2013 - WAEC Physics Past Questions and Answers - page 1
1
The rising of a liquid in an open and ended glass tube of narrow bore is
A
osmosis
B
adhesion
C
capillarity
D
surface tension
correct option: c
Users' Answers & Comments2
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
Users' Answers & Comments4
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
Users' Answers & Comments5
The slope of a linear distance-times graph represents
A
acceleration
B
displacement
C
speed
D
velocity
correct option: c
Users' Answers & Comments6
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
7
The total area under a force-velocity graph represents
A
energy
B
momentum
C
power
D
pressure
correct option: a
Users' Answers & Comments8
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
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 = \(\frac{1}{2} \times 0.02 \times 6^2 = 0.36J\)
Users' Answers & Comments= \(\frac{1}{2}\)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
Users' Answers & Comments