# Evan's Space

## Wonders of Physics ## 2016 Pure Physics P1 Q40 – Voltage and Power of Bulb Y when X is switched off Solutions: Option C
[There is another similar question in 2013 Nov Pure Physics P1Q40. The answer is Option C too. Refer to the last section of this post]

Refer to the 2 videos below. The 1st one is faster if you know that the bigger the resistance, by proportion, the bigger the potential difference of the component as it will take a larger portion of the emf wrt to the other components in series. This method is especially useful for MCQ.

If you are still unsure, you may put in values to find the I, V and P across the components. This will be more tedious and time-consuming.

Another similar question is 2013 Nov Pure Physics P1 Q40. Solutions: Option C

## Identifying what lens, focal length and image from 2 rays – PP2010P1Q23 and SP2014P1Q11

These 2 questions are actually the same. Q23 is from 2010 Pure Physics P1 while Q11 is from 2014 Sci Physics P1. Take a look at these 2 questions. If you are not sure, view the video below for the explanation.  If you do not know how to answer these 2 questions, view this video and also refer to the lens summary below. ## 2013SPp1q17 What is the new P when V across resistor is doubled? 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.

## N2010 P2Q12 Fuel gauge using variable resistor 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. 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.

## 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? 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

## N2010 P1 MCQ Q12 Sci Physics 2010 O-Level

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