JEE Physics Mechanical Properties Of Fluids MCQs Set 38

Physics Objective Questions and Answers: Mechanical Properties Of Fluids

Access targeted multiple-choice questions for Mechanical Properties Of Fluids designed to align with the latest JEE academic syllabus for JEE Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Download Mechanical Properties Of Fluids MCQs with Answers

Access the complete set of multiple-choice questions for Mechanical Properties Of Fluids below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: For the container above (open top, hole A/100 near the bottom, on a smooth floor, liquid of density \( \rho \) and initial mass \( m_0 \)), the velocity of efflux when 75% of the liquid has drained out is
(a) \( \sqrt{\dfrac{gm_0}{A\rho}} \)
(b) \( \sqrt{\dfrac{2gm_0}{A\rho}} \)
(c) \( \sqrt{\dfrac{gm_0}{4A\rho}} \)
(d) \( \sqrt{\dfrac{gm_0}{2A\rho}} \)
Show Answer & Explanation

Answer: (d) \( \sqrt{\dfrac{gm_0}{2A\rho}} \)

Explanation:
1. The remaining mass is \( 0.25m_0 \), so the height is \( h=\dfrac{0.25m_0}{\rho A} \).
2. \( v=\sqrt{2gh}=\sqrt{\dfrac{gm_0}{2A\rho}} \).

Teacher's Note:
1. Use the pressure formula \( P = P0 \) + rho \( g h \) for liquids at rest and the equation of continuity \( A1 v1 = A2 v2 \) with Bernoulli's equation for flow.
2. Check that the forces or pressures balance at a chosen reference level, or that the energy per unit volume is conserved.

Question 2: A container of large uniform cross-section A resting on a horizontal surface holds two immiscible, non-viscous and incompressible liquids of densities d and 2d, each of height H/2 (the denser liquid at the bottom, which is open to the atmosphere of pressure \( p_0 \)). A tiny hole of area s (s much less than A) is punched on the vertical side of the container at a height h (h < H/2). The initial speed of efflux of the liquid at the hole is
(a) \( \sqrt{\dfrac{(3H-4h)g}2} \)
(b) \( \sqrt{\dfrac{(3H-2h)g}2} \)
(c) \( \sqrt{(3H-4h)g} \)
(d) \( \sqrt{\dfrac{(H-2h)g}2} \)
Show Answer & Explanation

Answer: (a) \( \sqrt{\dfrac{(3H-4h)g}2} \)

Explanation:
1. Gauge pressure at the hole \( =g\left[d\cdot\dfrac H2+2d\left(\dfrac H2-h\right)\right]=gd\left(\dfrac{3H}2-2h\right) \).
2. The liquid at the hole has density \( 2d \): \( \tfrac12(2d)v^2=gd\left(\dfrac{3H}2-2h\right)\Rightarrow v=\sqrt{\dfrac{g(3H-4h)}2} \).

Teacher's Note:
1. Use the pressure formula \( P = P0 \) + rho \( g h \) for liquids at rest and the equation of continuity \( A1 v1 = A2 v2 \) with Bernoulli's equation for flow.
2. Check that the forces or pressures balance at a chosen reference level, or that the energy per unit volume is conserved.

Question 3: For the same container (liquids d and 2d, each of height H/2, hole at height h), the horizontal distance x travelled by the liquid initially is
(a) \( \sqrt{2h(3H-4h)} \)
(b) \( \sqrt{h(3H-4h)} \)
(c) \( 2\sqrt{h(3H-4h)} \)
(d) \( \sqrt{h(3H-2h)} \)
Show Answer & Explanation

Answer: (b) \( \sqrt{h(3H-4h)} \)

Explanation:
1. Efflux speed \( v=\sqrt{\dfrac{g(3H-4h)}2} \); fall time \( t=\sqrt{\dfrac{2h}g} \).
2. \( x=vt=\sqrt{h(3H-4h)} \).

Teacher's Note:
1. Use the pressure formula \( P = P0 \) + rho \( g h \) for liquids at rest and the equation of continuity \( A1 v1 = A2 v2 \) with Bernoulli's equation for flow.
2. Check that the forces or pressures balance at a chosen reference level, or that the energy per unit volume is conserved.

Question 4: For the same container (liquids d and 2d, each of height H/2), the hole should be punched at a height \( h_m \) so that the liquid travels the maximum distance \( x_m \) initially. The values of \( h_m \) and \( x_m \) are
(a) \( h_m=\dfrac{H}4,\ x_m=\dfrac{3H}4 \)
(b) \( h_m=\dfrac{3H}8,\ x_m=\dfrac{3H}2 \)
(c) \( h_m=\dfrac{3H}8,\ x_m=\dfrac{3H}4 \)
(d) \( h_m=\dfrac{H}2,\ x_m=\dfrac H2 \)
Show Answer & Explanation

Answer: (c) \( h_m=\dfrac{3H}8,\ x_m=\dfrac{3H}4 \)

Explanation:
1. Maximise \( x^2=h(3H-4h) \): \( \dfrac{d}{dh}=3H-8h=0\Rightarrow h_m=\dfrac{3H}8 \).
2. \( x_m^2=\dfrac{3H}8\left(3H-\dfrac{3H}2\right)=\dfrac{9H^2}{16}\Rightarrow x_m=\dfrac{3H}4 \).

Teacher's Note:
1. Use the pressure formula \( P = P0 \) + rho \( g h \) for liquids at rest and the equation of continuity \( A1 v1 = A2 v2 \) with Bernoulli's equation for flow.
2. Check that the forces or pressures balance at a chosen reference level, or that the energy per unit volume is conserved.

Question 5: The ratio of surface tensions of mercury and water is given to be 7.5 while the ratio of their densities is 13.6. Their contact angles with glass are close to \( 135^\circ \) and \( 0^\circ \) respectively. It is observed that mercury gets depressed by an amount h in a capillary tube of radius \( r_1 \), while water rises by the same amount h in a capillary tube of radius \( r_2 \). The ratio \( r_1/r_2 \) is then close to
(a) 3/5
(b) 2/3
(c) 4/5
(d) 2/5
Show Answer & Explanation

Answer: (d) 2/5

Explanation:
1. \( h=\dfrac{2T\cos\theta}{r\rho g}\Rightarrow r=\dfrac{2T\cos\theta}{h\rho g} \) (magnitude).
2. \( \dfrac{r_1}{r_2}=\dfrac{7.5\times\cos135^\circ}{13.6\times\cos0^\circ}\to\dfrac{7.5(0.707)}{13.6}=0.39\approx\dfrac25 \).

Teacher's Note:
1. Surface tension acts along the surface; excess pressure is \( 2T/r \) in a drop and \( 4T/r \) in a soap bubble.
2. In capillary rise h = 2T cos(theta)/(r rho \( g \)), and terminal speed in a viscous fluid scales as \( r^2 \) (rho - sigma)/eta.

Question 6: If M is the mass of water that rises in a capillary tube of radius r, then the mass of water which will rise in a capillary tube of radius 2r is
(a) 2M
(b) 4M
(c) M/2
(d) M
Show Answer & Explanation

Answer: (a) 2M

Explanation:
1. \( h\propto\dfrac1r \) and \( m=\rho\pi r^2h\propto r \).
2. So doubling the radius doubles the mass: \( 2M \).

Teacher's Note:
1. Surface tension acts along the surface; excess pressure is \( 2T/r \) in a drop and \( 4T/r \) in a soap bubble.
2. In capillary rise h = 2T cos(theta)/(r rho \( g \)), and terminal speed in a viscous fluid scales as \( r^2 \) (rho - sigma)/eta.

Question 7: In a capillary-rise experiment the diameter of the capillary is \( d=1.25\times10^{-2} \) m and the rise of water is \( h=1.45\times10^{-2} \) m. Using g = 9.80 m/s² and \( T=\dfrac{rhg}2\times10^3 \) N/m, the possible error in surface tension is closest to
(a) 2.4%
(b) 1.5%
(c) 10%
(d) 0.15%
Show Answer & Explanation

Answer: (b) 1.5%

Explanation:
1. Take the last digit as the uncertainty: \( \dfrac{\Delta r}r=\dfrac{0.01}{1.25}=0.8\% \), \( \dfrac{\Delta h}h=\dfrac{0.01}{1.45}=0.69\% \).
2. \( \dfrac{\Delta T}T=\dfrac{\Delta r}r+\dfrac{\Delta h}h\approx1.5\% \).

Teacher's Note:
1. Surface tension acts along the surface; excess pressure is \( 2T/r \) in a drop and \( 4T/r \) in a soap bubble.
2. In capillary rise h = 2T cos(theta)/(r rho \( g \)), and terminal speed in a viscous fluid scales as \( r^2 \) (rho - sigma)/eta.

Question 8: Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, density of liquid is \( \rho \) and L is its latent heat of vaporisation.
(a) \( \dfrac{\rho L}{T} \)
(b) \( \sqrt{\dfrac T{\rho L}} \)
(c) \( \dfrac{2T}{\rho L} \)
(d) \( \dfrac{T}{\rho L} \)
Show Answer & Explanation

Answer: (c) \( \dfrac{2T}{\rho L} \)

Explanation:
1. Let the radius shrink by dR: surface energy released \( =T\cdot8\pi R\,dR \).
2. Heat needed to evaporate the mass \( \rho4\pi R^2dR \) is \( L\rho4\pi R^2dR \). Equate: \( 8\pi RT=4\pi R^2\rho L\Rightarrow R=\dfrac{2T}{\rho L} \).

Teacher's Note:
1. Surface tension acts along the surface; excess pressure is \( 2T/r \) in a drop and \( 4T/r \) in a soap bubble.
2. In capillary rise h = 2T cos(theta)/(r rho \( g \)), and terminal speed in a viscous fluid scales as \( r^2 \) (rho - sigma)/eta.

Question 9: A glass tube of uniform internal radius r has a valve separating the two identical ends. Initially the valve is in a tightly closed position. End 1 has a hemispherical soap bubble of radius r. End 2 has a sub-hemispherical soap bubble (radius of curvature larger than r) as shown. Just after opening the valve,
(a) air from end 1 flows towards end 2, with no change in the volume of the soap bubbles
(b) air from end 2 flows towards end 1, and the volume of the soap bubble at end 1 increases
(c) no change occurs
(d) air from end 1 flows towards end 2, and the volume of the soap bubble at end 1 decreases
Show Answer & Explanation

Answer: (d) air from end 1 flows towards end 2, and the volume of the soap bubble at end 1 decreases

Explanation:
1. Excess pressure in a soap bubble is \( \dfrac{4T}R \). The hemispherical bubble has the smallest radius of curvature \( (r) \), hence the larger pressure.
2. Air flows from the higher pressure (end 1) to the lower pressure (end 2), so the bubble at end 1 shrinks and the other grows.

Teacher's Note:
1. Surface tension acts along the surface; excess pressure is \( 2T/r \) in a drop and \( 4T/r \) in a soap bubble.
2. In capillary rise h = 2T cos(theta)/(r rho \( g \)), and terminal speed in a viscous fluid scales as \( r^2 \) (rho - sigma)/eta.

Question 10: A uniform capillary tube of inner radius r is dipped vertically into a beaker filled with water. The water rises to a height h in the capillary tube above the water surface in the beaker. The surface tension of water is \( \sigma \) and the angle of contact is \( \theta \). Ignore the mass of water in the meniscus. Which statements are true? (a) For a given material of the capillary tube, h decreases with increase in r. (b) For a given material of the capillary tube, h is independent of \( \sigma \). (c) If this experiment is performed in a lift going up with a constant acceleration, then h decreases. (d) h is proportional to the contact angle \( \theta \).
(a) (a) and (c) only
(b) (a) and (b) only
(c) (c) and (d) only
(d) (b) and (d) only
Show Answer & Explanation

Answer: (a) (a) and (c) only

Explanation:
1. \( h=\dfrac{2\sigma\cos\theta}{r\rho g} \): \( h \) decreases with r (a true), depends on \( \sigma \) (\( b \) false), is proportional to \( \cos\theta \), not \( \theta \) (\( d \) false).
2. In an upward-accelerating lift the effective gravity is \( g+a \), so \( h \) decreases: \( (c) \) true.

Teacher's Note:
1. Surface tension acts along the surface; excess pressure is \( 2T/r \) in a drop and \( 4T/r \) in a soap bubble.
2. In capillary rise h = 2T cos(theta)/(r rho \( g \)), and terminal speed in a viscous fluid scales as \( r^2 \) (rho - sigma)/eta.

Question 11: When liquid medicine of density \( \rho \) is put in the eye with a dropper, a drop forms at the opening of radius r and is assumed spherical with radius R (r much smaller than R). The net vertical force due to the surface tension T on the drop of radius R is
(a) \( 2\pi rT \)
(b) \( \dfrac{2\pi r^2T}{R} \)
(c) \( 2\pi RT \)
(d) \( \dfrac{2\pi R^2T}{r} \)
Show Answer & Explanation

Answer: (b) \( \dfrac{2\pi r^2T}{R} \)

Explanation:
1. The surface tension force acts along the circle of radius \( r \) at an angle to the vertical with \( \sin\theta=\dfrac rR \).
2. Vertical force \( =T(2\pi r)\sin\theta=\dfrac{2\pi r^2T}R \).

Teacher's Note:
1. Surface tension acts along the surface; excess pressure is \( 2T/r \) in a drop and \( 4T/r \) in a soap bubble.
2. In capillary rise h = 2T cos(theta)/(r rho \( g \)), and terminal speed in a viscous fluid scales as \( r^2 \) (rho - sigma)/eta.

Question 12: For the dropper, \( r=5\times10^{-4} \) m, \( \rho=10^3 \) kg m\(^{-3}\), g = 10 m s\(^{-2}\) and \( T=0.11 \) N m\(^{-1}\). The radius of the drop when it detaches from the dropper is approximately
(a) \( 3.3\times10^{-3} \) m
(b) \( 2.0\times10^{-3} \) m
(c) \( 1.4\times10^{-3} \) m
(d) \( 4.1\times10^{-3} \) m
Show Answer & Explanation

Answer: (c) \( 1.4\times10^{-3} \) m

Explanation:
1. Detachment when weight equals the surface-tension force: \( \tfrac43\pi R^3\rho g=\dfrac{2\pi r^2T}R \).
2. \( R^4=\dfrac{3r^2T}{2\rho g}=\dfrac{3(25\times10^{-8})(0.11)}{2\times10^4}=4.1\times10^{-12} \), so \( R\approx1.4\times10^{-3} m \).

Teacher's Note:
1. Surface tension acts along the surface; excess pressure is \( 2T/r \) in a drop and \( 4T/r \) in a soap bubble.
2. In capillary rise h = 2T cos(theta)/(r rho \( g \)), and terminal speed in a viscous fluid scales as \( r^2 \) (rho - sigma)/eta.

Question 13: After the drop (radius \( R=1.4\times10^{-3} \) m, surface tension 0.11 N/m) detaches from the dropper, its surface energy is
(a) \( 1.4\times10^{-6} \) J
(b) \( 5.4\times10^{-6} \) J
(c) \( 8.1\times10^{-9} \) J
(d) \( 2.7\times10^{-6} \) J
Show Answer & Explanation

Answer: (d) \( 2.7\times10^{-6} \) J

Explanation:
1. Surface energy \( =4\pi R^2T=4\pi(1.43\times10^{-3})^2(0.11) \).
2. \( =2.8\times10^{-6}\approx2.7\times10^{-6} J \).

Teacher's Note:
1. Surface tension acts along the surface; excess pressure is \( 2T/r \) in a drop and \( 4T/r \) in a soap bubble.
2. In capillary rise h = 2T cos(theta)/(r rho \( g \)), and terminal speed in a viscous fluid scales as \( r^2 \) (rho - sigma)/eta.

Question 14: A drop of liquid of radius \( R=10^{-2} \) m having surface tension \( S=\dfrac{0.1}{4\pi} \) N m\(^{-1}\) divides itself into K identical drops. In this process the total change in the surface energy is \( \Delta U=10^{-3} \) J. If \( K=10^\alpha \), the value of \( \alpha \) is
(a) 6
(b) 3
(c) 4
(d) 9
Show Answer & Explanation

Answer: (a) 6

Explanation:
1. \( \Delta U=S\,4\pi R^2\left(K^{1/3}-1\right)=\dfrac{0.1}{4\pi}\cdot4\pi\cdot10^{-4}\left(K^{1/3}-1\right)=10^{-5}\left(K^{1/3}-1\right) \).
2. \( 10^{-3}=10^{-5}(K^{1/3}-1)\Rightarrow K^{1/3}=101\Rightarrow K\approx10^6 \), so \( \alpha=6 \).

Teacher's Note:
1. Surface tension acts along the surface; excess pressure is \( 2T/r \) in a drop and \( 4T/r \) in a soap bubble.
2. In capillary rise h = 2T cos(theta)/(r rho \( g \)), and terminal speed in a viscous fluid scales as \( r^2 \) (rho - sigma)/eta.

Question 15: Consider two solid spheres P and Q each of density 8 g cm\(^{-3}\) and diameters 1 cm and 0.5 cm respectively. Sphere P is dropped into a liquid of density 0.8 g cm\(^{-3}\) and viscosity \( \eta=3 \) poiseulles. Sphere Q is dropped into a liquid of density 1.6 g cm\(^{-3}\) and viscosity \( \eta=2 \) poiseulles. The ratio of the terminal velocities of P and Q is
(a) 1.5
(b) 3
(c) 6
(d) 2
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. \( v_t=\dfrac{2r^2(\rho-\sigma)g}{9\eta}\propto\dfrac{r^2(\rho-\sigma)}{\eta} \).
2. \( P \): \( \dfrac{(0.5)^2(7.2)}3=0.6 \); \( Q \): \( \dfrac{(0.25)^2(6.4)}2=0.2 \). Ratio \( =3 \).

Teacher's Note:
1. Surface tension acts along the surface; excess pressure is \( 2T/r \) in a drop and \( 4T/r \) in a soap bubble.
2. In capillary rise h = 2T cos(theta)/(r rho \( g \)), and terminal speed in a viscous fluid scales as \( r^2 \) (rho - sigma)/eta.

Question 16: Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure 8 N m\(^{-2}\). The radii of bubbles A and B are 2 cm and 4 cm respectively. The surface tension of the soap-water used to make bubbles is 0.04 N m\(^{-1}\). The ratio \( n_B/n_A \), where \( n_A \) and \( n_B \) are the number of moles of air in bubbles A and B respectively, is
(a) 2
(b) 3
(c) 6
(d) 8
Show Answer & Explanation

Answer: (c) 6

Explanation:
1. \( p_{in}=8+\dfrac{4T}r \): \( A \): \( 8+\dfrac{0.16}{0.02}=16 \); \( B \): \( 8+\dfrac{0.16}{0.04}=12 N m ^{-2} \).
2. \( n\propto pV\propto pr^3 \): \( \dfrac{n_B}{n_A}=\dfrac{12\times64}{16\times8}=6 \).

Teacher's Note:
1. Surface tension acts along the surface; excess pressure is \( 2T/r \) in a drop and \( 4T/r \) in a soap bubble.
2. In capillary rise h = 2T cos(theta)/(r rho \( g \)), and terminal speed in a viscous fluid scales as \( r^2 \) (rho - sigma)/eta.

Question 17: A small sphere falls from rest in a viscous liquid. Due to friction, heat is produced. The rate of production of heat when the sphere has attained the terminal velocity is proportional to
(a) \( r^3 \)
(b) \( r^4 \)
(c) \( r^2 \)
(d) \( r^5 \)
Show Answer & Explanation

Answer: (d) \( r^5 \)

Explanation:
1. Rate of heat \( =F\,v_t=6\pi\eta rv_t^2 \) and \( v_t\propto r^2 \).
2. So the rate \( \propto r\cdot r^4=r^5 \).

Teacher's Note:
1. Surface tension acts along the surface; excess pressure is \( 2T/r \) in a drop and \( 4T/r \) in a soap bubble.
2. In capillary rise h = 2T cos(theta)/(r rho \( g \)), and terminal speed in a viscous fluid scales as \( r^2 \) (rho - sigma)/eta.

Question 18: A container of width 2a is filled with a liquid. A thin wire of weight per unit length \( \lambda \) is gently placed over the liquid surface in the middle of the container. As a result, the liquid surface is depressed by a distance y (y much less than a). The surface tension of the liquid is
(a) \( \dfrac{\lambda a}{2y} \)
(b) \( \dfrac{\lambda y}{2a} \)
(c) \( \dfrac{2\lambda a}{y} \)
(d) \( \dfrac{\lambda a}{y} \)
Show Answer & Explanation

Answer: (a) \( \dfrac{\lambda a}{2y} \)

Explanation:
1. On each side the surface makes a small angle \( \theta\approx\dfrac ya \) with the horizontal. Per unit length of wire: \( 2T\sin\theta=\lambda \).
2. \( T=\dfrac{\lambda}{2\theta}=\dfrac{\lambda a}{2y} \).

Teacher's Note:
1. Surface tension acts along the surface; excess pressure is \( 2T/r \) in a drop and \( 4T/r \) in a soap bubble.
2. In capillary rise h = 2T cos(theta)/(r rho \( g \)), and terminal speed in a viscous fluid scales as \( r^2 \) (rho - sigma)/eta.

Question 19: A soap bubble is being blown at the end of a very narrow tube of radius b. Air (density \( \rho \)) moves with a velocity v inside the tube and comes to rest inside the bubble. The surface tension of the soap solution is T. After some time the bubble, having grown to radius r, separates from the tube. Assuming \( r\gg b \), so that the air can be considered to be falling normally on the bubble's surface, the value of r is
(a) \( \dfrac{2T}{\rho v^2} \)
(b) \( \dfrac{4T}{\rho v^2} \)
(c) \( \dfrac{T}{\rho v^2} \)
(d) \( \dfrac{8T}{\rho v^2} \)
Show Answer & Explanation

Answer: (b) \( \dfrac{4T}{\rho v^2} \)

Explanation:
1. The air stream's dynamic pressure \( \rho v^2 \) (momentum destroyed per unit area per second) pushes the bubble off.
2. The bubble separates when this exceeds the excess pressure \( \dfrac{4T}r \): \( r=\dfrac{4T}{\rho v^2} \).

Teacher's Note:
1. Surface tension acts along the surface; excess pressure is \( 2T/r \) in a drop and \( 4T/r \) in a soap bubble.
2. In capillary rise h = 2T cos(theta)/(r rho \( g \)), and terminal speed in a viscous fluid scales as \( r^2 \) (rho - sigma)/eta.

Question 20: Water from a pipe is coming at a rate of 100 litres per minute. If the radius of the pipe is 5 cm, the Reynolds number for the flow is of the order of (density of water = 1000 kg/m³, coefficient of viscosity of water = 1 mPa s)
(a) \( 10^3 \)
(b) \( 10^2 \)
(c) \( 10^4 \)
(d) \( 10^6 \)
Show Answer & Explanation

Answer: (c) \( 10^4 \)

Explanation:
1. Flow \( =\dfrac{0.1}{60}=1.67\times10^{-3} m^{3}/s \); \( v=\dfrac{1.67\times10^{-3}}{\pi(0.05)^2}=0.21 m/s \).
2. \( Re=\dfrac{\rho vD}\eta=\dfrac{1000(0.21)(0.1)}{10^{-3}}=2\times10^4 \), of order \( 10^4 \).

Teacher's Note:
1. Surface tension acts along the surface; excess pressure is \( 2T/r \) in a drop and \( 4T/r \) in a soap bubble.
2. In capillary rise h = 2T cos(theta)/(r rho \( g \)), and terminal speed in a viscous fluid scales as \( r^2 \) (rho - sigma)/eta.

Question 21: A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains the same, the stress in the leg will change by a factor of
(a) 1/9
(b) 81
(c) 1/81
(d) 9
Show Answer & Explanation

Answer: (d) 9

Explanation:
1. Weight \( \propto L^3 \); leg cross-section \( \propto L^2 \).
2. Stress \( \propto\dfrac{L^3}{L^2}=L \), so it increases 9 times.

Teacher's Note:
1. Write \( Y \) = (F/A)/(dl/l) and apply it to each wire or rod, keeping track of which quantities are equal (force, length or extension).
2. Convert all lengths to metres and areas to square metres before substituting.

Question 22: A person in a lift is holding a water jar which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance d of 1.2 m from the person. In each of the following situations, how does the distance at which the jet hits the floor compare with 1.2 m? (A) Lift accelerating vertically up. (B) Lift accelerating vertically down with an acceleration less than g. (C) Lift moving vertically up with constant speed. (D) Lift falling freely.
(a) A: d = 1.2 m; B: d = 1.2 m; C: d = 1.2 m; D: no water leaks
(b) A: d > 1.2 m; B: d < 1.2 m; C: d = 1.2 m; D: no water leaks
(c) A: d < 1.2 m; B: d > 1.2 m; C: d = 1.2 m; D: d = 1.2 m
(d) A: d = 1.2 m; B: d = 1.2 m; C: d = 1.2 m; D: d = 1.2 m
Show Answer & Explanation

Answer: (a) A: d = 1.2 m; B: d = 1.2 m; C: d = 1.2 m; D: no water leaks

Explanation:
1. With effective gravity \( g_e \): \( v=\sqrt{2g_eh} \) and time of fall \( t=\sqrt{\dfrac{2H}{g_e}} \), so \( x=vt=2\sqrt{hH} \) is independent of \( g_e \).
2. So \( d = 1.2 m \) in cases \( A, B \) and \( C \). In free fall \( g_e=0 \): there is no pressure difference and no water leaks.

Teacher's Note:
1. Use the pressure formula \( P = P0 \) + rho \( g h \) for liquids at rest and the equation of continuity \( A1 v1 = A2 v2 \) with Bernoulli's equation for flow.
2. Check that the forces or pressures balance at a chosen reference level, or that the energy per unit volume is conserved.

Question 23: The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by \( 100^\circ \)C is (for steel, Young's modulus is \( 2\times10^{11} \) N m\(^{-2}\) and coefficient of thermal expansion is \( 1.1\times10^{-5} \) K\(^{-1}\))
(a) \( 2.2\times10^9 \) Pa
(b) \( 2.2\times10^8 \) Pa
(c) \( 2.2\times10^7 \) Pa
(d) \( 2.2\times10^6 \) Pa
Show Answer & Explanation

Answer: (b) \( 2.2\times10^8 \) Pa

Explanation:
1. Required pressure \( =Y\alpha\Delta T \).
2. \( =2\times10^{11}\times1.1\times10^{-5}\times100=2.2\times10^8 \) Pa.

Teacher's Note:
1. Write \( Y \) = (F/A)/(dl/l) and apply it to each wire or rod, keeping track of which quantities are equal (force, length or extension).
2. Convert all lengths to metres and areas to square metres before substituting.

Question 24: An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by an additional 46 cm. What will be the length of the air column above mercury in the tube now? (Atmospheric pressure = 76 cm of Hg)
(a) 22 cm
(b) 38 cm
(c) 16 cm
(d) 6 cm
Show Answer & Explanation

Answer: (c) 16 cm

Explanation:
1. Let the mercury inside stand \( h \) cm above the outside level. The air column is \( 8+46-h=54-h \) with pressure \( 76-h \).
2. Boyle: \( 76(8)=(76-h)(54-h)\Rightarrow h^2-130h+3496=0\Rightarrow h=38 \).
3. Air column \( =54-38=16 \) cm.

Teacher's Note:
1. Use the pressure formula \( P = P0 \) + rho \( g h \) for liquids at rest and the equation of continuity \( A1 v1 = A2 v2 \) with Bernoulli's equation for flow.
2. Check that the forces or pressures balance at a chosen reference level, or that the energy per unit volume is conserved.

Question 25: A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring such that it is half-submerged in a liquid of density \( \sigma \) at equilibrium position. The extension \( x_0 \) of the spring when it is in equilibrium is
(a) \( \dfrac{Mg}k \)
(b) \( \dfrac{Mg}k\left(1-\dfrac{LA\sigma}M\right) \)
(c) \( \dfrac{Mg}k\left(1+\dfrac{LA\sigma}M\right) \)
(d) \( \dfrac{Mg}k\left(1-\dfrac{LA\sigma}{2M}\right) \)
Show Answer & Explanation

Answer: (d) \( \dfrac{Mg}k\left(1-\dfrac{LA\sigma}{2M}\right) \)

Explanation:
1. Equilibrium: \( kx_0+\text{buoyancy}=Mg \), with buoyancy \( =\dfrac{LA}2\sigma g \).
2. \( x_0=\dfrac{Mg}k\left(1-\dfrac{LA\sigma}{2M}\right) \).

Teacher's Note:
1. Use the pressure formula \( P = P0 \) + rho \( g h \) for liquids at rest and the equation of continuity \( A1 v1 = A2 v2 \) with Bernoulli's equation for flow.
2. Check that the forces or pressures balance at a chosen reference level, or that the energy per unit volume is conserved.

Question 26: When a block of iron floats in mercury at \( 0^\circ \)C, a fraction \( k_1 \) of its volume is submerged, while at the temperature of \( 60^\circ \)C, a fraction \( k_2 \) is seen to be submerged. If the coefficient of volume expansion of iron is \( \gamma_{Fe} \) and that of mercury is \( \gamma_{Hg} \), then the ratio \( k_1/k_2 \) can be expressed as
(a) \( \dfrac{1+60\gamma_{Fe}}{1+60\gamma_{Hg}} \)
(b) \( \dfrac{1-60\gamma_{Fe}}{1+60\gamma_{Hg}} \)
(c) \( \dfrac{1+60\gamma_{Fe}}{1-60\gamma_{Hg}} \)
(d) \( \dfrac{1+60\gamma_{Hg}}{1+60\gamma_{Fe}} \)
Show Answer & Explanation

Answer: (a) \( \dfrac{1+60\gamma_{Fe}}{1+60\gamma_{Hg}} \)

Explanation:
1. Fraction submerged \( =\dfrac{\rho_{Fe}}{\rho_{Hg}} \) (the densities decrease as \( \dfrac{1}{1+\gamma\Delta T} \)).
2. \( \dfrac{k_2}{k_1}=\dfrac{1+60\gamma_{Hg}}{1+60\gamma_{Fe}} \), so \( \dfrac{k_1}{k_2}=\dfrac{1+60\gamma_{Fe}}{1+60\gamma_{Hg}} \).

Teacher's Note:
1. Use the pressure formula \( P = P0 \) + rho \( g h \) for liquids at rest and the equation of continuity \( A1 v1 = A2 v2 \) with Bernoulli's equation for flow.
2. Check that the forces or pressures balance at a chosen reference level, or that the energy per unit volume is conserved.

Question 27: A closed compartment containing gas is moving with some acceleration in the horizontal direction (neglect the effect of gravity). Then the pressure in the compartment is
(a) the same everywhere
(b) lower in front side
(c) lower in the rear side
(d) lower in the upper side
Show Answer & Explanation

Answer: (b) lower in front side

Explanation:
1. Gas elements are accelerated forward by a pressure gradient: \( \dfrac{dp}{dx}=-\rho a \).
2. So the pressure is higher at the rear and lower in front.

Teacher's Note:
1. Use the pressure formula \( P = P0 \) + rho \( g h \) for liquids at rest and the equation of continuity \( A1 v1 = A2 v2 \) with Bernoulli's equation for flow.
2. Check that the forces or pressures balance at a chosen reference level, or that the energy per unit volume is conserved.

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