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Wonders of Physics


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N2010 P2Q12 Fuel gauge using variable resistor

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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&gt; on <a href=”https://vimeo.com”>Vimeo</a&gt;.</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.

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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. Current flowing through the circuit increases.

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.


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N2007P1Q9 Pure Physics – Energy needed to pump water up a building

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?

GPE up 10 m

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


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N2010 P1 MCQ Q12 Sci Physics 2010 O-Level

Which distance is equal to the focal length of a lens?
A. The distance between a distant object and its image.
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.
Solution: Option C.

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.

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


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N2004 P1 Q15 – Pressure P1V1 = P2V2

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 cm3.

Bubbles

 

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

 

A) 2 cm3          B) 3 cm3         C) 12 cm3      D) 18 cm3

 

 

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

                  PAVA = PBVB

(Patm + Pwater) VA = (Patm) VB

        (10 + 20) VA = 10 x 6

                      VA = 60 / 30

                           = 2 cm3

 


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N2005/P2/A1 – Tension in Spring

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

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(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.

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(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.

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