ysics 1. Tree identical cubes are arranged as in tree different (I, II and II) positions on a orizontal surface. Te pressure applied on te orizontal surface by te cubes in arrangement I is.. Solid objects K and L are placed on a orizontal surface in two different arrangements as sown in Figure-1 and Figure-. RESSURE Solid & Liquid ressures 10 ressure & Buoyancy Workseet : Solid & Liquid ressure 1. Solid objects X, Y and Z are made of te same substance. Te pressures applied by X, Y, Z on te orizontal surface are X, Y, Z and te magnitudes of te forces applied by tem on te orizontal surface are FX, FY, FZ. ( represents te base areas of te objects.). Wat are te pressures applied on te orizontal surface by te cubes in arrangement II and III in terms of? Read te following statements. If te statement is BSOLUTELY ORRET print "", if it is OSSIBLE print "" or if it is WRONG ten print "W". You ave to correct te wrong statements by using an appropriate word(s) of prase(s). a) ompare X, Y and Z. Z > Y > X b) ompare FX, FY and FZ. FZ > FY > FX Te weigt of object K is equal to te weigt of object L.. Solid objects K and L are placed on a orizontal surface in two different arrangements as sown in Figure-1 and Figure-. Te pressure applied by te objects in 4. Figure- is greater tan te pressure Two cylinders X andapplied Y are placed a orizontal by te on objects in Figure-1. surface as sown in te figure. Te radius of cylinder Y is twice te radiuste of cylinder X and teir by eigts total force applied te objects on are equal. Te pressure applied by cylinder X on te is te orizontal surface in Figure-1 orizontal surface is and te pressure applied by equal to te total force applied by te cylinder Y on te orizontal surface is. objects on te orizontal surface in Figure-. 1 1
ysics. 3. 10 ressure & Buoyancy Workseet : Solid & Liquid ressure Solid objects K and L are placed on a orizontal 3. surface in two different arrangements as sown in Tree identical cubes are arranged as in tree Figure-1 and Figure-. different (I, II and II) positions on a orizontal surface. Te pressure applied on te orizontal surface by te cubes in arrangement I is. 4. 4. Two cylinders X and Y are placed on a orizontal surface as sown in te figure. Te radius of cylinder Y is twice te radius of cylinder X and teir eigts are equal. Te pressure applied by cylinder X on te orizontal surface is and te pressure applied by cylinder Y on te orizontal surface is. Wat are te pressures applied on te orizontal surface te cubes instatements. arrangement II and III in Readby te following If te statement is terms of? BSOLUTELY ORRET print "", if it is OSSIBLE print "" or if it is WRONG ten print "W". You ave to 3G correct te wrong statements by = using an appropriate word(s) of prase(s). Wat would te pressure be, in terms, on te orizontal surface if cylinder X is placed on te top of cylinder Y? Te weigt of object K is equal to te weigt 3Gof object G L. II = 3 = II = 3 Te pressure applied by te objects in Figure- is greater tan te pressure applied by te objects in Figure-1. 3G III = III = Te total force applied by te objects on te orizontal surface in Figure-1 is equal to te total force applied by te objects on te orizontal surface in Figure-. 4. Two cylinders X and Y are placed on a orizontal surface as sown in te figure. Te radius of cylinder 1 GX = GX = π r πr GY = G = 8 π r Y 4π r GX + GY (π r ) + (8π r ) final = = 4π r 4π r 9 final = 4
5. 5. Tree containers K, L, M are placed on a orizontal surface as sown in te figure. Te forces applied by te liquids on te bases of containers K, L and M are equal. 6. liquid is poured into a container tat is placed on a orizontal surface as sown in te figure. Te pressure applied by te liquid on te base of te container is, te force applied by te liquid on te base of te container is F and te force applied by te container on te orizontal surface is F. Read te following statements. If te statement is BSOLUTELY ORRET print "", if it is OSSIBLE print "" or if it is WRONG ten print "W". You ave to correct te wrong statements by using an appropriate word(s) of prase(s). How would, F and F cange if te container were placed on its surface aving base area of? Te liquid pressures at te bottom of te containers are equal. Te base areas of te containers are equal. Te densities of te liquids in te containers are related as d L >d K >d M. Te eigts of te liquids in te containers are related as M > K > L. Te masses of te liquids in te containers are related as m L >m K >m M. 3 increases : decreases F : remains te same F :
7. 7. ontainers X and Y are identical. X contains some liquid as in te figure. Te liquid pressure on te base of container X is, te force applied by te liquid on te base of container X is F and te force applied by te container on te orizontal surface is F. (Figure is composed of identical squares.) 8. container is fulfilled wit a liquid as sown in te figure. Te areas of te lateral surfaces X and Y are "3" and "" respectively. Te force applied by te liquid on lateral surface X is "F". Read te following statements. If te statement is ORRET print "" or if it is WRONG ten print "W". You ave to correct te wrong statements by using an appropriate word(s) of prase(s). W If te liquid in container X was poured into container Y, te liquid pressure on te base of container Y would be /. If te liquid in container X were poured into container Y, te force applied by te liquid on te base of container Y would be F/. 3F/ If te liquid in container X were poured into container Y, te force applied by container Y would be F. 4 Wat is te force applied by te liquid on te lateral surface Y in terms of "F"? F X = F = 3 F Y = 5 F Y = 5F 9.d.g.3 = 9dg.d.g. = 5dg
9. 9. Two immiscible liquids X and Y are in equilibrium in a closed container as sown in te figure. 10. Te dimensions of a rectangular container are 15 cm, 0 cm and 30 cm. It is completely filled wit a liquid. Te force applied by te liquid on te lateral surface K is "F K " and te force applied by te liquid on te base of te container is "F B ". Read te following statements. If te statement is ORRET print "" or if it is WRONG ten print "W". You ave to correct te wrong statements by using an appropriate word(s) of prase(s). W If te container was made upside-down, te liquid pressure at te bottom of te container would not cange. Wat is te ratio of F K to F B? increase B = 15.0 = 300 cm If te container was made upside-down, te force applied by te liquids at te base of te container would be less. If te container was made upside-down, te pressure applied by te container on te orizontal surface would be greater. K = 0.30 = 600 cm F K = 15.d.g.600 F B = 30.d.g.300 F K F B = 1 5
11. Two containers are filled wit two liquids X and Y as in figure. Te weigt of liquid X is G 1 and te weigt of liquid Y is G. Te force applied by liquid X on te base of its container is F 1 and te force applied by liquid Y on te base of its container is F. Te ratio of F 1 to F is 1/. 1. 1. Two immiscible liquids of densities d and 3d are poured into a container as sown in te figure. Te surface areas of lateral surfaces are and. Te force exerted by te liquid on te lateral surface aving area of is F. Wat is te ratio of G 1 to G? V Y = 3V V X = 7V = y = F 1 = 1 F =.d.g. X d X = 1.d Y.g. d Y G 1 G = d X.V X.g d Y.V Y.g = 1.7V.3V = 7 6 6 alculate te forces exerted by te liquid on te lateral surface aving te area of and on te base of te container in terms of F? B = 15.0 = 300 cm K = 0.30 = 600 cm F K = 15.d.g.600 F B = 30.d.g.300 F K F B = 1
13. 13. Four containers X, Y, Z and T are placed on a orizontal surface. ontainers X, Y and Z are filled wit liquids and container T is empty. Te liquid pressures at te bottoms of te containers X, Y and Z are all equal to "". (Te liquids in te container are immiscible.) = 4d X g d X = 3d = d Y g d Y = 6d = 1dg = 3d z g d Z = 4d If te liquids were poured into container T, wat would be te total liquid pressure at te bottom of container T in terms of ""? T = ( 6dg)+ 5 4.4d.g + 3dg ( ) T 3d 4d 6d 5/4 T = 14dg T = 7 6 7
14. Te ydraulic system is in equilibrium as sown in te figure. Te piston is supposed to be weigtless and frictionless. Te pressure at point X is. ( and represent te cross-sectional areas of te columns of te combined container.) 15. 15. Two miscible liquids aving densities of d and d are in equilibrium as in te figure. Te liquid pressure at point X is. ( and represent te cross-sectional areas of te columns of te combined container.) Wat would be te pressure at point X (in terms of ), if te piston were pused from level K to level L? Wat will be te liquid pressure at point X in terms of after te valve is opened and te final equilibrium ad reaced? V initial = V final 4+ = /.3 / = d mix = (d.4v) + (d.v) 4V + V = 4d 3 Xinitial = = dg dg = Xfinal = 5 dg X final = 5 4 8 = dg / =. 4d 3.g / = 4 3
16. 16. Te ydraulic system is in equilibrium as sown in te figure. istons are supposed to be weigtless and frictionless. 1 and are te base areas of te pistons. Te density of liquid is d. 17. 17. Te flow rates of te valves M and N are te same. Tey are open at te same time and te final equilibrium is reaced. Te liquid pressures at te points X, Y and Z are X, Y and Z. (ll te columns of te combined container ave te same cross-sectional area. Dotted lines are equally spaced.) Read te following statements. If te statement is ORRET print "" or if it is WRONG ten print "W". You ave to correct te wrong statements by using an appropriate word(s) of prase(s). Te liquid pressure at point X is equal to.d.g. Te liquid pressures at point X and Y are equal. How can X, Y and Z be compared? fter te valves are opened M N Te total pressure at point X is greater tan te liquid pressure at point Y. X Y Z 9 X = Y < Z
18. 18. Liquid of density "d" and liquid B of density "3d" are in equilibrium in a combined container as sown in te figure. (tmosperic pressure is ignored.) 19. 19. combined container is in equilibrium wen an object of mass "m" is placed on piston K and object of mass "m" is placed on piston L as sown in te figure. K and L are weigtless and frictionless pistons. Teir cross-sectional areas are "" and "3" respectively. Te vertical distance between te levels of te pistons is "". K L Wat is te pressure at point X in terms of "dg"? K = L = 4dg+ 9dg = 13dg L = 13dg = 7.3d.g+ X X = 5dg 10 If te places of te objects were intercanged, wat would be te vertical distance between te levels of te pistons in terms of ""? 1 st condition; mg = mg 3 +dg m = 3d nd condition; mg = mg 3 +/ dg 5m 3 = / d / = 5
0. 0. olumn K of a combined container is filled wit a liquid as sown in te figure. Te liquid pressure on te bottom of column K is wen te valve is closed. ( and represent te cross-sectional areas of te columns K and L of te combined container.) V initial = V final 6 = ( + 3). / / = Te valve is opened and te final equilibrium is reaced. Wat will be te liquid pressure in terms of at te bottom of column K? 5 Te valve is closed; K L orizontal = 6dg dg = 6 Te valve is opened; 11 / = dg / = 3
1. Two immiscible liquids X and Y are poured into a container and teir equilibrium condition is given in te figure. Te density of liquid X is 3d and te density of liquid Y is d. Te liquid pressure at point L is two times te liquid pressure at point K... Liquid X aving density of d and weigt of G is poured into te container as sown in te figure. Te liquid pressure at te bottom of te container is. Te empty part of te container is filled wit liquid Y aving weigt of G. (ontainer is composed of identical cubic parts.) Wat is te ratio of 1 to? K = 1.d.g L = ( 1.d.g) + (.3d.g) K L = 1 If liquids X and Y are miscible, wat will be te liquid pressure at te bottom of te container in terms of? V X = V V Y = V G = d.v.g G = d Y.V.g d Y = 4d (d.v) + (4d.V) d mix = = d V + V 4 1 = 1 + 3 1 = 3 1 = dg / = 4 / =.d.g = 4dg
3. 3. Te liquids of densities d, d and liquid X are in 4. equilibrium in a U-tube as sown in te figure. 4. Mercury is poured into a U-tube as given in Figure-1. Te left arm of te U-tube as a cross-sectional area of 10 cm, and te rigt arm as a cross-sectional area of 5 cm. One undred grams of water are ten poured into te rigt arm, as sown in Figure-. (Te density of water is 1 g/cm 3.) =x Wat is te density of liquid X in terms of d? x B x = B (3.d.g) + (.d X.g) = 4.d.g (3d) + (d X ) = 8d d X = 5d 13 a) Wat is te lengt of te water column in te rigt arm of te U-tube in cm? V water = 100 cm 3 water = 100 5 = 0 cm b) If te density of mercury is 13,6 g/cm 3, wat distance () will te mercury rise in te left arm? = B 3x.1.g = 0.13,6.g x = 0,49 cm