Unit-1. Pressure & Buoyancy. Solutions 1.2 Buoyancy page a) F K =F L =F M. b) F K <F L =F M. c) F K =F M. d) F M >F L

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1 1.2 page Three solid objects (K, L and M) are in equilibrium in a liquid as shown in the figure. a) F K =F L =F M Answer the following questions: b) F K <F L =F M a) If the masses of the objects are the same then compare the buoyant forces acting on them. b) If the volumes of the objects are the same than compare the buoyant forces acting on them. c) F K =F M c) If the weight of the liquid displaced by K and M are the same then compare the weight of the objects K and M. d) F M >F L d) If the volume of the object L is half the volume of object M, then compare the buoyant forces acting on L and M. e) If sugar is added to the liquid to increase its density how would the buoyant force acting on K change? e) It would not change, because it is equal to the weight of object K.

2 1.2 page - 50 P : incresase 2. The system given in the figure is in equilibrium. The pressure of the gas inside the elastic balloon is P and the tension in the rope is T. When liquid is added to the container. the height of liquid increases. Hence, the gas pressure inside the balloon increases to balance the liquid pressure. T : decresase If the same liquid is added to the level L, how would P and T change? P : T : The volume of the balloon decreases. This cause a decrease in the buoyant force on the balloon. Hence, Tension decreases. F B = T + G balloon

3 1.2 page An object is in equilibrium in a liquid of density 2d as shown in the figure. The flow rate of the tap is constant. After the tap is opened, a liquid of density d fills the rest of the container and a homogenous mixture is obtained. a) It would not change. Half of solid object is submerged. Hence, it density is d. The density of the mixture is 3d/2. Solid object floats. The buoyant force on the object is equal to its weight. Hence, it will be the same. Answer the following questions. a) How would the buoyant force applied on the object change? b) How would the submerged volume of the object change? b) It would increase. The buoyant force on the object is the same. Volume of the submerged part of the object will increase. Because density of the liquid decreases. 4.

4 1.2 page Three different equilibrium conditions of objects K, L and M in the same liquid (its density is d ) are given in the figures. (Object K has five equal divisons.) mg = 2V.d.g! # # mg+m L.g = 3V.d.g " # # mg+m L.g+m M.g = 5 V.d.g $ # If the mass of object K is m what are the masses of objects L and M in terms of m? m L = Vdg m L = m 2 m M = 2Vdg m M = m

5 1.2 page K, L and M are three identical prisms. The objects X, Y and Z are placed on these prisms as shown in the drawing. The masses of these objects are m X, m Y and m Z. Compare the masses m X, m Y and m Z. m X +m = V.d! # m Y +m Z +m = 3V.d" m Y = V.d # m Z +m = 2V.d $ m = V.d m X m < V.d m X +m = V.d!# " m Z +m = 2V.d$# m > m Z X m < V.d m Z > V.d m Z > m y > m X

6 1.2 page The density of plastic is 2d and the density of liquid is d. A ball is produced by using plastic. When the ball is placed into the liquid, half of the ball submerges into the liquid as shown in the figure. V ball = V V submerged = V 2 F B = G V 2.d.g = V plastic.2d.g V plastic = V 4 V space = 3V 4 What percent of the ball is empty space? 75% is empty space

7 1.2 page F X + F Y = G (V.d X.g)+(3V.3.g)=(4V.2,5.g) A solid object having four equal divisions is in equilibrium in two immiscible liquids X and Y as shown in the figure. The density of the object is 2,5 g/cm 3 and the density of liquid Y is 3 g/cm 3. Calculate the density of liquid X in g/cm 3. d X + 9 = 10 d X = 1 g/cm 3

8 1.2 page The weight of object A is G and its two different equilibrium conditions in the same liquid (having density of d ) are given in the figures. Figure I; G + T 1 = F B1 F B1 = 2G 2G = 3V.d.g V.d.g = 2G 3 Figure II; G = T 2 + F B1 G = T 2 + V.d.g If the tension T 1 is G in Figure-I, what is the tension T 2 in Figure-II in terms of G? T 2 + 2G 3 = G T 2 = G 3

9 1.2 page Two immiscible liquids of densities d 1 =2 g/cm 3 and d 2 =6 g/cm 3 are poured into a container. Object K that has four equal divisions is in equlibrium in the container as shown in the figure. a) G K =F 1 +F 2 4V.d K.g=(3V.2.g)+(V.6.g) d K =3 g / cm 3 a) Calculate the density of the object K in g/cm 3? b) If the tap is opened and the liquid having density of 2 g/cm 3 is evacuated, draw the new equilibrium condition of the object K in the liquid having density 6 g/cm 3 on the following figure. Explain briefly. b) d K =3 g / cm 3 half of K submerges. d 2 =6 g / cm 3 tap K d2

10 1.2 page A container is filled with a liquid as shown in figure. Object X is dropped into the liquid. The mass of object X is 200 g and the density of liquid is 2 g/cm 3. The increase in the mass of container B is 50 grams. a) F B = 50.g ΔG G K = 200.g A = 200.g - 50.g= 150.g Δm A = 150 grams a) Calculate the increase in the mass of container A in grams. b) 200.g=V x.d x.g d 50.g=V x.d L.g x = 8 g / cm 3 b) Calculate the density of the object X.

11 1.2 page Two solid objects K and L are in equilibrium as shown in Figure-I. The solid objects X and Y are placed on objects K and L respectively and the new equilibrium condition is given in Figure-II. Read the following statements. If they are ABSOLUTELY CORRECT print "A", if they are POSSIBLE print "P" or if they are WRONG then print "W" to the spaces supplied. You have to correct the wrong statements with appropriate word(s) or phrase(s). a) The density of K is smaller than the density of L. A a) The density of K is smaller than the density of L. W b) The weight of X is equal to the weight of Y. A c) The weight of L is equal to the weight of Y. W smaller than d) The weight of K is greater than the weight of X. smaller b) The weight of X is equal to the weight of Y. c) The weight of L is equal to the weight of Y. d) The weight of K is greater than the weight of X.

12 1.2 page - 53 T 1 +V.d.g=V.2d.g 12. The volumes of objects Y and Z are equal and their densities are "d" and "4d" respectively. Objects are in equilibrium as shown in the figure. The tensions in the ropes are T 1 and T 2. T 1 =Vdg T 2 +V.2d.g=V.4d.g T 2 = 2Vdg Calculate the ratio of T 1 to T 2. T 1 = Vdg T 2 2Vdg T 1 T 2 = 1 2

13 1.2 page Two different equilibrium conditions of three object X, Y and Z in the same liquid are given in the following figure. Read the following statements. If they are ABSOLUTELY CORRECT print "A", if they are POSSIBLE print "P" or if they are WRONG then print "W" to the spaces supplied. You have to correct the wrong statements with appropriate word(s) or phrase(s). a) The volume of object X is the sum of the volumes of objects Y and Z. b) Object X is heavier than object Y. c) Object Z is lighter than object Y. Container at left; G X +G Y +T =V Y.d liquid.g G Z = T +V Z.d liquid.g G X +G Y +G Z =(V Y +V Z ).d liquid.g Container at right; G X +G Y +G Z =V X.d liquid.g V X = V Y +V Z A a) The volume of object X is the sum of the volumes of objects Y and Z. P b) Object X is heavier than object Y. P c) Object Z is lighter than object Y.

14 1.2 page Solid objects X, Y and Z are made of the same material and have the same volume. When they are dropped into a liquid, they are in equilibrium as shown in the figure. All the objects are floating; So, their weights are equal to the buoyant forces applied on them. The volumes of the objects are the same. The relation among the submerged volumes of the objects are a) What is the relation among the masses of the objects? V Y >V X >V Z then the buoyant forces on them F Y >F X >F Z b) What is the relation among the buoyant forces acting on the objects? c) What is the relation among the volumes of spaces in the objects? Objects are made of the same subsstance. Their densities are the same. The one having greater mass has smalle air space in it. So, the relation among the volume of air spaces in the objects are V Z >V X >V Y

15 1.2 page A container is filled with a liquid having density of 2 g/cm 3. Objects X and Y are dropped into the liquid as shown in Figure-I. Half of object X submerges in the liquid and object Y sinks as shown in the Figure-II. The volumes of objects X and Y are both 200 cm 3 and the density of object Y is 4 g/cm 3. a) F X = G X V sıb X.d liquid.g=m X.g =m X m X = 200 g X Y X liquid liquid Y Figure-I Figure-II a) What is the mass of the object X in grams? b) V sub-total = V sub-x + V sub-y V sub-total = V sub-total = 300 cm 3 b) What is the total volume of overflown liquid in cm 3? c) How many grams does the mass of container increase after the objects are dropped c) Δm container =(200.4)-(200.2) Δm container = Δm container = 400 g