Physics 2C Summer Session I - 2009
Quiz #1
Multiple Choice. Choose the answer that best completes the
statement or answers the question.
1. Assuming that one atmosphere of pressure is equal to 105 P a; the
density of sea water is
= 1:03gm=cm3 ; and the acceleration of gravity is
2
g = 9:8m= sec ; calculate the depth of water that is required to produce an
additional one atmosphere of pressure.
A) 9:91m
B) 10:2 103 m
C) 10:2m
The change in pressure is given by
P
=
h
=
D) 9:91
103 m
E) 9:8m.
gh = 105 = 1030 9:8h
105
= 9:907m ! A
1030 9:8
2. A tank is …lled with water, = 1gm=cm3 ; to a depth of 2m. At
the bottom there is a hatch 1m high and 2m wide. Assuming an atmospheric
pressure of P0 = 105 P a and the acceleration due to gravity of g = 10m=s2 ;
determine the net force on the hatch. Hint, the atmospheric pressure acts on
both sides of the hatch.
Problem 2
A) 2 104 N
B) 40N
C) 3 104 N
D) 4 104 N
E) 30N
Since the atmospheric pressure cancels, the net force on the hatch is due to
the pressure of the water, which is given by
P = g (1) + gh;
1
where h is measured from the top of the hatch. The total force is
Z
Z 1
F =
P dA =
( g (1) + gh) W dh
0
F
=
F
=
3
1
gW + gW =
gW
2
2
3
1000 10 2 = 3 104 N ! C
2
3. A rectangular glass beaker measures 10cm 5cm 5cm: It ‡oats in
water with 4cm out of its 10cm height submerged. How many 10gm rocks can
you add to the beaker before it sinks?
A) 15
B) 25
C) 50
D) 150
E) 50
The volume of the beaker above waterline is V = 6cm 5cm 5cm = 150cm3 :
The mass of water for this volume is w V = 150gm: Hence you can add 15 of
the 10gm rocks to the beaker at which point the beaker is barely ‡oating! A:
4. At what temperature are Celsius and Fahrenheit equal?
A) 72 C
B) 40 C
C) 80 C
D) 40 C
E) 80 C
The relation between Celsius and Fahrenheit temperatures is TC =
When TC = TF = T we …nd
9T = 5T
160 ! T =
40 C or T =
5
9
(TF
32) :
40 F ! B
5. A tank of height of h = 10m is full of water, see …gure. At what
height of the water level, y, should a small hole be cut in the tank so that
initially the water goes as far horizontally as it does vertically?
Problem 5
A) 2m
B) 4m
C) 5m
D) 6m
E) 8m
From Bernoulli’s equation the horizontal velocity of the water as it exits the
tank is
p
v 2 = 2g (10 y) ! v = 2g (10 y):
2
p
The time is takes the water to fall to ground level is t = 2y=g: During this
time the water travels a horizontal distance x = vt: From the condition x = y
we …nd
p
p
p
2g (10 y) 2y=g = 2 y (10 y)
y =
y 2 = 4 10y y 2 ! 5y 2 40y = 0
y = 0m; 8m ! E
Note that we ignored the uninteresting solution y = 0:
6. A 500gm piece of Cu-Ag alloy weighs 4:403N when submerged
in water. Given the densities wat = 1gm=cm3 ; Cu = 8:9gm=cm3 , Ag =
10:5gm=cm3 ; the percentage of Cu (by volume) in the alloy is (keep 4 signi…cant …gures during your calculation)
A) 30%
B) 40%
W
= mg
=
=
C) 50%
w gV
D) 60%
!V =
mg
m
= w
=
V
mg W
1
9:8592
mg
E) 70%
W
wg
w
W=mg
=
1
1
4:403=4:9
Since
=
Cu
+ (1
Ag
=
Cu
Ag
) Ag !
Ag =
Cu
9:8592 10:5
=
= :4 ! B
8:9 10:5
Ag
7. A constant volume thermometer is …lled with Ar gas whose pressure
is 1000P a at the triple point of water (:01 C)?. What will be the pressure at
the boiling point of water, 100 C?
A) 1316P a
B) 1136P a
C) 31666P a
D) 1361P a
E) 1366P a
The temperature at the triple point of water is T = 273:16K: Since P=T =
P3 =T3 we …nd
1000
P =
373:16 = 1366P a ! E
273:16
8. A :05kg of metal is heated to T = 200 C and then dropped into
a calorimeter containing .4kg of water initially at T = 20 C: The …nal temperature of the system is T = 22:4 C:Given that the speci…c heat of water is
c = 4:2kJ=kg K; what is the speci…c heat of the metal (in units J=kg K)?
A) 1132
B) .817
C) 454
D) 367
3
E) .454
At equilibrium the heat transfered from the metal is equal to the heat absorbed by the water. The expression for this heat transfer is
mmet cmet (200
:05 cmet
22:4) = mw cw (22:4 20)
177:6 = :4 4200 2:4
cmet = 454J=kg K ! C
9. At low temperatures the speci…c heat of a solid is approximately
proportional to its temperature cubed, T 3 : For this case the heat capacity is
found by integrating the expression dQ=dT = mc: For copper at low tempera3
tures c = 31 (T =343) J=gm K where T is the temperature in K. Find the heat
required to heat 40gm of copper from 10K to 25K.
A) 7:31J
B) :150J
C) 11:7J
Integrating dQ=dT = mc results in
Q = 40
Z
25
31
10
D) 2:92J
40 31 1
T3
dT =
254
3
343
3433 4
E) :094J
104 = 2:92J ! D:
10. One end of an iron rod (thermal conductivity k = 80:4W= (m K)),
which is 40cm long and 3cm in diameter is immersed in ice water. The other
end is immersed in boiling water. The rod is well insulated so that there is no
heat loss out the sides. What is the heat ‡ow, H, along the rod?
A) 5:68kW
B) 56:8W
C) 1:42kW
Using the expression for heat ‡ow we …nd
H=
kA
T
=
x
80:4
2
4
(:03)
D) 14:2W
E) 4:52W
100
= 14:2W ! D:
:4
Possible equations of interest
9
1
TC + 32; P + v 2 + gh = const:; FB =
5
2
dQ
T
Q = mc T or
= mc; H = kA
dT
x
P = P0 + gh; TF =
P=T = P3 =T3 :;
4
w gVdisp