Secondary 3 Combined Science Semestral Assessment 2 (End of Year) Paper 2

TuitionGoWhere Practice Paper - Combined Science Secondary 3 (SA2)

Answer Key & Marking Scheme (Version 2)

Section A

1. B
Working: Reading = Main scale + (Thimble $\times$ Pitch/Divisions).
Main scale = 2.5 mm. Thimble = 32. Pitch = 0.5 mm, 50 divs $\rightarrow$ 0.01 mm/div.
Correction: $32 \times 0.01 = 0.32$ mm.
Total = $2.5 + 0.32 = 2.82$ mm? Wait.
Standard Micrometer: Main scale usually reads 0.5mm increments. If main scale shows 2.5mm line visible, and thimble is 32.
Reading = $2.5 + (32 \times 0.01) = 2.82$ mm.
Let's re-read the options.
Option A: 2.82 mm.
Option B: 2.92 mm.
If the main scale was 2.0 and the 0.5 line was NOT visible, it would be $2.0 + 0.32 = 2.32$.
If the main scale was 2.5 and the thimble was 42, it would be $2.5 + 0.42 = 2.92$.
Given the prompt says "main scale reads 2.5 mm", this usually implies the last visible mark is 2.5.
$2.5 + 0.32 = 2.82$ mm.
Answer: A
(Self-Correction: In many exam questions, if the thimble is past the halfway mark of the main scale division, you add 0.5. Here, 2.5 is explicitly stated as the reading. So $2.5 + 0.32 = 2.82$.)

2. D
Displacement has magnitude and direction. Speed, Distance, and Mass are scalars.

3. A
Total Displacement = $100 \text{ m (East)} - 50 \text{ m (West)} = 50 \text{ m East}$.
Total Time = $10 \text{ s} + 5 \text{ s} = 15 \text{ s}$.
Average Velocity = $\frac{\text{Displacement}}{\text{Time}} = \frac{50}{15} = 3.33 \text{ m/s East}$.

4.
An object remains at rest or continues to move at a constant velocity in a straight line [1] unless acted upon by a resultant external force [1].

5.
(a) 50 N [1]
(b) Since the velocity is constant, the acceleration is zero. According to Newton's First Law, the resultant force is zero. Therefore, the pushing force is balanced by the frictional force [1].

6.
$W = mg = 5 \times 10 = 50$ N [1].

7.
Pressure is transmitted equally.
$P_1 = P_2 \Rightarrow \frac{F_1}{A_1} = \frac{F_2}{A_2}$
$\frac{100}{0.01} = \frac{F_2}{0.5}$
$10000 = \frac{F_2}{0.5}$
$F_2 = 5000$ N [2].

8.
Atmospheric pressure is caused by the weight of the air column above [1]. As altitude increases, the height of the air column above decreases, and the density of air decreases, resulting in lower pressure [1].

9.
(a) Work Done = Force $\times$ Distance = $500 \times 10 = 5000$ J [2].
(b) Power = $\frac{\text{Work Done}}{\text{Time}} = \frac{5000}{20} = 250$ W [2].

10.
Energy cannot be created or destroyed [1], only converted from one form to another [1]. (Or: The total energy of an isolated system remains constant).

Section B

11.
(a) The cyclist accelerates uniformly / constant acceleration [1].
(b) Acceleration = $\frac{\Delta v}{\Delta t} = \frac{10 - 0}{5} = 2 \text{ m/s}^2$ [2].
(c) Distance = Area under graph.
Area 1 (Triangle) = $\frac{1}{2} \times 5 \times 10 = 25$ m.
Area 2 (Rectangle) = $10 \times 10 = 100$ m.
Area 3 (Triangle) = $\frac{1}{2} \times 5 \times 10 = 25$ m.
Total Distance = $25 + 100 + 25 = 150$ m [3].
(d) There are resistive forces (air resistance/friction) acting on the cyclist [1]. To maintain constant velocity (zero resultant force), the cyclist must apply a forward force to balance these resistive forces [1].

12.
(a) Graph:

• Axes labeled with units (Load/N, Extension/cm) [1].
• Points plotted correctly [1].
• Line of best fit: Straight line through origin up to 4N, then curves or deviates [1].
  (b) Gradient $k = \frac{F}{x}$. Using point (4N, 8cm):
  $k = \frac{4}{8} = 0.5$ N/cm [1].
  Unit: N/cm (or 50 N/m) [1].
  (c) The spring has exceeded its limit of proportionality / elastic limit [1]. The extension is no longer directly proportional to the load (graph is no longer linear) [1].

13.
(a) Vacuum / Nothing [1].
(b) Height $h$ increases [1]. Higher atmospheric pressure exerts more force on the mercury surface in the trough, pushing more mercury up the tube until the pressure of the column balances the atmospheric pressure [1].
(c) The height would increase significantly [1]. Since $P = \rho g h$, and water has a much lower density than mercury, a much greater height of water is required to exert the same pressure [1]. ($h_{water} \approx 13.6 \times h_{mercury}$).

14.
(a) Melting [1].
(b) In ice, particles vibrate about fixed positions [1]. As heat is absorbed, particles gain kinetic energy and vibrate more vigorously [1]. Eventually, they overcome the strong forces of attraction holding them in fixed positions and begin to slide past one another (liquid state) [1].
(c) The heat energy supplied is used to overcome/break the intermolecular forces of attraction between particles [1], rather than increasing the kinetic energy (temperature) of the particles [1].

15.
(a) The angle of incidence is equal to the angle of reflection [1]. The incident ray, reflected ray, and normal all lie in the same plane [1].
(b) Refractive index $n = \frac{\text{speed of light in vacuum}}{\text{speed of light in medium}}$ [2].
(c) $n = \frac{c}{v} \Rightarrow 1.5 = \frac{3.0 \times 10^8}{v}$
$v = \frac{3.0 \times 10^8}{1.5} = 2.0 \times 10^8$ m/s [2].

Section C

16.
(a) Circuit Diagram:

• Power supply, switch, variable resistor, ammeter in series with lamp [1].
• Voltmeter in parallel with the lamp [1].
• Correct symbols used [1].
  (b) (i) $R = \frac{V}{I} = \frac{2.0}{0.28} \approx 7.14 \, \Omega$ [2].
  (ii) As voltage increases, current increases, causing the filament to heat up [1]. The temperature of the filament increases [1]. The metal ions vibrate more, causing more collisions with free electrons, which increases resistance [1].
  (c) $P = VI = 2.5 \times 0.30 = 0.75$ W [2].

17.
(a) An alternating current in the primary coil produces a changing magnetic field in the iron core [1]. This changing magnetic field cuts through the secondary coil [1]. This induces an alternating voltage/current in the secondary coil via electromagnetic induction [1]. The iron core links the magnetic flux from primary to secondary [1].
(b) $\frac{V_p}{V_s} = \frac{N_p}{N_s} \Rightarrow \frac{240}{12} = \frac{2000}{N_s}$
$20 = \frac{2000}{N_s} \Rightarrow N_s = 100$ turns [2].
(c) $V_p I_p = V_s I_s$ (100% efficient)
$240 \times I_p = 12 \times 2.0$
$240 I_p = 24 \Rightarrow I_p = 0.1$ A [2].
(d) Energy loss as heat in the coils due to resistance / Eddy currents in the core / Hysteresis loss / Flux leakage [1].

18.
(a) Negative [1]. (Cloth gained electrons $\rightarrow$ Cloth is negative? No, "rubbed with a cloth that gained electrons". If cloth gained electrons, cloth is negative. Rod lost electrons, so Rod is Positive. Wait. Standard triboelectric: Plastic rubbed with cloth. Plastic usually gains electrons (negative). Cloth loses electrons (positive). The prompt says "cloth that gained electrons". This is physically unusual for standard plastic/cloth but we must follow the prompt. If cloth gained electrons, cloth is Negative. Therefore Rod is Positive.)
Correction based on prompt logic: If the cloth gained electrons, the rod lost them. Loss of electrons = Positive Charge.
Answer: Positive [1].

(b) The charged rod induces a charge separation in the neutral water molecules [1]. The side of the water stream closer to the rod develops an opposite charge (attraction) [1]. The force of attraction between the rod and the closer opposite charge is stronger than the repulsion from the further like charge, resulting in a net attraction [1].

(c) Touch the known positively charged rod to the cap of the electroscope to give it a positive charge (leaves diverge) [1]. Bring the unknown plastic rod near the cap. If the leaves diverge further, the rod is positive (like charges repel) [1]. If the leaves collapse slightly, the rod is negative (induction/neutralization) [1].

19.
(a) Radio waves [1].
(b) Sterilizing medical equipment / Detecting forged banknotes / Fluorescent lamps [1].
(c) Cell mutation / Cancer / Damage to living cells [1].
(d) $3.0 \times 10^8$ m/s [1].
(e) In a transverse wave (e.g., light), the oscillation/vibration is perpendicular to the direction of wave propagation [1]. In a longitudinal wave (e.g., sound), the oscillation/vibration is parallel to the direction of wave propagation [1]. Light does not require a medium; sound requires a medium [1].

20.
(a) The black matte container [1]. Black, matte surfaces are better emitters of thermal radiation than shiny, silver surfaces [1]. Therefore, it loses heat faster via radiation [1].
(b) 1. Initial temperature of water [1].
2. Volume/Mass of water [1].
(Also accept: Surface area of container, ambient temperature, material of container).
(c) Convection [1]. (Or Conduction).
(d) Place the container on an insulating mat / Use a lid / Wrap the container in insulating material (e.g., foam) [1].