# 2016 - WAEC Physics Past Questions and Answers - page 1

1

The dimensions of momentum are

A

MLT

B

ML

^{-1}T^{-1}C

MLT

^{-1}D

ML

^{-1}T**Answer & Comments**

2

A student measures the volume of a liquid using a measuring cylinder. What else needs to be measured by the student in order to determine the density of the liquid?

A

depth of the liquid in the cylinder

B

mass of the cylinder

C

mass of the liquid

D

tempersture of the liquid

**Answer & Comments**

**correct option:**c

3

The diagram given, represents the graph of stress against strain for an elastic wire. The point Q on the graph is the

A

elastic linit

B

breaking point

C

yield point

D

proportional limit

**Answer & Comments**

**correct option:**b

4

A ball is dropped from the top of a tower. Due to air resistance,it reaches terminal velocity. Which of the following statement(s) about its motion is/are correct? i. The acceleration of the ball is zero. ii. The net force on the ball is zero. iii. The velocity of the ball increases.

A

i only

B

i and ii only

C

ii and iii only

D

i, ii and iii

**Answer & Comments**

**correct option:**a

5

Which of the following substances lowers the surface tension of water?

A

metal

B

sand

C

detergent

D

paper

**Answer & Comments**

**correct option:**c

6

An object of volume 400cm

^{3}and 2.5gcm^{-3}is suspended from a spring balance with half its volume immersed in water. Determine the reading on the spring balance. (Density of water = 1gcm^{-3})A

1200g

B

1000g

C

800g

D

400g

**Answer & Comments**

**correct option:**c

Mass = density x volume

400 x 2.5 = 1000g

Loss in mass = volume x density

= \(\frac{400}{2}\) x 1 = 200g

Reading on spring balance = 1000 - 200g = 800g

Users' Answers & Comments400 x 2.5 = 1000g

Loss in mass = volume x density

= \(\frac{400}{2}\) x 1 = 200g

Reading on spring balance = 1000 - 200g = 800g

7

The diagram given illustrates a force-distance graph for the motion of a wooden block. Determine the work done on the block when moved through a distance of 5m.

A

4J

B

15J

C

5J

D

100J

**Answer & Comments**

8

The two person of a body undergoing a uniformly accelerated motion are (10s, 10ms

^{-1}) and (30s, 50ms^{-1}). On the velocity-time graph pf the body.A

0.5ms

^{-2}B

2.0ms

^{-2}C

10.0ms

^{-2}D

40.0ms

^{-2}**Answer & Comments**

**correct option:**b

Slope = a = \(\frac{\deltav}{\deltat}\)

= \(\frac{50 - 10}{30 - 10}\)

= 2.0ms

Users' Answers & Comments= \(\frac{50 - 10}{30 - 10}\)

= 2.0ms

^{-2}9

At a birthday party, the celebrant pops a corked fruit wine. If the cork shoots out of the bottle at an angle of 40

^{o}to the horizontal and travels a horizontal distance of 4.50m in 1.25s. Calculate the initial speed of the cork.A

4.2ms

^{-1}B

4.7ms

^{-1}C

5.6ms

^{-1}D

7.1ms

^{-1}**Answer & Comments**

**correct option:**b

R = Ucos\(\theta\) x t

U = \(\frac{R}{Cos \theta \times t}\)

= \(\frac{4.50}{Cos40N.25}\)

= 4.699m/s

Users' Answers & CommentsU = \(\frac{R}{Cos \theta \times t}\)

= \(\frac{4.50}{Cos40N.25}\)

= 4.699m/s

10

A uniform metre rule is balanced on a fulcrum placed at the 35cm mark by suspending a mass of 120g at the 10cm mark. Calculate the mass of the metre rule.

A

60g

B

80g

C

120g

D

200g

**Answer & Comments**

**correct option:**d

Clock-wise moment = Anti-clock wise

15 x m = 5 x 10

m = 200g

Users' Answers & Comments15 x m = 5 x 10

m = 200g