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Physics IX · 2025 · E-Marking Notes (Questions & Solved Better Responses)

AGA KHAN UNIVERSITY EXAMINATION BOARD
Notes from E-Marking Centre on SSC-I Physics Annual Examinations 2025
Digital Edition: Questions, Marking Schemes & Solved Better Responses (2025 Syllabus Aligned)
Q.1Constructed Response Questions (CRQs)
2025 Syllabus Outcome (1.6.1): identify sources of errors (systematic and random) in the measurement;
Question Text
i.How many significant figures are there in 0.0102030?
ii.List the rules of significant figures applied on the number given part (i).
Checking Hints (Marking Scheme • 2 Marks)
  • 1 mark for writing the correct number of significant figures (6 significant figures)
  • 1 mark for writing any ONE rule of significant figures
✓ BETTER RESPONSE (SOLVED MODEL ANSWER)Full Marks (2/2)
(i) Identification of Measuring Instrument (1 Mark):
•Vernier Callipers (used to measure internal/external diameter and depth up to 0.1 mm0.1\text{ mm} or 0.01 cm0.01\text{ cm}).
(ii) Least Count of Vernier Callipers (1 Mark):
Least Count (L.C.)=Smallest reading on main scaleTotal number of divisions on Vernier scale=0.1 mm=0.01 cm\text{Least Count (L.C.)} = \frac{\text{Smallest reading on main scale}}{\text{Total number of divisions on Vernier scale}} = 0.1\text{ mm} = 0.01\text{ cm}
Q.2Constructed Response Questions (CRQs)
2025 Syllabus Outcome (4.4.4): explain different conditions of equilibrium with examples;
Question Text
Describe the THREE states of equilibrium on the basis of the centre of gravity.
Checking Hints (Marking Scheme • 3 Marks)
  • 1 mark for describing each state of equilibrium (3 required). Note: No mark will be awarded for mentioning the name only.
✓ BETTER RESPONSE (SOLVED MODEL ANSWER)Full Marks (3/3)
Effect of Lowering Centre of Gravity on Stability (3 Marks):
1.Condition for Stability: An object's stability increases when its **centre of gravity (GG) is kept as low as possible and its base area** is made as wide as possible.
2.Reason: Lowering the centre of gravity increases the tilt angle required before the vertical line of action of weight (W=mgW = mg) falls outside the base of support.
3.Application: Racing cars and double-decker buses are designed with heavy chassis/engines at the bottom and wide wheelbases so that the restoring moment returns them to stable equilibrium without toppling.
Q.3Constructed Response Questions (CRQs)
Out of 2025 Syllabus2 Marks • Level U
Question Text
The orbits of comets are large and elliptical as compared to the orbits of planets which are smaller and circular. Describe the difference in the shapes of orbits with the help of TWO reasons.
Checking Hints (Marking Scheme • 2 Marks)
  • 1 mark for describing each reason (Any 2 required)
✓ BETTER RESPONSE (SOLVED MODEL ANSWER)Full Marks (2/2)
Given Data:
•Mass of first car (m1m_1) = 1000 kg1000\text{ kg}
•Mass of second car (m2m_2) = 1000 kg1000\text{ kg}
•Distance between centres (dd) = 2 m2\text{ m}
•Universal gravitational constant (GG) = 6.673×10−11 N⋅m2/kg26.673 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2
Formula:
F=Gm1m2d2F = G\frac{m_1 m_2}{d^2}
Substitution & Calculation:
F=(6.673×10−11)×1000×1000(2)2=6.673×10−54F = \frac{(6.673 \times 10^{-11}) \times 1000 \times 1000}{(2)^2} = \frac{6.673 \times 10^{-5}}{4}
F=1.67×10−5 NF = 1.67 \times 10^{-5}\text{ N}
Q.4Constructed Response Questions (CRQs)
2025 Syllabus Outcome (7.3.1): explain atmospheric pressure;
Question Text
Write any THREE differences between evaporation and boiling.
Checking Hints (Marking Scheme • 3 Marks)
  • 1 mark for EACH difference (Any 3 required)
✓ BETTER RESPONSE (SOLVED MODEL ANSWER)Full Marks (3/3)
Dependence of Liquid Pressure on Depth (3 Marks):
1.Hydrostatic Pressure Formula: The pressure exerted by a liquid column at depth hh is given by:
P=ρghP = \rho g h
where ρ\rho is the density of the liquid, gg is acceleration due to gravity, and hh is depth below the free surface.
2.**Direct Proportionality (P∝hP \propto h):** Since ρ\rho and gg are constant for a given liquid, hydrostatic pressure increases linearly with depth (P∝hP \propto h).
3.Consequence: Water emerges with the greatest horizontal speed and range from the lowest hole of a container (and dam walls are built thickest at the bottom) to withstand the maximum pressure at the greatest depth.
Q.5Constructed Response Questions (CRQs)
2025 Syllabus Outcome (8.2.2): explain the thermal expansion of liquids as the: a. real expansion, b. apparent expansion;
Question Text
A copper rod is 20 m long with a cross-sectional area of 4 m². If its two ends are kept at temperatures 200°C and 300°C respectively, calculate the amount of heat flowing across the rod in one second. (Note: Take the co-efficient of thermal conductivity as 386.4 J/m.K.s.)
Checking Hints (Marking Scheme • 3 Marks)
  • 1 mark for using the correct formula: Q/t = kA(T₁ - T₂)/L
  • 1 mark for the correct substitution: (386.4 × 4 × 100) / 20
  • 1 mark for writing the correct answer with SI unit: 7728 J (or J/s)
✓ BETTER RESPONSE (SOLVED MODEL ANSWER)Full Marks (3/3)
Given Data:
•Mass of water (mm) = 2 kg2\text{ kg}
•Initial temperature (T1T_1) = 20 ∘C20\ ^\circ\text{C}
•Final temperature (T2T_2) = 100 ∘C100\ ^\circ\text{C}
•Change in temperature (ΔT=T2−T1\Delta T = T_2 - T_1) = 100−20=80 ∘C=80 K100 - 20 = 80\ ^\circ\text{C} = 80\text{ K}
•Specific heat capacity of water (cc) = 4200 J⋅kg−1⋅K−14200\text{ J}\cdot\text{kg}^{-1}\cdot\text{K}^{-1}
Formula:
ΔQ=mcΔT\Delta Q = m c \Delta T
Calculation:
ΔQ=2 kg×4200 J⋅kg−1⋅K−1×80 K=672,000 J=6.72×105 J (or 672 kJ)\Delta Q = 2\text{ kg} \times 4200\text{ J}\cdot\text{kg}^{-1}\cdot\text{K}^{-1} \times 80\text{ K} = 672,000\text{ J} = 6.72 \times 10^5\text{ J}\text{ (or }672\text{ kJ)}
Q.6(a)Extended Response Questions (ERQs)
Out of 2025 Syllabus6 Marks • Level A
Question Text
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.)
Checking Hints (Marking Scheme • 6 Marks)
  • 1 mark for extracting the correct data
  • 1 mark for writing the correct formula (2gh = vf² - vi²)
  • 1 mark for substituting values for the dropped ball (vi = 0, vf = 25 m/s)
  • 1 mark for finding the height h = 31.89 m
  • 1 mark for substituting values for the thrown ball (vi = 25 m/s, h = 31.89 m)
  • 1 mark for calculating the final velocity vf = 35.35 m/s
✓ BETTER RESPONSE (SOLVED MODEL ANSWER)Full Marks (6/6)
Better responses applied 2gh = vf² - vi² first with vi = 0 m/s to find height (31.89 m) and then with vi = 25 m/s to find the new final velocity (35.35 m/s).
Q.6(b)Extended Response Questions (ERQs)
Out of 2025 Syllabus6 Marks • Level U
Question Text
State and explain Newton's 2nd law of motion and prove that the rate of change of momentum is equal to force.
Checking Hints (Marking Scheme • 6 Marks)
  • 1 mark for each logical/mathematical step from F = ma and momentum P = mv to F = (Pf - Pi)/t = ΔP/Δt (6 marks total)
✓ BETTER RESPONSE (SOLVED MODEL ANSWER)Full Marks (6/6)
Better responses stated F = ma, defined initial momentum Pi = mvi and final momentum Pf = mvf, wrote ΔP = m(vf - vi), and divided by time t to prove F = ΔP/t.
Q.7(a)Extended Response Questions (ERQs)
Out of 2025 Syllabus6 Marks • Level A
Question Text
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.
Checking Hints (Marking Scheme • 6 Marks)
  • 1 mark for listing given data (m = 30 kg, vi = 2 m/s, vf = 4 m/s, t = 5 s)
  • 1 mark for calculating acceleration a = (4 - 2)/5 = 0.4 m/s²
  • 1 mark for calculating force F = ma = 30 × 0.4 = 12 N
  • 1 mark for calculating distance S = vit + ½at² = 15 m (or work via ΔKE = ½m(vf² - vi²) = 180 J)
  • 1 mark for calculating work W = F × S = 180 J
  • 1 mark for calculating power P = W/t = 180/5 = 36 W
✓ BETTER RESPONSE (SOLVED MODEL ANSWER)Full Marks (6/6)
Better responses systematically calculated acceleration (0.4 m/s²), force (12 N), distance (15 m), work (180 J), and power (36 W) with proper SI units.
Q.7(b)Extended Response Questions (ERQs)
Out of 2025 Syllabus6 Marks • Level A
Question Text
i.Explain why the walls of a dam filled with water are usually wedge-shaped. Relate this to the pressure of liquid at different depths.
ii.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.)
Checking Hints (Marking Scheme • 6 Marks)
  • Part i: 1 mark for each reason relating depth, hydrostatic pressure P = ρgh, and thicker base wall strength (3 marks)
  • Part ii: 1 mark for formula P = ρgh, 1 mark for substitution 100,000 = 1000 × 9.8 × h, 1 mark for answer h = 10.2 m (3 marks)
✓ BETTER RESPONSE (SOLVED MODEL ANSWER)Full Marks (6/6)
Better responses explained that liquid pressure increases directly with depth (P = ρgh), requiring a wider, stronger base to withstand higher lateral forces, and calculated h = 100,000 / (1000 × 9.8) = 10.2 m.