2005 - JAMB Physics Past Questions and Answers - page 4
31
The electric field intensity in a place where a charge of 10-10c experiences a force of 0.4N is?
A
4.0 x 107 NC-1
B
8.0 x 109 NC-1
C
8.O x 10-12 NC-1
D
4.0 x 109 NC-1
correct option: d
The electric force experienced in an electric field of intensity E is given by:
F = qE: where q = the test charge.
E = F/Q = 0.4/10-10 = 0.4 X 10-10
= 4.0 X 109 NC-1
Users' Answers & CommentsF = qE: where q = the test charge.
E = F/Q = 0.4/10-10 = 0.4 X 10-10
= 4.0 X 109 NC-1
32
Two charged particles are projected into a region where there is a magnetic field perpendicular to their velocities. if the charges are deflected in opposite directions, which of the following statements is true of the charge?
A
the charge must be of opposite sign
B
positive charges are more in number
C
the charges are electrically neutral
D
the charges are of the same sign
correct option: a
The charge must be opposite in sign, since the positive charges will deflect to the negative plate, and the negative charge to the positive plate
Users' Answers & Comments33
The resistance of a pieces of wire of lenght 20m are crossed-sectional area 8 x 10 6m2is?
[Resistivity of the wire = 4 x 10-7 Ω m]
[Resistivity of the wire = 4 x 10-7 Ω m]
A
5.0 Ω
B
0.5 Ω
C
10.0 Ω
D
1.0 Ω
correct option: d
since resistance R = p (l)/a
= (4 x 10-7 x 20)/8.0 x 10-6
R = 1.0 Ω
Users' Answers & Comments= (4 x 10-7 x 20)/8.0 x 10-6
R = 1.0 Ω
34
A
20o
B
18o
C
35o
D
25o
correct option: d
Users' Answers & Comments35
A
10N
B
12N
C
20N
D
27N
36
A
5 ms-2
B
15 ms-2
C
3 ms-2
D
9 ms-2
correct option: a
Normal reaction R = 3 kg x 10 = 30N
coefficient of friction , μ = 0.5
∴ acting frictional force = μR = 30 x 0.5
Pulling/sliding force F = 6kg x 10 = 60N.
∴ Net downward force = ma = 60 - 15.0
=> (3 + 6)a = 45
∴ 9a = 45
a = 5m/s2
Users' Answers & Commentscoefficient of friction , μ = 0.5
∴ acting frictional force = μR = 30 x 0.5
Pulling/sliding force F = 6kg x 10 = 60N.
∴ Net downward force = ma = 60 - 15.0
=> (3 + 6)a = 45
∴ 9a = 45
a = 5m/s2
37
A
11N
B
9N
C
4N
D
6N
correct option: b
If the string is in equilibrium, the the tension in the string Q = 4N + 5N = 9N
Users' Answers & Comments38
A
800J
B
600J
C
100J
D
400J
correct option: d
Since the force is inclined at an angle 60o to the horizontal, the effective component of te force along the horizontal is given by F x cos 60o
= 100 x 0.5 = 50N.
∴ work done = F x cos 60 x displacement
= 50 x 8.0m
= 400.0J
Users' Answers & Comments= 100 x 0.5 = 50N.
∴ work done = F x cos 60 x displacement
= 50 x 8.0m
= 400.0J
39
A
EMITTER, BASE and COLLECTOR
B
COLLECTOR, EMITTER and BASE
C
BASE, COLLECTOR and EMITTER
D
BASE, ENITTER and COLLECTOR
correct option: a
The fig above shows a conventional transistor, showing the three basic features, the Base, the Emitter and the collector. Thus comparing this with the figure in the question it implies that P = Emitter, Q = Base and R = collector
Users' Answers & Comments40
\( ^{235} _{92}U + ^1 _0n → ^{144} _{56}Ba + ^{90} _{36}Kr +2X \)
In the reaction above, X is
In the reaction above, X is
A
electron
B
proton
C
neutrino
D
neutron
correct option: d
235U + 1n → 144Ba + 90Kr + 2x9205636
In the reaction above, balancing the left with the right, show that X is neutron
Users' Answers & CommentsIn the reaction above, balancing the left with the right, show that X is neutron