2003 - JAMB Physics Past Questions and Answers - page 3
21
A 2000W electric heater is used to heat a metal object of mass 5kg initially at 10oC. If a temperature rise of 30oC is obtained after 10min, the heat capacity of the material is
A
6.0 x 104JoC-1
B
4.0 x 104JoC-1
C
1.2 x 104JoC-1
D
8.0 x 104JoC-1
correct option: b
2000W heater → 2000J/S
=> Heat supplied by the heater within 10 min
(600 seconds) = 2000 x 600
= 1200000J
∴ Heat capacity MC = 1200000/30
= 4000J/C
Heat capacity = 4.0 x 104J/C
Users' Answers & Comments=> Heat supplied by the heater within 10 min
(600 seconds) = 2000 x 600
= 1200000J
∴ Heat capacity MC = 1200000/30
= 4000J/C
Heat capacity = 4.0 x 104J/C
22
If 1.2 x 106J of heat energy is given off in 1 sec from a vessel maintained at a temperature gradient of 30Km-1, the surface area of the vessel is
A
1.0 x 102m2
B
9. 0 x 102m2
C
1.0 x 103m2
D
9.0 x 104m2
correct option: a
Generally in thermal/heat conduction, if 1.2 x 106 J is the heat flow per seconds; 30KM-1 is the temperature gradient of the materials; and 400Wm-1K-1 the thermal conductivity of the material, then we have that:
heat flow per seconds per unit area = thermal conductivity x temp. gradient
i.e = 1.2 x 106Area
= 400 x 30
=> 400 x 30 x Area = 1.2 x 106
∴ Area = 1.2 x 106400 x 30
Area = 100m2 or 1.0 x 102m2
Users' Answers & Commentsheat flow per seconds per unit area = thermal conductivity x temp. gradient
i.e = 1.2 x 106Area
= 400 x 30
=> 400 x 30 x Area = 1.2 x 106
∴ Area = 1.2 x 106400 x 30
Area = 100m2 or 1.0 x 102m2
23
A positive charged rod X is brought near as uncharged metal sphere Y and is then touched by a finger with X still in place. When the finger is removed, the result is that y has
A
no charge and a zero potential
B
a positive charge and zero potential
C
a negative charge and positive potential
D
a negative charge and a negative potential
correct option: d
Users' Answers & Comments24
A
27W
B
18W
C
9W
D
5W
correct option: a
From the above diagram, if each of the resistors can dissipate a maximum of 18W, the from the relation power, P = I2R, the current in the 2Ω series resistor is given by:
I2 = P/R = 18/2 = 9 => I = √9 = 3A
The effective parallel arrangement of the two 2Ω resistors: 1/R = 1/2 + 1/2 = 1Ω
∴ Total resistance in the circuit = 1 + 2 = 3Ω
current flowing the circuit = 3A
∴ Maxi. power = P = I2R = 32 x 3 = 27W
Users' Answers & CommentsI2 = P/R = 18/2 = 9 => I = √9 = 3A
The effective parallel arrangement of the two 2Ω resistors: 1/R = 1/2 + 1/2 = 1Ω
∴ Total resistance in the circuit = 1 + 2 = 3Ω
current flowing the circuit = 3A
∴ Maxi. power = P = I2R = 32 x 3 = 27W
25
The operation of an optical fibre is based on the principal of
A
dispersion of light
B
interference of light
C
refraction of light
D
polarization of light
correct option: c
Users' Answers & Comments26
A
It will increase because the equivalent resistance will increase
B
It will decrease if R2 is greater than R1
C
It will increase because the effective resistance will decrease
D
It will decrease if R2 is less than R1
correct option: c
Users' Answers & Comments27
A wire of 5Ω resistance is drawn out so that its new length is two times the original length. If the resistivity of the wire remains the same and the cross-sectional area is halved, the new resistance is
A
40Ω
B
20Ω
C
10Ω
D
5Ω
correct option: b
Let the original length = L1
Let the original resistance = R1 = 5Ω
Let the original resistivity = P1
Let the original area = a1
Let the new length = L2 = 2L1
let the new area = a2 = 1/(2a2)
Let the new resistance = R2
Let the new resistivity = P2
But since the resistivity remains the same,
=> P1 = P2
But a2 = 1/2a1; and L2 = 2L1
= 20Ω
Users' Answers & CommentsLet the original resistance = R1 = 5Ω
Let the original resistivity = P1
Let the original area = a1
Let the new length = L2 = 2L1
let the new area = a2 = 1/(2a2)
Let the new resistance = R2
Let the new resistivity = P2
But since the resistivity remains the same,
=> P1 = P2
| ∴ P1 = | R1 a1 |
| _L1 |
| = P2 = | R2 a2 |
| _L2 |
But a2 = 1/2a1; and L2 = 2L1
| ∴ | R1 a1 |
| _L1 |
| = | R2 2a1/2 |
| _2L1 |
| => | 5 x a1 |
| _L1 |
| = | R2 x a1 |
| _4L1 |
| ∴R2 = | 5 x a1 x 4L1 |
| _a1 x L1 |
= 20Ω
28
An electron of charge 1.6 x 10-19C is accelerated between two metal plates. If the kinetics energy of the electron is 4.8 x 10-17J, the potential difference between the plate is
A
30V
B
40V
C
300V
D
400V
correct option: c
In general, if a charged particle is accelerated between two metal plates/a region by potential difference V volts, the work done on the charge, given by the product of the charge and the potential difference, is equal to the kinetic energy of the charge
Thus eV = K.E
1.6 x 10-19 x V = 4.8 x 10-17
= 300V
Users' Answers & CommentsThus eV = K.E
1.6 x 10-19 x V = 4.8 x 10-17
| ∴ V = | 4.8 x 10-17 |
| 1.6 x 10-19 |
= 300V
29
A magnetic field is said to exist at a point if a force is
A
exerted on a stationary charge at the point
B
exerted on a moving charge at the point
C
deflected at the point
D
strengthened at the point
correct option: b
A charge in motion always experienced a force on entering a magnetic field
Users' Answers & Comments30
I It retains its magnetic much longer than steel.
II It is more easily magnetized than steel
III It is more easily demagnetized than steel
IV It produces a stronger magnet than steel
Which combination of the above makes iron preferable to steel in the making of electromagnets?
II It is more easily magnetized than steel
III It is more easily demagnetized than steel
IV It produces a stronger magnet than steel
Which combination of the above makes iron preferable to steel in the making of electromagnets?
A
I and II only
B
II and III only
C
I, III, and IV only
D
II, III and IV only
correct option: b
Iron (spft iron) is usually preferred to steel in the making of electromagnets because it is more easily magnetized and equally more easily demagnetized than [II and III]
Users' Answers & Comments