Water from the capillary tube of a diameter of ½f to a height above surface level, if the density of water is g and p is the acceleration due to gravity, then the surface tension is
a) δ = ρglf/2 b) δ = ρglf/4 cosθ c) δ = ρglf/4 b) δ = ρglf/g
b. δ = ρglf/4 cosθ
How many molecule are in a 1.00kg bottle of water? The molar mass of the water is 18.0 g/mol
a) 6.02 × 1023 b) 1.08 × 1023 c) 3.35 × 1025 d) 3.35 × 1023
Molar mass of H2O = 1 × 2 + 16 = 18 g/mol (H=1, O=16)
No of moles = mass/molar mass
If 1 mole of H2O contains 6.022 × 1023 molecules
0.0556 or 1/18 moles will contain x molecules
Therefore x = 6.022 × 1023 × 0.0556 = 3.348×1022 molecules
The following are similarities between coulombs law and gravitational law except
[i] Both equation has an inverse square relationship.
[ii] Both equation has the same proportionality constant value
[iii] Both equation show that force is proportional to the product the quantity causes the force.
[iv] Both the attractive and repulsive forces
a) i only b) i and iii only c) ii and iv only d) all of the above
c.
Buoyant force is a property of
a) pressure law b) Archimedes principles c) Hook’s law d) None of the above
b.
Heat is added into a 500.0g Solid sample at the rate of 10.0 Kj/min, while recording its temperature as a function of time, you plot your data and obtain the graph show in fig 1. What is the latent heat of the fusion for this solid?
Vertical axis is temperature in degree Celsius and the horizontal axis is time in mins
a) 50 Kj/kg b) 25 Kj/kg c) 15 Kj/kg d) 30 Kj/kg
d.
Q~ = mlf also Q~ = Q/t therefore Q = Q~ × t where t from the graph it
t = t2 – t1 = 2.5 -1 = 1.5 mins
lf = Q~/m = (Q~ × t)/m = (10 × 103 × 1.5)/ 0.5
lf = 30Kj/Kg
Metal rod that is 4 m long and 0.5 cm2 in cross sectional area is found to stretch 0.20 cm under a tension of 5000N. The strain in the material is;
a) 5 × 10-5 b) 5 × 10-2 c) 5 × 10-7 d) 5 × 10-4
c.
ε = Δl/l = 0.2cm/4000cm = 5 × 10-7
Palm oil flow slugglishly from a bottle, but if heated, it flow more freely because
a) Surface tension increases b) Adhesion force decreases. c) Viscosity increases d) Viscosity decreases.
c.
A 500.0 g chunk of an unknown metal, which has been in boiling water for several minutes, is quickly dropped into an insulating Styrofoam beaker containing 1.00 kg of water at room temperature (20oC), after waiting and gently stirring for 5.00 minutes, you observe that the water temperature has reached constant value of 22.0o Assuming that the Styrofoam absorb a negligibly small amount of heat and thus no heat was lost to the surrounding, what is the specific heat of the metal
a) 16.80 Kj/kg b) 0.1 Kj/kg c) 0.22 Kj/kg d) 8.40 Kj/kg
c.
The heat lost by the chunk metal is equal to the heat gain by the water. Therefore
Qm = – Qw
mm cm ΔT = – mw cw ΔT
cm = – (mw cw ΔT)/( mm ΔT) = – [1 × 4200 × (220 -20)]/[0.5 × (22 -100)]
= – 8400/-39 =215.38j/kgK = 0.22Kj/kgK
Shear strain, bulk strain and normal strain have unit of;
a) m b) m2 c) dimensionless d) N
c.
What is the pressure (gauge) required to raise or water supply to raise a water supply system by 1,500.cm
a)110.3 mmHg b) 88.3KN/m2 c) 1.45 atm d) None of the above
c.
Pguage = ρgh = 1000 × 9.81 × 1500 × 10-2 = 1.4715 × 105 N/m2
1 atm = 101325 N/m2 therefore 1.4715 × 105 is equal to 1.45 atm
An ice cube tray of negligible mass contains 0.350kg of water at 18.00 How much heat must be removed to cool the water at 0oC and freeze it.
a)137.7 Kj b) 26.5 Kj/kg c) 117.3 Kj/kg d) 410.6 Kj/kg
b.
Q = m cm ΔT = 0.35 × 4200 × (0-18) = -26.46Kj heat is loss
The following are properties of liquid except.
a) Definite shape b) definite volume c)it cannot be compressed d)it takes the shape of the containing vessel.
a.
The radius of water droplet is 2cm what is the excess pressure within the droplet if the surface tension is T N/cm.
a) T N/cm2 b) 2T N/cm2 c) 4T N/cm2 d) 8T N/cm2
a.
You propose a new temperature scale with temperatures given in oM, You define 0.0oM to be the normal melting point of mercury and 100.0oM to be the normal boiling point of a mercury. If the melting point and boiling point of mercury -39oC and 357oC respectively, what is the normal boiling point of water in o
a) 35.1 oM b) 100 oM c) 38.9 oM d) 43.7 oM
a.
0o M = -39o C
100o M = 357o C
(357 – 100)/(100 – x) = (357 + 39)/(100 – 0)
25700 = 396(100 – x)
396x = 39600 – 25700
x = 35.1o M
A tank of uniform cross-sectional area of 1.5 m2 is filled with water to the height of 3.2 m. what is the absolute pressure at the base of the tank if the atmospheric pressure 1 atm?
a) 1.50 × 105 N/m2 b) 1.33 × 105 N/m2 c) 11.0 × 105 N/m2 d) 1.24 × 105 N/m2
b.
Pabsolute = Patm + Pguage(ρgh)
= 101325 +1000 × 9.81 × 3.2 = 1.33 × 105 N/m2
A 20.0 L tank contains 0.225 kg of helium at 18.0o the molar mass of helium is 4.00 g/mol. What is the pressure in the tank, in pascal?
a) 6.8 × 106 Pa b) 4.2 × 105 Pa c) 2.7 × 104 Pa d) 6.8 × 103 Pa
d.
Pv =nRT = mass(m)/ molar mass(M)
ρ = m/v = 0.225/20 × 10-3 = 11.25 kg/m3
Pv =nRT = mRT/vM = ρRT/M
=(11.25 × 8.314 × 291)/4 = 6.8 × 103 Pa
Two liters of water when subjected to a pressure of 10 atm are compressed by 1.013 × 10-6. The compressibility of water is:
a) 5 × 10-10 m2/N b) 2 × 109 N/m2 c) 9 × 10-6 d) 1 × 1020 N
a.
Β = δv/Vδp = 1.013× 10-6 /(2× 10-3 × 10 × 101325) = 5 × 10-10 m2/N
For a fluid flowing through a pipe, if the Reynold’s number is <2000, then the flow is
a) Steady b) Laminar c) Streamline d) Turbulent
b.
Helium gas is with a volume of 2.60 L under a pressure of 1.30 atm and at a temperature 41.0oC, is warmed until both pressure and volume are doubled. What is the final temperature?
a) 437oC b) 437 K c) 164oC d) 983oC
d.
P1V1/T1 = P2V2/T2
T2 = P2V2 T1/P1V1 = 2P1 × 2V1 × 314/P1V1
4 × 314 = 1256 K
ToC = 1256 – 273 = 983oC
The force of attraction between molecules of the same substance is called?
a) Combination b) Fusion c) Cohesion d) Adhesion.
c.
An apple losses 4.5 Kj of heat as it cools per oC drop in its temperature. The amount of heat loss from the apple is per oF drop in its temperature is;
a) 1.25 KJ b) 2.50 KJ c) 5.0 KJ d) 8.1 KJ
b.
T(oF) = 1.8T + 32
ΔQ/ΔT = 4.5Kj/oC
= 4.5/1.8 = 2.50Kj
Two balloons with charges of +6.74μC and -16.42μC attract each other with a force of 0.1252N. the separation distance between the two balloons is; k= 9 × 109 Nm2/C3
a) 1.99m b) 0.56 c) 2.82m d) 2.01m
c.
F = KQ1Q2/r2
r2 = KQ1Q2/F = (9 × 109 × 6.74 × 10-6 × 16.42 × 10-6)/0.1252
r = 2.82m
In a capillary tube of diameter 0.3mm, water at 4oC rises 2.6 cm above the surface level. What will be the surface tension?
a) 26.87 × 10-3 N/m b) 6.54 × 10-3 N/m c) 65.43 × 10-3 N/m d) 19.1 × 10-3 N/m
d.
T = ρgdh/4 cosθ = (1000 × 9.81 × 0.3 × 10-3× 2.6 × 10-2)/ 4 cos 0
= 19.13 × 10-3 N/m
At which temperature will celsius and fereinheit thermometers give the same reading?
a) -40 b) 25.6 c) -72 d) 40
a.
Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.75 m long and 1.50 cm in diameter. The combination is subjected to a tensile force with magnitude 4000N. Young modulus of steel and copper are 20 × 1010 N/m2 and 11 × 1010 N/m2 respectively
The total extension of the rod is
a) 2.39 × 10-4 m b) 23.76 × 10-4 m c) 10-4 m d) 2.39 × 10-5 m
a.
young modulus (Y) = stress(σ)/strain(ε)
σ = Force(F)/Area(A)
ε = extension(e)/length (l)
σsteel =σcu = F/A = 4000 [π(1.5 × 10-2)/4] = 2.2635 × 107 N/m2
εsteel = σ/Ys = 2.2635 × 107 /20 × 1010 = 1.132 ×10-4
εcu = σ/Ys = 2.2635 × 107 /11 × 1010 = 2.058 ×10-4
esteel = εsteel × l =1.132 ×10-4 × 0.75 = 8.49 ×10-5 m
ecu = εcu × l =2.058 ×10-4× 0.75 = 1.543 ×10-4 m
eT = esteel + ecu =8.49 ×10-5 + 1.543 ×10-5 = 2.389 ×10-4 m
The total final length of the rod is;
a) 0.8901 m b) 0.7502 m c) 0.2345 d) 0.6374 m
b.
lT = original length + total extension
lT = 0.75 + 2.389 × 10-4 = 0.75024m
A steel bridge has a length of 1410 m. Calculate the change in length of the steel deck when the temperature increases from -5.0oC to 18.0oC, if the coefficient of the linear expansion of steel is 1.2 × 10-5 K-1
a) 0.39m b) 0.22m c) 0.93 m d) 0.0039 m
A steel bridge has a length of 1410 m. Calculate the change in length of the steel deck when the temperature increases from -5.0oC to 18.0oC, if the coefficient of the linear expansion of steel is 1.2 × 10-5 K-1
a) 0.39m b) 0.22m c) 0.93 m d) 0.0039 m
a.
Δl/l = αl ΔT T2 = 273 + 18 = 291K T1 = 237 + (-5) = 268K
Δl = αl ΔT l = 1.2 × 10-5 × 1410 × (291 -268) = 0.3892m
Water flows through a pipe whose cross-sectional area 35 cm2 into another pipe with cross-sectional area of 20 cm2. Both pipes are horizontal, if the velocity of larger pipe is 3.6 m/s, then the velocity of small pipe is
a) 4.21 m/s b) 6.3 m/s c) 2.46 m/s d) 6.83m/s
b.
A1V1 = A2V2
V2 = A1V1 / A2
(35 × 3.6)/20 = 6.3 m/s
In an effort to stay awake for an all night study session, a student makes a cup of coffee by first placing a 200 W electric immersion heater in 0.320 kg of water. How much time is required to raise it’s temperature from 20.0oC to 80oC assuming that all the heaters power goes into heating the water.
a) 7.6 mins b) 11.2 mins c) 6.7 mins d) 21.1 mins
c.
P = Q/t t = Q/P = mcm ΔT/P = [0.32 × 4200 × (353 – 343)]/200 = 403.2 sec
60 sec = 1 min
403.2 sec = x min
Therefore x = 403.2/60 = 6.72 mins
For a streamline motion of liquid, the equation of continuity is given by
a) A1A2=ρ1ρ2 b) A1A2=v1v2 c) A1ρ1= A2ρ2 d) A1v2= A2v2
d.
You are given a sample of metal and ask to determine its specific heat. You weight the sample and find that its weight is 28.4 N. You carefully added 1.25×104 J of heat energy to the sample’s and find that its temperature rises 18.0C°. What is the sample’s specific heat?
a) 240 j/kgK b) 4318 J/kgK c) 15 J/kgK d) 240 J/kgK
a.
Q = mcm ΔT F = mg m = f/g
Therefore c = Q/[ (f/g) × ΔT]
= 1.25 × 104 /[(28.4/9.81) × 18] = 239.88 j/kgK
What pressure should be applied to a lead block to reduce its volume by 10%. Compressibility for lead 1.667 × 10-10m2/N
a) 30 × 109N/m2 b) 6 × 106m2/N c) 90×107N/m2 d) 60×107 N/m2
d.
β = δv/Vδp
δp = δv/βv = 0.1/ (1.667 × 10-10 × 1) =59.988 × 107 N/m
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