# Chapter 7 - Worked Solutions

### to selected Complex Reasoning questions from the OUP text New Century Senior Physics - Concepts in Context  by Walding, Rapkins and Rossiter

Question 31
(a) SG = SG(liquid) x fraction submerged
\ fraction submerged = 0.58
(b) Rises (receives some upthrust from the water).

Question 32
(a) P = r gh = 1025 x 10 x 100 = 1.025 MPa
F = PA = 1.025 x 106 x 1.2 x 0.6 = 7.38 x 105 N
(b) m = F/g = 7.38 x 104 kg = 74 tonnes

Question 33
NOTE: the diagram is missing from the textbook - it should look like this: (a) container (ii); (b) container (ii); (c) container (ii); (d) all the same (pressure depends on the depth of the liquid, not the size of the base).

Question 34
Firstly, calculate the time to strike ground ( s = ut + ½ at2) and then calculate the horizontal velocity using vh = Ö 2gh, then calculate the horizontal range of the liquid in that time using sh = vht: eg for hole at a height of 30 cm (0.30 m):
0.30 = 0 + ½ x 10 x t2
t = 0.245 s
vh = Ö (2gh) = Ö (2 x 10 x 0.30) = 1.41 ms-1
sh = vht = 1.41 x 0.245 = 0.346 m
Full set of results:

 Height (m) t (s) vh = Ö 2gh sh = vht 0.30 0.245 1.41 0.346 0.20 0.200 2.00 0.400 0.10 0.141 2.45 0.345

The 30 cm and 10 cm holes have spouts that land in the same spot (0.346 m)

Question 35
By Archimedes' Principle, the upthrust equals the weight of water displaced. The weight of the ice cubes = weight of the water displaced which is the volume of water below the water line. When the ice melts it will have a volume that fits into the space occupied by the ice that was underwater: Question 36
Hydrogen balloon:
Upthrust = V r = 400000 x 1.3/1000 x 10 = 5200 N
Weight of balloon = 200 x 10 = 2000 N
Weight of gas = V r = 400000 x 0.09/1000 x 10 = 360 N
Result: Nett Force = 5200 - 2000 - 360 = 2840 N

Helium balloon:
Upthrust = V r = 400000 x 1.3/1000 x 10 = 5200 N
Weight of balloon = 200 x 10 = 2000 N
Weight of gas = V r = 400000 x 0.18/1000 x 10 = 720 N
Result: Nett Force = 5200 - 2000 - 720 = 2480 N

Conclusion: Hydrogen balloon lifts only 1.14 as much as the helium balloon.

Question 37
Equal. The weight of the wood equals the upthrust which equals the weight of water displaced. The displaced water flows over the side of the bucket so the weight of the bucket is the same as before.

Question 38
Nothing. The bucket and the water and the cork all fall at the same rate.

Question 39
P = 1187 kgf /cm2 = 11870 N/cm2 = 1.187 x 108 N/m2 (= Pa).
P = r gh
r = P/gh = 1.187 x 108 / (10 x 10916) = 1087 kg m-3

Question 40
P = F/A = mg/A = r Vg/A = r hAg/A = r hg (the A cancels out)
P = 1.06 x 103 x 1.83 x 10 = 19400 Pa (19.4 kPa).

Question 41
P = 1/20 x 101.3 kPa = 5065 Pa
P = r gh
h = P/r g = 5065/(1000 x 10) = 0.5 m

Question 42
SG = SG(fluid) x fraction submerged
0.98 = SG x 2/3
SG = 1.47 g cm-3

Question 43
Water level is still higher The weight of water still has to equal the weight of mercury above the other level of mercury:
Pwater = Pmercury
r w g hw = r m g hm (cancel the g)
1 x hw = hm x 13.6
The water column is 13.6 times the height of the mercury.

Question 44

For both balloons: Upthrust = weight of air displaced;
Weight of air = mass of air x g
Mass of air = density of air x volume of balloon = DairV
Weight of air (upthrust) = DairV x g

Weight of balloon and contents = (density of gas inside x volume x g) + (mass of balloon structure x g)
Weight (hydrogen balloon) = DHV g + mg
Weight (helium balloon) = DHeV g + mg
Lift = Upthrust - weight
Lift (hydrogen) = DHV g + mg - DairV x g
Lift (helium) = DHeV g + mg - DairV x g = 2DHV g + mg - DairV x g
Ratio: difference in lift/lift of helium
= ((-mg -DHVg + DairVg - (-mg -2DHVg + DairVg))/(-mg -2DHVg + DairVg)
= (-DHVg +2 DHVg)/(-mg - 2DHVg = DairVg)
= (DHV)/(-m - 2DH + Dair) x 100%
8% = (0.089 x 100)/(-m - 0.178 +1.20)%
8 = 8.9/(-m + 1.022)
8m = 0.724
m = 0.09 kg
NOTE: answer in back of book says 2.13 kg but this is wrong. My mistake.

Question 45