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- 1. Q. a cube of mass 1kg with each side of 1cm is lying on the table. find thepressure exerted by the block onthe table. take g=10 m/s2Ans: Pressure is given as force/area so, Force, F = mg = 1000 X 981 gm.m/s2and area, A = 1x1 cm2 = 1cm2Thus, the pressure exerted would be P = (981 X 1000) / 1 orP = 9.81 X 10 5 paQ. The mass of a solid iron cube of side 3cm is to be determined usig a spring balance. If the of iron isapproximately 8.5 g/cm3, the best suited spring balance for determining weight of the solid would be of1. range 0-250gwt ; least count 1gwt2. range 0-250gwt ; least count 5gwt3. range 0-1000gwt ; least count 5gwt4. range 0-1000gwt ; least count 10gwtAns: Edge=3 cm , Density=8.5 g/cm3Mass= density x volume = 8.5 x(3x3)=229.5gwt. Therefore secondspring balance of range 0-250 gwt with least count 5gwt will be suitable.Q. The density of turpentine oil is 840 Kg/ m3. What will be its relativedensity. (Density of water at 4degree C is 10 cube kg minus cube)Ans:Relative Density = Density of Substance/ Density of water at 4 0cDensity of turpentine oil = 840 kg/m3 ( given).Density of water at 4 0c = 1000 kg/ m3Relative density of turpentine oil = Density ofturpentine oil / Density of water at 4 0c= (840 / 1000 ) kg m-3/ kg m-3 = 0.84Since, the relative density ofthe turpentine oil is less than 1, therefore it will float in water.Q. A solid body of mass 150 g and volume 250cm3 is put in water . will following substance float or sinkif the density of water is 1 gm-3?Ans:Q. a cube of mass 1kg with each side of 1cm is lying on the table. find the pressure exerted by the blockon the table. take g=10 m/s2Ans: Pressure is given as force/area so, Force, F = mg = 1000 X 981 gm.m/s2and area, A = 1x1 cm2 = 1 cm2Thus, the pressure exerted would be P = (981 X 1000) / 1 or P = 9.81 X 10 5 paQ. The mass of a solid iron cube of side 3cm is to be determined usig a spring balance. If the of iron isapproximately 8.5 g/cm3, the best suited spring balance for determining weight of the solid would be of1. range 0-250gwt ; least count 1gwt 2. range 0-250gwt ; least count 5gwt3. range 0-1000gwt ; least count 5gwt 4. range 0-1000gwt ; least count 10gwtAns: Edge=3 cm , Density=8.5 g/cm3Mass= density x volume = 8.5 x(3x3)=229.5gwtTherefore second spring balance of range 0-250 gwt with least count 5gwt will be suitable.
- 2. Q. The density of turpentine oil is 840 Kg/ m3. What will be its relative density. (Density of water at 4degree C is 10 cube kg minus cube)Ans:Relative Density = Density of Substance/ Density of water at 4 0cDensity of turpentine oil = 840 kg/ m3 ( given).Density of water at 4 0c = 1000 kg/ m3Relative density of turpentine oil = Density of turpentine oil / Density of water at 4 0c= (840 / 1000 ) kg m-3/ kg m-3 = 0.84Since, the relative density of the turpentine oil is less than 1, therefore it will float in water.Q. A solid body of mass 150 g and volume 250cm3 is put in water . will following substance float or sinkif the density of water is 1 gm-3?Ans: The substance will float if its density is less than water and will sink if its greater.so, density of solid body is d = mass/volumeor d = 150/250 = 0.6 gm/cm3 which is less than the density of water (1 gm/3).So, the solid body will float on water.Q. A body weighs 50 N in air and when immersed in water it weighs only 40 N. Find itsrelative density.Ans: the relative density would be ratio of the density of the body with respect to air and the density ofthe body with respect to water.so, F1 = 50 N F2 = 40 Nso, F1/F2 = 50/40 or relative masses m1/m2 = 5/4 anddensity = mass/volume and as volume remains constant,Relative density = d1/d2 = 5/4Q. A ball of relative density 0.8 falls into water from a height of 2m. find the depth to which the ball willsink ?Ans: Speed of the ballV = 2gh = 2x10x2 = 6.32 m/sBuoyancy force by water try to stop the ball.Buoyancy force = weight of displaced water = dx Vxgwhere d = density of water V = volume of the ball , g = 10 m/s2deceleration of the body by buoyancy force, a = (dVg)/ mwhere m= dV d = density of blocka = dVg/(d V) = dg/d =(d/d)*g =g/(0.8)= 10/0.8 (Given, d/d = 0.8)= 12.5 m/s2Net deceleration of ball,a = a-g = 2.5 m/s2Final speed of ball v = 0 Use v 2 = v2 + 2as s= depth of ball in the water=> 40 = 0 + 2x2.5xs => s = 8m
- 3. Q. Equal masses of water and a liquid of relative density 2 are mixed together. Then, the mixture hasa relative density of (in g/cm3) a)2/3 b)4/3 c)3/2 d)3Ans: he masses of two liquids are equal, let it be m.Let the relative densities of water and liquid be ρ1 and ρ2 respectively.The volume of the two be V1 and V2, of water and liquid respectively.The volume of the mixture would be, V = V1 + V2 (1)also, volume = mass/densitythus,2m/ρ (V) = m/ρ1 (V 1 ) + m/ρ2 (V 2 )here ρ1 = 1,ρ2 = 2 and ρ is the relative density of the mixture.now,2/ρ = 1/ρ1 + 1/ρ2by substituting the values, weρ/2 = 2/3or, the relative density of the combined liquid will be, ρ=4/3 Density Problems1) What is the volume of a tank that can hold 18 754 Kg of methanol whose density is 0.788g/cm3?2) What is the density of a board whose dimensions are 5.54 cm x 10.6 cm X 199 cm and whose mass is 28.6 Kg?3) CaCl2 is used as a de-icer on roads in the winter. It has a density of 2.50 g/cm3. What is the mass of 15.0 L this substance?4) In the Macys Thanksgiving day parade, 2000, there was a new float, the Cheerios bee, Buzzbee. The jar of honey it was holding contained 36,763 L of helium. The honey jar was 1/10 of the body size. What is the volume of the entire float? If the density of helium is 0.000178 g/cm3, what is the mass of the helium for the entire float in Kg?5) My gas tank in my car has a total volume of 68 L. The manual says the gas gauge light will come on when there are only 5 L remaining in the tank and that the car will not be able to draw on the last 2 L in the tank. Practically speaking, what is the functional volume of the gas tank? If I ran out of gas, the car wont go, and I put in 5 L of gas, what is the total volume of gas in my tank? If 5 L of gas has a mass of 450 Kg, what is the mass of the gas in my tank?
- 4. What is the density of the gas?6) The density of paper is 1.20 g/cm3. What is the mass of the paper in a notebook that is 76 mm thick, 215.9 mm wide, and 279.4mm long? Express your answer in significant digits.7) In pottery class, you throw a pot from a lump of wet clay. Your pots mass is 5.5 Kg. After the pot is fired, its mass is 4.9 Kg. The density of wet clay is 1.60 g/cm3 and the density of fired clay is 1.36 g/cm3. What was the volume of your pot before it was fired? What was the volume of the pot after it was fired? Express your answer in significant digits.8) The Italian government is giving Michelangelos David a bath. In the newspaper, there was a photo of a person working on the statue. Their hand was half the size of the statues hand. If the statue is 4.9m3 and the density of marble is 2.76 g/cm3 what is the mass? Express your answer in significant digits.9) The volume of the aquarium in our classroom is 1890 L. The density of seawater is 1.03g/cm3. What is the mass of the water in our tank? Express your answer in significant digits.10) Spanish mahogany is a red, lustrous wood. It is prized over Honduras mahogany, which does have the characteristic red color and similar grain, but is the wood is not as compact and has a looser appearance than the Spanish mahogany. A hope chest made of Honduras mahogany has a volume of .648 m3 and a mass of 349 Kg. An identical hope chest made of Spanish mahogany has a mass of 550.8 Kg. What is the density of Honduras mahogany? What is the density of Spanish mahogany? Express your answer in significant digits.Solutions to Density ProblemsDensity = mass volume1) What is the volume of a tank that can hold 18 754 Kg of methanol whose density is 0.788g/cm3? D = m/v v = m/D v = [18754 Kg][103 g][ cm3] [ 1 Kg][0.788 g] v = 2.38 x 107 cm32) What is the density of a board whose dimensions are 5.54 cm x 10.6 cm X 199 cm and whose mass is 28.6 Kg?
- 5. D = m/v D = [28.6 Kg][103 g] [ 1 Kg][5.54cm x 10.6 cm x 199 cm] D = [28.6 Kg][103 g] [ 1 Kg][11686 cm3] D = 2.45 g cm33) CaCl2 is used as a de-icer on roads in the winter. It has a density of 2.50 g/cm3. What is the mass of 15.0 L this substance? D = m/v m = Dv m = [2.50 g][15.0 L][103 mL][1cm3] [ cm3] [ 1 L ][ 1mL] m = 3.75 x 104 g4) In the Macys Thanksgiving day parade, 2000, there was a new float, the Cheerios bee, Buzzbee. The jar of honey it was holding contained 36,763 L of helium. The honey jar was 1/10 of the body size. What is the volume of the entire float? If the density of helium is 0.000178 g/cm3, what is the mass ofthe helium for the entire float in Kg? D = m/v m = Dv v = 36,763 L He x 10 = 367,630 L He m = [0.000178 g][367,630 L][103 mL][1cm3][ 1 Kg] [ cm3] [ 1 L ][ 1mL ][103 g] m = 65.4 Kg5) My gas tank in my car has a total volume of 68 L. The manual says the gas gauge light will come on when there are only 5 L remaining in the tank and that the car will not be able to draw on the last 2 L in the tank. Practically speaking, what is the functional volume of the gas tank? If I ran out of gas, the car wont go, and I put in 5 L of gas, what is the total volume of gas in my tank? If 5 L of gas has amass of 450 Kg, what is the mass of the gas in my tank? What is the density of the gas? D = m/v functional volume of tank is 68L - 2 L = 66 L gas "empty tank" + 5 L = 7 L gas
- 6. m = [7 L][450 Kg] [ 5 L] m = 630 Kg D = [630 Kg][103 g] [ 1 L ][ 1 mL] [ 1 Kg][7 L][103 mL][1 [cm3] D = 630 g 7 cm3 D = 9.0 x 101 g cm3Problem Example 1Ordinary commercial nitric acid is a liquid having a density of 1.42 g mL–1, and contains 69.8% HNO3 byweight. a) Calculate the mass of HNO3in 800 ml of nitric acid. b) What volume of acid will contain 100 g ofHNO3?Solution: The mass of 800 mL of the acid is (1.42 g mL–1) × (800 mL) = 1140 g. The weight of acid thatcontains 100 g of HNO3 is (100 g) / (0.698) = 143 g and will have a volume of (143 g) / (1.42 g mL–1) =101 mL.Problem Example 2A glass bulb weighs 66.3915 g when evacuated, and 66.6539 g when filled with xenon gas at 25°C. Thebulb can hold 50.0 mL of water. Find the density and specific volume of xenon under these conditions.Solution: The mass of xenon is found by difference: (66.6539 – 66.3915)g = 0.2624 g. The density ρ = m/V =(0.2624 g)/(0.050 L) = 5.248 g L–1. The specific volume is 1/(5.248 g L–1 = 0.190 L g–1.Problem Example 3Suppose that you place 1000 mL of pure water at 25°C in the refrigerator and that it freezes, producing iceat 0C. What will be the volume of the ice?Solution: From the graph above, the density of water at 25°C is 0.9997 kg L–1, and that of ice at 0°C 0.917 gL–1.Problem Example 4An object weighs 36 g in air and has a volume of 8.0 cm3. What will be its apparent weight whenimmersed in water?Solution: When immersed in water, the object is buoyed up by the mass of the water it displaces, which ofcourse is the mass of 8 cm3 of water. Taking the density of water as unity, the upward (buoyancy) force isjust 8 g.The apparent weight will be (36 g) – (8 g) = 28 g.
- 7. Problem Example 5A balloon having a volume of 5.000 L is placed on a sensitive balance which registers a weight of 2.833 g. What isthe "true weight" of the balloon if the density of the air is 1.294 g L–1?Solution: The mass of air displaced by the balloon exerts a buoyancy force of(5.000 L) × (1.294 g L –1) = 6.470 g. Thus the true weight of the balloon is this much greater than the apparentweight: (2.833 + 6.470) g = 9.303 g.Problem Example 6A piece of metal weighs 9.25 g in air, 8.20 g in water, and 8.36 g when immersed in gasoline. a) What is the densityof the metal? b) What is the density of the gasoline?Solution: When immersed in water, the metal object displaces (9.25 – 8.20) g = 1.05 g of water whose volume is(1.05 g) / (1.00 g cm–3) = 1.05 cm3. The density of the metal is thus (9.25 g) / (1.05 cm3) = 8.81 g cm–3.The metal object displaces (9.25 - 8.36) g = 0.89 g of gasoline, whose density must therefore be (0.89 g) / (1.05 cm3)= 0.85 g cm–3.Problem Example 7A cube of ice that is 10 cm on each side floats in water. How many cm does the top of the cube extend above thewater level? (Density of ice = 0.917 g cm–3.)Solution: The volume of the ice is (10 cm)3 = 1000 cm3 and its mass is(1000 cm3) x (0.917 g cm–3) = 917 g. The ice is supported by an upward force equivalent to this mass of displacedwater whose volume is (917 g) / (1.00 g cm–3) = 917 cm3 . Since the cross section of the ice cube is 100-cm2, itmust sink by 9.17 cm in order to displace 917 cm3 of water. Thus the height of cube above the water is(10 cm – 9.17 cm) = 0.83 cm.... hence the expression, “the tip of the iceberg”, implying that 90% of its volume is hidden under the surface of thewater.Problem Example 8If the weight of the crown when measured in air was 4.876 kg and its weight in water was 4.575 kg, what was thedensity of the crown?Solution: The volume of the crown can be found from the mass of water it displaced, and thus from its buoyancy:(4876 – 4575) g / (1.00 g cm–3) = 301 cm3. The density is then(4876 g) / (301 cm3) = 16.2 g cm–3The densities of the pure metals: silver = 10.5, gold = 19.3 g cm–3,Problem Example 9Estimate the diameter of the neon atom from the following information:Density of liquid neon: 1.204 g cm–3; molar mass of neon: 20.18 g.
- 8. Solution: This problem can be divided into two steps. 1 - Estimate the volume occupied by each atom. One mole (6.02E23 atoms) of neon occupy a volume of (20.18 g) / (1.204 g cm–3) = 16.76 cm3. If this space is divided up equally into tiny boxes, each just large enough to contain one atom, then the volume allocated to each atom is given by: (16.76 cm3 mol–1) / (6.02E23 atom mol–1) = 2.78E–23 cm3 atom–1.2 - Find the length of each box, and thus the atomic diameter. Each atom of neon has a volume of about 2.8E–23 cm3. If we re-express this volume as 28E–24 cm3 and fudge the “28” a bit, we can come up with a reasonablygood approximationof the diameter of the neon atom without even using a calculator. Taking the volume as 27E–24 cm3 allows us tofind the cube root, 3.0E–8 cm = 3.0E–10 m = 300 pm, which corresponds to the length of the box and thus to thediameter of the atom it encloses.The accepted [van der Waals] atomic radius of neon is 154 pm, corresponding to a diameter of about 310 pm. Thisestimate is suprisingly good, since the atoms of a liquid are not really confined to orderly little boxes in the liquid.SUMMARYUse these questions to evaluate how well you have achieved the goals of this chapter.The answers to these questions are given at the end of this summary with the number ofthe section where you can find related content material.Definitions1. Any object can be deformed by applying a force to the object. An object is elastic if________ after the force is removed. However, if the object has been deformedbeyond its________, it will remain in a deformed state.2. The stress on an object is defined as the ratio of the________ to the_________.3. The strain of an object is defined as the ratio of the________ to the_________.4. The generalized statement of Hookes law states that________ is proportionalto________ of the system.5. Youngs modulus is a constant with units of _________, which characterizes theresponse of a solid to _________ and which has a magnitude equal to ________.6. The bulk modulus of a material is a constant with units of ________, whichcharacterizes the response of a material to _________.7. The shear modulus, also called the________, is a constant with units of _________,
- 9. which characterize the response of a material to_________ and which has amagnitude equal to _________.8. A steel rod 2 m long and 0.5 cm2in area stretches 0.24 cm under a tension 12,000N.a. What is the stress of the rod?b. What is the strain of the rod?9. What is the compressibility of water?Elasticity Problems10. What is Youngs modulus for the steel rod in question 8 above?11. If the volume of an iron sphere is normally 100 cm3and the sphere is subjected toa uniform pressure of 108N/m2, what is its change in volume?Answers1. It returns to its original state, elastic limit (Section 13.3)2. Applied force, cross-sectional area (Section 13.2)3. Change in a spatial variable, original value of that variable (Section 13.2)4. Strain, stress (Section 13.3)5. N/m2, forces applied to change its length, FL/AΔL (Section 13.4)6. N/m2
- 10. , forces applied to change its volume without changing its shape,FV/AΔL (Section 13.5)7. Modulus of rigidity, N/m2force applied to change its shape without changingits volume, (F/A)φ (Section 13.6)8. a. 2.4 x 108N/m2b. 12 x 10-4(Section 13.4)9. 4.5 x 10-10m2/N (Section 13.5)10. 20 x 1010N/m2(Section 13.4)11. 6.3 x 10-2cm3(Section 13.5)Problems
- 11. 1. If the stress produced in stretching a wire is 5.00 x 106N/m2by an applied force of10.0 N, what is the cross-sectional area of the wire?2. If the strain for the wire above is 0.100 percent, what is the length of the wire that willhave an elongation of 1.00 mm?3. In an experiment one finds that a force of 160 N produces a stretch of 8.00 cm in agiven spring. What is the spring constant of the spring?4. What is Youngs modulus for the wire described in problems 1 and 2?5. A 5.00-cm cube of gelatin has its upper surface displaced 1.00 cm by a tangential force0.500 N. What is shear modulus of this substance?6. What is the potential energy stored in the spring in problem 3 when it is stretched 8cm?Answers1. 2.00 x 10-6m22. 1.00 m3. 2.00 x 103N/m4. 5.0 x 109N/m2
- 12. 5. 400 N/m26. 6.4 J

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