2\pi bS\cos\left(\theta+\tfrac\alpha2\right)=\pi b^{2}h\rho g \).
3. \( h=\dfrac{2S}{b\rho g}\cos\left(\theta+\dfrac\alpha2\right) \).

Teacher's Note:
1. The tilted wall changes the angle between the liquid surface and the vertical.
2. Balance the vertical component of tension with the weight of the column.

Answer: (a) (ii) and (iii) only

Explanation:
1. For a uniform self-gravitating fluid sphere, \( P(r)=\dfrac{2\pi G\rho^{2}}{3}\left(R^{2}-r^{2}\right) \), so \( P(0) \) is the maximum and not zero (i false).
2. (ii): \( \dfrac{1-9/16}{1-4/9}=\dfrac{7/16}{5/9}=\dfrac{63}{80} \), true.
3. (iii): \( \dfrac{1-9/25}{1-4/25}=\dfrac{16}{21} \), true. (iv): \( \dfrac{1-1/4}{1-1/9}=\dfrac{27}{32}\neq\dfrac{20}{27} \), false.

Teacher's Note:
1. Integrate \( dP=-\rho g(r)\,dr \) with \( g(r)\propto r \).
2. Pressure falls to zero only at the surface.

Answer: (d) 3

Explanation:
1. \( v_T\propto\dfrac{r^{2}(\rho-\sigma)}{\eta} \).
2. \( \dfrac{v_P}{v_Q}=\dfrac{(0.5)^{2}(8-0.8)/3}{(0.25)^{2}(8-1.6)/2}=\dfrac{0.6}{0.2}=3 \).

Teacher's Note:
1. Use Stokes' terminal velocity formula and take ratios.
2. Radius is half the diameter.

Answer: (c) 3

Explanation:
1. Electric forces are of order \( \dfrac{q^{2}}{4\pi\varepsilon_0a^{2}} \) and the film tension force is of order \( \gamma a \).
2. Balance gives \( a^{3}\sim\dfrac{q^{2}}{\gamma} \) apart from numerical constants.
3. So \( a\propto\left(\dfrac{q^{2}}{\gamma}\right)^{1/3} \) and \( N=3 \).

Teacher's Note:
1. Equate the scaling of the electrostatic force with the surface-tension force.
2. \( q^{2}/a^{2}\sim\gamma a \) gives the power directly.

Answer: (a) \( 6\pi\times10^{-3} \) J

Explanation:
1. A bubble has two surfaces.
2. \( W=2T\times4\pi\left(R_2^{2}-R_1^{2}\right)=2\times25\times10^{-3}\times4\pi(0.04-0.01)=6\pi\times10^{-3} \) J.

Teacher's Note:
1. Two surfaces mean the area increase is doubled.

Answer: (b) 0.6 g

Explanation:
1. Both film surfaces pull on the straw: \( mg=2Tl=2\times3\times10^{-2}\times0.1=6\times10^{-3} \) N.
2. \( m=6\times10^{-4} \) kg \( =0.6 \) g.

Teacher's Note:
1. The film has two surfaces, so count \( 2Tl \).

Answer: (b) \( 4\pi R^{3}T\left(\dfrac1r-\dfrac1R\right) \)

Explanation:
1. \( n=\dfrac{R^{3}}{r^{3}} \).
2. Energy released \( =T\times4\pi\left(nr^{2}-R^{2}\right)=4\pi T\left(\dfrac{R^{3}}{r}-R^{2}\right)=4\pi R^{3}T\left(\dfrac1r-\dfrac1R\right) \).

Teacher's Note:
1. The released energy equals the decrease in total surface area times \( T \).

Answer: (b) 4.76 mm

Explanation:
1. \( h=\dfrac{2T}{\rho g}\left(\dfrac1{r_1}-\dfrac1{r_2}\right)=\dfrac{2\times0.07}{1000\times9.8}\left(1000-666.7\right)=4.76\times10^{-3} \) m.

Teacher's Note:
1. The narrower tube holds a higher column.
2. Subtract the two capillary rises.

Answer: (b) 2.5 cm

Explanation:
1. \( \Delta h=\dfrac{4T}{\rho g}\left(\dfrac1{d_1}-\dfrac1{d_2}\right)=\dfrac{4\times0.07}{10^{4}}(1000-100)=2.52\times10^{-2} \) m \( \approx2.5 \) cm.

Teacher's Note:
1. Capillary rise is \( 4T/(\rho gd) \) in terms of diameter.

Answer: (d) 20 cm

Explanation:
1. In free fall the effective gravity is zero, so there is no weight to balance and the surface tension pulls water up the whole tube.
2. The water column fills the entire 20 cm.

Teacher's Note:
1. The capillary rise \( h\propto1/g_{eff} \) becomes unbounded as \( g_{eff}\to0 \).
2. The tube length limits it.

Answer: (a) 20.4 m

Explanation:
1. Terminal velocity in water \( v=\dfrac{2r^{2}(\rho-\sigma)g}{9\eta}=\dfrac{2\times10^{-8}\times9000\times9.8}{9\times9.8\times10^{-6}}=20 \) m/s.
2. The free-fall speed must equal this: \( h=\dfrac{v^{2}}{2g}=\dfrac{400}{19.6}=20.4 \) m.

Teacher's Note:
1. The ball enters at its terminal velocity, so drag balances net weight.
2. Use \( v^{2}=2gh \) for the fall in air.

Answer: (b) 4.5 m

Explanation:
1. At the surface the balloon's mean density is \( \tfrac23\rho_w \).
2. For neutral buoyancy its density must double by a factor \( \tfrac32 \), meaning its volume must shrink by \( \tfrac23 \).
3. Boyle's law: \( 9\,V=(9+d)\tfrac23V \), so \( 9+d=13.5 \) and \( d=4.5 \) m.

Teacher's Note:
1. Neutral buoyancy needs the mean density to equal that of water.
2. The volume of the balloon varies inversely with the pressure.

Answer: (a) 3

Explanation:
1. With constant velocity: buoyancy \( =\) weight \( +\) drag.
2. Buoyancy \( =4mg \), so drag \( =3mg \), which is 3 times the weight.

Teacher's Note:
1. Net force is zero at constant velocity.
2. Buoyancy is four times the weight here.

Answer: (b) 10.9 m/s

Explanation:
1. \( v=\dfrac{2r^{2}\rho g}{9\eta}=\dfrac{2\times(3\times10^{-4})^{2}\times1000\times9.8}{9\times1.8\times10^{-5}}\approx10.9 \) m/s.

Teacher's Note:
1. Use Stokes' law with the air density neglected.

Answer: (d) 152 Pa s

Explanation:
1. For an air bubble (density negligible) \( v=\dfrac{2r^{2}\sigma g}{9\eta} \).
2. \( \eta=\dfrac{2\times10^{-4}\times1470\times9.8}{9\times2.1\times10^{-3}}\approx152 \) Pa s (about \( 1.52\times10^{3} \) poise).

Teacher's Note:
1. The liquid density is the dominant density for a rising bubble.

Answer: (c) 0.1 m/s

Explanation:
1. \( v_T\propto(\rho-\sigma) \) for the same size and liquid.
2. \( v=0.2\times\dfrac{10.5-1.5}{19.5-1.5}=0.2\times\dfrac{9}{18}=0.1 \) m/s.

Teacher's Note:
1. Only the density difference with the liquid matters here.

Answer: (d) \( \sqrt{\dfrac{Vg(\rho_1-\rho_2)}{k}} \)

Explanation:
1. At terminal speed, weight minus buoyancy equals drag: \( Vg(\rho_1-\rho_2)=kv^{2} \).
2. \( v=\sqrt{\dfrac{Vg(\rho_1-\rho_2)}{k}} \).

Teacher's Note:
1. A quadratic drag law gives a square-root dependence.

Answer: (b) \( 0.4\pi \) mJ

Explanation:
1. \( W=2T\times4\pi\left(r_2^{2}-r_1^{2}\right)=2\times0.03\times4\pi\left(25-9\right)\times10^{-4} \) J.
2. \( =3.84\pi\times10^{-4}\approx0.4\pi\times10^{-3} \) J \( =0.4\pi \) mJ.

Teacher's Note:
1. A bubble has two surfaces, so the area change is doubled.

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

Explanation:
1. \( v\propto\dfrac{\rho-\sigma}{\eta} \).
2. \( v'=10\times\dfrac{7.8-1.2}{7.8-1.0}\times\dfrac{8.5\times10^{-4}}{13.2}=10\times0.9706\times6.44\times10^{-5}=6.25\times10^{-4} \) cm/s.

Teacher's Note:
1. Compare the two liquids by the ratio of \( (\rho-\sigma)/\eta \).

Answer: (d) \( 2^{8/3}\pi r^{2}T \)

Explanation:
1. \( R=2^{1/3}r \).
2. Surface energy \( =T\times4\pi R^{2}=4\pi r^{2}T\times2^{2/3}=2^{8/3}\pi r^{2}T \).

Teacher's Note:
1. The volume doubles, so the radius grows by \( 2^{1/3} \).

Answer: (d) \( \dfrac{2T}{\rho L} \)

Explanation:
1. Evaporating a thin layer of mass \( dm=\rho\,4\pi r^{2}dr \) needs heat \( L\,dm \).
2. The surface energy released as the radius shrinks by \( dr \) is \( T\times8\pi r\,dr \).
3. Equating: \( 8\pi rT=4\pi r^{2}\rho L \), so \( r=\dfrac{2T}{\rho L} \).

Teacher's Note:
1. The energy for evaporation comes from the drop's own surface energy.
2. Equate the surface energy lost to the latent heat needed.

Answer: (c) \( R^{2}\sqrt{\dfrac{2\rho_wg}{3T}} \)

Explanation:
1. Surface tension pulls the bubble down along the circle of contact with vertical component \( T\,2\pi r\sin\theta \) where \( \sin\theta\approx\dfrac rR \): \( \dfrac{2\pi r^{2}T}{R} \).
2. At detachment the buoyancy balances it: \( \dfrac43\pi R^{3}\rho_wg=\dfrac{2\pi r^{2}T}{R} \).
3. \( r^{2}=\dfrac{2\rho_wgR^{4}}{3T} \), so \( r=R^{2}\sqrt{\dfrac{2\rho_wg}{3T}} \).

Teacher's Note:
1. Only the vertical component of tension holds the bubble down.
2. Use \( \sin\theta=r/R \) for small \( r \).