# 2014 - WAEC Physics Past Questions and Answers - page 1

1

A ball falling through a viscous liquid is acted upon by

A

upthrust only

B

upthrust and the ball's weight

C

the ball's weight and viscous force

D

upthrust, the ball's weight and viscous force

**correct option:**d

2

An object of mass m moves with a uniform speed v round a circular path of radius. If its angular speed is \(\omega\), the magnitude of the centripetal force acting on it is

A

\(m \omega^2 r\)

B

\(\frac{mv^2}{r}\)

C

\(m \omega r\)

D

\(\frac{mv^2}{r^2}\)

**correct option:**b

3

When an elastic material is stretched by a force, the energy stored in it is

A

kinetic

B

potential

C

thermal

D

electrical

**correct option:**b

4

When the linear momentum of a body is constant, the net force acting on it

A

is zero

B

increases

C

decreases

D

remains constant

**correct option:**a

5

If no net force acts on an object, the object maintains a state of rest or constant speed in a straight line. The above is a statement of Newton's

A

first law of motion

B

second law of motion

C

third law of motion

D

law of universal gravitation

**correct option:**a

6

Which of the following pairs of physical quantities is made up of vectors?

A

speed and displacement

B

mass and force

C

dislacement and acceleration

D

momentum and length

**correct option:**c

7

The knowledge of surface tension is applied in the following areas except in

A

manufacturing of rain coats

B

production of palm oil

C

floating of a needle on water

D

washing of plates with soapy water

**correct option:**b

8

A body is projected vertically upwards with a speed of 10 ms

^{-1}from a point 2m above the ground. Calculate the total time taken for the body to reach the ground. [g = 10ms^{-2}]A

1.00s

B

2.00s

C

2.18s

D

3.00s

9

A car starts from rest and covers a distance of 40m in 10s. Calculate the magnitude of its acceleration

A

0.25ms

^{-2}B

0.80ms

^{-2}C

3.20ms

^{-2}D

4.00ms

^{-2}**correct option:**b

S = ut + \(\frac{1}{2} at^2\)

40 = 0 x 10 + \(\frac{1}{2}\) x a x 10

50a = 40

a = \(\frac{40}{50}\)

= 0.8ms

Users' Answers & Comments40 = 0 x 10 + \(\frac{1}{2}\) x a x 10

^{2}50a = 40

a = \(\frac{40}{50}\)

= 0.8ms

^{-2}10

The diagram above illustrates a uniform metre rule which is balanced on a pivot by some masses. Calculate the value of M

A

500.0g

B

400.0g

C

44.4g

D

25.0g

**correct option:**d

clockwise moment = Anti-clock wise moment

= M x 40 = 100 x 10

M = \(\frac{1000}{40}\)

= 25g

Users' Answers & Comments= M x 40 = 100 x 10

M = \(\frac{1000}{40}\)

= 25g