**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.**

Which distance is equal to the focal length of a lens?

(A) the distance between a distant object and the 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

**Solutions: Option C**

**Focal length f** is the **distance between** the **focal point** and the** centre of the lens (optical centre)**.

Note that only when **parallel rays of light** enter a converging lens, the rays will converge to a point. That point is considered to be focal point F (principal focus). The distance between focal point F and the optical centre is the focal length f. Refer to below.

As none of the options is similar to the above definition, you have to consider that the rays from a **distant (far away) object** are considered **parallel**. Hence the sharp image formed on the screen is considered the forcal point F of the lens and the distance between the image and the optical centre is the focal length f.

If it is a **close object,** the rays entering are not considered parallel. Hence even if the rays converged to a point, that point is NOT focal point F and the distance between this converged point and the optical centre is NOT focal length f.

(a) (i) A bigger force than F can be obtained due to the **level system** and **hydraulic system**.**Level system:** Applying principle of moments,** the anticlockwise moment by the F is equal to the clockwise moment by the force on piston A** (note that the handle is pushing the piston down, but the piston A is pushing on the handle upwards – action = reaction).

Hence these two systems allow the force on piston B to be greater than F at the handle.

(a) (ii) Both liquid and gas molecules are in a continuously random motion. But in liquid, the molecules are closely packed together and able to slide around one another. There is very little empty space between the molecules hence liquid is not compressible. Gas molecules are far apart from one another, hence gas can be easily compressed.

(b) P = F/A = 12000 / 0.060 = 200 000 Pa

(c) Velocity is the vector quantity while speed is a scalar. As the car goes round the bend, the direction of the car changes. Hence velocity is changing even though speed is constant.

Likewise, as the velocity is changing, the car is considered to have an acceleration (not in the sense of increasing speed though).

Solutions: Option B

As **lamp X is thicker and shorter**, these 2 factors make the **resistance of X lower than Y**.

*Recall: Length increases, R increases & cross-sectional area increases, R decreases. *

*[R = pL/A]*

As both bulbs are connected to the same mains, (assume 240V), the potential difference across the bulbs are 240 V.

**Brightness of the bulb depends on power**, P = IV or V2/R, since V is constant, I across X is higher due to lower R,

hence **power of X is greater than power of Y, hence X is brighter**.