# 2013 - WAEC Physics Past Questions & Answers - page 1

1
The rising of a liquid in an open and ended glass tube of narrow bore is
A
osmosis
B
C
capillarity
D
surface tension
CORRECT OPTION: c
2

Which of the following units is equivalent to the watt?

A
kgms-2
B
kgm2s-3
C
kgm2s-2
D
kgm2s-1
CORRECT OPTION: b

$$Power = \frac{\text{Work done}}{time}$$

= $$\frac{Force \times distance}{time}$$

= $$\frac{mass \times acceleration \times distance}{time}$$

= $$\frac{kg \times m \times m}{s^{2} \times s}$$

= $$kgm^{2} s^{-3}$$

3
Which of the following statements about pressure in a liquid is correct?
A
the pressure in a liquid increases with depth
B
the higher the density of a liquid, the lower the pressure it exerts
C
pressure in liquid act only in a direction perpendicular to the sides of the containing vessel
D
pressure is independent of the acceleration due to gravity
CORRECT OPTION: a
4
Solid friction, like viscosity is
A
independent of the surface area in contact
B
independent of the relative motion between layers
C
dependent on normal reaction
D
in opposition to motion
CORRECT OPTION: d
5
The slope of a linear distance-times graph represents
A
acceleration
B
displacement
C
speed
D
velocity
CORRECT OPTION: c
6
A body of mass 2kg is released from rest on a plane inclined at an angle of 60o to the horizontal. Calculate the acceleration of the body down the plane.[g = 10ms-2]
A
3.1ms-2
B
5.2ms-2
C
6.0ms-2
D
8.7ms-2
CORRECT OPTION: d
a = g sin $$\theta$$

= 10 sin 60

= 8.7 ms-2
7
The total area under a force-velocity graph represents
A
energy
B
momentum
C
power
D
pressure
CORRECT OPTION: a
8
A body of mass 20g projected vertically upwards in vacuum returns to the point of projection after 1.2s. [g = 10ms-2]. Calculate the speed of projection
A
1.2ms-1
B
6.0ms-1
C
12.0ms-1
D
0.6ms-1
CORRECT OPTION: b
V = u - gt

u = v + gt

= 0 + 10 x 0.6

= 6ms-2
9
A body of mass 20g projected vertically upwards in vacuum returns to the point of projection after 1.2s. [g = 10ms-2]. Determine the potential energy of the body at the maximum height of its motion
A
0.36J
B
0.72J
C
360.00J
D
720.00J
CORRECT OPTION: a
P.E at max = K.E on projection

= $$\frac{1}{2}$$mv2 = $$\frac{1}{2} \times 0.02 \times 6^2 = 0.36J$$
10
The maximum displacement on either side of the equilibrium position of an object in simple harmonic motion represents
A
period
B
amplitude
C
wavelength
D
frequency
CORRECT OPTION: b
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