JEE Physics Mechanical Properties Of Fluids MCQs Set 31

Answer: (a) 1018 J

Explanation:
1. Take a horizontal slice at depth \( y \) below the rim with radius \( r \), where \( r^{2}=R^{2}-y^{2} \) (centre of the sphere at the rim).
2. Work to lift the slice \( dW=\rho g\,\pi(R^{2}-y^{2})\,y\,dy \).
3. \( W=\rho g\pi\int_0^R(R^{2}y-y^{3})dy=\rho g\pi\dfrac{R^{4}}{4}=1000\times10\times\pi\times\dfrac{0.1296}{4}\approx1018 \) J.

Teacher's Note:
1. Each layer is lifted by its own depth, so integrate.
2. Equivalently, work = weight of all the water times the depth of its centre of mass \( \tfrac38R \).

Answer: (a) 54.4 cm

Explanation:
1. The mercury falls 2 cm in the water limb and rises 2 cm in the other, so the mercury levels differ by 4 cm.
2. Pressure balance at the lower mercury level: \( h_w\rho_wg=0.04\,\rho_{Hg}g \).
3. \( h_w=4\times13.6=54.4 \) cm.

Teacher's Note:
1. Equal volume displacement means equal level changes in a uniform tube.
2. The water column must balance the mercury column of height difference.

Answer: (b) 19.6 m

Explanation:
1. Speed at the water surface: \( v^{2}=2g(19.6) \).
2. In water the net upward force gives an acceleration upward of \( a=\dfrac{(\rho_w-\rho)g}{\rho}=\dfrac{0.5g}{0.5}=g \).
3. Depth \( =\dfrac{v^{2}}{2a}=\dfrac{2g(19.6)}{2g}=19.6 \) m.

Teacher's Note:
1. The deceleration in water equals \( g \) here, the same magnitude as the acceleration during the fall.
2. So the ball goes as deep as it fell.

Answer: (d) 2.8 g/cm³

Explanation:
1. At exactly half full (250 g of water inside) the beaker just floats while totally submerged to its rim: displaced water mass equals beaker mass plus contents \( =390+250=640 \) g.
2. Displaced volume \( =V_{glass}+500 \) cm³ \( =640 \) cm³, so \( V_{glass}=140 \) cm³.
3. Density \( =\dfrac{390}{140}\approx2.8 \) g/cm³.

Teacher's Note:
1. The critical condition is that the whole beaker (glass plus interior) is just immersed.
2. Density is mass divided by the volume of the glass alone.

Answer: (b) 100 kg

Explanation:
1. The surface tilts with \( \tan\theta=\dfrac ag=0.2 \), so across the 1 m length the level drops by 0.2 m.
2. Spilled volume \( =\tfrac12\times0.2\times1\times1=0.1\ \text{m}^3 \).
3. Mass \( =1000\times0.1=100 \) kg.

Teacher's Note:
1. The spilled water forms a wedge of height \( \tan\theta\times\text{length} \).
2. Use the volume of the wedge and the density of water.

Answer: (c) 5

Explanation:
1. In the container frame the effective gravity has magnitude \( g_{eff}=\sqrt{g^{2}+a^{2}}=\sqrt{100+44}=12\ \text{m/s}^2 \).
2. The ball is acted on by buoyancy and weight, so its acceleration is \( \left(\dfrac{\rho_w}{\rho}-1\right)g_{eff}=\left(\dfrac{17}{12}-1\right)\times12 \) in the direction opposite to \( g_{eff} \).
3. Acceleration \( =5\ \text{m/s}^2 \).

Teacher's Note:
1. Treat the container frame as having an effective gravity at an angle to the vertical.
2. The net upthrust per unit mass is \( (\rho_w/\rho-1)g_{eff} \).

Answer: (b) 4 m/s²

Explanation:
1. With the surface tilted by \( \tan\theta=\dfrac ag \), the level at the rear rises by \( \dfrac{a}{g}\cdot\dfrac52 \) above the mean level of 2 m.
2. No spilling means \( 2+\dfrac{5a}{2g}\le3 \), so \( a\le\dfrac{2g}{5}=4\ \text{m/s}^2 \).

Teacher's Note:
1. The mean water level stays at its original height as the surface tilts.
2. The rear wall limits the rise to 1 m.

Answer: (d) 10%

Explanation:
1. New acceleration \( =1.2\times4=4.8\ \text{m/s}^2 \), so \( \tan\theta=0.48 \) and the level drops by \( 0.48\times5=2.4 \) m across the tank.
2. With the rear at the rim (3 m), the front level is \( 3-2.4=0.6 \) m, still above the bottom. Remaining side-view area \( =5\times\dfrac{3+0.6}{2}=9\ \text{m}^2 \) compared with \( 5\times2=10\ \text{m}^2 \) initially.
3. Spilled fraction \( =\dfrac{1}{10}=10\% \).

Teacher's Note:
1. After spilling, the surface passes through the rear rim.
2. Compare the trapezoidal cross-section with the original rectangle.

Answer: (a) \( \rho_1<\rho_3<\rho_2 \)

Explanation:
1. Liquid 1 floats on liquid 2, so \( \rho_1<\rho_2 \).
2. A ball resting at the interface sinks through liquid 1 but floats on liquid 2, so \( \rho_1<\rho_3<\rho_2 \).

Teacher's Note:
1. A body settles at the interface when its density lies between the two liquid densities.
2. The lighter liquid is always on top.

Answer: (a) 2

Explanation:
1. In air: \( \tan\theta=\dfrac{F_e}{mg} \). In the liquid: \( \tan\theta=\dfrac{F_e/K}{mg-\text{buoyancy}} \).
2. Buoyancy \( =V\rho_lg=\dfrac{0.8}{1.6}mg=\dfrac{mg}{2} \).
3. Equal angles: \( \dfrac{F_e}{mg}=\dfrac{F_e/K}{mg/2} \), so \( K=2 \).

Teacher's Note:
1. The liquid reduces both the weight (buoyancy) and the electric force (dielectric).
2. Equal angles require the ratio to stay the same.

Answer: (a) \( 3.6\times10^{-3} \) m

Explanation:
1. Speed after falling 0.2 m: \( v^{2}=0.4^{2}+2\times9.8\times0.2=4.08 \), so \( v\approx2.02 \) m/s.
2. Continuity: \( \dfrac{d_2}{d_1}=\sqrt{\dfrac{v_1}{v_2}}=\sqrt{\dfrac{0.4}{2.02}}\approx0.445 \).
3. \( d_2\approx8\times10^{-3}\times0.445\approx3.6\times10^{-3} \) m.

Teacher's Note:
1. The diameter scales as the inverse square root of the speed.
2. Use \( v^{2}=u^{2}+2gh \) for the falling water.

Answer: (b) 50

Explanation:
1. Depth of the hole \( =3-0.525=2.475 \) m.
2. Bernoulli with the falling surface speed \( V=\dfrac aA\,v \): \( v^{2}-V^{2}=2gh \).
3. \( v^{2}(1-0.01)=2\times10\times2.475=49.5 \), so \( v^{2}=50 \).

Teacher's Note:
1. The level in the container falls slowly, but it is not negligible when \( a/A=0.1 \).
2. Combine continuity with Bernoulli's equation.

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. Held upward, the water slows as it rises and, by \( Av= \) constant, the stream widens; held downward, it speeds up and the stream narrows.
2. The equation of continuity (constant volume flow rate) is exactly the reason for both effects.
3. Both statements are true and the second explains the first.

Teacher's Note:
1. Continuity links the speed of a stream to its cross-sectional area.
2. Gravity changes the speed, and continuity changes the area.

Answer: (d) the force due to gravity and the force due to viscosity of the liquid

Explanation:
1. The buoyancy force is itself the resultant of the pressure forces of the liquid on the bubble.
2. So, apart from buoyancy, only gravity and the viscous drag from the liquid act on the bubble.

Teacher's Note:
1. Do not count the pressure force separately from buoyancy.
2. A moving bubble experiences viscous drag opposing its motion.

Answer: (c) more than half-filled if \( \rho_c \) is less than 0.5

Explanation:
1. Let the outer volume be \( V \), shell material volume \( V_s \), inner volume \( V-V_s \) and water volume \( V_w \). Half-submerged floating gives \( \rho_cV_s+V_w=\dfrac V2 \) (in water-density units).
2. Then \( V_w-\dfrac{V-V_s}{2}=V_s\left(\dfrac12-\rho_c\right) \).
3. This is positive when \( \rho_c<0.5 \), so the shell is more than half-filled in that case.

Teacher's Note:
1. Compare the actual water volume with half of the inner volume.
2. The sign is decided by \( \tfrac12-\rho_c \).

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

Explanation:
1. The heavy sphere (density \( 3\rho \)) hangs below in the liquid: net downward force \( (3\rho-2\rho)Vg=\rho Vg \), which the spring supports, so \( kx=\rho Vg=\dfrac{4\pi R^{3}\rho g}{3} \) and \( x=\dfrac{4\pi R^{3}\rho g}{3k} \).
2. The light sphere has net upward force \( (2\rho-\rho)Vg=\rho Vg \) when fully submerged, exactly balanced by the spring pull downward.
3. So the light sphere is completely submerged: (i) and (iv).

Teacher's Note:
1. Treat each sphere separately with weight, buoyancy and spring force.
2. Both spheres feel the same spring force of magnitude \( \rho Vg \).

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

Explanation:
1. A is pulled down by the string and tends to float up, so \( d_A<d_F \) (i). B is pulled up by the string and tends to sink, so \( d_B>d_F \) (ii).
2. The tension satisfies \( T=(d_F-d_A)Vg=(d_B-d_F)Vg \), so \( d_A+d_B=2d_F \) (iv).
3. (iii) contradicts (i).

Teacher's Note:
1. Write the force balance on each sphere with the same tension.
2. Equal volumes mean equal buoyancy.

Answer: (d) 2 m/s

Explanation:
1. Continuity: \( A_1v_1=A_2v_2 \).
2. \( v_2=5\times\left(\dfrac{20}{1}\right)^{2}=5\times400=2000 \) mm/s \( =2 \) m/s.

Teacher's Note:
1. The flow speed scales inversely with the area, which goes as the radius squared.

Answer: (d) \( 2RT \)

Explanation:
1. The film outside pulls on each half of the loop with force \( 2T \) per unit length (two surfaces), over a length \( 2R \): total \( 4RT \).
2. This is shared by the two cut ends of the thread, so \( 2\tau=4RT \) and \( \tau=2RT \).

Teacher's Note:
1. Consider half of the ring and balance forces along the diameter.
2. A soap film has two surfaces, so the pull per unit length is \( 2T \).

Answer: (c) \( 1.44\times10^{-5} \) J

Explanation:
1. Increase in area of one face \( =0.1\times0.001=10^{-4}\ \text{m}^2 \), and the film has two faces.
2. \( W=T\,\Delta A=72\times10^{-3}\times2\times10^{-4}=1.44\times10^{-5} \) J.

Teacher's Note:
1. Count both surfaces of the film.
2. Work equals surface tension times the increase in area.

Answer: (d) \( 7.2\times10^{-2} \) N/m

Explanation:
1. The liquid film clings to both the inner and outer circumference, so the pull is \( T\,\pi(d_1+d_2)=T\pi(8.5+8.7)\times10^{-2} \) m \( =T\times0.5404 \) m.
2. This equals \( 3.97\times10^{-3}\times9.8=0.0389 \) N.
3. \( T=0.072 \) N/m.

Teacher's Note:
1. Both the inner and the outer edges of the ring are in contact with water.
2. Total contact length is \( \pi(d_{in}+d_{out}) \).

Answer: (b) 60 dyn

Explanation:
1. The net force is the difference of tensions times the length: \( (50-40)\times6=60 \) dyn, directed towards the side of higher tension.

Teacher's Note:
1. Each side pulls with force \( T\times\text{length} \).
2. The resultant is directed toward the larger surface tension.

Answer: (d) \( 10^{5} \) dyn

Explanation:
1. The meniscus excess pressure is \( \dfrac{2T}{d/2}\cdot\dfrac12=\dfrac{2T}{d} \) (a cylindrical meniscus of radius \( d/2 \)).
2. Force \( =\dfrac{2TA}{d}=\dfrac{2\times75\times8}{0.012}=10^{5} \) dyn.

Teacher's Note:
1. The thin water film between plates has a curved edge of radius half the separation.
2. Force equals pressure difference times wetted area.

Answer: (d) \( \dfrac{r_1r_2}{r_1-r_2} \)

Explanation:
1. Pressures: \( P_2-P_0=\dfrac{4T}{r_2} \), \( P_1-P_0=\dfrac{4T}{r_1} \), and across the common film \( P_2-P_1=\dfrac{4T}{r} \).
2. So \( \dfrac1r=\dfrac1{r_2}-\dfrac1{r_1} \), giving \( r=\dfrac{r_1r_2}{r_1-r_2} \).

Teacher's Note:
1. The smaller bubble has the higher pressure, so the common film bulges into the larger bubble.
2. Use the pressure difference across the common film.

Answer: (a) air flows from the smaller bubble to the bigger one

Explanation:
1. The excess pressure \( 4T/r \) is greater in the smaller bubble.
2. Air therefore flows from the smaller to the bigger bubble, and the smaller keeps shrinking.

Teacher's Note:
1. A smaller radius means a higher pressure.
2. The flow continues until the small bubble collapses into the tube end.