WAEC Past Questions and Answers - Page 3076

15,376.

The diagram above illustrates a beaker of water being heated. It is observed that as the bubbles rise they get bigger in size.The reason for this observation is

A.

water pressure on the bubbles decreases

B.

density of water increases with rise in temperature

C.

volume of water increases with rise in temperature

D.

atmospheric pressure on the bubbles decreases

Correct answer is A

When a bubble is deep under water, it has both atmospheric pressure and water pressure (giving it depth), acting on it but as it rises, the water pressure decreases. At this point, the basic pressre acting on the bubble is the atmospheric pressure. Recall that pressure is inversely proportional to volume (Boyles' law), hence, the less the pressure, the more the volume

15,377.

A plane is inclined at an angle \(\theta\) to the horizontal. Its velocity ratio is

A.

\(sin\theta\)

B.

\(tan\theta\)

C.

\(\frac{1}{sin\theta}\)

D.

\(\frac{1}{tan\theta}\)

Correct answer is C

Velocity ratio = \(\frac{Effort distance}{Load distance}\)

On an inclined plane, it is the inverse of \(sin\theta\) = \(\frac{1}{sin\theta}\)

15,378.

Ice of mass 50g at 0°C was completely melted by 70% of the total heat energy supplied. Calculate the total heat energy supplied  [Specific latent heat of fusion of ice = 336\(Jg^{-1}\)]

A.

2400J

B.

6800J

C.

7251J

D.

24000J

Correct answer is D

The quantity of heat needed to melt ice = mL (where m is the mass of ice and L is the specific latent heat of fusion of ice)

Q = \(50 \times 336 = 16800J\) but this is 70% of the total heat supplied so that we have,

70% total heat = 16800J

\(\frac{70}{100}\) of total heat = 16800J

⇒ Total heat supplied = \(\frac{16800\times100}{70} = 24000J\)

15,379.

The energy stored in a spring of stiffness constant k = 2000\(Nm^{-1}\) when extended 4cm is

A.

0.16J

B.

1.60J

C.

16.00J

D.

160.00J

Correct answer is B

The spring constant = the measure of stiffness = 2000\(Nm^{-1}\)

Energy stored in a spring = Work done to stretch it through a distance = change in potential energy

= \(\frac{kx^{2}}{2}\) = \(\frac{2000 \times (0.04)^{2}}{2}\) (converting 4cm to metres)

= \(\frac{3.2}{2} = 1.60J\)