JEE Physics Mechanical Properties Of Fluids MCQs Set 24

Answer: (d) 3.98 × 10⁻² J

Explanation:
1. Volume conservation: \( r=\frac{R}{100}=10^{-4}\ \text{m} \).
2. \( \Delta A=4\pi\left(10^6r^2-R^2\right)=4\pi(10^{-2}-10^{-4})=4\pi\times9.9\times10^{-3}\ \text{m}^2 \)
3. \( E=T\,\Delta A=0.32\times0.1244=3.98\times10^{-2}\ \text{J} \)

Teacher's Note:
1. Energy needed to increase the surface of a liquid is surface tension multiplied by the increase in area.
2. Use volume conservation to find the droplet radius first: n r³ = R³.

Answer: (b) 1.44 × 10⁻⁵ J

Explanation:
1. The film has two surfaces, so the area added is \( \Delta A=2\times0.10\times0.001=2\times10^{-4}\ \text{m}^2 \).
2. \( W=T\,\Delta A=72\times10^{-3}\times2\times10^{-4}=1.44\times10^{-5}\ \text{J} \)

Teacher's Note:
1. A free liquid film has a front and a back surface, so the area is counted twice.
2. Work done against surface tension equals T times the extra area.

Answer: (b) \( \frac{2T}{\rho g r} \)

Explanation:
1. Water at depth h has pressure excess \( \rho g h \) over the air above it.
2. The water surface in the hole bulges inwards as a hemisphere of radius r and can balance a pressure difference at most \( \frac{2T}{r} \).
3. \( \rho g h=\frac{2T}{r}\Rightarrow h=\frac{2T}{\rho g r} \)

Teacher's Note:
1. A thin film across a small hole can support a pressure difference of 2T/r, the excess pressure of a hemispherical meniscus.
2. Beyond this depth, hydrostatic pressure beats surface tension and water enters.

Answer: (b) full length of the tube

Explanation:
1. In a satellite the effective gravity is zero (free fall), so \( h=\frac{2T\cos\theta}{r\rho g}\to\infty \).
2. Water therefore fills the entire tube.

Teacher's Note:
1. Capillary rise is inversely proportional to g.
2. In orbit, apparent weight is zero, so there is no hydrostatic pressure to oppose surface tension.

Answer: (c) \( 2\pi rT\cos\theta+W \)

Explanation:
1. Upward forces: buoyancy W plus the vertical component of surface tension along the perimeter, \( 2\pi rT\cos\theta \).
2. For equilibrium the weight of the disc equals this total: \( mg=W+2\pi rT\cos\theta \).

Teacher's Note:
1. Surface tension acts along the liquid surface; only its vertical component supports the disc.
2. The contact length is the circumference, 2πr.

Answer: (c) \( \frac{\rho g h\,r_1r_2}{2(r_2-r_1)} \)

Explanation:
1. Capillary rise in each arm: \( h_1=\frac{2T}{r_1\rho g},\ h_2=\frac{2T}{r_2\rho g} \).
2. \( h=h_1-h_2=\frac{2T}{\rho g}\left(\frac1{r_1}-\frac1{r_2}\right)=\frac{2T(r_2-r_1)}{\rho g\,r_1r_2} \).
3. \( T=\frac{\rho g h\,r_1r_2}{2(r_2-r_1)} \)

Teacher's Note:
1. The narrower arm has the higher meniscus, so the level difference is the difference of the two capillary rises.
2. Check dimensions of the final expression.

Answer: (c) 1/2 of that for a smaller droplet

Explanation:
1. Volume conservation: \( R^3=8r^3\Rightarrow R=2r \).
2. \( \Delta P\propto\frac1{\text{radius}}\Rightarrow \frac{\Delta P_{big}}{\Delta P_{small}}=\frac{r}{R}=\frac12 \)

Teacher's Note:
1. Excess pressure 2T/r is inversely proportional to the radius.
2. Bigger drop, smaller excess pressure.

Answer: (b) \( P+hdg+\frac{2T}{r} \)

Explanation:
1. Pressure in the water at depth h: \( P+hdg \).
2. The bubble is a cavity of air with a single liquid surface, so the air pressure exceeds that of the water by \( \frac{2T}{r} \).

Teacher's Note:
1. Pressure inside the concave side of a curved surface is higher than outside.
2. An air bubble in liquid has only one surface, so the factor is 2T/r, not 4T/r.

Answer: (d) 5 cm

Explanation:
1. In vacuum the pressure inside a bubble is \( P=\frac{4T}{r} \).
2. Isothermal: \( P_1V_1+P_2V_2=PV \) gives \( \frac{4T}{r_1}r_1^3+\frac{4T}{r_2}r_2^3=\frac{4T}{R}R^3 \).
3. \( R^2=r_1^2+r_2^2=9+16=25\Rightarrow R=5\ \text{cm} \)

Teacher's Note:
1. With no outside pressure the amount of gas is proportional to r² for a soap bubble.
2. The result R² = r₁² + r₂² also means the surface areas add.

Answer: (d) \( 8p_1+\frac{24\sigma}{r} \)

Explanation:
1. Gas inside the bubble: initial pressure \( p_1+\frac{4\sigma}{r} \), volume \( \frac43\pi r^3 \).
2. Final: pressure \( p_2+\frac{8\sigma}{r} \) at radius r/2, volume one eighth.
3. \( \left(p_1+\frac{4\sigma}{r}\right)=\frac18\left(p_2+\frac{8\sigma}{r}\right)\Rightarrow p_2=8p_1+\frac{24\sigma}{r} \)

Teacher's Note:
1. Apply Boyle's law to the air inside the bubble, including the 4σ/r excess pressure.
2. Halving the radius reduces the volume by a factor 8 and doubles the excess pressure.

Answer: (d) 28.0 cm

Explanation:
1. The meniscus in a hole of radius r = 0.05 mm supports a pressure \( \frac{2T}{r} \).
2. \( h\rho g=\frac{2T}{r}\Rightarrow h=\frac{2\times0.07}{5\times10^{-5}\times10^3\times10}=0.28\ \text{m}=28\ \text{cm} \)

Teacher's Note:
1. Water that does not wet the bottom forms a convex meniscus below the hole, which holds back the liquid.
2. Convert 70 dyne/cm to 0.07 N/m.

Answer: (b) \( \sqrt{\frac{6T}{\rho}\left(\frac1a-\frac1b\right)} \)

Explanation:
1. Number of drops \( n=\frac{b^3}{a^3} \). Energy released \( E=T\cdot4\pi(na^2-b^2)=4\pi T\,b^3\left(\frac1a-\frac1b\right) \).
2. Kinetic energy \( \frac12\left(\frac43\pi b^3\rho\right)v^2=E \).
3. \( v=\sqrt{\frac{6T}{\rho}\left(\frac1a-\frac1b\right)} \)

Teacher's Note:
1. Find n from volume conservation, then the area lost.
2. Equate the surface energy released to the kinetic energy of the big drop, whose mass is (4/3)πb³ρ.

Answer: (a) 58.8 dyne/cm

Explanation:
1. Excess pressure \( =0.2\times0.8\times980=156.8\ \text{dyne/cm}^2 \).
2. \( \frac{4T}{r}=156.8\Rightarrow T=\frac{156.8\times1.5}{4}=58.8\ \text{dyne/cm} \)

Teacher's Note:
1. A soap bubble has two surfaces, so the excess pressure is 4T/r.
2. Use the radius (1.5 cm), not the diameter.

Answer: (b) 3

Explanation:
1. The column has two concave menisci (top and bottom), each supporting a pressure \( \frac{2T}{r} \).
2. \( h\rho g=\frac{4T}{r}\Rightarrow h=\frac{4\times72}{0.1\times1\times1000}=2.88\ \text{cm}\approx3\ \text{cm} \)

Teacher's Note:
1. With two free surfaces, the supportable pressure doubles to 4T/r.
2. Compare this with one meniscus, where the height would be 2T/(rρg) = 1.44 cm.

Answer: (b) 6

Explanation:
1. Upward force = surface tension on both boundaries: \( T\,\pi(d_1+d_2) \).
2. Weight of liquid: \( \frac{\pi}{4}(d_2^2-d_1^2)h\rho g \).
3. \( h=\frac{4T}{(d_2-d_1)\rho g}=\frac{4\times0.073}{0.5\times10^{-3}\times1000\times9.8}=0.0596\ \text{m}\approx6\ \text{cm} \)

Teacher's Note:
1. The annular gap acts like two parallel plates a distance (d₂ − d₁)/2 apart.
2. Equate the surface tension pull on inner and outer wall to the weight of the liquid ring.

Answer: (a) The free liquid surface inside the tube will no longer be hemispherical, and the liquid will fill the tube without coming out of its upper end

Explanation:
1. The height is fixed by \( h=\frac{2T\cos\theta}{r\rho g} \) only when the meniscus has radius of curvature R/cosθ.
2. When the tube is shorter than h, the liquid reaches the top and the meniscus adjusts: its radius of curvature increases so that the pressure balance holds with the smaller height. The contact angle is a property of the pair of materials, and the surface just becomes less curved.
3. Energy cannot be created from surface tension, so no fountain forms.

Teacher's Note:
1. The meniscus curvature is adjustable; the contact angle at a fixed wall is not only fixed at the rim of the tube top.
2. A liquid cannot flow out of the tube without a source of energy.

Answer: (b) 3.60 mm of Hg

Explanation:
1. Excess pressure beneath a convex mercury meniscus is \( \frac{2T}{r} \) in each limb.
2. For equal levels, \( \Delta P=2T\left(\frac1{r_1}-\frac1{r_2}\right)=2\times480\left(\frac1{0.1}-\frac1{0.2}\right)=4800\ \text{dyne/cm}^2 \).
3. \( h=\frac{4800}{13.6\times980}=0.36\ \text{cm}=3.6\ \text{mm of Hg} \)

Teacher's Note:
1. Convert the pressure difference into a mercury column using P = hρg.
2. The narrower limb has the greater excess pressure.

Answer: (a) 6

Explanation:
1. \( P_A=8+\frac{4\times0.04}{0.02}=16\ \text{N/m}^2,\quad P_B=8+\frac{4\times0.04}{0.04}=12\ \text{N/m}^2 \).
2. \( \frac{n_B}{n_A}=\frac{P_BV_B}{P_AV_A}=\frac{12\times4^3}{16\times2^3}=6 \)

Teacher's Note:
1. Use PV = nRT at the same temperature for both bubbles.
2. Include the excess pressure 4T/r for each soap bubble.

Answer: (c) \( \frac{2\pi r^2T}{R} \)

Explanation:
1. Surface tension acts along the circle of contact of length \( 2\pi r \).
2. The surface there makes an angle θ with the vertical where \( \sin\theta=\frac{r}{R} \).
3. Vertical component: \( 2\pi rT\sin\theta=\frac{2\pi r^2T}{R} \)

Teacher's Note:
1. Only the vertical component of the surface tension pull supports the drop.
2. The geometry gives sinθ = r/R.

Answer: (d) 1.4 × 10⁻³ m

Explanation:
1. \( \frac{2\pi r^2T}{R}=\frac43\pi R^3\rho g\Rightarrow R^4=\frac{3r^2T}{2\rho g} \).
2. \( R^4=\frac{3\times(5\times10^{-4})^2\times0.11}{2\times10^3\times10}=4.13\times10^{-12}\Rightarrow R\approx1.4\times10^{-3}\ \text{m} \)

Teacher's Note:
1. Set the supporting force equal to the weight of the spherical drop.
2. Take the fourth root at the end.

Answer: (a) \( \frac{2S}{b\rho g}\cos(\theta+\alpha/2) \)

Explanation:
1. The wall makes an angle α/2 with the axis, widening upwards, so the meniscus normal differs from the axis by \( \theta+\frac{\alpha}{2} \).
2. Radius of curvature of the meniscus: \( R_c=\frac{b}{\cos(\theta+\alpha/2)} \).
3. \( h\rho g=\frac{2S}{R_c}\Rightarrow h=\frac{2S}{b\rho g}\cos\left(\theta+\frac{\alpha}{2}\right) \)

Teacher's Note:
1. For a vertical tube R_c = b/cosθ; here the wall is tilted by α/2, which adds to the contact angle.
2. The semi-apex angle, not the apex angle, enters the formula.

Answer: (a) 6

Explanation:
1. \( r=\frac{R}{K^{1/3}},\quad \Delta U=S\,4\pi R^2\left(K^{1/3}-1\right) \).
2. \( \frac{0.1}{4\pi}\times4\pi\times10^{-4}\left(K^{1/3}-1\right)=10^{-3}\Rightarrow K^{1/3}-1=100 \).
3. \( K=101^3\approx1.03\times10^6\Rightarrow\alpha\approx6 \)

Teacher's Note:
1. Volume conservation gives the radius of the small drops.
2. The −1 is negligible against K^(1/3) here, so K is close to 10⁶.

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

Explanation:
1. \( h=\frac{2\sigma\cos\theta}{r\rho g} \).
2. (i) true: h is proportional to 1/r. (ii) false: h is proportional to σ. (iii) true: in an upward-accelerating lift g_eff = g + a, so h decreases. (iv) false: h depends on cosθ, not on θ.

Teacher's Note:
1. Write Jurin's law and test each statement against it.
2. Upward acceleration increases the effective gravity.

Answer: (b) \( \frac{2T}{\rho L} \)

Explanation:
1. Decrease of radius by dr releases surface energy \( T\,d(4\pi r^2)=8\pi rT\,dr \).
2. Heat needed to evaporate the mass lost: \( 4\pi r^2\rho\,dr\cdot L \).
3. \( 8\pi rT=4\pi r^2\rho L\Rightarrow r=\frac{2T}{\rho L} \)

Teacher's Note:
1. Equate the energy released by shrinking to the latent heat needed for the evaporated mass.
2. Smaller drops release more energy per unit mass.

Answer: (c) 1860 N/m²

Explanation:
1. Hydrostatic difference \( \rho g h=10^3\times9.8\times0.2=1960\ \text{N/m}^2 \).
2. Meniscus corrections: \( 2T\left(\frac1{r_2}-\frac1{r_1}\right)=2\times0.072\left(\frac{1}{7.2\times10^{-4}}-\frac1{1.44\times10^{-3}}\right)=100\ \text{N/m}^2 \).
3. The narrow tube's meniscus already supports 100 N/m² more, so the air pressure difference is \( 1960-100=1860\ \text{N/m}^2 \).

Teacher's Note:
1. The raised level in a narrow tube is partly due to capillarity, so the real air pressure difference is smaller than ρgh.
2. Subtract the difference of the two meniscus excess pressures.