JEE Physics Mechanical Properties Of Fluids MCQs Set 32

Answer: (b) \( \left(\dfrac{R}{24d}\right)^{1/3} \)

Explanation:
1. Volume of the film \( 4\pi R^{2}d=\dfrac43\pi r^{3} \), so \( r=(3R^{2}d)^{1/3} \).
2. Drop: \( \dfrac{2T}{r} \); bubble: \( \dfrac{4T}{R} \).
3. Ratio \( =\dfrac{R}{2r}=\left(\dfrac{R^{3}}{8\cdot3R^{2}d}\right)^{1/3}=\left(\dfrac{R}{24d}\right)^{1/3} \).

Teacher's Note:
1. The liquid volume is conserved.
2. A bubble has two surfaces and a drop has one.

Answer: (d) \( \dfrac{3T}{\rho s}\left(\dfrac1r-\dfrac1R\right) \)

Explanation:
1. Energy released \( =4\pi T(nr^{2}-R^{2}) \) with \( n=R^{3}/r^{3} \), which equals \( 4\pi TR^{3}\left(\dfrac1r-\dfrac1R\right) \).
2. Mass \( =\dfrac43\pi R^{3}\rho \), so \( \Delta\theta=\dfrac{4\pi TR^{3}(1/r-1/R)}{\frac43\pi R^{3}\rho s}=\dfrac{3T}{\rho s}\left(\dfrac1r-\dfrac1R\right) \).

Teacher's Note:
1. The surface energy released heats the drop.
2. Equate the released energy to \( ms\Delta\theta \).

Answer: (d) \( P_1=P_2 \)

Explanation:
1. An air bubble in water has a single surface: excess pressure \( \dfrac{2T}{r} \).
2. A drop in air also has one surface: \( \dfrac{2T}{r} \).
3. They are equal.

Teacher's Note:
1. Count the surfaces: one for each, since neither is a soap film.

Answer: (c) 1.5 cm

Explanation:
1. Water is held by two menisci, one at each end, each giving an upward force \( 2\pi rT \).
2. \( \rho g\pi r^{2}h=2\times2\pi rT \), so \( h=\dfrac{4T}{\rho gr}=\dfrac{4\times73.5\times10^{-3}}{1000\times9.8\times2\times10^{-3}}=1.5\times10^{-2} \) m.

Teacher's Note:
1. An open-ended tube has two curved surfaces.
2. Total surface tension force is twice that of a single meniscus.

Answer: (c) 1860 N/m²

Explanation:
1. The capillary rise difference is \( \dfrac{2T}{\rho g}\left(\dfrac1{r_2}-\dfrac1{r_1}\right)=\dfrac{0.144}{9800}\times694.4\approx0.0102 \) m.
2. The applied pressure difference supports the rest of the height difference: \( \Delta P=\rho g(0.2-0.0102) \).
3. \( \Delta P\approx1860 \) N/m².

Teacher's Note:
1. Part of the level difference comes from capillarity.
2. Only the remainder is due to the applied pressure difference.

Answer: (a) 4.5 cm/s

Explanation:
1. For a Newtonian fluid, the viscous force is proportional to the velocity gradient, hence \( F\propto v \).
2. \( v'=1.5\times\dfrac{2400}{800}=4.5 \) cm/s.

Teacher's Note:
1. Newtonian means viscous force is linear in speed.
2. Triple the force gives triple the speed.

Answer: (a) \( \dfrac{3\rho agt}{5v} \)

Explanation:
1. Constant velocity means \( mg\sin37^{\circ}=\eta A\dfrac vt \) with \( A=a^{2} \) and \( m=\rho a^{3} \).
2. \( \eta=\dfrac{\rho a^{3}g\times\frac35\,t}{a^{2}v}=\dfrac{3\rho agt}{5v} \).

Teacher's Note:
1. The viscous force on the film is \( \eta A v/t \).
2. Balance it against the component of weight along the incline.

Answer: (a) \( r^{2} \)

Explanation:
1. From Stokes' law \( v_T=\dfrac{2r^{2}(\rho-\sigma)g}{9\eta} \), so \( v_T\propto r^{2} \).

Teacher's Note:
1. Stokes' drag is linear in both \( r \) and \( v \), while weight goes as \( r^{3} \).

Answer: (c) 15

Explanation:
1. Pressure at the lower end must exceed the hydrostatic pressure of the water: \( 12 \) cm.
2. It must also overcome the excess pressure of the meniscus, which equals the capillary rise of 3 cm of water.
3. Required \( x=12+3=15 \) cm.

Teacher's Note:
1. Blowing the bubble needs the hydrostatic pressure plus the surface-tension excess pressure \( 2T/r \).
2. The latter equals \( h\rho g \) for the capillary rise \( h \).

Answer: (a) 0.5 g

Explanation:
1. The two surfaces of the film pull the wire up with \( 2Tl \).
2. \( Wg=2\times25\times10^{-3}\times0.1=5\times10^{-3} \) N, so \( W=5\times10^{-4} \) kg \( =0.5 \) g.

Teacher's Note:
1. The film has two faces.
2. Balance the total pull with the weight of the load.

Answer: (c) \( \dfrac1{14} \) mm

Explanation:
1. Water enters when \( \rho gh=\dfrac{2T}{r} \).
2. \( r=\dfrac{2\times0.07}{1000\times10\times0.4}=3.5\times10^{-5} \) m.
3. Diameter \( =7\times10^{-5} \) m \( =\dfrac1{14} \) mm.

Teacher's Note:
1. At the threshold the hydrostatic pressure equals the excess pressure of the meniscus.
2. Convert radius to diameter at the end.

Answer: (b) \( 2.8\times10^{-4} \) m

Explanation:
1. Surface tension acts along the perimeter \( 4\times0.1=0.4 \) m: force \( =0.07\times0.4=0.028 \) N.
2. Extra displaced volume \( =\dfrac{0.028}{1000\times9.8}=2.86\times10^{-6}\ \text{m}^{3} \).
3. Over the base area \( 0.01\ \text{m}^2 \): \( \Delta h=2.86\times10^{-4} \) m (the cube floats this much deeper, since the film pulls it down).

Teacher's Note:
1. The film pulls the cube down along its perimeter.
2. Extra buoyancy needed equals the vertical surface-tension force.

Answer: (d) \( \dfrac{2Tl}{g} \)

Explanation:
1. The depressed surface pulls on both sides of the needle, each side with force \( Tl \), acting vertically at best.
2. \( mg=2Tl \), so \( m=\dfrac{2Tl}{g} \).

Teacher's Note:
1. There are two long edges of contact.
2. The maximum is when the film pulls vertically upward on both.

Answer: (d) \( 8.08\times10^{5} \) N/m²

Explanation:
1. Initial air pressure inside the bubble \( =10^{5}+\dfrac{4\times0.08}{2.4\times10^{-4}}=1.01333\times10^{5} \).
2. Volume becomes \( \tfrac18 \), so \( P'=8\times1.01333\times10^{5}=8.107\times10^{5} \).
3. Cylinder pressure \( =8.107\times10^{5}-\dfrac{4\times0.08}{1.2\times10^{-4}}=8.107\times10^{5}-2667\approx8.08\times10^{5} \).

Teacher's Note:
1. Apply Boyle's law to the air inside the bubble.
2. The cylinder pressure is the inside pressure minus \( 4T/r \).

Answer: (c) 1 : 1

Explanation:
1. Drop: \( \dfrac{2T_w}{R} \).
2. Bubble: \( \dfrac{4T_s}{R}=\dfrac{4(T_w/2)}{R}=\dfrac{2T_w}{R} \).
3. Ratio \( 1:1 \).

Teacher's Note:
1. The bubble has two surfaces while the drop has one.
2. Half the tension doubled gives the same total.

Answer: (a) \( 2:\pi \)

Explanation:
1. For a semicircle of radius \( r \), the flat side has length \( 2r \) and the curved part \( \pi r \).
2. Surface-tension force is \( T\times\text{length} \), so the ratio is \( 2r:\pi r=2:\pi \).

Teacher's Note:
1. Force equals tension times contact length.
2. Compare the two lengths.

Answer: (c) water will stay at a height of 12 cm in the capillary

Explanation:
1. The water rises only as high as the tube allows, so it reaches the cut end at 12 cm.
2. The meniscus flattens (its radius of curvature increases) to \( r'=\dfrac{2T}{h\rho g} \) with \( h=12 \) cm, so the pressure balance still holds.
3. No fountain forms and the water stays at 12 cm.

Teacher's Note:
1. Capillary rise is a maximum height.
2. A shorter tube is simply filled up to its top with a less curved meniscus.

Answer: (b) \( \dfrac{\beta d}{4} \)

Explanation:
1. \( \dfrac{dv}{dr}=-\dfrac{\beta r}{2\mu} \).
2. At the wall \( r=d/2 \): \( \left|\dfrac{dv}{dr}\right|=\dfrac{\beta d}{4\mu} \).
3. Shear stress \( =\mu\left|\dfrac{dv}{dr}\right|=\dfrac{\beta d}{4} \).

Teacher's Note:
1. Shear stress is viscosity times the velocity gradient.
2. Differentiate the profile and evaluate at the wall.

Answer: (a) the initial acceleration is \( g\left(\dfrac{\rho-\sigma}{\rho}\right) \)

Explanation:
1. At the instant of release there is no drag, so \( ma=Vg(\rho-\sigma) \) giving \( a=g\dfrac{\rho-\sigma}{\rho} \).
2. At terminal speed the net force is zero, so (c) is false, and the time to reach it depends on \( \sigma \) through the net weight, so (b) is false.

Teacher's Note:
1. The initial acceleration has weight and buoyancy but no viscous force.
2. Net force vanishes at terminal speed.

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

Explanation:
1. Initial area of one face \( =0.1\times0.005=5\times10^{-4} \), final \( =0.1\times0.006=6\times10^{-4}\ \text{m}^{2} \); two faces give \( 12\times10^{-4}\ \text{m}^{2} \), so (ii) is true.
2. \( W=T\times2\times\Delta A=72\times10^{-3}\times2\times10^{-4}=1.44\times10^{-5}=144\times10^{-7} \) J, so (i) is true and (iii) is false.

Teacher's Note:
1. Count both faces of the film.
2. Work is surface tension times the increase of total area.

Answer: (a) all four

Explanation:
1. A drop in air, a spherical meniscus, and an air bubble in water each have a single curved surface of radius \( R \), giving \( 2T/R \).
2. A cylindrical soap bubble in air has two surfaces, each with excess \( T/R \) for a cylinder, which again totals \( 2T/R \).
3. All four can give \( 2T/R \).

Teacher's Note:
1. Count the surfaces and the principal radii of curvature.
2. For a cylinder one radius is infinite, so each surface contributes \( T/R \).

Answer: (a) (1) and (4) only

Explanation:
1. A bubble has an excess pressure \( 2T/r \) over the liquid, so (1) is true.
2. As the bubble rises, the outside pressure falls and the bubble expands, so the internal pressure decreases: (4).

Teacher's Note:
1. The inside pressure equals the liquid pressure plus \( 2T/r \).
2. Liquid pressure drops with depth.

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

Explanation:
1. Final area \( =n^{2/3}\times \) single area, so the final surface energy is \( n^{2/3}E \).
2. Released energy \( =nE-n^{2/3}E=E\left(n-n^{2/3}\right)>0 \).
3. (i) and (iii) are true.

Teacher's Note:
1. Volume conservation gives \( R=n^{1/3}r \).
2. Surface energy scales with area.

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

Explanation:
1. The liquid can only rise up to the top of the shorter tube.
2. Its meniscus flattens (radius of curvature increases) so that \( \dfrac{2T\cos\theta}{r'}=h'\rho g \) with the smaller \( h' \).
3. So the tube is filled without overflow, and the surface is no longer hemispherical.

Teacher's Note:
1. Capillary rise is limited by the tube height.
2. The meniscus radius adjusts to balance the smaller column.

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

Explanation:
1. Viscous force depends on the velocity gradient (so on speed) and on temperature, while (kinetic) friction is nearly independent of speed and temperature.
2. Viscous force depends on the area of the layers, unlike what (iv) says.

Teacher's Note:
1. Viscous drag grows with speed and falls steeply with temperature in liquids.
2. Friction is nearly constant in both respects.