1998 - JAMB Physics Past Questions and Answers - page 2
11
In a wheel and axle mechanism the diameters of the wheel and axle are 40cm and 8cm respectively. Given that the machine is 80% efficient, what effort is required to lift a load of 100N?
A
20N
B
25N
C
50N
D
80N
correct option: b
Efficiency = \(\frac{\text{work done by hand}}{\text{work done by effort}}\)
80 = \(\frac{100{\times 8}}{E {\times} 40}\) x \(\frac{100}{1}\)
E = 25N
Users' Answers & Comments80 = \(\frac{100{\times 8}}{E {\times} 40}\) x \(\frac{100}{1}\)
E = 25N
12
The tendon in a man's leg is 0.01m long. If a force of 5N stretches the tendon by 2.0 x 10-5m, calculate the strain on the muscle.
A
5 x 106
B
5 x 102
C
2 x 10-3
D
2 x 10-7
correct option: c
Strain = \(\frac{extension}{\text{original length}}\)
\(\frac{2 {\times} 10^{-5}}{0.01}\)
= 2 x 10-3
Users' Answers & Comments\(\frac{2 {\times} 10^{-5}}{0.01}\)
= 2 x 10-3
13
Which of these statements are correct for the pressures in liquids? i. pressure in a liquid at a point acts equally in all directions. ii. pressure increases with depth. iii. pressure at a depth depends on the shape of the container. iv. pressure at the same depth in different liquids are proportional to the density at different liquids.
A
i, ii, iii only
B
i, ii, iv only
C
i, iii, iv only
D
ii, iii, iv only
correct option: b
Users' Answers & Comments14
The atmospheric pressure due to water is 1.3 x 106Nm-2. What is the total pressure at the bottom of an ocean 10m deep? [Density of water = 1000kgm-3, g = 10ms-2]
A
1.3 x 107Nm-2
B
1.4 x 106Nm-2
C
1.4 x 105Nm-2
D
1.0 x 10 5Nm-2
correct option: b
Pressure = heg = 1000 x 10 x 10 = 105Nm-2
Total pressure = (1.3 x 106) + (1 x 105)
= 1.4 x 106Nm-2
Users' Answers & CommentsTotal pressure = (1.3 x 106) + (1 x 105)
= 1.4 x 106Nm-2
15
A solid of weight 0.600N is totally immersed in oil and water respectively. If the upthrust of oil is 0.210N and the relative density of oil is 0.875. Find the upthrust in water.
A
0.600N
B
0.360N
C
0.240N
D
0.180N
correct option: c
Rd = \(\frac{\text{upthrust in oil}}{\text{upthrust in }H_2O}\)
thus; 0.875 = \(\frac{0.21}{\text{upthrust in }H_2O}\)
upthrust in H2O = 0.24N
Users' Answers & Commentsthus; 0.875 = \(\frac{0.21}{\text{upthrust in }H_2O}\)
upthrust in H2O = 0.24N
16
A platinum resistance thermometer records 3.0\(\Omega\) at 0oC and 8.0\(\Omega\) at 100oC. If it records 6.0\(\Omega\) in a certain environment, The temperature of the medium is
A
80oC
B
60oC
C
50oC
D
30oC
correct option: b
8\(\Omega\).....100oC
6\(\Omega\).....\(\theta\)oC
3\(\Omega\).....0oC
\(\frac{6 - 3}{8 - 3}\) = \(\frac{\theta - 0}{100 - 0}\)
therefore, \(\theta\) = 60oC
Users' Answers & Comments6\(\Omega\).....\(\theta\)oC
3\(\Omega\).....0oC
\(\frac{6 - 3}{8 - 3}\) = \(\frac{\theta - 0}{100 - 0}\)
therefore, \(\theta\) = 60oC
17
The linear expansivity of brass is 2 x 10-5C-1. If the volume of a piece of brass is 15.00cm3 at 00C, What is the volume at 100oC?
A
16.03cm3
B
16.00cm3
C
15.09cm3
D
15cm3
correct option: c
linear expansivity = \(\frac {V_2 - V_1}{V_1 {\times} {\Delta}{\theta}}\)
3 x 2 x 10-5 =\(\frac {V_2 - 15}{15 {\times} (100 - 0)}\)
V2 = 15.09cm3
Users' Answers & Comments3 x 2 x 10-5 =\(\frac {V_2 - 15}{15 {\times} (100 - 0)}\)
V2 = 15.09cm3
18
A gas has a volume of 100cm3 at 27oC. If it is heated to temperature T until a final volume of 120cm3 is attained, calculate T
A
33oC
B
60oC
C
87oC
D
114oC
correct option: c
\(\frac {V_1}{T_1}\) = \(\frac {V_2}{T_2}\)
\(\frac {100}{27 + 273}\) = \(\frac {120}{T + 273}\)
T = 87oC
Users' Answers & Comments\(\frac {100}{27 + 273}\) = \(\frac {120}{T + 273}\)
T = 87oC
19
A 500W heater is used to heat 0.6kg of water from 25oC to 100oC in t2 seconds. If another 1000W heater is used to heat 0.2kg of water from 10oC to 100oC in t, seconds, find \(\frac {t_1}{t_2}\)
A
50
B
5
C
5/4
D
1/5
correct option: b
Heat energy = Electrical energy
mc\(\theta\) = power x time
0.6 x c x (100 - 25) = 500 x t1
and, 0.2 x c x (100 - 10)= 100 x t2
\(\frac {t_1}{t_2}\) = 5
Users' Answers & Commentsmc\(\theta\) = power x time
0.6 x c x (100 - 25) = 500 x t1
and, 0.2 x c x (100 - 10)= 100 x t2
\(\frac {t_1}{t_2}\) = 5
20
How many grams of water at 17oC must be added to 42g of ice at 0oC to melt the ice completely? [specific latent heat of fusion of ice = 3.4 x 105Jkg-1, specific heat capacity of water = 4200Jkg-1k-1]
A
200g
B
300g
C
320g
D
400g
correct option: a
Heat lost = Heat gained
mc\(\theta\) = mL
m x 4200 x 17 = 42 x 3.4 x 105
therefore, m = 200g
Users' Answers & Commentsmc\(\theta\) = mL
m x 4200 x 17 = 42 x 3.4 x 105
therefore, m = 200g