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

TuitionGoWhere Practice Paper - Combined Science Secondary 3 (Answer Key)

Assessment: SA2 Practice Paper (Version 3 of 5)
Subject: Combined Science (Physical Sciences Focus)

Section A: Multiple Choice & Short Structured Questions

1. A
Reasoning: Length = End reading - Start reading = $14.5 - 2.0 = 12.5$ cm.

2. D
Reasoning: Displacement has both magnitude and direction. Speed, distance, and mass are scalars.

3. 250 m
Working:
Distance at constant speed = $20 \times 10 = 200$ m.
Distance during deceleration = Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 20 = 50$ m.
Total distance = $200 + 50 = 250$ m.

4. Energy cannot be created or destroyed [1]; it can only be converted from one form to another [1].
(Alternative: The total energy of an isolated system remains constant.)

5.
(a) 200 N [1]
(b) Since the velocity is constant, the acceleration is zero. According to Newton's First Law, the net force is zero, so the pushing force equals the frictional force [1].

6. 800 Pa
Working:
$P = \frac{F}{A}$
$P = \frac{400}{0.5}$
$P = 800$ Pa [2]
(1 mark for formula/substitution, 1 mark for answer with unit)

7.
(a) 0 Pa (or Zero) [1]
(The space above the mercury column is a vacuum.)
(b) It remains the same / No change [1]
(The height depends on atmospheric pressure, not the depth of the tube in the trough, provided the top remains above the mercury level in the trough.)

8. 400 $\text{J}/(\text{kg}^\circ\text{C})$
Working:
$E = mc\Delta\theta$
$2000 = 0.5 \times c \times (30 - 20)$
$2000 = 0.5 \times c \times 10$
$2000 = 5c$
$c = 400$ $\text{J}/(\text{kg}^\circ\text{C})$ [2]

9. Metals contain free electrons [1] that can move freely through the lattice structure, transferring kinetic energy rapidly from the hot end to the cold end [1].

10. 200 m/s
Working:
$v = f \lambda$
$v = 50 \times 4$
$v = 200$ m/s [2]

Section B: Structured Questions

11.
(a) Graph:

• Axes labeled correctly with units (Time/s, Distance/m) [1].
• Points plotted correctly [1].
• Smooth curve drawn through points [1].
  (Note: The curve should be parabolic, starting flat and getting steeper.)

(b) The trolley is accelerating / speeding up [1].
(Reason: The gradient of the distance-time graph is increasing.)

(c) 1.0 m/s
Working:
$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$
$\text{Speed} = \frac{2.0 \text{ m}}{2.0 \text{ s}} = 1.0$ m/s [2]

12.
(a) For an object in equilibrium, the sum of clockwise moments about any pivot is equal to the sum of anticlockwise moments about that same pivot [2].

(b) 3 N
Working:
Pivot at 50 cm.
Force 1: 4 N at 20 cm. Distance from pivot = $50 - 20 = 30$ cm = 0.3 m.
Moment 1 (Anticlockwise) = $4 \times 0.3 = 1.2$ Nm.
Force 2: $W$ at 80 cm. Distance from pivot = $80 - 50 = 30$ cm = 0.3 m.
Moment 2 (Clockwise) = $W \times 0.3$.
Equilibrium: $1.2 = W \times 0.3$
$W = \frac{1.2}{0.3} = 4$ N? Wait, let's re-read the diagram description.
Correction based on standard template logic:
If 4N is at 20cm (30cm from pivot) and W is at 80cm (30cm from pivot), W=4N.
Let's adjust the question logic for variety in Version 3:
Let's assume the question meant 4N at 10cm mark (40cm from pivot) to make it distinct from V1/V2.
Re-calculating based on typical exam variation:
If 4N is at 20cm (dist 30cm) and W is at 90cm (dist 40cm):
$4 \times 30 = W \times 40 \rightarrow 120 = 40W \rightarrow W = 3$ N.
Let's stick to the text provided in Q12:
4N at 20cm (dist 30cm). W at 80cm (dist 30cm).
$4 \times 30 = W \times 30 \rightarrow W = 4$ N.
However, to ensure distinctness from simple symmetry, let's assume the standard "trap" where students forget to subtract the pivot position.
Let's provide the answer for the text as written:
Anticlockwise Moment = $4 \text{ N} \times (50-20)\text{ cm} = 4 \times 30 = 120$ Ncm.
Clockwise Moment = $W \times (80-50)\text{ cm} = W \times 30$ Ncm.
$120 = 30W \rightarrow W = 4$ N.
(Self-Correction: To make this a "Version 3" distinct question, I will provide the answer for a slightly modified scenario often seen: If the weight was at 10cm. But I must answer the question AS WRITTEN in the prompt. As written, W=4N.)
Answer: 4 N [3]
(1 mark for correct distances, 1 mark for equation, 1 mark for answer)

(c) The 4 N weight must be moved closer to the pivot [1].
Reason: Moving W closer reduces its moment. To balance, the moment of the 4 N weight must also decrease, which requires decreasing its distance from the pivot [1].

13.
(a) 30,000 J
Working:
$\text{Work Done} = \text{Force} \times \text{Distance}$
$\text{Force} = \text{Weight} = mg = 200 \times 10 = 2000$ N.
$\text{Work} = 2000 \times 15 = 30,000$ J [3]

(b) 3,000 W
Working:
$\text{Power} = \frac{\text{Work Done}}{\text{Time}}$
$\text{Power} = \frac{30,000}{10} = 3,000$ W [2]

(c) 75%
Working:
$\text{Efficiency} = \frac{\text{Useful Energy Output}}{\text{Total Energy Input}} \times 100\%$
$\text{Efficiency} = \frac{30,000}{40,000} \times 100\% = 75\%$ [2]

14.
(a) Refractive index $n$ is the ratio of the speed of light in vacuum (or air) to the speed of light in the medium [1].
$n = \frac{c}{v}$ or $n = \frac{\sin i}{\sin r}$ [1].

(b) 1.52 (or 1.5)
Working:
$n = \frac{\sin i}{\sin r}$
$n = \frac{\sin 40^\circ}{\sin 25^\circ}$
$n = \frac{0.6428}{0.4226} \approx 1.52$ [2]

(c) The image is virtual / upright / laterally inverted / same size as object / same distance behind mirror as object is in front. [1]
(Any one correct property)

15.
(a) Circuit Diagram:

• Power supply symbol [1].
• Resistor, Ammeter, Rheostat in series [1].
• Voltmeter in parallel across the fixed resistor [1].

(b) 5 $\Omega$
Working:
$R = \frac{V}{I}$
Using any pair, e.g., $V=2.0, I=0.4$:
$R = \frac{2.0}{0.4} = 5$ $\Omega$ [2]

Section C: Free Response & Application

16.
(a) 2 m/s²
Working:
$\text{Acceleration} = \frac{\text{Change in Velocity}}{\text{Time Taken}}$
$a = \frac{10 - 0}{5} = 2$ m/s² [2]

(b) 150 m
Working:
Distance = Area under graph.
Area 1 (Triangle, 0-5s): $\frac{1}{2} \times 5 \times 10 = 25$ m.
Area 2 (Rectangle, 5-15s): $10 \times 10 = 100$ m.
Area 3 (Triangle, 15-20s): $\frac{1}{2} \times 5 \times 10 = 25$ m.
Total Distance = $25 + 100 + 25 = 150$ m [3]

(c) The forward driving force is equal in magnitude and opposite in direction to the resistive forces (friction/air resistance) [1]. Therefore, the resultant force is zero, and there is no acceleration [1].

17.
(a) The silvered surfaces are good reflectors of infrared radiation (heat) [1]. They reflect radiant heat back into the liquid (or back in from the surroundings), reducing heat loss by radiation [1].

(b) The vacuum contains no particles/molecules [1]. Therefore, heat cannot be transferred by conduction or convection, as these methods require a medium [1].

(c) Plastic and cork are poor conductors of heat (insulators) [1]. This reduces heat loss by conduction through the stopper.

18.
(a) 3 $\Omega$
Working:
Total Voltage $V = 12$ V. Total Current $I = 2$ A.
Total Resistance $R_T = \frac{V}{I} = \frac{12}{2} = 6$ $\Omega$.
Since lamps are identical and in series, $R_T = R_1 + R_2 = 2R$.
$2R = 6 \rightarrow R = 3$ $\Omega$ [3]

(b) Lamp L2 will go out / not light up [1].
Reason: In a series circuit, there is only one path for current. If L1 breaks, the circuit is open, and current cannot flow through L2 [1].

(c) 12 V [1]
(In parallel, voltage across each branch is equal to the source voltage.)

19.
(a) The galvanometer needle deflects (moves) to one side as the magnet enters the coil [1], returns to zero when the magnet is fully inside/stationary [1], and deflects to the opposite side as the magnet leaves the coil [1].
(Explanation: Changing magnetic field lines cutting the coil induces an EMF/current. Direction changes as motion relative to coil changes.)

(b) Any two of:

1. Use a stronger magnet [1].
2. Increase the number of turns on the coil [1].
3. Move the magnet faster [1].

20.
(a) Kinetic energy (or Mechanical energy) to Electrical energy [1].

(b) As the coil rotates, the sides of the coil cut the magnetic field lines in opposite directions during each half-turn [1]. This causes the direction of the induced current to reverse every half-rotation, producing an alternating voltage [1].

(c) Any one of:

• Increase the speed of rotation [1].
• Increase the strength of the magnetic field [1].
• Increase the number of turns on the coil [1].