# 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

2

Which of the following units is equivalent to the watt?

A

kgms

^{-2}B

kgm

^{2}s^{-3}C

kgm

^{2}s^{-2}D

kgm

^{2}s^{-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 60

^{o}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

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 projectionA

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 motionA

0.36J

B

0.72J

C

360.00J

D

720.00J

**correct option:**a

P.E at max = K.E on projection

= \(\frac{1}{2}\)mv

Users' Answers & Comments= \(\frac{1}{2}\)mv

^{2}= \(\frac{1}{2} \times 0.02 \times 6^2 = 0.36J\)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