JEE Physics Mechanical Properties Of Fluids MCQs Set 39

Answer: (c) less than \( t \)

Explanation:
1. Breaking into droplets increases the total surface area, so the surface energy increases.
2. The system is isolated, so this energy must come from the internal (thermal) energy of the liquid.
3. Hence the temperature falls and the droplets are colder than \( t \).

Teacher's Note:
1. Surface energy is stored in the surface; creating more surface requires work.
2. With no outside energy supply, the liquid cools.

Answer: (b) \( r=\frac{r_1r_2}{r_1-r_2} \)

Explanation:
1. Excess pressure inside bubble 1 is \( 4S/r_1 \) and inside bubble 2 is \( 4S/r_2 \). The excess across the common film is \( 4S/r \).
2. So \( \dfrac4{r}=\dfrac4{r_2}-\dfrac4{r_1} \), giving \( \dfrac1r=\dfrac1{r_2}-\dfrac1{r_1} \).
3. \( r=\dfrac{r_1r_2}{r_1-r_2} \).

Teacher's Note:
1. The film bulges into the larger bubble because the smaller bubble has the higher pressure.
2. The common surface has the pressure difference of the two bubbles.

Answer: (c) A will become smaller and B will become larger

Explanation:
1. Pressure inside a bubble is \( p_0+4S/r \), higher for the smaller bubble.
2. Air flows from the higher pressure \( (A) \) to the lower pressure \( (B) \).
3. A shrinks and \( B \) grows; as \( A \) shrinks its pressure rises further, so the process continues until \( A \) is a small cap.

Teacher's Note:
1. A smaller bubble has greater excess pressure, so it empties into the larger one.
2. This is an unstable equilibrium, unlike a liquid drop pair of equal size.

Answer: (d) none of these (it is dimensionless)

Explanation:
1. \( [\rho]=ML^{-3} \), \( [v]=LT^{-1} \), \( [D]=L \), \( [\eta]=ML^{-1}T^{-1} \).
2. \( [R]=\dfrac{ML^{-3}\cdot LT^{-1}\cdot L}{ML^{-1}T^{-1}}=M^0L^0T^0 \).
3. So \( R \) (the Reynolds number) is dimensionless.

Teacher's Note:
1. A quantity that decides between two flow regimes is usually a dimensionless ratio.
2. Check dimensions by writing each factor in \( M, L, T \).

Answer: (b) viscosity

Explanation:
1. Viscosity of a liquid decreases with temperature.
2. Lower viscosity means less internal friction between layers, so the oil flows more easily.

Teacher's Note:
1. Gases become more viscous when heated; liquids become less viscous.
2. Surface tension also falls with temperature, but flow resistance is governed by viscosity.

Answer: (b) \( \frac{Ft}{A} \)

Explanation:
1. \( \eta=\dfrac{F}{A\,(dv/dx)} \), and \( dv/dx \) has dimension \( T^{-1} \).
2. So \( [\eta]=\dfrac{[F]}{[A]}T=\dfrac{Ft}{A} \).

Teacher's Note:
1. Velocity gradient has dimension of inverse time.
2. The SI unit of viscosity, \( N s m^{-2} \) (Pa \( s \)), matches Ft/A.

Answer: (c) \( \eta_w \) only

Explanation:
1. The viscous drag acts on the drop due to the water flowing past it.
2. Only the surrounding liquid's viscosity enters the drag formula, so the velocity depends on \( \eta_w \) only.

Teacher's Note:
1. The drag on a body depends on the viscosity of the medium around it.
2. The drop's own viscosity matters only for small internal circulation effects, which are neglected here.

Answer: (d) Water flowing out through a hole in the side of a tank

Explanation:
1. Flow out of a hole in a tank is described by Torricelli's law, which is based on energy conservation and ignores viscosity.
2. The other situations involve friction between layers or against walls and are strongly affected by viscosity.

Teacher's Note:
1. Ideal (non-viscous) flow is the Bernoulli picture.
2. Short, fast flow through a wide opening hardly feels viscosity.

Answer: (b) \( r^2 \)

Explanation:
1. At terminal speed, weight equals viscous drag: \( \tfrac43\pi r^3(\rho-\sigma)g=6\pi\eta rv \).
2. So \( v=\dfrac{2r^2(\rho-\sigma)g}{9\eta}\propto r^2 \).

Teacher's Note:
1. Weight grows as \( r^3 \) but drag only as \( r \), giving \( r^2 \).
2. Bigger drops fall faster.

Answer: (d) \( 32p \)

Explanation:
1. Terminal velocity \( \propto r^2 \), so it becomes 4 times.
2. Mass \( \propto r^3 \), so it becomes 8 times.
3. Momentum \( =mv \) becomes \( 8\times4=32 \) times, i.e. \( 32p \).

Teacher's Note:
1. Combine the scaling of mass and terminal speed.
2. The momentum scales as \( r^5 \).

Answer: (d) In air flowing through a tube of variable cross-section, the pressure becomes different at different points

Explanation:
1. The first three involve drag or friction with air (viscosity).
2. Pressure variation along a tube of changing cross-section follows from Bernoulli's principle for ideal flow and does not need viscosity.

Teacher's Note:
1. Bernoulli's principle assumes no viscosity.
2. The sideways motion of a spinning ball (Magnus effect) relies on air being dragged round by the surface.

Answer: (a) \( \sin\theta=\frac12\sqrt{\rho_0/\rho} \)

Explanation:
1. Rod length \( L \), area \( A \). Submerged length \( l=\dfrac{L/2}{\sin\theta} \).
2. Torques about the lower end: weight \( \rho ALg\cdot\dfrac L2\cos\theta \) equals buoyancy \( \rho_0Alg\cdot\dfrac l2\cos\theta \).
3. \( \rho\dfrac{L^2}{2}=\rho_0\dfrac{L^2}{8\sin^2\theta} \), so \( \sin\theta=\dfrac12\sqrt{\rho_0/\rho} \).

Teacher's Note:
1. Take torques about the contact point so the normal reaction drops out.
2. Buoyancy acts at the centre of the submerged part.

Answer: (b) \( h\rho(g+a) \)

Explanation:
1. In the accelerating frame the effective gravity is \( g+a \).
2. Pressure \( =h\rho g_{eff}=h\rho(g+a) \).

Teacher's Note:
1. Upward acceleration increases the effective weight of the liquid.
2. Downward acceleration would reduce it to \( g-a \).

Answer: (d) remain in the same position relative to the water

Explanation:
1. Both buoyancy and weight scale with the effective gravity \( g\pm a \).
2. The floating condition (submerged fraction equals density ratio) does not involve \( g \), so the depth stays unchanged.

Teacher's Note:
1. Effective gravity cancels in the floating condition \( \rho_{wood}V=\rho_{liquid}V_{sub} \).
2. The same applies to a lift in motion.

Answer: (a) \( ad/g \)

Explanation:
1. The free surface tilts so that \( \tan\theta=\dfrac ag \).
2. The level difference over the horizontal distance \( d \) is \( d\tan\theta=\dfrac{ad}{g} \).

Teacher's Note:
1. The liquid surface is perpendicular to the effective gravity \( \vec g-\vec a \).
2. Check the pressure at the bottom: \( \rho g\Delta h=\rho ad \).

Answer: (a) conservation of energy only

Explanation:
1. Bernoulli's equation \( p+\tfrac12\rho v^2+\rho gh=\text{constant} \) is the work-energy theorem applied to an ideal flowing fluid.
2. It is therefore a statement of conservation of energy.

Teacher's Note:
1. It holds for steady, incompressible, non-viscous flow along a streamline.
2. The continuity equation (mass conservation) is a separate relation.

Answer: (c) The mass of water flowing past any cross-section must remain constant; since water is almost incompressible, the rate of volume flow must remain constant, so as velocity increases the area decreases

Explanation:
1. The equation of continuity gives \( Av=\text{constant} \) for an incompressible liquid.
2. The falling water speeds up under gravity, so the cross-section must shrink.

Teacher's Note:
1. Continuity is conservation of mass in flow.
2. Surface tension does matter at very small radii, but it is not the main cause.

Answer: (b) \( 1.5\ \text{cm}^2 \)

Explanation:
1. Speed after falling \( 0.8 m \): \( v^2=3^2+2\times10\times0.8=25 \), so \( v=5 m/s \).
2. Continuity: \( A_2=\dfrac{2.5\times3}{5}=1.5\ \text{cm}^2 \).

Teacher's Note:
1. First find the new speed by kinematics, then apply \( A_1v_1=A_2v_2 \).
2. The column narrows as the water speeds up.