2006 - JAMB Physics Past Questions and Answers - page 5
41
A 40 kW electric cable is used to transmit electricity through a resistor of resistance 2.0Ω at 800 V. The power loss as internal energy is?
A
4.0 x 102W
B
5.0 x 102W
C
4.0 x 103W
D
5.0 x 103W
correct option: d
In general, power = IV; 40kw = IV.
∴ 40,000 = 1 x 800
1 = (40.000)/800 = 50A. tune the current through resistor = 50A.
∴ power loss = 12R = 502 X 2 = 2500 X 2 = 5.0 X 103W
Users' Answers & Comments∴ 40,000 = 1 x 800
1 = (40.000)/800 = 50A. tune the current through resistor = 50A.
∴ power loss = 12R = 502 X 2 = 2500 X 2 = 5.0 X 103W
42
The instrument that measures both a.c and d.c is?
A
a current balance
B
a moving current ammeter
C
a moving iron ammeter
D
an inverter
correct option: c
Users' Answers & Comments43
A 120 V, 60 W lamp is to be operated on 220 V a.c. the supply mains. calculate the values of non-inductive resistance that would be required to ensure that the lamp is run on current value.
A
100 Ω
B
200 Ω
C
300 Ω
D
500 Ω
correct option: b
Power of the lamp = 60W = (V2)/R and the voltage rate = 120v.
R = (V2)/60 = (120 X 120)/60 = 240 Ω, the fuse wire that should ensure current does not exceed this value should be rated 200 Ω
Users' Answers & CommentsR = (V2)/60 = (120 X 120)/60 = 240 Ω, the fuse wire that should ensure current does not exceed this value should be rated 200 Ω
44
A radioactive substance has a half-life of 20 days what fraction the original radiation nuclei will remain after 80 days?
A
(1)/32
B
(1)/16
C
(1)/ 8
D
(1)/4
correct option: b
80 days = 4 half lives (i.e 4 x 2odqs = 80 days)
But the fraction always left by any radioactive material after 4half lives is (1/2)4 = (1)/16
Users' Answers & CommentsBut the fraction always left by any radioactive material after 4half lives is (1/2)4 = (1)/16
45
A
Breaking point
B
Yield point
C
Elastic limit
D
Proportional limit
correct option: a
In the stress - strain graph shown above, X represents the breaking point of the wire
Users' Answers & Comments46
A
86 cm of mercury
B
66 cm of mercury
C
96 cm of mercury
D
76 cm mercury
47
A mirror of weight 75N is hung by a cord from a hook on the wall. If the cord makes an angle of 30o with horizontal, the tension in the cord is
A
100N
B
1500N
C
250N
D
150N
correct option: d
For vertical equilibrium, we have
2Ty = 75N
But Ty/T = adj/hyp = cos 60o
∴ Ty = T Cos 60o
=> 2Ty = 2Tcos 60o = 75N
2 x T x 0.5 = 75 => T = 75N
∴ Total tension in the rope = 2T.
= 2 x 75 = 150N
Users' Answers & Comments2Ty = 75N
But Ty/T = adj/hyp = cos 60o
∴ Ty = T Cos 60o
=> 2Ty = 2Tcos 60o = 75N
2 x T x 0.5 = 75 => T = 75N
∴ Total tension in the rope = 2T.
= 2 x 75 = 150N
48
In a gas experiment, if the volume of the gas is plotted against the reciprocal of the pressure, the unit of the slope of the resulting curve is
A
force
B
temperature
C
power
D
work
correct option: d
IF the volume of gas is plotted against the reciprocal of pressure, the slope will be given as
Slope = V/1/P = P x V
thus the unit of the slope will be;
= Nm-2, = V = M3 = NM.
The unit NM (Newton - Meter) is the practical unit of work, which is the product of force (N) and Displacement (M).
Note: It should be noted that the practical unit of work is Newton-Meter, whereas its S.I unit is Joules, just as energy.
Users' Answers & CommentsSlope = V/1/P = P x V
thus the unit of the slope will be;
| P = | Force |
| Area |
= Nm-2, = V = M3 = NM.
The unit NM (Newton - Meter) is the practical unit of work, which is the product of force (N) and Displacement (M).
Note: It should be noted that the practical unit of work is Newton-Meter, whereas its S.I unit is Joules, just as energy.
49
A shooter wants to fire a bullet in such a way that its horizontal range is equal to three times its maximum height. At the what angle should he fire the bullet to achieve this?
A
68o
B
53o
C
30o
D
45o
correct option: b
Horizontal range = 3 x maxi. height
=> (U2Sin2θ) / g = (3 x U2sin2θ) / 2g
∴ 4sinθcosθ = 3sin2θ
=> 4cosθ = 3sinθ
∴ sinθ/cosθ = 4/3; => tanθ = 4/3 = 1.3333
θ = tan-1 (1.3333)
= 53o
Users' Answers & Comments=> (U2Sin2θ) / g = (3 x U2sin2θ) / 2g
| ∴sin2θ = | 3sin2θ |
| 2 |
| => 2sinθcosθ = | 3sin2θ |
| 2 |
∴ 4sinθcosθ = 3sin2θ
=> 4cosθ = 3sinθ
∴ sinθ/cosθ = 4/3; => tanθ = 4/3 = 1.3333
θ = tan-1 (1.3333)
= 53o
50
An engine of a car of power 80 kW moves on a rough road with a velocity of 32 ms-1. The forces required to bring it to rest is
A
2.56 x 106N
B
2.50 x 106N
C
2.80 x 103N
D
2.50 x 103N
correct option: d
In terms of dimension: power , P, = ML2T3
Force F1 = MLT-2; Vel, V, LT-1
From the above, => P = F x V
i.e, ML2T-3 = MLT-2 x LT-1
ML2T-3
=> 80,000 = F x 32
= 2500N
= 2.50 x 103N
Users' Answers & CommentsForce F1 = MLT-2; Vel, V, LT-1
From the above, => P = F x V
i.e, ML2T-3 = MLT-2 x LT-1
ML2T-3
=> 80,000 = F x 32
| ∴ F = | 80,000 |
| 32 |
= 2500N
= 2.50 x 103N