JEE Physics Mechanical Properties Of Fluids MCQs Set 20

Answer: (d) zero

Explanation:
1. Both P and S lie on free liquid surfaces that are open to the atmosphere.
2. So each is at atmospheric pressure, and the difference between them is zero.
3. The 10 cm and 20 cm heights affect the pressure inside the tube, not the pressure at the free surfaces.

Teacher's Note:
1. The pressure at any open free surface is atmospheric.
2. Siphon flow is driven by the height difference between the free surfaces.

Answer: (a) (H − h)ρg

Explanation:
1. The depth of P below the free surface is H − h.
2. Pressure at P is \( (H-h)\rho g \).
3. The thrust per unit area on the wall is this pressure acting normal to the wall, whatever its inclination.

Teacher's Note:
1. Depth is measured from the free surface, not from the bottom.
2. Hydrostatic pressure acts normal to every wall element; no cos θ factor appears.

Answer: (d) FA = FB and WA > WB

Explanation:
1. The force on the base is \( \rho g h\times\text{base area} \) in both vessels, so FA = FB.
2. Vessel A holds more water because it is wider above the base, so WA > WB.
3. The extra weight in A is supported by the sloping walls.

Teacher's Note:
1. The force on the base depends only on depth, density and base area (hydrostatic paradox).
2. Compare volumes of liquid to compare weights.

Answer: (c) zero

Explanation:
1. In free fall the effective gravity is zero, so the pressure in the liquid has no depth variation.
2. With no pressure difference between the top and bottom of the body, there is no buoyant force.

Teacher's Note:
1. Buoyancy exists because of the pressure increase with depth, which needs effective gravity.
2. In a freely falling frame all bodies are weightless and buoyancy disappears.

Answer: (a) \( 2\sqrt{D(H-D)} \)

Explanation:
1. Speed of efflux: \( v=\sqrt{2gD} \).
2. The hole is at height H − D above the ground, so the fall time is \( t=\sqrt{\frac{2(H-D)}{g}} \).
3. Range \( x=vt=2\sqrt{D(H-D)} \).

Teacher's Note:
1. Torricelli's speed plus projectile motion from a horizontal launch.
2. Maximum range 2·√(D(H−D)) is H when D = H/2.

Answer: (c) 1.0 m/s

Explanation:
1. Continuity: flow in = flow out.
2. \( A(3.0)=A(1.5)+1.5A\,v \).
3. So \( 1.5=1.5v \) and v = 1.0 m/s.

Teacher's Note:
1. For an incompressible liquid the volume flow rate is conserved at a junction.
2. Add the flow rates of the branches.

Answer: (a) \( \sqrt2\,\rho VL \)

Explanation:
1. Mass flow rate = ρL.
2. Velocity changes from V upwards to V horizontal, a change of magnitude \( \sqrt2V \).
3. Force on water = rate of change of momentum = \( \sqrt2\rho LV \), and the water exerts an equal and opposite force on the corner.

Teacher's Note:
1. Because the cross-section is uniform, the speed is the same before and after the bend.
2. The momentum change is a vector difference of two perpendicular vectors.

Answer: (a) ρgℓA₀

Explanation:
1. The liquid stands at the top of the limb at A, 2ℓ above the horizontal section.
2. B is at height ℓ, so its depth below the free surface is \( 2\ell-\ell=\ell \).
3. Pressure at B = ρgℓ, so the force on the closed end is \( \rho g\ell A_0 \).

Teacher's Note:
1. The pressure at a point depends on its vertical depth below the free surface, not on the path of the tube.
2. Force = pressure × area.

Answer: (a) B

Explanation:
1. In the frame of the container the fluid is in a uniform pseudo-gravity of \( -\vec a \), i.e. along (−î + ĵ − k̂).
2. Pressure increases in the direction of effective gravity, so it is greatest at the vertex that is furthest along (−x, +y, −z): the left, top, back vertex.
3. That vertex is B.

Teacher's Note:
1. Effective gravity points opposite to the acceleration of the container.
2. Pressure is greatest at the 'bottom' with respect to effective gravity.

Answer: (d) H

Explanation:
1. Pressure decreases along the direction of the acceleration, so it is least at the vertex furthest along (+x, −y, +z): the right, bottom, front vertex.
2. That vertex is H.

Teacher's Note:
1. The minimum is opposite to the maximum.
2. Opposite vertices of the cube (B and H) hold the extreme values.

Answer: (d) picture (1): the level is high in the tube over the wide part and low in the tube over the neck

Explanation:
1. By continuity the liquid moves faster in the narrow neck.
2. By Bernoulli's theorem the pressure is lower where the speed is higher, so the pressure in the neck is less than in the wide part.
3. A lower pressure supports a shorter liquid column, so the level in the neck tube is lower than in the wide-part tube: picture (1).

Teacher's Note:
1. Higher speed means lower pressure (Venturi effect).
2. The same level in both tubes would apply only to a liquid at rest.

Answer: (c) h

Explanation:
1. The jet from a hole of area a at depth d exerts a reaction \( \rho a v^2=2\rho g a d \).
2. The two reactions act in opposite horizontal directions, so the net force is \( 2\rho g a(d_1-d_2)=2\rho g a h \).
3. This is proportional to h.

Teacher's Note:
1. Each jet's reaction force is proportional to the depth of its hole.
2. Opposite jets subtract.

Answer: (b) Statement-1 is true, Statement-2 is true and Statement-2 is a correct explanation for Statement-1

Explanation:
1. A floating body displaces water equal to its own weight (Archimedes).
2. In the second container the ice replaces its own weight of water, so with the same final level the total weight is unchanged.
3. Statement-2 states exactly this reason, so both statements are true and Statement-2 explains Statement-1.

Teacher's Note:
1. Floating means weight = weight of displaced liquid.
2. The level is the same, so the volume of liquid-plus-submerged-ice is the same.