2007 - JAMB Physics Past Questions and Answers - page 3
21
A charge 50 µc has electric field strength of 360 NC-1 at a certain point. The electric field strength due to another charge 120 µc kept at the same distance apart and in the same medium is?
A
864 NC-1
B
150 NC-1
C
144 NC-1
D
18 NC-1
correct option: b
In general, electric field strength E = (F)/q
(qq)/4RE0 V2
∴if E1 = 360 NC-1 F = q1 E1 = q2E2
E2 = (q1E1)/q2
∴ E2 = (50 X 10-6 X 360)/120 X 10-6 = 150 NC -1
Users' Answers & Comments(qq)/4RE0 V2
∴if E1 = 360 NC-1 F = q1 E1 = q2E2
E2 = (q1E1)/q2
∴ E2 = (50 X 10-6 X 360)/120 X 10-6 = 150 NC -1
22
The energy E of a photon and it's wavelength are related by Eλ = X. the numerical value of X is?
[h = 6.63 x 10-34js, c = 3 x 108 ms-1]
[h = 6.63 x 10-34js, c = 3 x 108 ms-1]
A
6.6 X 10-28
B
1.99 X 10-27
C
6.60 x 10-26
D
1.99 x 10-25
correct option: d
E = HF = (hc)/λ
Eλ = hc
∴ X = hc ==6.63 x 10-34 x 108
= 1.989 x 10-25
=1.99 x 10-25
Users' Answers & CommentsEλ = hc
∴ X = hc ==6.63 x 10-34 x 108
= 1.989 x 10-25
=1.99 x 10-25
23
The bond that forms a semiconductor is?
A
covalent
B
electrovalent
C
ionic
D
metallic
correct option: a
Users' Answers & Comments24
Silicon doped with aluminum and germanium doped with arsenic become?
A
p- and n- types respectively
B
n- en p- type respectively
C
p- type semiconductor
D
n- type semiconductors
correct option: a
When silicon, a four valency element, is doped with aluminum, a three valency element, a p-type semiconductor is produced where the major carriers are the electrons.
Thus, the correct option is p-type and n-type respectively
Users' Answers & CommentsThus, the correct option is p-type and n-type respectively
25
A
5N
B
5√ 3N
C
10N
D
10√ 3N
correct option: c
Resolution of force:
Total vertical upward force:
= 3√3 + 10 cos 60o
= 3√3 + 5.
Total downward force
= 6 cos 30o
= 6 x √(3/2) = 3√3
Total horizontal forces towards right;
= 3N + cos 30o
= 3N + 10 x √(3/2)
= 3N + 5√3
Total horizontal forces towards left:
= 6 cos 60o
= 6 x 0.5
= 3.0N
Summary:
Vertical resultant forces upward:
= 3√3 + 5 - 3√3
= 5N
Horizontal resultant forces towards right:
= 3N + 5√3 - 3√3
= 5N
Horizontal resultant force towards right
= 3N + 5√3 - 3N
= 5√3N
∴ R2 = 52 + (5√3)2
= 25 + 25 x 3
= 100
∴ R = √100
= 10N
Users' Answers & CommentsTotal vertical upward force:
= 3√3 + 10 cos 60o
= 3√3 + 5.
Total downward force
= 6 cos 30o
= 6 x √(3/2) = 3√3
Total horizontal forces towards right;
= 3N + cos 30o
= 3N + 10 x √(3/2)
= 3N + 5√3
Total horizontal forces towards left:
= 6 cos 60o
= 6 x 0.5
= 3.0N
Summary:
Vertical resultant forces upward:
= 3√3 + 5 - 3√3
= 5N
Horizontal resultant forces towards right:
= 3N + 5√3 - 3√3
= 5N
Horizontal resultant force towards right
= 3N + 5√3 - 3N
= 5√3N
∴ R2 = 52 + (5√3)2
= 25 + 25 x 3
= 100
∴ R = √100
= 10N
26
A
X
B
Y
C
Z
D
W
correct option: a
The ball has its fastest speed at its equilibrium position when its displacement X = O; at X
Users' Answers & Comments27
A
40N
B
80N
C
120N
D
200N
correct option: b
Being a uniform rod, the height W should be acting at the mid point 25cm mark.
∴ Taking moment about the fulcrum, F,
200 x 10 = W x 10 + 40 x 30
2000 = 10W + 1200
=> 10W = 2000 - 1200
= 800
w = 800/10 = 80N
Users' Answers & Comments∴ Taking moment about the fulcrum, F,
200 x 10 = W x 10 + 40 x 30
2000 = 10W + 1200
=> 10W = 2000 - 1200
= 800
w = 800/10 = 80N
28
A boy drags a bag of rice along a smooth horizontal floor with a force of 2N applied at the angle of 60o to the floor. The work done after a distance of 3m is
A
3 J
B
4 J
C
5 J
D
6 J
correct option: a
Effective component of the force along the horizontal = F cos 60o
Work done by this force : 60
F cos 60 x X
= 2 x 0.5 x 3
= 3J
Users' Answers & CommentsWork done by this force : 60
F cos 60 x X
= 2 x 0.5 x 3
= 3J
29
A blacksmith heated a metal whose cubic expansivity is 6.3 x 10-6K-1. The areal expansivity is
A
2.0 x 10-6K-1
B
2.1 x 10-6K-1
C
4.2 x 10-6K-1
D
6.3 x 10-6K-1
correct option: c
If the lineal expansivity = ∝
=> area expansivity β= 2∝.
And cubic expansivity δ = 3∝.
=> 6.3 x 10-6 = 3∝
∴ ∝ =
∴ Area = 2∝ = 2 x 2.1x10-6
= 4.2 x 10-6K-1
Users' Answers & Comments=> area expansivity β= 2∝.
And cubic expansivity δ = 3∝.
=> 6.3 x 10-6 = 3∝
∴ ∝ =
| ∴ ∝ = | 6.3 x 10-6 | = 2.1x 10-6 |
| __3 |
∴ Area = 2∝ = 2 x 2.1x10-6
= 4.2 x 10-6K-1
30
The ratio of the coefficient of linear expansion of two metals ∝1/∝2 is 3:4. If, when heated through the same temperature change, the ratio of the increase in length of the two metals, e1/e2 is 1:2, the ratio of the original lengths l1/l2 is
A
8/3
B
3/2
C
2/3
D
3/8
correct option: c
Ratio of their linear expansion = ∝1/∝2= 3:4.
When heated to the same temperature range, the ratio of their increase in length e1/e2 = 1:2
But the increase in length of
1 = e1 = ∝1l1Δθ and increase i length of
2 = e2 = ∝2l2Δθ
=> 6l1 = 4l2
Users' Answers & CommentsWhen heated to the same temperature range, the ratio of their increase in length e1/e2 = 1:2
But the increase in length of
1 = e1 = ∝1l1Δθ and increase i length of
2 = e2 = ∝2l2Δθ
| => | e1 | = | ∝1l1Δθ | = | ∝1l1 | = | 1 |
| e2 | ∝2l2Δθ | ∝2l2 | 2 |
| But | ∝1 | = | 3 | , ∴ | 3 | x | l1 | = | 1 |
| ∝2 | 4 | 4 | l2 | 2 |
=> 6l1 = 4l2
| ∴ | l1 | = | 4 | = | 2 | OR 2 : 3 | l2 | 6 | 3 |