2007 - JAMB Physics Past Questions and Answers - page 5
41
The difference observed in solids, liquids and gases can be accounted for by
A
The different molecules in each of them
B
The spaces and forces acting between the molecules
C
Thier melting points
D
Their relative massess
correct option: b
Users' Answers & Comments42
If two charged plates are maintained at a potential difference of 3 kv, the work done in taking a charge of 600µc across the field is
A
0.8j
B
1.8j
C
9.0j
D
18.0j
correct option: b
Workdone = product of the charge Q, and the p.d V.
i.e Work done = QV
= 600 * 10-6 *2.1 * 103
=1.8j
Users' Answers & Commentsi.e Work done = QV
= 600 * 10-6 *2.1 * 103
=1.8j
43
A conductor has a diameter of 1.00mm and length of 2.00m. If the resistance of the material is 0.1Ω, the resistivity is
A
2.55 * 105 Ωm
B
2.55 * 103 Ωm
C
3.93 * 10-6 Ωm
D
3.93 * 10-8 Ωm
correct option: d
\( R = \frac{1}{a} p \implies p = \frac{R.a}{1} = \frac{0.1 \times 3.14}{2} \left ( \frac{10^{-3}}{2} \right)^2 \\
\text{Therefore } p = \frac{0.1 \times 3.14 \times 10^{-6}}{2 \times 4} \\
= 0.3925 \times 10^{-7} \text{ OR } -3.930 \times 10 ^{-8}Ωm \)
Users' Answers & Comments\text{Therefore } p = \frac{0.1 \times 3.14 \times 10^{-6}}{2 \times 4} \\
= 0.3925 \times 10^{-7} \text{ OR } -3.930 \times 10 ^{-8}Ωm \)
44
A 40KW electric cable was uses to transmit electricity through a resistor of resistance 2.00Ω at 800V. 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; \( \implies 40KW = IV \\
\text{Therefore } 40000 = 1 \times 800\\
\implies I = \frac{40000}{800} = 50A,\text{tune the current through} \)
Resistor = 50A
power loss= I2R = 502 x 2
= 2500 x 2 = 5.0 x 103W
Users' Answers & Comments\implies I = \frac{40000}{800} = 50A,\text{tune the current through} \)
Resistor = 50A
power loss= I2R = 502 x 2
= 2500 x 2 = 5.0 x 103W
45
The instrument that measures both ac and dc is called
A
A current balance
B
a moving coil ammenter
C
a moving iron ammeter
D
an inverter
correct option: c
Users' Answers & Comments46
two long parallel wires X and Y carry currents 3A and 5A each. if the force experienced by unit length is 5 x 10-5N. the force per unit length experienced by wire Y is
A
3 x 10-5
B
3 x 10-6
C
5 x 105
D
5 x 10-5
correct option: d
since the wires are parallel to each other, the forces which they exert on each other are equal and opposite, since they must agree with newton's law of action an reaction. Hence the force exerted by the wire is equally
5 x 10-5
Users' Answers & Comments5 x 10-5
47
Lenz's law is a law of conservation of
A
electric charge
B
electric current
C
Energy
D
Momentum
correct option: c
Lenz's lew of electrical magnetic induction is usefully referred to as the law of conservation of energy
Users' Answers & Comments48
A 120V, 60W lamp is to be operated on 220V ac supply mains. calculate the value of non inductive resistance that would be required to ensure that the lamp is run on correct value
A
100Ω
B
200Ω
C
300Ω
D
500Ω
correct option: b
Power of the lamp = 60W = \( \frac{v^2}{R} \) and the voltage rate = 120V
\( \implies R = \frac{V^2}{60} = \frac{120 \times 120}{60} = 240Ω \)
Now, if the resistance of the lamp is 240Ω the fuse wire that should ensure current does not exceed this value should be rated 200Ω.
Users' Answers & Comments\( \implies R = \frac{V^2}{60} = \frac{120 \times 120}{60} = 240Ω \)
Now, if the resistance of the lamp is 240Ω the fuse wire that should ensure current does not exceed this value should be rated 200Ω.
49
A radioactive substance has a half life of 20 days. What fraction of the original radioactive nuclei will remain after 80 days?
A
\( \frac{1}{32} \)
B
\( \frac{1}{16} \)
C
\( \frac{1}{8} \)
D
\( \frac{1}{4} \)
correct option: b
80 days = 4 half lives (i.e 4 x 20dqs = 80dq)
But the fraction always left by any radioactive material after 4 half lives is
\( \left ( \frac{1}{2} \right)^2 = \frac{1}{16} \)
Users' Answers & CommentsBut the fraction always left by any radioactive material after 4 half lives is
\( \left ( \frac{1}{2} \right)^2 = \frac{1}{16} \)