2001 - JAMB Physics Past Questions and Answers - page 3
21
The cost of running five 60 W lamps and four 100 W lamps for 20 hours if electrical energy cost #10.00 per kwh is
A
280.00
B
160.00
C
120.00
D
140.00
correct option: d
Total power in watt = 60*5 + 4*100
= 300 + 400
= 700w
Total power in kilo watt = 700/100
= 0.7 killo w
Unit of electricity in killo watt-hour
= 0.7 * 20hrs
= 14 okwh
i.e If I kwh costs # 10.00
14kwh = 10.00 * 14 = #140
Users' Answers & Comments= 300 + 400
= 700w
Total power in kilo watt = 700/100
= 0.7 killo w
Unit of electricity in killo watt-hour
= 0.7 * 20hrs
= 14 okwh
i.e If I kwh costs # 10.00
14kwh = 10.00 * 14 = #140
22
In a Daniel cell, the depolarizer, positive and negative electrodes are respectively
A
copper sulphate, copper and zinc
B
manganese dioxide, carbon and zinc
C
sulphuric acid, lead oxide and lead
D
potassium hydroxide, nickel and iron.
correct option: a
In the Daniel cell,
i. Depolarize = CuSO4 solution
ii. Positive electrode = Copper container
iii Negative electrode = zinc rod/plate
Users' Answers & Commentsi. Depolarize = CuSO4 solution
ii. Positive electrode = Copper container
iii Negative electrode = zinc rod/plate
23
when a piece of rectangular glass block is inserted between two parallel capacitor,at constant plate area and distance of separation, the capacitance of the capacitor will
A
increase
B
decrease
C
decrease then increase
D
remain constant
correct option: a
Users' Answers & Comments24
the ratio of electrostatic force FE to gravitational force FG between two protons each of charge e and mass m, at a distance d is
A
e/4πԐoGm
B
e2/Gm2
C
Gm2/4πԐoe2
D
e2/4πԐoeGm2
correct option: d
Electrostatic force = FE = e2/4λEod2
Gravitational force = FG = GM2/d2
Users' Answers & CommentsGravitational force = FG = GM2/d2
25
A cell can supply current of 0.4 A and 0.2 A through a 4.0Ω and 10.0Ω resistors respectively.
This internal resistance of the cell is
This internal resistance of the cell is
A
2.0Ω
B
1.0Ω
C
2.5Ω
D
1.5Ω
correct option: a
let the internal resistance of the cell = r,
therefore E = I (R+r)
(i) E = 0.4(4+r), (ii) E = 0.2 (10+r)
therefore 0.4 (4+r)= 0.2 (10+r)
1.6 + 0.4r = 2.0 * 0.2r
0.4r - 0.2r = 2.0 - 1.6
therefore 0.2r = 0.4 => r = 0.4/0.2 = 2Ω
therefore r = 2Ω
Users' Answers & Commentstherefore E = I (R+r)
(i) E = 0.4(4+r), (ii) E = 0.2 (10+r)
therefore 0.4 (4+r)= 0.2 (10+r)
1.6 + 0.4r = 2.0 * 0.2r
0.4r - 0.2r = 2.0 - 1.6
therefore 0.2r = 0.4 => r = 0.4/0.2 = 2Ω
therefore r = 2Ω
26
The particles emitted when \( ^{39}_{19}K \text{ decay to } ^{39}_{19}K \) is
A
gamma
B
beta
C
electron
D
alpha
correct option: a
3919K 3919K.
Since neither the mass number (39), nor the atomic number (19) is affected by the emission,the particles emitted can only be gamma ray, which is not a charge particle, but simply a radiation.
Users' Answers & CommentsSince neither the mass number (39), nor the atomic number (19) is affected by the emission,the particles emitted can only be gamma ray, which is not a charge particle, but simply a radiation.
27
I. Low pressure.
II. High pressure.
III. High p.d.
IV. Low p.d.
Which combination of the above is true of the conduction of electricity through gases?
II. High pressure.
III. High p.d.
IV. Low p.d.
Which combination of the above is true of the conduction of electricity through gases?
A
I and IV only
B
I and III only
C
II and IV only
D
II and III only
correct option: b
Electricity conduction through gases is usually done under low pressure (I) and high voltage (III). Thus /B/
Users' Answers & Comments28
The current through a resistor in an a.c circuit is given as 2 sin wt. Determine the d.c. equivalent of the current
A
A/√2
B
2/√A
C
2 A
D
√2 A
correct option: b
If I = 2 Sin wt; but in a.c. circuit current I = I0 Sin wt; thus comparing the two equations, it implies that I0 = 2A; Again, the root means square (I(rms)) of an a.c. current is the value of the d.c. current that will dissipate the same amount of heat in a given resistance as the a.c.
and I(rms) = I0/√2 = 2.0A/√2
Users' Answers & Commentsand I(rms) = I0/√2 = 2.0A/√2
29
The primary coil of a transformer has N turns and is connected to a 120 V a.c. power line. If the secondary coils has 1 000 turns and a terminal voltage of 1 200 volts, what is the value of N?
A
120
B
100
C
1 000
D
1 200
correct option: b
For the transformer, V ∝ ∩
=> V s ∝ ∩s; Vp ∝ ʌp
=> Vp/Vv = ∩p/∩s, therefore 120/1200 = ∩p/1000
therefore ∩p * 1200 = 120 * 1000
∩p = 120*1000/1200 = 100
Users' Answers & Comments=> V s ∝ ∩s; Vp ∝ ʌp
=> Vp/Vv = ∩p/∩s, therefore 120/1200 = ∩p/1000
therefore ∩p * 1200 = 120 * 1000
∩p = 120*1000/1200 = 100
30
Which of the following metals will provide the greatest shield against ionizing radiation
A
Iron
B
Manganese
C
Alminium
D
Lead
correct option: d
Users' Answers & Comments