JEE Physics Mechanical Properties Of Fluids MCQs Set 30

Answer: (a) 0.051 m

Explanation:
1. In steady state outflow \( av=10^{-4}\ \text{m}^3/\text{s} \) gives \( v=1 \) m/s.
2. \( h=\dfrac{v^{2}}{2g}=\dfrac{1}{19.6}\approx0.051 \) m.

Teacher's Note:
1. Steady level means inflow rate equals outflow rate.
2. Use Torricelli's theorem for the efflux speed.

Answer: (c) density of liquid

Explanation:
1. Efflux speed \( v=\sqrt{2gh} \) and discharge \( =av \) per second.
2. The discharge depends on the orifice area, the height \( h \) and \( g \), but not on the density.

Teacher's Note:
1. In Torricelli's law the density cancels.
2. Pressure and weight both scale with density.

Answer: (c) \( \dfrac{h_2+h_1}{2} \)

Explanation:
1. With water height \( H \), a hole at height \( h \) has range \( R=2\sqrt{h(H-h)} \).
2. Equal ranges for \( h_1 \) and \( h_2 \) mean they are symmetric about \( H/2 \): \( h_1+h_2=H \).
3. The range is maximum at \( h=\dfrac H2=\dfrac{h_1+h_2}{2} \).

Teacher's Note:
1. The range is maximum for a hole at half of the liquid height.
2. Holes equidistant from the midpoint have equal ranges.

Answer: (a) \( F_1=F_2 \)

Explanation:
1. For a vessel with vertical walls and base area \( A \), the force on the base is \( \rho gh\,A=\rho g\,(hA)=\rho gV=mg \).
2. The same liquid has the same weight in both vessels.
3. So \( F_1=F_2 \).

Teacher's Note:
1. With vertical walls, the force on the base equals the weight of the liquid.
2. The different base areas change the depth, but not the force.

Answer: (c) \( \sin\theta=\dfrac12\sqrt{\rho_0/\rho} \)

Explanation:
1. Let the rod have length \( l \) and area \( A \). The immersed length is \( \dfrac{l/2}{\sin\theta} \).
2. Torque about the lower end: weight \( \rho Alg \) at \( \dfrac l2 \) gives \( \rho Alg\dfrac l2\cos\theta \); buoyancy \( \rho_0A\dfrac{l}{2\sin\theta}g \) acts at \( \dfrac{l}{4\sin\theta} \), giving \( \rho_0A\dfrac{l}{2\sin\theta}g\dfrac{l}{4\sin\theta}\cos\theta \).
3. Equating: \( \rho=\dfrac{\rho_0}{4\sin^{2}\theta} \), so \( \sin\theta=\dfrac12\sqrt{\dfrac{\rho_0}{\rho}} \).

Teacher's Note:
1. Take torques about the bottom end, where the normal force acts.
2. Buoyancy acts at the centre of the immersed part.

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

Explanation:
1. In a horizontally accelerating liquid, pressure rises in the direction opposite to the acceleration: \( P_C-P_A=\rho al \).
2. Vertically below C: \( P_B-P_C=\rho gh \).
3. \( P_B-P_A=\rho gh+\rho al \).

Teacher's Note:
1. Break the path from A to B into a horizontal part and a vertical part.
2. The horizontal pressure gradient equals \( \rho a \).

Answer: (a) \( 4m \)

Explanation:
1. Initially \( M=\rho\cdot\tfrac45a^{3} \).
2. After placing the coin, \( M+m=\rho a^{3} \).
3. Subtracting: \( m=\rho a^{3}-\tfrac45\rho a^{3}=\tfrac15\rho a^{3} \), so \( M=\tfrac45\rho a^{3}=4m \).

Teacher's Note:
1. The extra load equals the extra buoyancy from the extra submerged volume.
2. One-fifth of the volume corresponds to the coin.

Answer: (d) \( d_1>\dfrac34d \)

Explanation:
1. Floating: total weight \( =A L(d_1+d_2)g \) equals the buoyancy of the submerged length \( 2L-\tfrac L2=\tfrac{3L}{2} \): \( d_1+d_2=\tfrac32d \).
2. Since \( d_1>d_2 \), \( d_1>\tfrac12\cdot\tfrac32d=\tfrac34d \).
3. So \( d_1>\dfrac{3d}{4} \).

Teacher's Note:
1. The larger of two numbers with a fixed sum exceeds half of the sum.
2. Weight equals buoyancy at equilibrium.

Answer: (b) \( V(g+a)(d-\rho) \)

Explanation:
1. In the accelerating frame the effective gravity is \( g+a \).
2. Buoyant force \( =Vd(g+a) \) upward, weight \( =V\rho(g+a) \) downward.
3. \( T=V(g+a)(d-\rho) \).

Teacher's Note:
1. Replace \( g \) by \( g+a \) for an upward-accelerating container.
2. The string supplies the difference between buoyancy and weight.

Answer: (c) falls

Explanation:
1. Initially the ice with the metal displaces water equal in weight to the ice plus the metal, which is larger than the volume of the metal.
2. After melting, the ice turns into water of volume equal to the volume the ice-water mass displaced, and the metal displaces only its own volume.
3. The metal's displaced volume is smaller than the water volume it displaced while embedded in the ice, so the level falls.

Teacher's Note:
1. A dense object displaces more water when floating (by weight) than when sunk (by volume).
2. The melted ice itself does not change the level.

Answer: (b) \( 2g/3 \)

Explanation:
1. The free surface tilts with \( \tan\theta=\dfrac ag \). Let the cube side be \( L \) and the surface drop by \( x \) across the width.
2. Spilled volume \( =\tfrac12xL\cdot L=\tfrac13L^{3} \), so \( x=\dfrac23L \).
3. \( \tan\theta=\dfrac{x}{L}=\dfrac23 \), so \( a=\dfrac{2g}{3} \).

Teacher's Note:
1. The spilled water is a wedge-shaped prism.
2. The slope of the surface is \( a/g \).

Answer: (a) 4.9 litre/min

Explanation:
1. Continuity: \( A_1v_1=A_2v_2 \), so \( v_2=2v_1 \).
2. Falling 10 cm: \( v_2^{2}=v_1^{2}+2gh \), so \( 3v_1^{2}=2\times980\times10 \) and \( v_1\approx80.8 \) cm/s.
3. Flow rate \( =A_1v_1\approx80.8\ \text{cm}^3/\text{s}=4850\ \text{cm}^3/\text{min}\approx4.9 \) litre/min.

Teacher's Note:
1. The stream narrows because falling water speeds up.
2. Convert cm³/s to litres per minute at the end.

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

Explanation:
1. In the accelerating frame the effective gravity is \( g+a \) for upward acceleration.
2. Pressure due to the liquid \( =\rho(g+a)h \).

Teacher's Note:
1. Upward acceleration makes the liquid seem heavier.
2. Downward acceleration would reduce it to \( \rho(g-a)h \).

Answer: (d) 20

Explanation:
1. Equation of continuity: \( A_Av_A=A_Bv_B \).
2. \( v_B=\dfrac{1\times10}{0.5}=20 \) cm/s.

Teacher's Note:
1. The height difference does not affect the speed in an incompressible steady flow through a given tube.
2. Use continuity, not Bernoulli, for the speed ratio.

Answer: (d) The mass of water flowing past any cross-section must remain constant. Also, water is almost incompressible. Hence, the rate of volume flow must remain constant. As this is equal to velocity \( \times \) area, the area decreases as velocity increases

Explanation:
1. In steady flow of an incompressible fluid, \( Av= \) constant (equation of continuity).
2. Falling water speeds up, so the area must decrease.
3. Surface tension and terminal velocity are not the cause.

Teacher's Note:
1. Continuity is the direct explanation.
2. The atmosphere does not squeeze the stream.

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

Explanation:
1. In the accelerating frame the effective gravity is \( g_{eff}=g\pm a \).
2. Both the weight of the wood and the buoyant force scale with \( g_{eff} \), so the fraction submerged \( \dfrac{\rho_{wood}}{\rho_{water}} \) is unchanged.
3. The wood stays at the same position relative to the water.

Teacher's Note:
1. Buoyancy equilibrium depends only on the density ratio.
2. A uniform change of \( g \) cancels from both sides.

Answer: (c) 3 atm

Explanation:
1. Range \( R=v\,t \) with the same fall time, so doubling the range needs double the efflux speed, and \( v^{2} \) becomes 4 times.
2. \( v^{2}=2\left(gh+\dfrac{\Delta P}{\rho}\right) \): \( gh+\dfrac{\Delta P}{\rho}=4gh \), so \( \Delta P=3\rho gh \).
3. \( \Delta P=3\times10^{3}\times10\times10=3\times10^{5} \) Pa \( =3 \) atm.

Teacher's Note:
1. The extra pressure acts like extra head: \( \Delta P/\rho g \).
2. The time of fall is unchanged since the hole height is the same.

Answer: (c) \( \dfrac1{\sqrt2} \)

Explanation:
1. Hole 1 is in the lighter liquid at depth \( h/2 \): \( v_1=\sqrt{2g\dfrac h2}=\sqrt{gh} \).
2. Hole 2 is in the heavier liquid, with pressure \( \rho gh+2\rho g\dfrac h2=2\rho gh \), so \( \tfrac12(2\rho)v_2^{2}=2\rho gh \) and \( v_2=\sqrt{2gh} \).
3. \( \dfrac{v_1}{v_2}=\dfrac{1}{\sqrt2} \).

Teacher's Note:
1. Equate the pressure at the hole to \( \tfrac12\rho v^{2} \) using the density of the liquid at that hole.
2. Add the contributions of both layers for the lower hole.

Answer: (a) \( \sqrt{\dfrac{2\mu M}{\pi\rho H}} \)

Explanation:
1. Efflux speed \( v=\sqrt{2gH} \). Thrust \( F=\rho av^{2}=\rho\dfrac{\pi d^{2}}{4}\cdot2gH=\dfrac{\pi\rho gHd^{2}}{2} \).
2. The vessel starts moving when \( F=\mu Mg \) (limiting friction).
3. \( d^{2}=\dfrac{2\mu M}{\pi\rho H} \), so \( d=\sqrt{\dfrac{2\mu M}{\pi\rho H}} \).

Teacher's Note:
1. The jet exerts a reaction thrust of \( \rho av^{2} \) on the vessel.
2. Limiting friction is \( \mu \) times the normal force \( Mg \).

Answer: (d) \( x \) is maximum for \( y=H/2 \)

Explanation:
1. Range \( x=2\sqrt{y(H-y)} \) (hole at height \( H-y \) above the ground).
2. It rises from 0, is maximum at \( y=\dfrac H2 \) with \( x_{max}=H \), and falls to 0 at \( y=H \).
3. So only the statement that \( x \) is maximum for \( y=H/2 \) is correct; the maximum value does not depend on density.

Teacher's Note:
1. The product \( y(H-y) \) is largest at \( y=H/2 \).
2. The density cancels in Torricelli's law.

Answer: (a) The manometer at P shows greater pressure than at Q

Explanation:
1. By continuity the water is slower in the wide section X and faster in the narrow section Y, so (a) and (c) are false.
2. By Bernoulli, the slower flow at X has the higher pressure.
3. So the manometer at P (wide part) reads a higher pressure than at Q.

Teacher's Note:
1. Narrow section means fast flow and low pressure.
2. Compare pressures using Bernoulli's equation for a horizontal tube.

Answer: (d) \( v_0/2^{1/4} \)

Explanation:
1. Let the side be \( H \). Initially \( v_0=\sqrt{2gH} \).
2. When half full and tilted by 45° with the spout at the lowest corner, the free surface passes through the two side corners, so the depth of the spout below it is half the face diagonal: \( \dfrac{\sqrt2H}{2}=\dfrac{H}{\sqrt2} \).
3. \( v=\sqrt{2g\dfrac{H}{\sqrt2}}=\dfrac{v_0}{2^{1/4}} \).

Teacher's Note:
1. Torricelli's speed depends on the vertical depth of the liquid above the opening.
2. Find the depth from the geometry of the tilted cube.

Answer: (c) \( w_2=w_1 \)

Explanation:
1. The balloon is filled with air of the same density as the surrounding air, so the air inside is neutrally buoyant.
2. The weight \( w \) of the enclosed air is exactly cancelled by the buoyant force on the balloon.
3. The spring balance reading stays the same: \( w_2=w_1 \).

Teacher's Note:
1. An object with the same density as the surrounding fluid has zero apparent weight in it.
2. Buoyancy acts on the whole balloon volume.

Answer: (d) \( x_m=2h \) and \( y=h \)

Explanation:
1. A hole at depth \( y \) is at height \( 2h-y \) above the ground, so \( x=2\sqrt{y(2h-y)} \).
2. This is maximum when \( y=h \), where \( x=2\sqrt{h\cdot h}=2h \).
3. So \( x_m=2h \) and \( y=h \).

Teacher's Note:
1. The range is maximum when the depth equals the height above the ground.
2. Substitute the total height \( 2h \) above the ground.

Answer: (b) (2) and (4) only

Explanation:
1. Thrust \( =\dfrac{dp}{dt}=\rho Av^{2} \), so doubling \( v \) gives four times the thrust.
2. Energy lost per second \( =\tfrac12\rho Av\cdot v^{2}=\tfrac12\rho Av^{3} \), which becomes eight times.
3. So statements (2) and (4) are correct.

Teacher's Note:
1. Mass flow rate is proportional to \( v \) and momentum per kg to \( v \): thrust goes as \( v^{2} \).
2. The kinetic energy per kg goes as \( v^{2} \), giving \( v^{3} \) per second.