2009 - JAMB Physics Past Questions and Answers - page 2
11
When a pentavalent impurity is added to a pure semiconductor, it becomes
A
a p-type semiconductor
B
an n-type semiconductor
C
an insulator
D
an intrinsic semi-conductor
correct option: b
When a pentavalent (5 outer electron)atom is added to a pure semiconductor, there is an extra electrons which does not take part in the covalent bonding resulting in an n-type semiconductor.
Users' Answers & Comments12
The value of x, y and z respectively in the expression MxLyTz for the universal gravitational constant G are
A
2, -3, -2
B
-1, 3, -2
C
-1, 2, -3
D
-2, -1, 3
correct option: b
| F ∝ | M1M2 |
| __r2 |
| => F = | GM1M2 |
| __r2 |
where F is the force of attraction
M1 and M2 are masses of the two bodies, r is the distance between them, and G is the universal gravitational constant
| ∴ G = | _Fr2 |
| M1M2 |
| ∴ dimension for force, F = | ML | = MLT-2 |
| T.T |
| The dimension for G = | MLT-2L2 | = | L3T-2 |
| _MM | _M |
= M-1L3T-2
In comparison with MxLyTz, the values for x, y and z = -1, 3, -2
13
An air craft attempt too fly due north at 100 kmh-1. If the wind blows against it from east to west at 60 kmh-1, its resultant velocity is
A
117 kmh-1, N31oE
B
127 kmh-1, N31oE
C
117 kmh-1, N31oW
D
127 kmh-1, N31oW
correct option: c
R2 = 1002 + 602
10000 + 3600 = 13600
R = √13600 = 117km
60kml & = tan = 60/100
tan-1 0.6
= 31o west of the North
117km N31ow
Users' Answers & Comments10000 + 3600 = 13600
R = √13600 = 117km
60kml & = tan = 60/100
tan-1 0.6
= 31o west of the North
117km N31ow
14
A car accelerate uniformly from rest at 4 ms-2. How far will it travel in the fifth complete second?
A
100 m
B
50m
C
32 m
D
18 m
correct option: d
From S = ut + 1/2 at2
S = 1/2 ut2, since U = 0
Let S5 = distance in 5 seconds
= 1/2 x 4 x 52 = 2 x 25 = 50m
S4 = distance covered in 4 seconds
= 1/2 x 4 x 42 = 2 x 16 = 32m
Distance covered in the fifth second
S5 - S4 = 50m - 32m = 18m
Users' Answers & CommentsS = 1/2 ut2, since U = 0
Let S5 = distance in 5 seconds
= 1/2 x 4 x 52 = 2 x 25 = 50m
S4 = distance covered in 4 seconds
= 1/2 x 4 x 42 = 2 x 16 = 32m
Distance covered in the fifth second
S5 - S4 = 50m - 32m = 18m
15
A body of mass 12kg traveling at 4.2ms-1 collides with a second body of mass 18kg at rest. Calculate their common velocity if the two bodies coalesce after collision
A
1.5 ms-1
B
1.4 ms-1
C
2.1 ms-1
D
1.7 ms-1
correct option: d
Sum of momentum before collision
M1 U1 + M2M2
12 x 4.2 + 18 x 0
12 x 4.2
= 50.4
Sum of momentum after collision
(m1 + m2)v
= (12 + 18)v
= 30v
V = 50.4/30
= 1.7ms-1
Users' Answers & CommentsM1 U1 + M2M2
12 x 4.2 + 18 x 0
12 x 4.2
= 50.4
Sum of momentum after collision
(m1 + m2)v
= (12 + 18)v
= 30v
V = 50.4/30
= 1.7ms-1
16
The terminal velocity of a ball-bearing falling through a viscous fluid reached when the
A
upthrust is equal to the weight of the ball
B
ball accelerate uniformly
C
upthrustis equal to the velocity of the ball
D
velocity is uniform
correct option: d
At vt = terminal velocity
= zero acceleration
= uniform velocity
Users' Answers & Comments= zero acceleration
= uniform velocity
17
The pressure of two moles of an ideal gas at a temperature of 27oC and volume 10-2m3 is
[R = 8.313 J mol-1K-1]
[R = 8.313 J mol-1K-1]
A
4.99 x 105 Nm-2
B
9.80 x 103 Nm-2
C
4.98 x 103 Nm-2
D
9.80 x 105 Nm-2
correct option: a
PV = nRT
= 4.987 x 105 Nm-2
Users' Answers & Comments| ∴ P = | nRT | = | 2 x 8.313 x 300 |
| _V | ___10-2 |
= 4.987 x 105 Nm-2
18
A
reflection
B
deflection
C
diffraction
D
refraction
correct option: c
Diffraction is spreading of wave after passing through a slit
Users' Answers & Comments19
A
20 Nm-1
B
40 Nm-1
C
30 Nm-1
D
10 Nm-1
20
If a light wave has a wavelength of 500nm in air what is the frequency of the wave?
[c = 3 x 108 ms-1]
[c = 3 x 108 ms-1]
A
3.0 x 1014 Hz
B
6.0 x 1014 Hz
C
6.0 x 1012 Hz
D
2.5 x 1014 Hz