Tiêu đề 11. Fluid mechanics Easy Ans.pdf ✅

1. (b)Pressure at bottom of the lake = P h g 0 +  Pressure at half the depth of a lake 0 2 h = + P g  According to given condition 0 0 1 2 ( ) 2 3 P h g P h g + = +    0 1 1 3 6 P h g =   5 0 3 2 2 10 20 10 10 P h m g  = = =  . 2. (c)Apparent weight ( ) ( ) m V g g      = − = − where m = mass of the body,  = density of the body  = density of water If two bodies are in equilibrium then their apparent weight must be equal.  1 2 1 2 1 2 ( ) ( ) m m       − = −  2 2 36 48 (9 1) ( 1) 9   − = − By solving we get 2  = 3. 3. (b)According to Boyle's law, pressure and volume are inversely proportional to each other i.e. 1 P V   PV PV 1 1 2 2 =  0 1 0 2 ( ) P h g V PV + = w 2 1 0 1 w h g V V P     = +      2 2 1 47.6 10 1 1000 1 70 13.6 1000 V V      = +       3 3 2  = + = V cm cm (1 5)50 300 . [As 2 0 P P cm = = 70 of Hg =   70 13.6 1000 ] 4. (b)Force acting on the base F P A hdgA =  = 3 0.4 900 10 2 10 7.2N − =     = 5. (c)As the both points are at the surface of liquid and these points are in the open atmosphere. So both point possess similar pressure and equal to 1 atm. Hence the pressure difference will be zero. 6. (b)Difference of pressure between sea level and the top of hill P ( ) ( ) 2 1 2 75 50 10 Hg Hg h h g g   − = −   = −    ...(i) and pressure difference due to h meter of air P = air h g    ...(ii) By equating (i) and (ii) we get ( ) 2 75 50 10 air Hg h g g   −   = −    P2V2 (P1 V1) h 
2 2 4 25 10 25 10 10 2500 Hg air h m   − −    =  =   =      Height of the hill = 2.5 km. 7. (c)Volume of ice M  = , volume of water M  = .  Change in volume M M 1 1 M       = − = −     8. (b)If two liquid of equal masses and different densities are mixed together then density of mixture 1 2 1 2 2 2 1 2 4 1 2 3        = = = + + 9. (d)Let M0 = mass of body in vacuum. Apparent weight of the body in air = Apparent weight of standard weights in air  Actual weight – upthrust due to displaced air = Actual weight – upthrust due to displaced air  0 0 1 2 M M M g dg Mg dg d d     − = −         2 0 1 1 1 d M d M d d   −    =     −     10. (c) P h g =  i.e. pressure does not depend upon the area of bottom surface. 11. (c) PV PV 1 1 2 2 =  ( ) 3 0 4 3 P h g r + +   = 3 0 4 (2 ) 3 P r   Where, h = depth of lake  0 h g P  = 7  7 7 . H g h H g   =  = 12. (c) PV PV 1 1 2 2 =  (P h g V 0 +  ) = 3 0 4 (2 ) 3 P r    0 h g P  = 2  2 75 13.6 13.6 10 g h g    =  = 15 m 13. (a) P h g =  1 h g  (P and  are constant) If value of g decreased by 2% then h will increase by 2%. 14. (d) P h g =  1 h g  . If lift moves upward with some acceleration then effective g increases. So the value of h decreases i.e. reading will be less than 76 cm. 15. (a) Due to acceleration towards right, there will be a pseudo force in a left direction. So the pressure will be more on rear side (Points A and B) in comparison with front side (Point D and C). Also due to height of liquid column pressure will be more at the bottom (points B and C) in comparison with top (point A and D). So overall maximum pressure will be at point B and minimum pressure will be at point D. 16. (b) Total pressure at (near) bottom of the liquid P P h g = +0  As air is continuously pumped out from jar (container), P0 decreases and hence P decreases. a A D B C
17. (a) cos60 h l  =  76 cos 60 1/2 h l = =   l cm =152 18. (b) Pressure at the bottom = h g  and pressure on the vertical surface = 1 2 h g  Now, according to problem Force at the bottom = Force on the vertical surface  2 1 2 2 h g r h g rh      =   h = r 19. (d) At the condition of equilibrium Pressure at point A = Pressure at point B P P A B =  10 1.3 0.8 (10 ) 13.6   =   + −   g h g h g By solving we get h = 9.7 cm 20. (b)Thrust on lamina = pressure at centroid × Area = 3 h g A   = 1 . 3 A gh  21. (c) 1 2 1 2 Total mass 2 2 Total volume 1 1 m m V V m    = = = +     +    1 2 1 2 2     = + 22. (a) ( 1 2 ) Total mass 1 2 Total volume 2 2 m m V V V    + + = = = 1 2 2   + = 23. (b) Bulk modulus, 0 0 p p B V V V V B   = −   = −   0 1 p V V B    = −      Density, 1 0 0 1 1 p p B B    −       = − = +         where, 0 0  = − = p p p h g  = pressure difference between depth and surface of ocean h 60° Wate l h Mercury Glycerine 10 cm Oil h A B 10–h
 0 0 1 gy B      = +     (As h = y) 24. (b)Since, with increase in temperature, volume of given body increases, while mass remains constant so that density will decrease. i.e. 0 0 0 0 0 / (1 ) / (1 ) m V V V m V V V r      = = = = −  +   0     = −  (1 ) 25. (b)  mixture = 1 2 3 ( 2 3 ) 3 3 m m m V d d d V V + + + + = = 2d. 26. (b)  mix = 1 2 3 3 3 2 3 m m V V V m m m d d d = + + + + = 3 6 18 11 11 d d  = 27. (d)Pressure = hg i.e. pressure at the bottom is independent of the area of the bottom of the tank. It depends on the height of water upto which the tank is filled with water. As in both the tanks, the levels of water are the same, pressure at the bottom is also the same. 28. (a) 29. (c)A torque is acting on the wall of the dam trying to make it topple. The bottom is made very broad so that the dam will be stable. 30. (c) 31. (d) 32. (c) Let the total volume of ice-berg is V and its density is . If this ice-berg floats in water with volume Vin inside it then V g V g in  =  V V in     =     [  = density of water] or V V V V out in      − = − =      1000 900 1 1000 10 Vout V      − − = = =      Vout =10% of V 33. (a) Volume of log of wood mass 120 density 600 V = = =0.2 m3 Let x weight that can be put on the log of wood. So weight of the body = (120 10 +  x N ) Weight of displaced liquid = 3 V g N  =   0.2 10 10 The body will just sink in liquid if the weight of the body will be equal to the weight of displaced liquid.  ( ) 3 120 10 0.2 10 10 +  =   x  + = 120 200 x  x = 80 kg 34. (c)Weight of the bowl = mg = 3 3 4 3 2 2 D d V g g          = −               where D = Outer diameter , d = Inner diameter  = Density of bowl Weight of the liquid displaced by the bowl 3 4 3 2 D V g g      = =     where  is the density of the liquid.