2009 - JAMB Physics Past Questions and Answers - page 3
21
A
12 V
B
8 V
C
2 V
D
36 V
correct option: b
1/R = 1/R1 + 1/R2 = 1/3 + 1/6
= 3/6
∴ R = 6/3 = 2
E = I(R + r) = 4(2 + 0) = 8v
Users' Answers & Comments= 3/6
∴ R = 6/3 = 2
E = I(R + r) = 4(2 + 0) = 8v
22
A
116.3 V
B
50.0 V
C
230.0 V
D
106.3 V
correct option: d
V =
√ (v1 + v2)2 + VR2
V =
√ (40 + 110)2 + 802
V =
√ (-70)2 + 802
=
√ 4900 + 6400
=
√ 11300
= 106.3v
Users' Answers & Comments√ (v1 + v2)2 + VR2
V =
√ (40 + 110)2 + 802
V =
√ (-70)2 + 802
=
√ 4900 + 6400
=
√ 11300
= 106.3v
23
A binary operation \(\otimes\) defined on the set of integers is such that m \(\otimes\) n = m + n + mn for all integers m and n. Find the inverse of -5 under this operation, if the identity element is 0
A
\(\frac{-5}{4}\)
B
\(\frac{-5}{6}\)
C
o
D
5
correct option: d
Users' Answers & Comments24
A progressive wave is represented by \( y = 10 \text{Sin} (1000\pi t - \frac{\pi x}{34}) \) . two layers of the wave separated by 153cm have a phase difference of
A
2700
B
450
C
900
D
1800
correct option: e
Correct Answer: Option E
Comparing y = 10Sin\( (1000\pi t - \frac{\pi x}{34} ) \\
\text{With } y = a \text{Sin }(net - kx)\\
K = \frac{\pi}{34}, \text{and phase difference}\\
J = KAC = \frac{\pi}{34} \times 1.53rad.\\
\text{But} \pi rad = 180^0 \\ \text{Therefore } \frac{1.53rad}{34} = \frac{1.53 \pi}{34} \times \frac{1}{\pi} \times 180^0 \\ =8.1^0 \)
Users' Answers & CommentsJ = KAC = \frac{\pi}{34} \times 1.53rad.\\
\text{But} \pi rad = 180^0 \\ \text{Therefore } \frac{1.53rad}{34} = \frac{1.53 \pi}{34} \times \frac{1}{\pi} \times 180^0 \\ =8.1^0 \)
25
A near sighted student has a near point of 0.1m and a focal length of 5.0cm. What is the student's far point?
A
0.200m
B
8.000m
C
0.125m
D
2.100m
correct option: e
Correct Answer: Option E
The person's image distance v will be at the distance of 0.1m = 10cm for him to see objects at a distance of u.F = 5.0cm
\( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies \frac{1}{10} + \frac{1}{u} = \frac{1}{5} \\
\frac{1}{u} = \frac{1}{5} - \frac{1}{10} = \frac{2-1}{10} = \frac{1}{10} \\
\text{therefore } u = 10cm = 0.1m \)
Users' Answers & Comments\( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies \frac{1}{10} + \frac{1}{u} = \frac{1}{5} \\
\frac{1}{u} = \frac{1}{5} - \frac{1}{10} = \frac{2-1}{10} = \frac{1}{10} \\
\text{therefore } u = 10cm = 0.1m \)
26
I. Diameter of a small ball bearing.
II. Thickness of a pieces of paper.
III. Diameter of a measuring cylinder.
IV. Length of a piece of wire.
Which of the above can be measured using a micrometer screw gauge?
II. Thickness of a pieces of paper.
III. Diameter of a measuring cylinder.
IV. Length of a piece of wire.
Which of the above can be measured using a micrometer screw gauge?
A
I, II and IV only .
B
I and II only
C
II and III only
D
III and IV only
correct option: b
Of all the instruments used in measuring length,the micrometer screw gauge has the highest reading accuracy (0.01 mm), and is therefore used in measuring very small length such as diameter of a small ball bearing, and thickness of a piece of paper.
Users' Answers & Comments27
What is the acceleration between two points on a velocity time graph which has coordinates
(10s, 15 ms-1)
and (20 s, 35 ms-1)?
(10s, 15 ms-1)
and (20 s, 35 ms-1)?
A
1.7 ms-2
B
3.50 ms-2
C
1.00 ms-2
D
2.00 ms-2
correct option: d
Acceleration = (AV)/At = (V2) - U1)/t2 - t1 = (35 - 15)/20 - 10 = (20)/10 = 2m5-2
Users' Answers & Comments28
A bullet fired vertically from upward from a gun held 2.0 m above the ground reaches its maximum height in 4.0s. calculate its initial velocity.
[g = 10ms-2]
[g = 10ms-2]
A
10 ms-1
B
8 ms-1
C
40 ms-1
D
20 ms-1
correct option: c
At maximum height, the final velocity, v = 0
thus from v = u + at
0 = u - at
∴ U = at = 10 x 4 = 40 mls
Users' Answers & Commentsthus from v = u + at
0 = u - at
∴ U = at = 10 x 4 = 40 mls
29
I. The motion of the tyres of a moving car.
II. The motion of a loaded test tube oscillated
vertically.
III. The beating of the heart.
IV. A stone tied to a string and whirled round.
V. The motion of the pistons in a gasoline engine.
Which of the motions above is simple harmonic?
II. The motion of a loaded test tube oscillated
vertically.
III. The beating of the heart.
IV. A stone tied to a string and whirled round.
V. The motion of the pistons in a gasoline engine.
Which of the motions above is simple harmonic?
A
III, IV and V only
B
II, III and V only
C
II, III and IV only
D
I,II and III only
correct option: b
Simple harmonic motion is the motion of a body whose acceleration is directed towards a fixed point and is directly proportional to its displacement from point.
Users' Answers & Comments30
If a cage containing a truck of coal weighing 750 kg is raised to a height of 90 m in a minute, what is the total power expended?
[g = 10 ms2]
[g = 10 ms2]
A
11.50 kw
B
12.60 kw
C
11.25 kw
D
12.10 kw
correct option: c
Power = (work done)/time taken = (mgh)/t
= (750 x 10 x 90)/60 = 11250w = 11.25kw
Users' Answers & Comments= (750 x 10 x 90)/60 = 11250w = 11.25kw