AKU-EBClass IX (SSC Part I)CHAPTER-WISE ERQS • EXTENDED RESPONSE

Physics — Chapter-Wise Extended Response Questions (ERQs)

Practice 4–6 mark structured derivations, graph interpretations, and multi-part problems from 2024 & 2025 Annual Paper II, Model Paper II, and curriculum banks with official marking schemes and E-Marking scripts.

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UNIT 1 • 1 EXTENDED QUESTIONS (ERQS)

Unit 1: Measurement

2025 Syllabus Outcome (1.6.1): identify sources of errors (systematic and random) in the measurement;
Differentiate between systematic errors and random errors in physical measurements, giving two causes and one method of reducing each.
✓ MODEL SOLUTION (Full Marks Guidance)
Systematic errors occur consistently in one direction due to zero error of instruments or faulty calibration; reduced by applying zero correction. Random errors are unpredictable fluctuations due to parallax or environmental changes; reduced by taking the average of multiple repeated readings.
UNIT 2 • 5 EXTENDED QUESTIONS (ERQS)

Unit 2: Kinematics

2025 Syllabus Outcome (2.4.1): interpret distance-time graph and speed-time graph;
The following graph shows the path ABCDEF on which a boy walks.
Using the above graph calculate the
i.distance between A to B, B to C, C to D, D to E and E to F. (4 Marks)
ii.total distance covered. (1 Mark)
iii.total displacement covered. (1 Mark)
(Note: One small division is equal to one metre.)
Diagram for Q.6.a
📋 Official Checking Hints (Marking Rubric • 6 Marks)
  • 1 mark for both stationary intervals: Distance covered during A to B = 0 m AND during C to D = 0 m.
  • 1 mark for distance covered during B to C: 18 m - 2 m = 16 m.
  • 1 mark for distance covered during D to E: 18 m - 10 m = 8 m.
  • 1 mark for distance covered during E to F: 14 m - 10 m = 4 m.
  • 1 mark for Total Distance covered = 0 + 16 + 0 + 8 + 4 = 28 m.
  • 1 mark for Total Displacement covered = 0 + 16 + 0 + (-8) + 4 = 12 m (or final position 14 m minus initial position 2 m = 12 m).
✓ MODEL SOLUTION (Full Marks Guidance)
(i) Distance covered in each segment:
•From A to B: Position remains at 2 m2\text{ m}, so dAB=∣2−2∣=0 md_{AB} = |2 - 2| = 0\text{ m}
•From B to C: Position changes from 2 m2\text{ m} to 18 m18\text{ m}, so dBC=∣18−2∣=16 md_{BC} = |18 - 2| = 16\text{ m}
•From C to D: Position remains at 18 m18\text{ m}, so dCD=∣18−18∣=0 md_{CD} = |18 - 18| = 0\text{ m}
•From D to E: Position changes from 18 m18\text{ m} to 10 m10\text{ m}, so dDE=∣10−18∣=8 md_{DE} = |10 - 18| = 8\text{ m}
•From E to F: Position changes from 10 m10\text{ m} to 14 m14\text{ m}, so dEF=∣14−10∣=4 md_{EF} = |14 - 10| = 4\text{ m}
(ii) Total distance covered:
Total Distance=dAB+dBC+dCD+dDE+dEF=0+16+0+8+4=28 m\text{Total Distance} = d_{AB} + d_{BC} + d_{CD} + d_{DE} + d_{EF} = 0 + 16 + 0 + 8 + 4 = 28\text{ m}
(iii) Total displacement covered:
Total Displacement=0+16+0+(−8)+4=12 m\text{Total Displacement} = 0 + 16 + 0 + (-8) + 4 = 12\text{ m} (or xF−xA=14 m−2 m=12 mx_F - x_A = 14\text{ m} - 2\text{ m} = 12\text{ m}).

Examiner Commentary (Better Responses): Better responses accurately read the vertical axis values for each segment (0 m, 16 m, 0 m, 8 m, 4 m), summed their magnitudes to get 28 m for total distance, and accounted for the negative direction from D to E (-8 m) to find 12 m for total displacement.

⚠ WEAKER RESPONSE (Common Misconceptions & Lost Marks)
Weaker responses misread the grid divisions (e.g., writing half-values 0 m, 8 m, 0 m, 4 m, 2 m to get 14 m) or confused total distance with total displacement.
Examiner Suggestion for Improvement: Teachers should focus on reading graph axes and scale notes carefully ('One small division is equal to one metre') and distinguishing scalar path length (distance) from vector net position change (displacement).
2025 Syllabus Outcome (2.5.2): solve word problems related to free-falling bodies using the relation g = Δv/ Δt.
Zohan dropped a ball from a certain height. Its final velocity becomes 25 m/s when it reaches the ground. Now from the same height, if he throws the ball vertically downward with an initial velocity of 25 m/s, then calculate its final velocity. (Note: Take the value of acceleration due to gravity as 9.8 m/s2.)
📋 Official Checking Hints (Marking Rubric • 6 Marks)
  • Using appropriate equation of motion: v_f^2 = v_i^2 + 2gh or v_f = gt
  • Correct substitution and final numerical answer with units (m/s).
✓ MODEL SOLUTION (Full Marks Guidance)
First scenario:
Initial velocity (viv_i) = 0 m/s
Final velocity (vfv_f) = 25 m/s
Acceleration (gg) = 9.8 m/s2\text{m/s}^2
Using third equation of motion:
vf2=vi2+2gsv_f^2 = v_i^2 + 2gs
(25)2=(0)2+2(9.8)s(25)^2 = (0)^2 + 2(9.8)s
625=19.6s625 = 19.6s
s=62519.6≈31.89s = \frac{625}{19.6} \approx 31.89 m
Second scenario:
Initial velocity (viv_i) = 25 m/s
Distance (ss) = 31.89 m
Acceleration (gg) = 9.8 m/s2\text{m/s}^2
Using third equation of motion:
vf2=vi2+2gsv_f^2 = v_i^2 + 2gs
vf2=(25)2+2(9.8)(31.89)v_f^2 = (25)^2 + 2(9.8)(31.89)
vf2=625+625=1250v_f^2 = 625 + 625 = 1250
vf=1250≈35.36v_f = \sqrt{1250} \approx 35.36 m/s
2025 Syllabus Outcome (2.3.4): describe the universal speed limit of any object in the universe that is approximately 3 × 108 m/s;
A sports car starts from rest. It covers a distance of 900 m to attain a speed of 80 m/s. Determine the acceleration of the car and the time required to reach this speed.
📋 Official Checking Hints (Marking Rubric • 6 Marks)
  • Identifying given data (v_i = 0, v_f = 80 m/s, S = 900 m)
  • Using third equation of motion to find acceleration
  • Correct value of acceleration with units
  • Using first or second equation of motion to find time
  • Correct value of time with units
✓ MODEL SOLUTION (Full Marks Guidance)
Given: v_i = 0 m/s, v_f = 80 m/s, S = 900 m. Using 2aS = v_f^2 - v_i^2 => 2a(900) = (80)^2 - 0 => 1800a = 6400 => a = 3.56 m/s^2. Using v_f = v_i + at => 80 = 0 + (3.56)t => t = 22.48 s.
2025 Syllabus Outcome (2.5.2): solve word problems related to free-falling bodies using the relation g = Δv/ Δt.
Zohan dropped a ball from a certain height and its final velocity reached 25 m/s at the ground. From the same height, if he throws the ball vertically downward with an initial velocity of 25 m/s, calculate its final velocity (g = 9.8 m/s²).
✓ MODEL SOLUTION (Full Marks Guidance)
Stage 1 (Dropped, v_i = 0): 2gh = v_f² => h = 25² / (2 × 9.8) = 31.89 m. Stage 2 (Thrown, v_i = 25 m/s): v_f² = v_i² + 2gh = 625 + 625 = 1250 => v_f = 35.35 m/s.
2025 Syllabus Outcome (2.3.4): describe the universal speed limit of any object in the universe that is approximately 3 × 108 m/s;
A sports car starts from rest and covers 900 m to attain a speed of 80 m/s. Determine its acceleration and the time taken.
✓ MODEL SOLUTION (Full Marks Guidance)
Using 2aS = v_f² - v_i²: a = 80² / (2 × 900) = 6400 / 1800 = 3.56 m/s². Time t = (v_f - v_i)/a = 80 / 3.56 = 22.5 s.
UNIT 3 • 3 EXTENDED QUESTIONS (ERQS)

Unit 3: Dynamics

2025 Syllabus Outcome (3.4.4): solve word problems related to momentum and impulse;
A sports bicycle acquires a velocity of 25 m/s25\text{ m/s} from the state of rest when a force of 100 N100\text{ N} acts on it for 5 s5\text{ s}. Find
i.the mass of the bicycle. (4 Marks)
ii.the momentum of the bicycle after 5 s5\text{ s}. (2 Marks)
(Note: Extract data from the question and provide your answers in SI units.)
📋 Official Checking Hints (Marking Rubric • 6 Marks)
  • 1 mark for extracting and writing the correct given data (v_i = 0 m/s, v_f = 25 m/s, F = 100 N, t = 5 s).
  • 1 mark for finding the correct acceleration using a = (v_f - v_i)/t = (25 - 0)/5 = 5 m/s².
  • 1 mark for writing the correct formula for mass: m = F / a (from F = ma).
  • 1 mark for calculating mass with correct SI unit: m = 100 / 5 = 20 kg.
  • 1 mark for writing the correct formula for momentum: P = m × v_f (or P = F × t).
  • 1 mark for calculating momentum with correct SI unit: P = 20 × 25 = 500 kg·m/s (or 500 N·s).
✓ MODEL SOLUTION (Full Marks Guidance)
Given Data:
Initial velocity (viv_i) = 0 m/s0\text{ m/s}
Final velocity (vfv_f) = 25 m/s25\text{ m/s}
Force (FF) = 100 N100\text{ N}
Time (tt) = 5 s5\text{ s}
**(i) Mass of the bicycle (mm):**
First, find acceleration from the first equation of motion:
a=vf−vit=25−05=5 m/s2a = \frac{v_f - v_i}{t} = \frac{25 - 0}{5} = 5\text{ m/s}^2
By Newton's second law (F=maF = ma):
m=Fa=100 N5 m/s2=20 kgm = \frac{F}{a} = \frac{100\text{ N}}{5\text{ m/s}^2} = 20\text{ kg}
**(ii) Momentum of the bicycle after 5 s5\text{ s} (PP):**
P=m×vf=20 kg×25 m/s=500 kg⋅m/sP = m \times v_f = 20\text{ kg} \times 25\text{ m/s} = 500\text{ kg}\cdot\text{m/s} (or 500 N⋅s500\text{ N}\cdot\text{s}).

Examiner Commentary (Better Responses): Better responses extracted all given data with SI units, calculated acceleration a = 5 m/s², found m = F/a = 20 kg, and computed P = mv = 500 kg·m/s clearly.

⚠ WEAKER RESPONSE (Common Misconceptions & Lost Marks)
Weaker responses divided force directly by velocity (M = 100 N / 25 m/s = 4 N·s/m) or wrote an invalid formula for momentum such as P = (v_i - v_f)/t.
Examiner Suggestion for Improvement: Teachers should emphasise extracting data with SI symbols first, finding acceleration before applying F = ma (or using F = ΔP/t directly), and verifying SI units at each step.
2025 Syllabus Outcome (3.4.2): explain the relationship between force and momentum;
State and explain Newton’s 2nd law of motion and prove that the rate of change of momentum is equal to force.
📋 Official Checking Hints (Marking Rubric • 6 Marks)
  • Statement: Acceleration is directly proportional to applied net force and inversely proportional to mass.
  • Mathematical derivation showing a ∝ F and a ∝ 1/m leading to F = ma.
  • Definition of unit force: 1 Newton is force required to produce 1 m/s² acceleration in 1 kg mass.
✓ MODEL SOLUTION (Full Marks Guidance)
Statement: When a net force acts on a body, it produces acceleration in the direction of the force. The acceleration is directly proportional to the net force and inversely proportional to the mass of the body.
Explanation and Proof:
Let a body of mass 'mm' move with initial velocity 'viv_i'. Let a force 'FF' act on it for time 'tt' so that its velocity becomes 'vfv_f' .
Initial momentum (PiP_i) = mvimv_i
Final momentum (PfP_f) = mvfmv_f
Change in momentum = Pf−Pi=mvf−mvi=m(vf−vi)P_f - P_i = mv_f - mv_i = m(v_f - v_i)
Rate of change of momentum = m(vf−vi)t\frac{m(v_f - v_i)}{t}
Since acceleration a=vf−vita = \frac{v_f - v_i}{t},
Rate of change of momentum = mama
According to Newton's second law, F=maF = ma.
Therefore, F=Rate of change of momentumF = \text{Rate of change of momentum}.
2025 Syllabus Outcome (3.4.2): explain the relationship between force and momentum;
State and explain Newton's 2nd law of motion and prove that the rate of change of momentum is equal to the applied net force.
✓ MODEL SOLUTION (Full Marks Guidance)
By Newton's 2nd law, F = ma. For a body of mass m changing velocity from v_i to v_f in time t, a = (v_f - v_i)/t. Substituting gives F = m(v_f - v_i)/t = (mv_f - mv_i)/t = (P_f - P_i)/t = ΔP/Δt.
UNIT 4 • 1 EXTENDED QUESTIONS (ERQS)

Unit 4: Turning Effect of Forces

2025 Syllabus Outcome (4.4.2): classify the different types of equilibrium by using examples from everyday life;
State the principle of moments and explain the two conditions required for a rigid body to remain in complete equilibrium.
✓ MODEL SOLUTION (Full Marks Guidance)
Principle of moments: Sum of clockwise moments about a pivot equals sum of anticlockwise moments. Conditions of equilibrium: First condition ΣF = 0 (no linear acceleration); Second condition Στ = 0 (no angular acceleration).
UNIT 5 • 1 EXTENDED QUESTIONS (ERQS)

Unit 5: Deformation of Solids

2025 Syllabus Outcome (5.1.3): define spring constant;
Describe an experiment to determine the spring constant k of a helical spring and interpret the elastic limit on its load–extension curve.
✓ MODEL SOLUTION (Full Marks Guidance)
Suspend a helical spring vertically next to a millimetre scale, record pointer readings for incremental slotted weights, and plot Load F against Extension x. The gradient F/x in the linear region gives the spring constant k.
UNIT 6 • 4 EXTENDED QUESTIONS (ERQS)

Unit 6: Work, Energy and Power

2025 Syllabus Outcome (6.3.3): solve word problems related to the kinetic and potential energy;
The diagram shows a block is dropped from a point A and moves to a point B.
i.Calculate the change in potential energy (P.E.) of the block at point B with respect to ground. (3 Marks)
ii.Calculate the velocity of the block when it reaches to the point B. (3 Marks)
(Note: Take the value of acceleration due to gravity 'gg' as 10 m/s210\text{ m/s}^2.)
Diagram for Q.7.a
📋 Official Checking Hints (Marking Rubric • 6 Marks)
  • 1 mark for extracting the correct data: m = 10 kg, g = 10 m/s², vertical drop h = 30 m - 10 m = 20 m.
  • 1 mark for the correct potential energy formula: P.E. = mgh.
  • 1 mark for the correct change in potential energy from A to B: ΔP.E. = 10 × 10 × 20 = 2000 J.
  • 1 mark for equating loss in P.E. to gain in K.E. (K.E. = P.E.) or using 2gh = v_f² - v_i².
  • 1 mark for writing K.E. = ½mv² (or substituting into 2gh = v_f²).
  • 1 mark for the correct velocity at point B: v = √(2 × 10 × 20) = √400 = 20 m/s.
✓ MODEL SOLUTION (Full Marks Guidance)
Given Data:
Mass of block (mm) = 10 kg10\text{ kg}
Height at point A (hAh_A) = 30 m30\text{ m}
Height at point B (hBh_B) = 10 m10\text{ m}
Vertical distance fallen from A to B (h=hA−hBh = h_A - h_B) = 30 m−10 m=20 m30\text{ m} - 10\text{ m} = 20\text{ m}
Acceleration due to gravity (gg) = 10 m/s210\text{ m/s}^2
**(i) Change in Potential Energy (ΔP.E.\Delta P.E.) of the block from A to B:**
ΔP.E.=mgh=10 kg×10 m/s2×(30−10) m=10×10×20=2000 J\Delta P.E. = mgh = 10\text{ kg} \times 10\text{ m/s}^2 \times (30 - 10)\text{ m} = 10 \times 10 \times 20 = 2000\text{ J}
*(Note: The potential energy remaining at point B with respect to the ground is P.E.B=mghB=10×10×10=1000 JP.E._B = mgh_B = 10 \times 10 \times 10 = 1000\text{ J}, and the loss/change in P.E. from A to B is 3000 J−1000 J=2000 J3000\text{ J} - 1000\text{ J} = 2000\text{ J}.)*
**(ii) Velocity of the block at point B (vfv_f):**
By conservation of energy (Gain in K.E.=Loss in P.E.\text{Gain in } K.E. = \text{Loss in } P.E.) or the third equation of motion (2gh=vf2−vi22gh = v_f^2 - v_i^2 with vi=0v_i = 0):
12mvf2=2000 J  ⟹  vf2=2×200010=400\frac{1}{2}mv_f^2 = 2000\text{ J} \implies v_f^2 = \frac{2 \times 2000}{10} = 400
vf=400=20 m/sv_f = \sqrt{400} = 20\text{ m/s}.

Examiner Commentary (Better Responses): Better responses subtracted the heights to find h = 30 - 10 = 20 m, calculated P.E. = mgh = 10(10)(20) = 2000 J, and applied 2gh = v_f² - v_i² (or ½mv² = P.E.) to obtain v_f = √400 = 20 m/s.

⚠ WEAKER RESPONSE (Common Misconceptions & Lost Marks)
Weaker responses used the full 30 m height without subtracting 10 m, or wrote invalid formulas such as P.E. = S/g = 30/10 = 3 m/s² and v = d × g = 30 × 10 = 300 m/s².
Examiner Suggestion for Improvement: Ensure students use the correct formulas for potential energy (P.E. = mgh) and kinetic energy (K.E. = ½mv²) consistently and determine the vertical displacement between points A and B carefully from the diagram.
2025 Syllabus Outcome (6.6.2): solve word problems related to the concept of power;
Sara has mass of 30 kg. While running in the playground, she accelerates from 2 m/s to 4 m/s in 5 s. Calculate the power consumed in the given situation.
📋 Official Checking Hints (Marking Rubric • 6 Marks)
  • Formula F = ma substituted with m = 30 kg.
  • Correct acceleration calculation and units.
✓ MODEL SOLUTION (Full Marks Guidance)
Given data:
Mass (mm) = 30 kg
Initial velocity (viv_i) = 2 m/s
Final velocity (vfv_f) = 4 m/s
Time (tt) = 5 s
Work done (WW) = Change in kinetic energy (ΔKE\Delta KE)
W=12mvf2−12mvi2W = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2
W=12(30)(4)2−12(30)(2)2W = \frac{1}{2} (30) (4)^2 - \frac{1}{2} (30) (2)^2
W=15(16)−15(4)W = 15(16) - 15(4)
W=240−60=180W = 240 - 60 = 180 J
Power (PP) = Wt\frac{W}{t}
P=180 J5 s=36P = \frac{180 \text{ J}}{5 \text{ s}} = 36 W
2025 Syllabus Outcome (6.4.3): discuss that perpetual machines do not work;
Describe renewable and non-renewable sources of energy and give any FOUR examples of each source.
📋 Official Checking Hints (Marking Rubric • 6 Marks)
  • Definition of renewable sources of energy
  • Four examples of renewable sources of energy
  • Definition of non-renewable sources of energy
  • Four examples of non-renewable sources of energy
✓ MODEL SOLUTION (Full Marks Guidance)
Renewable energy sources are naturally replenished in a short time (e.g., Solar, Wind, Hydroelectric, Geothermal). Non-renewable sources are finite and cannot be replenished easily once consumed (e.g., Coal, Petroleum, Natural Gas, Nuclear/Uranium).
2025 Syllabus Outcome (6.6.3): explore different innovative energy transfer devices used in transportation and communication.
Sara has a mass of 30 kg. While running in the playground, she accelerates from 2 m/s to 4 m/s in 5 s. Calculate the power consumed in the given situation.
✓ MODEL SOLUTION (Full Marks Guidance)
a = (4 - 2)/5 = 0.4 m/s²; F = ma = 30 × 0.4 = 12 N; S = v_i t + ½at² = 2(5) + 0.5(0.4)(25) = 15 m; W = F × S = 12 × 15 = 180 J; P = W/t = 180 / 5 = 36 W.
UNIT 7 • 5 EXTENDED QUESTIONS (ERQS)

Unit 7: Pressure

2025 Syllabus Outcome (7.4.2): discuss the use of Pascal’s law (including hydraulic press and hydraulic brakes on vehicles);
The given figure shows the brake system of a vehicle.
i.Name the law which is used in the given brake system. (1 Mark)
ii.Calculate the pressure exerted by the liquid at G and H and the force that is applied at X. (5 Marks)
Diagram for Q.7.b
📋 Official Checking Hints (Marking Rubric • 6 Marks)
  • 1 mark for correctly identifying Pascal's law.
  • 1 mark for calculating pressure at G: P_G = F₁ / A₁ = 1000 N / 2 m² = 500 N/m² (or 500 Pa).
  • 1 mark for stating the pressure at H: P_H = 500 N/m² (or 500 Pa) because pressure is transmitted equally throughout an enclosed fluid.
  • 1 mark for using the correct transmitted pressure value P = 500 N/m².
  • 1 mark for writing the correct formula for force: F = P × A (or F₂ = P_H × A₂).
  • 1 mark for calculating the correct force at X: F = 500 × 100 = 50,000 N.
✓ MODEL SOLUTION (Full Marks Guidance)
(i) Name of the law:
The law used in the hydraulic brake system of a vehicle is Pascal's law.
(ii) Pressure at G, Pressure at H, and Force at X:
•**Pressure exerted at piston G (PGP_G):**
PG=Force (F1)Area (A1)=1000 N2 m2=500 N/m2 (or 500 Pa)P_G = \frac{\text{Force }(F_1)}{\text{Area }(A_1)} = \frac{1000\text{ N}}{2\text{ m}^2} = 500\text{ N/m}^2\text{ (or }500\text{ Pa)}
•**Pressure exerted at piston H (PHP_H):**
According to Pascal's law, pressure applied to an enclosed liquid is transmitted undiminished to every portion of the liquid. Therefore:
PH=PG=500 N/m2 (or 500 Pa)P_H = P_G = 500\text{ N/m}^2\text{ (or }500\text{ Pa)}
•**Force applied at X (F2F_2):**
Since PH=F2A2P_H = \frac{F_2}{A_2}:
F2=PH×A2=500 N/m2×100 m2=50,000 NF_2 = P_H \times A_2 = 500\text{ N/m}^2 \times 100\text{ m}^2 = 50,000\text{ N}.

Examiner Commentary (Better Responses): Better responses named Pascal's law, computed P_G = 1000 / 2 = 500 N/m², explained that P_H = 500 N/m² by Pascal's law, and solved F = P × A = 500 × 100 = 50,000 N with proper units.

⚠ WEAKER RESPONSE (Common Misconceptions & Lost Marks)
Weaker responses misidentified the law as Archimedes' principle, Boyle's law, or Newton's law of gravitation, or used an inverted formula F = P / A.
Examiner Suggestion for Improvement: Emphasise clear definitions of fluid mechanics laws (distinguishing Pascal's law for hydraulic systems from Archimedes' principle for buoyancy) and practice rearranging P = F/A.
2025 Syllabus Outcome (7.4.3): explain pressure beneath a liquid surface increase with depth and depends on the density of the liquid, i.e., P = ρgh;
Explain why walls of a dam filled with water are usually wedge-shaped. Relate this with the pressure of liquid at different depths.
📋 Official Checking Hints (Marking Rubric • 3 Marks)
  • Liquid pressure increases linearly with depth: P = ρgh.
  • Wedge shape provides greater thickness and strength at the bottom to withstand immense hydrostatic pressure.
✓ MODEL SOLUTION (Full Marks Guidance)
Liquid pressure increases with depth due to the weight of the water column above (P=ρghP = \rho g h). Consequently, the pressure exerted by water at the bottom of a dam is much greater than at the surface. To withstand this tremendously high pressure at greater depths, the walls of a dam are made thicker at the bottom and thinner at the top, giving them a wedge-shaped profile.
2025 Syllabus Outcome (7.4.4): solve word problems related to the relationship between pressure, depth and liquid density;
If the density of water is 1,000 kg/m3, then calculate the depth of water in a dam when the pressure at the bottom is 1,00,000 Pa. (Note: Take the value of acceleration due to gravity as 9.8 m/s2.)
📋 Official Checking Hints (Marking Rubric • 3 Marks)
  • Formula: P = ρgh rearranged to h = P / (ρg)
  • Substitution: h = P / (1000 * 9.8)
  • Calculation of depth h with metres (m) unit.
✓ MODEL SOLUTION (Full Marks Guidance)
Given data:
Density of water (ρ\rho) = 1000 kg/m31000 \text{ kg/m}^3
Pressure (PP) = 100,000 Pa100,000 \text{ Pa}
Acceleration due to gravity (gg) = 9.8 m/s29.8 \text{ m/s}^2
Formula:
P=ρghP = \rho g h
Calculation:
100,000=1000×9.8×h100,000 = 1000 \times 9.8 \times h
100,000=9800×h100,000 = 9800 \times h
h=100,0009800h = \frac{100,000}{9800}
h≈10.2h \approx 10.2 m
2025 Syllabus Outcome (7.4.1): state Pascal’s law;
Describe the types of forces that act on a body in a fluid, if it is I. partially immersed. II. completely immersed.
📋 Official Checking Hints (Marking Rubric • 4 Marks)
  • Description of forces acting on a partially immersed body (Weight downwards, Upthrust upwards)
  • Description of forces acting on a completely immersed body (Weight downwards, Upthrust/Buoyant force upwards, and potentially tension/normal force)
✓ MODEL SOLUTION (Full Marks Guidance)
I.Partially immersed: Weight of the body acting downward and Upthrust acting upward. II. Completely immersed: Weight acting downward, Upthrust acting upward, and Net force determining sinking or floating.
2025 Syllabus Outcome (7.4.4): solve word problems related to the relationship between pressure, depth and liquid density;
i.Explain why the walls of a dam filled with water are usually wedge-shaped. ii. If the density of water is 1,000 kg/m³, calculate the depth of water in a dam when the pressure at the bottom is 100,000 Pa (g = 9.8 m/s²).
✓ MODEL SOLUTION (Full Marks Guidance)
i.Liquid pressure increases with depth (P = ρgh), so water exerts a much larger lateral force near the bottom of the reservoir; a wider wedge-shaped base provides the structural strength to withstand this pressure. ii. Using P = ρgh: h = 100,000 / (1000 × 9.8) = 10.2 m.
UNIT 8 • 1 EXTENDED QUESTIONS (ERQS)

Unit 8: Thermal Physics

2025 Syllabus Outcome (8.2.5): describe melting, solidification, boiling and condensation in terms of energy transfer without the change in temperature;
Explain the temperature–time heating curve when crushed ice at -10°C is slowly heated until it turns into steam at 100°C, identifying why the temperature remains constant during melting and boiling.
✓ MODEL SOLUTION (Full Marks Guidance)
During melting at 0°C (latent heat of fusion) and boiling at 100°C (latent heat of vaporisation), absorbed heat energy goes into overcoming intermolecular forces rather than increasing average translational kinetic energy, so temperature stays constant.
UNIT 9 • 1 EXTENDED QUESTIONS (ERQS)

Unit 9: Transfer of Thermal Energy

2025 Syllabus Outcome (9.1.4): solve word problems related to the thermal conductivity of solid conductors;
A copper rod is 20 m long with a cross-sectional area of 4 m². If its two ends are kept at 200°C and 300°C respectively, calculate the amount of heat flowing across the rod in one second (k = 386.4 J/m·K·s).
✓ MODEL SOLUTION (Full Marks Guidance)
Using Q/t = kA(T₁ - T₂)/L: Q/t = (386.4 × 4 × (300 - 200)) / 20 = (386.4 × 400) / 20 = 7728 J/s (7728 J in 1 s).
UNIT 10 • 1 EXTENDED QUESTIONS (ERQS)

Unit 10: Nature of Science

2025 Syllabus Outcome (10.4.1): describe the importance of Physics in science, technology and engineering.
Differentiate between science, technology, and engineering, and describe how physics principles enable modern medical and communication technologies.
✓ MODEL SOLUTION (Full Marks Guidance)
Science discovers natural laws through inquiry; technology applies scientific knowledge to practical tools; engineering designs and optimises systems under real-world constraints (e.g., optical fibres using total internal reflection, MRI using electromagnetism).