1999 - JAMB Physics Past Questions and Answers - page 2
11
The equation of a wave travelling along the positive x-direction is given by;
y = 0.25 x 10-3 sin (500t - 0.025x).
Determine the angular frequency of the wave motion.
y = 0.25 x 10-3 sin (500t - 0.025x).
Determine the angular frequency of the wave motion.
A
0.25 x 10-3rad s-1
B
0.25 x 10-1rad s-1
C
5.00 x 102rad s-1
D
2.50 x 103rad s-1
correct option: c
The general wave equation is given by;
y = A Sin(wt - 2∏x/Wavelenght)
where w = angular frequency
Thus comparing this equation with the equation
y = 0.25 x 10-3 Sin (500t - 0.025x)
That Amplitude = 0.25 x 10-3m
the angular frequency w = 500 (the coefficient of t in both equations)
∴ w = 500 rad s-1
= 5.00 x 102rad s-1
Users' Answers & Commentsy = A Sin(wt - 2∏x/Wavelenght)
where w = angular frequency
Thus comparing this equation with the equation
y = 0.25 x 10-3 Sin (500t - 0.025x)
That Amplitude = 0.25 x 10-3m
the angular frequency w = 500 (the coefficient of t in both equations)
∴ w = 500 rad s-1
= 5.00 x 102rad s-1
12
If a sound wave goes from a cold air region to a hot air region, its wavelength will
A
increase
B
decrease
C
decrease then increase
D
remain constant
correct option: b
If a sound wave goes from a cold air region to a hot air region (which will be less dense), the velocity decreases, and since the frequency is always constant, the wave length decreases with the velocity
Users' Answers & Comments13
The lowest note emitted by a stretched string has a frequency of 40Hz. How many overtones are there between 40hz and 180Hz?
A
4
B
3
C
2
D
1
correct option: b
The lowest note (fundamental note) emitted by any stretched string has a fundamental frequecy fo = V/2l = 40Hz
The 1st overtone = 2fo = 2 x 40 = 80Hz
2nd overtone = 3fo = 3 x 40 = 120Hz
3rd overtone = 4fo = 4 x 40 = 160Hz
4th overtone = 5fo = 5 x 40 = 200Hz
Overtones between 40Hz and 180Hz are 80Hz, 120Hz, 160Hz = 3 Overtones
Users' Answers & CommentsThe 1st overtone = 2fo = 2 x 40 = 80Hz
2nd overtone = 3fo = 3 x 40 = 120Hz
3rd overtone = 4fo = 4 x 40 = 160Hz
4th overtone = 5fo = 5 x 40 = 200Hz
Overtones between 40Hz and 180Hz are 80Hz, 120Hz, 160Hz = 3 Overtones
14
A man stands 4m in front of a plane mirror. If the mirror is moved 1m towards the man, the distance between him and his new image is
A
3m
B
5m
C
6m
D
10m
correct option: c
The image of an object placed in front of a plane mirror is always as far behind the mirror as the object is in front of the mirror. Thus when the object is 4m in front of the mirror, the image will be 4m behind the mirror. But when the mirror is moved in towards the man, (that is the man is now 3m from the mirror), the image distance will again decrease by 1m from behind the mirror. Thus the new image position is 3m behind the mirror. Therefore, the distance between the man and his new image is 6m.
Users' Answers & Comments15
The inside portion of part of a hollow metal sphere of diameter 20cm is polished. The portion will therefore form a
A
concave mirror of focal length 5cm
B
concave mirror of focal length 10cm
C
convex mirror of focal length 5cm
D
convex mirror of focal length 20cm
correct option: a
Users' Answers & Comments16
The velocities of light in air and glass are 3.0 x 108ms-1 and 2.0 x 108ms-1 respectively. If the angle of refraction is 30°, the sine of the angle of incidence is
A
0.33
B
0.50
C
0.67
D
0.75
correct option: d
Refractive index
= \(\frac{Sin i}{Sin r}\) = \(\frac{\text{Vel. of light in air}}{\text{vel. of light in glass}}\)
= \(\frac{Sin i}{Sin 30^o}\)= \(\frac{3.0 \times 10^8}{2.0 \times 10^8}\)
= Sin i = \(\frac{(3.0 \times 10^8 \times Sin 30^o)}{2.0 \times 10^8}\)
= Sin i = \(\frac{(3.0 \times 10^8 \times 0.5)}{2.0 \times 10^8}\)
= \(\frac{1.5}{2}\)
= 0.75
Users' Answers & Comments= \(\frac{Sin i}{Sin r}\) = \(\frac{\text{Vel. of light in air}}{\text{vel. of light in glass}}\)
= \(\frac{Sin i}{Sin 30^o}\)= \(\frac{3.0 \times 10^8}{2.0 \times 10^8}\)
= Sin i = \(\frac{(3.0 \times 10^8 \times Sin 30^o)}{2.0 \times 10^8}\)
= Sin i = \(\frac{(3.0 \times 10^8 \times 0.5)}{2.0 \times 10^8}\)
= \(\frac{1.5}{2}\)
= 0.75
17
An astronomical telescope is said to be in normal adjustment when the
A
eye is accommodated
B
focal length of objective lens is longer than that of eye piece
C
final image is at the near point of eye
D
final image is at infinity
correct option: d
Users' Answers & Comments18
Which of the following electromagnetic waves is least energetic?
A
Infra-red rays
B
X-rays
C
Ultra-violet rays
D
Gamma rays
correct option: a
Users' Answers & Comments19
Steel is more suitable for permanent magnet than iron because the former
A
is easily demagnetized by shaing vigorously
B
is an alloy of many metals
C
is easily magnetized by alternating current through one cycle
D
retains magnetism more than iron
correct option: d
Users' Answers & Comments20
Four cells each of e.m.f 1.5V and internal resistance of 4Ω are connected in parallel. What is the effective e.m.f. and internal resistance of the combination?
A
6.0V, 16Ω
B
6.0V, 1Ω
C
1.5V, 4Ω
D
1.5V, 1Ω
correct option: d
Since all the cells are connected in parallel, the effective e.m.f. is just that of one single cell. Which means e.m.f. = 1.5V
However, the effective internal resistance adds as parallel arrangement given as
1/r = 1/r1 + 1/r2 + 1/r3 + 1/r4
= 1/4 + 1/4 + 1/4 + 1/4
= 1Ω
∴ r = 1Ω
Answer = 1.5V, 1Ω
Users' Answers & CommentsHowever, the effective internal resistance adds as parallel arrangement given as
1/r = 1/r1 + 1/r2 + 1/r3 + 1/r4
= 1/4 + 1/4 + 1/4 + 1/4
= 1Ω
∴ r = 1Ω
Answer = 1.5V, 1Ω