Title 8.MECHANICAL-PROPERTIES-OF-FLUIDS.pdf ✅

Chapter 8 Mechanical Properties of Fluids Solutions SECTION - A Objective Type Questions (One option is correct) 1. Pressure has the dimensions (1) [ML−1 T −2 ] (2) [ML0 T −1 ] (3) [ML1 T −1 ] (4) [MLT−2 ] Sol. Answer (1) 2. The radius of one arm of a hydraulic lift is three times the radius of the other arm. What force should be applied on the narrow arm so as to lift 50 kg at the wider arm? (1) 60 N (2) 54.4 N (3) 26.7 N (4) 30 N Sol. Answer (2) Let radius of narrow arm = r1 Then radius of broader arm = r2 = 3r1 The mass to be lifted at broader arm = 50 kg, then from Pascal's law F1 πr1 2 = F2 πr2 2 F1 = ( r1 r2 ) 2 (50g) = 1 9 × 490 = 54.4 N 3. The pressure at two points in a liquid in pascal are P and 2P. The piston attached to the mouth of the liquid is given a push with pressure 2 Pa. The respective pressures at the two points now are in the ratio (1) P−2 2P−2 (2) P+2 2P+2 (3) P+1 2P+1 (4) P−1 2P−1 Sol. Answer (2) Pascal's law of transmission of fluid pressure. 4. An incompressible liquid of density ρ is enclosed with two frictionless pistons one of cross-sectional area A and the other of 4A. When the narrow piston moves in by a distance h, the wider piston moves out by a distance. (1) h (2) 4h (3) h 4 (4) h 5 Sol. Answer (3)
Since the fluid is incompressible, Volume of fluid pushed in = volume of fluid moved out. ⇒ Ah = 4A ⋅ h ′ ⇒ h ′ = h 4 5. The pressure at a point in water is 10 N/m2 . The depth below this point where the pressure becomes double is (Given density of water = 103 kg m−3 , g = 10 m s −2 ) (1) 1 mm (2) 1 cm (3) 1 m (4) 10 cm Sol. Answer (1) Let the pressure become double at a depth h below the given point. Thus 20 N/m2 = 10 N/m2 + ρgh ⇒ ρgh = 10 ⇒ h = 10 104 = 10−3 m = 1 mm [Given, ρ = 103 kg m−3 , g = 10 m s −2 ] 6. A block of density ρ floats in a liquid with its one third volume immersed. The density of the liquid is (1) ρ (2) ρ 3 (3) ρ 2 (4) 3ρ Sol. Answer (4) Let the volume of the block be V. Since the block floats in water Weight of the block = weight of the liquid displaced Let the density of liquid = σ Then, ⇒ Vρg = V 3 σg ⇒ σ = 3ρ 7. A piece of steel has a weight W in air, W1 when completely immersed in water and W2 when completely immersed in an unknown liquid. The relative density (specific gravity) of liquid is (1) W−W1 W−W2 (2) W−W2 W−W1 (3) W1−W2 W−W1 (4) W1−W2 W−W2 Sol. Answer (2)
w1 = w − ρwvg, w2 = w − ρlvg ⇒ ρ1 ρw = w2 − w w1 − w = w − w2 w − w1 8. A ball made up of a cork material of relative density 1 2 , is dropped from rest from a height 20 m into a lake. Neglecting all dissipative forces, calculate the maximum depth to which the body sinks before re- turning to float on the surface (poission ratio is zero) (1) 10 m (2) 40 m (3) 20 m (4) 5 m Sol. Answer (3) From work energy theorem, mg(h + 20) = ρl × V × g × h h + 20 = ρl × V × g × h ρB × Vg = 2h ⇒h = 20 m 9. What is the barometric height of a liquid of density 3.4 g cm−3 at a place, where that for mercury ba- rometer is 70 cm ? (1) 70 cm (2) 140 cm (3) 280 cm (4) 340 cm Sol. Answer (3) ρgh = ρ ′gh ′ h ′ = ρh ρ ′ = 13.6 × 70 3.4 = 280 cm 10. A wooden cube floats just inside the water, when a mass of x (in grams) is placed on it. If the mass is removed, the cube floats with a height x 100 ( cm) above the water surface. The length of the side of cube is (density of water is 1000 kg/m3 ) (1) 10 cm (2) 15 cm (3) 20 cm (4) 30 cm Sol. Answer (1) Extra depth submerged on placing the block on the wooden cube = x 100 cm ⇒ 10−3xg = ρ. l ( x × 10 100 ) g Where x = Mass of block (in g )
ρ = Density of water I = Side of cube ⇒ I = 1 10 m = 10 cm 11. When at rest a liquid stands at the same level in the tubes shown in figure. But as indicated a height dif- ference h occurs when the system is given an acceleration a towards the right. Here h is equal to (1) aL 2g (2) gL 2a (3) gL a (4) aL g Sol. Answer (4) tan θ = h L tan θ = a g h = aL g 12. A ball floats on the surface of water in a container exposed to the atmosphere. Volume V1 of its volume in inside the water. If the container is now covered and the air is pumped out. Now let V2 be the volume immersed in water, then (1) V1 = V2 (2) V1 > V2 (3) V2 > V1 (4) V2 = 0 Sol. Answer (1) The net force acting on the ball is independent of the pressure, since force experience by the ball is only buoyancy force & gravity force 13. If equal masses of two liquids of densities d1 and d2 are mixed together, the density of the mixture is (1) d1d2 (d1+d2 )