1991 - JAMB Physics Past Questions and Answers - page 2
11
A pilot records the atmospheric pressure outside its plane as 63cm of Hg while a ground observer records a reading of 75cm of Hg for the atmospheric pressure on the ground. Assuming that the density of the atmosphere is constant, calculate the height of the plane above the ground. (Relative density of Hg = 13.0 and that of air = 0.00013)
A
1200m
B
6300m
C
7500m
D
13800m
correct option: a
Difference in pressure = 75 - 63
= 12cmHg
pressure = hge
difference in height = \(\frac{12}{10 \times \frac{0.00013}{13}}\)
= 1200m
Users' Answers & Comments= 12cmHg
pressure = hge
difference in height = \(\frac{12}{10 \times \frac{0.00013}{13}}\)
= 1200m
12
In which of the following is surface tension important?
A
the floating of a ship in water
B
the floating of a dry needle in water
C
the floating of a baloon in air
D
the diffusion of sugar solution across a membrane
correct option: b
Users' Answers & Comments13
A thermometer with an arbitrary scale S, of equal division registers -30oS at the ice point and +90oS at the steam point. Calculate the Celsius temperature corresponding to 60oS
A
25.0oC
B
50.0oC
C
66.7oC
D
75.0oC
correct option: d
100oC ... 90oS
\(\theta\)oC ... 60oS
0oC ... 30oS
\(\frac{\theta - 0}{100 - 0}\) = \(\frac{60 - (-30)}{90 - (-30)}\)
\(\theta\) = 75oC
Users' Answers & Comments\(\theta\)oC ... 60oS
0oC ... 30oS
\(\frac{\theta - 0}{100 - 0}\) = \(\frac{60 - (-30)}{90 - (-30)}\)
\(\theta\) = 75oC
14
A brass rod is 2m long at a certain temperature. What is the length for a temperature rise of 100K, if the expansivity of brass is 18 x 10-6K-1?
A
2.0036m
B
2.0018cm
C
2.1800
D
2.0360m
correct option: a
\(\alpha\) = \(\frac{L_2 - L_1}{L_1 \times \Delta \theta}\)
18 x 10-6 = \(\frac{L_2 - 2}{2\times 100}\)
L = 2.0036m
Users' Answers & Comments18 x 10-6 = \(\frac{L_2 - 2}{2\times 100}\)
L = 2.0036m
15
What is the difference in the amount of heat given out by 4kg of steam and 4kg of water when both are cooled from 100oC to 80oC? [the specific latent heat of steam is 2,260,000Jkg-1, specific heat capacity of water is 4200Jkg-1K-1]
A
4200J
B
2,260,OOOJ
C
9,040,000J
D
9,380,000J
correct option: c
Heat by water = mc\(\theta\)
Heat by steam = mc\(\theta\) + mL
Difference = mL = 4 x 2,260,000
= 9,040,000J
Users' Answers & CommentsHeat by steam = mc\(\theta\) + mL
Difference = mL = 4 x 2,260,000
= 9,040,000J
16
How long does it take a 750W heater to raise the temperature of 1kg of water from 20oC to 50oC [specific heat capacity of water = 4200Jkg-1K-1]
A
84s
B
112s
C
168s
D
280s
correct option: c
Heat energy = Elect. energy
mc\(\theta\) = power x time
1 x 4200 x (50 - 20) = 750 x t
t = 168s
Users' Answers & Commentsmc\(\theta\) = power x time
1 x 4200 x (50 - 20) = 750 x t
t = 168s
17
The saturated vapour pressure of a liquid increases as the
A
volume of the liquid increases
B
volume of the liquid decreases
C
temperature of the liquid increases
D
temperature of the liquid decreases
correct option: c
Users' Answers & Comments18
The absolute temperature of a perfect gas is proportional to the average
A
potential energy of the molecules
B
seperation between the molecules
C
kinetic energy of the molecules
D
velocity of the molecules
correct option: c
Users' Answers & Comments19
A room is heated by means of a charcoal fire, An occupant of the room standing away from the fire is warmed mainly by
A
convection
B
radiation
C
conduction
D
reflection
correct option: b
Users' Answers & Comments20
A boy timed 30 oscillations of a certain pendulum thrice and obtained 1min. 10s, 1min. 12s and 1min. 7s respectively.The mean period of oscillation of the pendulum is
A
0.14s
B
0.43s
C
2.32s
D
6.97s
correct option: c
Total time = (3 x 60) + 29
= 209sec
period \(\frac{t}{n}\) = \(\frac{209}{90}\)
= 2.32sec
Users' Answers & Comments= 209sec
period \(\frac{t}{n}\) = \(\frac{209}{90}\)
= 2.32sec