JEE Physics Mechanical Properties Of Fluids MCQs Set 29

Answer: (b) 1 kPa

Explanation:
1. Let the bucket plus water have mass \( M \) and the counterweight \( M/2 \). Then \( Mg-T=Ma \) and \( T-\tfrac M2g=\tfrac M2a \), giving \( a=\dfrac g3 \) (bucket accelerating downward).
2. In a container accelerating downward the effective gravity is \( g-a=\dfrac{2g}{3} \).
3. \( P=\rho(g-a)h=1000\times\dfrac{20}{3}\times0.15=1000 \) Pa \( =1 \) kPa.

Teacher's Note:
1. In an accelerating container the pressure at depth \( h \) is \( \rho\,g_{eff}\,h \).
2. Downward acceleration reduces the effective gravity.

Answer: (d) 35 cc

Explanation:
1. All 60 cm of water must be in the other vertical arm (the horizontal arm is full of the heavy liquid), so the water column there is 60 cm tall.
2. The pressure at the level of the horizontal arm must balance: \( 4\times h=1\times60 \), so the heavy liquid column in its vertical arm is \( h=15 \) cm.
3. Volume of heavy liquid \( =20+15=35 \) cc.

Teacher's Note:
1. Balance the pressures at the level of the horizontal arm.
2. Add the horizontal arm to the vertical column to get the volume.

Answer: (d) \( \dfrac{13}{5}P \)

Explanation:
1. Gauge pressure at the bottom \( \rho gh=3P-P=2P \).
2. After lowering the level by one-fifth, the depth is \( \tfrac45h \), so \( \rho gh'=\tfrac85P \).
3. Absolute pressure \( =P+\tfrac85P=\dfrac{13}{5}P \).

Teacher's Note:
1. Add the atmospheric pressure to the gauge pressure.
2. The gauge pressure is proportional to depth.

Answer: (a) 2/1

Explanation:
1. The heavy liquid settles at the bottom of its arm. Let the water column above it be \( w \) cm, so \( h_1=5+w \).
2. Pressure balance at the bottom: \( 4\times5+w=h_2 \). Water volume is conserved: \( w+h_2=40 \). So \( w=10 \) cm and \( h_2=30 \) cm.
3. \( h_1=15 \) cm, so \( \dfrac{h_2}{h_1}=\dfrac{30}{15}=2 \).

Teacher's Note:
1. Use pressure balance at the bottom of the tube and conservation of water volume.
2. Heights are measured from the bottom of the tube.

Answer: (d) 15 cm

Explanation:
1. Extra pressure under the piston \( =\dfrac{12\times10}{800\times10^{-4}}=1500 \) Pa.
2. This balances the water column: \( \rho gh=1500 \).
3. \( h=\dfrac{1500}{1000\times10}=0.15 \) m \( =15 \) cm.

Teacher's Note:
1. Convert 800 cm² to m² before dividing.
2. The level difference depends on the pressure and not on the tube widths.

Answer: (b) 25 cm

Explanation:
1. Total weight \( =(600+1150)\times10=17500 \) N.
2. If the lighter cube projects by \( x \) m, the displaced volume is \( 1+(1-x) \) m³ and the buoyant force is \( 1000\,(2-x)\times10 \).
3. Equating: \( 2-x=1.75 \), so \( x=0.25 \) m \( =25 \) cm.

Teacher's Note:
1. Treat the two cubes and the wire as a single floating body.
2. Weight equals the weight of water displaced.

Answer: (d) \( 2\pi\sqrt{\dfrac{da}{g}} \)

Explanation:
1. Restoring force per unit displacement \( k=\rho_wbc\,g \) (extra buoyancy from the base area \( bc \)).
2. Mass \( m=d\rho_w\,abc \).
3. \( T=2\pi\sqrt{\dfrac mk}=2\pi\sqrt{\dfrac{d\,abc}{bc\,g}}=2\pi\sqrt{\dfrac{da}{g}} \).

Teacher's Note:
1. A floating body behaves like a spring with constant \( \rho_wAg \), where \( A \) is the horizontal area at the waterline.
2. The vertical side \( a \) and relative density \( d \) set the period.

Answer: (b) the same as before

Explanation:
1. The boy carries the bucket, water and fish in both cases.
2. Placing the fish in the water just changes how its weight is supported (partly by buoyancy, which loads the water and hence the bucket).
3. The total weight carried is unchanged.

Teacher's Note:
1. Treat the bucket, water and fish together as one system.
2. Internal forces such as buoyancy cancel out.

Answer: (d) \( \dfrac{W-W_2}{W-W_1} \)

Explanation:
1. Buoyant force in water \( =W-W_1=V\rho_wg \) and in the liquid \( =W-W_2=V\rho_lg \).
2. Relative density \( =\dfrac{\rho_l}{\rho_w}=\dfrac{W-W_2}{W-W_1} \).

Teacher's Note:
1. Loss of weight in a fluid equals the weight of the displaced fluid.
2. The same volume is displaced in both liquids.

Answer: (a) \( \dfrac{W}{L^{2}\rho_Mg} \)

Explanation:
1. The iron floats because mercury is denser.
2. Weight equals buoyancy: \( W=\rho_MgL^{2}d \).
3. \( d=\dfrac{W}{L^{2}\rho_Mg} \).

Teacher's Note:
1. A floating body displaces its own weight of liquid.
2. The depth uses the density of the liquid, not of the solid.

Answer: (d) 16%

Explanation:
1. Loss of weight in water gives the ball's volume: \( V=\dfrac{1.5}{1000}=1.5\times10^{-3}\ \text{m}^3 \).
2. Volume of metal \( =\dfrac{9.8}{7800}=1.256\times10^{-3}\ \text{m}^3 \).
3. Cavity fraction \( =\dfrac{1.5-1.256}{1.5}\approx0.16 \), that is 16%.

Teacher's Note:
1. Buoyancy gives the total volume including the cavities.
2. Mass divided by density gives only the solid metal volume.

Answer: (a) 2.5 kg

Explanation:
1. Volume of the stone \( =\dfrac{0.5}{500}=10^{-3}\ \text{m}^3 \), so it displaces 1 kg of water.
2. The stone is held from outside, so the water exerts a buoyant force on it and the stone exerts an equal reaction of 1 kg-weight on the water.
3. Reading \( =1.5+1=2.5 \) kg.

Teacher's Note:
1. By Newton's third law the stone pushes down on the water with the buoyant force.
2. The stone is supported externally, so its own weight is not added.

Answer: (c) 86.4 N

Explanation:
1. Mass per second \( =\rho Av=1000\times6\times10^{-4}\times12=7.2 \) kg/s.
2. Change in normal momentum per kg \( =2v\cos60^\circ=12 \) m/s (the tangential component is unchanged).
3. \( F=7.2\times12=86.4 \) N.

Teacher's Note:
1. Only the component normal to the wall changes in an elastic rebound.
2. Force equals mass flow rate times change in velocity.

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

Explanation:
1. A liquid at rest cannot sustain a tangential (shear) stress, so any tangential component of the effective gravity would make it flow.
2. So at rest the surface must be perpendicular to \( \vec g_{eff} \).
3. Both statements are true and the second explains the first.

Teacher's Note:
1. Fluids deform continuously under shear.
2. This is why the free surface is horizontal in a stationary container.

Answer: (d) 4 times that of Q

Explanation:
1. Equation of continuity: \( A_Pv_P=A_Qv_Q \).
2. \( \dfrac{v_P}{v_Q}=\dfrac{A_Q}{A_P}=\left(\dfrac{4}{2}\right)^{2}=4 \).

Teacher's Note:
1. Area is proportional to the square of the diameter.
2. The narrower pipe carries the faster flow.

Answer: (c) \( v_1=v_2,\ p_2>p_1 \)

Explanation:
1. Uniform cross-section means equal speeds by continuity: \( v_1=v_2 \).
2. By Bernoulli, \( p_1+\rho gh_1=p_2+\rho gh_2 \) with \( h_1>h_2 \) (CD lower).
3. So \( p_2>p_1 \).

Teacher's Note:
1. Pressure increases as the liquid flows to a lower level when the speed is unchanged.
2. Check continuity before Bernoulli.

Answer: (d) velocity of water will be maximum and pressure minimum

Explanation:
1. By continuity, the narrowest section has the highest velocity.
2. By Bernoulli's theorem for a horizontal pipe, higher velocity means lower pressure.
3. So velocity is maximum and pressure minimum at the most contracted place.

Teacher's Note:
1. Speed and pressure change in opposite senses in horizontal flow.
2. This is the principle of the Venturi meter.

Answer: (d) 4 : 9

Explanation:
1. Continuity: \( \pi r_1^{2}v_1=\pi r_2^{2}v_2 \).
2. \( \dfrac{v_1}{v_2}=\left(\dfrac{r_2}{r_1}\right)^{2}=\left(\dfrac23\right)^{2}=\dfrac49 \).

Teacher's Note:
1. The speed is inversely proportional to the square of the radius.
2. The wider entrance has the slower flow.

Answer: (c) energy

Explanation:
1. Bernoulli's equation \( P+\tfrac12\rho v^{2}+\rho gh= \) const is the work-energy theorem for a steady, non-viscous, incompressible flow.
2. Conservation of mass gives the equation of continuity instead.

Teacher's Note:
1. Each term of Bernoulli's equation is an energy per unit volume.
2. Continuity is the mass conservation statement.

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

Explanation:
1. Mass flow rate \( =\rho L \). Water enters moving upward with speed \( V \) (uniform cross-section) and leaves horizontally with speed \( V \).
2. Change in momentum per second: vertical component \( \rho LV \) and horizontal component \( \rho LV \).
3. Force magnitude \( =\sqrt{(\rho LV)^{2}+(\rho LV)^{2}}=\sqrt2\,\rho VL \).

Teacher's Note:
1. The force is the rate of change of momentum, a vector difference.
2. The two components are at right angles.

Answer: (c) \( \dfrac{R^{2}}{4h} \)

Explanation:
1. Efflux speed at depth \( d \): \( v=\sqrt{2gd} \). Fall time from the table height: \( t=\sqrt{2h/g} \).
2. \( R=vt=2\sqrt{dh} \).
3. \( d=\dfrac{R^{2}}{4h} \).

Teacher's Note:
1. Torricelli's speed combines with projectile motion.
2. The hole is at the base of the barrel, which is at height \( h \) above the ground.

Answer: (c) 3.4 m/s

Explanation:
1. Pressure due to the piston \( =\dfrac{10\times10}{0.1}=1000 \) Pa, equivalent to \( \dfrac{1000}{1000\times10}=0.1 \) m of water.
2. Effective head \( =0.5+0.1=0.6 \) m.
3. \( v=\sqrt{2g(0.6)}=\sqrt{12}\approx3.4 \) m/s.

Teacher's Note:
1. Extra pressure on the free surface adds to the head of water.
2. The small hole means the surface velocity is negligible.

Answer: (c) 565.7 N

Explanation:
1. Mass flow rate \( =\rho Av=1000\times10^{-3}\times20=20 \) kg/s.
2. Velocity change on a 90° turn has magnitude \( \sqrt2\,v \).
3. \( F=20\times\sqrt2\times20\approx565.7 \) N.

Teacher's Note:
1. A 90° turn changes the velocity vector by \( \sqrt2 v \) in magnitude.
2. The force is directed along the line bisecting the change.

Answer: (a) 48

Explanation:
1. Continuity: \( A_{pipe}v_{pipe}=100\,A_{hole}v_{hole} \).
2. \( v_{hole}=\dfrac{(2)^{2}\times3}{100\times(0.05)^{2}}=\dfrac{12}{0.25}=48 \) m/s.

Teacher's Note:
1. The total area of 100 holes is much smaller than the pipe area.
2. Use diameters squared since the area ratio is the same as the ratio of squared diameters.

Answer: (c) 5 cm

Explanation:
1. In steady state, inflow equals outflow: \( av=100 \), so \( v=\dfrac{100}{1}=100 \) cm/s.
2. Torricelli: \( v=\sqrt{2gh} \), so \( h=\dfrac{v^{2}}{2g}=\dfrac{10^{4}}{2000}=5 \) cm.

Teacher's Note:
1. Steady state means the water level does not change.
2. Keep all quantities in cm and s.