JEE Physics Mechanical Properties Of Fluids MCQs Set 17

Answer: (b) the ball will finally stop

Explanation:
1. With gravity: weight = buoyancy + drag at terminal speed. Buoyancy is half the weight (air density is half the ball's), so the drag at speed v is \( \frac{mg}{2} \).
2. In gravity-free space there is no weight and no buoyancy, only the viscous drag, which opposes the motion.
3. The initial deceleration is \( \frac{mg/2}{m}=\frac g2 \), not 2g, and the speed keeps falling, so the ball finally stops. There is no terminal velocity.

Teacher's Note:
1. Buoyancy exists only because of gravity.
2. A resisting force alone keeps slowing the ball down.

Answer: (c) 1.5 H

Explanation:
1. Depth of the hole below the free surface \( =3H-x \) (the surface is at 3H above the ground).
2. Speed of efflux \( v=\sqrt{2g(3H-x)} \), fall time \( t=\sqrt{\frac{2x}{g}} \).
3. Range \( R=vt=2\sqrt{(3H-x)x} \) is maximum when \( x=3H-x\Rightarrow x=1.5H \).

Teacher's Note:
1. The product of the two heights is largest when they are equal.
2. Measure both depth and fall height from the correct references.

Answer: (d) \( \frac{h_2+h_1}{2} \)

Explanation:
1. For liquid height H and a hole at height h: \( R=2\sqrt{h(H-h)} \).
2. Equal ranges for h1 and h2 means they are symmetric: \( h_1+h_2=H \).
3. Range is maximum at \( h=\frac H2=\frac{h_1+h_2}{2} \).

Teacher's Note:
1. The range expression is symmetric about H/2.
2. Two holes with equal range are equally spaced about the mid-height.

Answer: (d) None of these

Explanation:
1. Terminal speed \( v\propto r^2 \) (Stokes' law).
2. Volume doubles on coalescing: \( R^3=2r^3\Rightarrow R=2^{1/3}r \).
3. \( v'=2^{2/3}v\approx1.59v \), which is none of the given values.

Teacher's Note:
1. Stokes' law: terminal speed is proportional to the square of the radius.
2. Volume is conserved when the drops merge.

Answer: (a) 8 m

Explanation:
1. Speed on reaching the water \( v^2=2g(2)=4g \).
2. In water, buoyancy \( =\frac{mg}{0.8}=1.25mg \) upwards, so the net upward force is \( 0.25mg \), a deceleration of \( 0.25g \).
3. Depth \( d=\frac{v^2}{2(0.25g)}=\frac{4g}{0.5g}=8\ \text{m} \).

Teacher's Note:
1. The volume of the ball is m/(0.8ρ), so buoyancy is 1.25 times the weight.
2. Treat the motion in water as uniform deceleration.

Answer: (d) \( \frac{\ell_1T_2-\ell_2T_1}{T_2-T_1} \)

Explanation:
1. Hooke's law: extension is proportional to tension: \( \frac{\ell_1-\ell_0}{\ell_2-\ell_0}=\frac{T_1}{T_2} \).
2. \( T_2\ell_1-T_2\ell_0=T_1\ell_2-T_1\ell_0 \).
3. \( \ell_0=\frac{\ell_1T_2-\ell_2T_1}{T_2-T_1} \).

Teacher's Note:
1. Extension = length − unstretched length.
2. Divide the two Hooke's law equations to eliminate the constant.

Answer: (c) \( 9.8\times10^{10}\ \text{N/m}^2 \)

Explanation:
1. Pressure at the bottom \( P=\rho gh=1000\times9.8\times1000=9.8\times10^6\ \text{N/m}^2 \).
2. Volume strain \( \frac{\Delta V}{V}=0.01\%=10^{-4} \).
3. \( B=\frac{P}{\Delta V/V}=\frac{9.8\times10^6}{10^{-4}}=9.8\times10^{10}\ \text{N/m}^2 \).

Teacher's Note:
1. Bulk modulus = pressure change / volume strain.
2. Convert the density 1 g/cm³ to 1000 kg/m³.

Answer: (b) the cohesive force between mercury molecules is greater than the adhesive force between mercury and glass

Explanation:
1. The liquid surface in a capillary is depressed when the angle of contact is obtuse.
2. The angle of contact is obtuse for mercury on glass because the mercury-mercury cohesion is stronger than the mercury-glass adhesion.
3. Hence the mercury is depressed (statement (b)).

Teacher's Note:
1. Rise happens when adhesion exceeds cohesion; depression when cohesion exceeds adhesion.
2. Viscosity has no effect on the final level.

Answer: (b) the angle of contact may be 90°

Explanation:
1. Capillary rise \( h=\frac{2T\cos\theta}{r\rho g} \).
2. For \( h=0 \) we need \( \cos\theta=0 \), that is, \( \theta=90^\circ \) (or T = 0, which is not required).
3. So the angle of contact may be 90°.

Teacher's Note:
1. h depends on cos θ; it vanishes at 90°.
2. An angle of 0° gives the maximum rise.

Answer: (b) \( 2.0014\times10^5\ \text{N/m}^2 \)

Explanation:
1. Pressure in the water at the depth: \( P_0+\rho gh=10^5+10^3\times10\times10=2\times10^5\ \text{N/m}^2 \).
2. Excess pressure inside an air bubble (one surface): \( \frac{2T}{r}=\frac{2\times0.07}{10^{-3}}=140\ \text{N/m}^2 \).
3. Total \( =2\times10^5+140=2.0014\times10^5\ \text{N/m}^2 \).

Teacher's Note:
1. A bubble inside a liquid has only one surface, so the excess pressure is 2T/r.
2. Add the hydrostatic pressure to the atmospheric pressure.

Answer: (a) \( 480\ \text{N/m}^2 \)

Explanation:
1. Pressure needed = hydrostatic pressure at the depth + excess pressure of the bubble.
2. Hydrostatic: \( \rho gh=10^3\times10\times0.02=200\ \text{N/m}^2 \).
3. Excess pressure: \( \frac{2T}{r}=\frac{2\times0.07}{5\times10^{-4}}=280\ \text{N/m}^2 \).
4. Total \( =480\ \text{N/m}^2 \).

Teacher's Note:
1. Gauge pressure ignores atmospheric pressure.
2. Bubble formation at the tube mouth involves the radius of the tube.

Answer: (d) \( \frac{2T}{\rho gd} \)

Explanation:
1. Per unit length of the plates, the upward surface tension force is \( 2T \) (one on each plate).
2. Weight of liquid raised per unit length \( =\rho g\,d\,h \).
3. \( 2T=\rho gdh\Rightarrow h=\frac{2T}{\rho gd} \).

Teacher's Note:
1. Take a unit length perpendicular to the plane of the plates.
2. Two liquid-plate contact lines each contribute T.

Answer: (c) \( RT \)

Explanation:
1. The film outside pulls on each element of the thread outwards with force \( T\,dl \) (\( dl=R\,d\theta \)).
2. Take a small element subtending angle dθ: balance of the outward pull and the inward components of the thread tension \( \tau \): \( \tau\,d\theta=T\,R\,d\theta \).
3. \( \tau=RT \).

Teacher's Note:
1. Thread tension is balanced by the film's outward pull per unit length.
2. For a small element the inward component of the tension is τ dθ.

Answer: (b) \( 81(4\pi)^{1/3}T\ \text{erg} \)

Explanation:
1. Volume \( \frac43\pi R^3=9\Rightarrow R^2=\left(\frac{27}{4\pi}\right)^{2/3} \) (in cm). Small drops have radius \( r=\frac R{10} \).
2. Work \( =T\Delta A=T\,4\pi(1000r^2-R^2)=4\pi TR^2(10-1)=36\pi TR^2 \).
3. \( 36\pi\left(\frac{27}{4\pi}\right)^{2/3}=9\times4\pi\times\frac{9}{(4\pi)^{2/3}}=81(4\pi)^{1/3} \), so \( W=81(4\pi)^{1/3}T \) erg.

Teacher's Note:
1. Work done equals surface tension times the increase in surface area.
2. Use the conservation of volume to find r.

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

Explanation:
1. Mercury does not wet the vessel, so it bulges out of the hole as a part of a sphere of radius r.
2. The excess pressure from surface tension supports the liquid column: \( \rho gh=\frac{2T}{r} \).
3. \( h=\frac{2T}{r\rho g} \).

Teacher's Note:
1. The surface acts as a membrane with a single curved surface.
2. Pressure from the column equals the excess pressure 2T/r.