**Solutions: C**

Alternatively, you may put values to work out using the 2 basic formulae of P = IV and V = IR. Refer the video below.

**Solutions: C**

Alternatively, you may put values to work out using the 2 basic formulae of P = IV and V = IR. Refer the video below.

**Solutions:**

View the video on how the gauge works.

<p><a href=”https://vimeo.com/138991946″>fuel guage using ammeter and variable resistor</a> from <a href=”https://vimeo.com/user10931667″>evantoh</a> on <a href=”https://vimeo.com”>Vimeo</a>.</p>

As the fuel level drops, the float which stays on the fuel surface will descend. The rod which is attached to the float will turn clockwise about the pivot X. As the rod turns, the resistance on the variable resistor increases. This increases the resistance of the circuit. Hence the current flowing through the circuit will decreases, causing the needle to deflect more to the left, indicating towards E (empty). Thus the reading on the fuel gauge decreases.

**Solutions:**

(i) Both the fixed resistor and sensor are in series.

Total effective resistance Re = 5000 + 1000 = 6000 ohms

V = IR

12 = I x 6000

I = 0.0020 A

Hence potential across Y, V = IR (where I is constant in a series circuit)

= 0.0020 x 1000

= 2.0 V

(ii) When the temperature increases and the resistance of sensor Y decreases,** the total effective resistance of the circuit decreases**.

Since V = IR, where the R of the fixed resistor is a constant 5000 ohms, as current I increases, **the potential difference across the 5000 ohms resistor will increase.**

An apartment block receives water from a nearby reservoir. A pump is necessary to lift the water into a storage tank at the top of the building.

The gravitational field strength is 10 N / kg. How much energy does the pump supply to lift each kilogram of water into the tank?

**Solutions: 100J**

Water will always find its own level. Hence with or without the pump, the water level in the pipe in the building will be of the same level as the reservoir. Hence, the pump is only needed to pump water up a height of 10m only (instead of the 50m).

Energy, E = mgh = 1 kg x 10 N/kg x 10m =** 100 J**

B. The distance between the image of a close object and the centre of the lens.

C. The distance between the image of a distant object and the centre of the lens.

D. The distance between two principal foci.

Recall that only reflected rays from distant object (object placed far away) are considered parallel. When these parallel light rays pass through the converging lens, they converge to a point, Focal Point (F).

Only then, the distance between the (sharp) image of distant object to the centre of the lens, optical centre (C), is known as **focal length (f)**.

Refer to the diagrams below.

Also refer to another question on lens: http://evantoh.posterous.com/2010/11/09/converging-lens-important-concepts

Water if depth 10 m exerts a pressure equal to atmospheric pressure. An air bubble rises to the surface of a lake which is 20 m deep. When the bubble reaches the surface, its volume is 6 cm^{3}.

What was the volume of the air bubble at the bottom of the lake?

A) 2 cm^{3 } B) 3 cm^{3 } C) 12 cm^{3 } D) 18 cm^{3}

Since the air bubble is enclosed, PV at A is equal to PV at B.

P_{A}V_{A} = P_{B}V_{B}

(P_{atm} + P_{water}) V_{A} = (P_{atm}) V_{B}

(10 + 20) V_{A} = 10 x 6

V_{A} = 60 / 30

= 2 cm^{3}

The figure below shows a 0.40 kg mass hanging at rest from a spring.

(a) State what is meant by the mass of an object.

Mass is the amount of matter in the body. SI unit is kg.

(b) (i) On the figure above, draw an arrow showing the ling of action and the direction for each of the two forces that act on the mass. Write the name of the force next to each arrow.

(ii) The gravitational field strength is 10 N/kg. Calculate the values of the two forces you have drawn in (i).

W = mg = 0.40 x 10 = 4 N

Weight = Tension = 4 N (Newton’s 1st law, net force = 0N)

(c) The mass is pulled downwards and then released. Explain, in terms of any changes in the forces acting on the mass, why the mass accelerates upwards.

As the spring is pulled downwards, there is a gain in elastic potential energy. The tension in the spring increases compared to previously.

Since Tension > Weight, there is an resultant force acting upwards, hence there is an acceleration upwards.

<p><a href=”http://vimeo.com/61785429″>closed loop triangle mpeg4</a> from <a href=”http://vimeo.com/user10931667″>evantoh</a> on <a href=”http://vimeo.com”>Vimeo</a>.</p>