AKU-EBClass X (SSC Part II)CHAPTER-WISE NUMERICALS & WORKED PROBLEMS

Physics — Chapter-Wise Physics Numericals & Formula Applications

Step-by-step solved numerical word problems from 2024 & 2025 Annual Papers I & II, Model Papers I & II, and chapter formula banks, mapped to 2025 syllabus SLOs with SI units and KaTeX equations.

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UNIT 11 • 6 NUMERICAL PROBLEMS

Unit 11: Wave Motion — Worked Numericals

2025 Syllabus Outcome (11.1.3): solve word problems related to time period of a simple pendulum;
A simple pendulum is attached with 4 m long string.
i.Find the time period of the simple pendulum if it is placed in an elevator moving upwards with an acceleration of 8 m/s^2.
ii.If the elevator is moving downwards with an acceleration of 8 m/s^2, then what will be its time period?
(Note: The value of acceleration due to gravity 'g' is 9.8 m/s^2.)
📋 Official Checking Hints (Marking Scheme):
  • Formula for time period of simple pendulum
  • Calculating effective 'g' for upward acceleration (g + a)
  • Correct calculation and unit for upward movement time period
  • Calculating effective 'g' for downward acceleration (g - a)
  • Correct calculation and unit for downward movement time period
  • Proper substitution and rounding in calculations
Given: Length L = 4 m, g = 9.8 m/s^2, a = 8 m/s^2
i.Elevator moving upwards:
Effective gravity g' = g + a = 9.8 + 8 = 17.8 m/s^2
T = 2 * pi * sqrt(L / g') = 2 * 3.14 * sqrt(4 / 17.8) = 2.97 seconds.
ii.Elevator moving downwards:
Effective gravity g' = g - a = 9.8 - 8 = 1.8 m/s^2
T = 2 * pi * sqrt(L / g') = 2 * 3.14 * sqrt(4 / 1.8) = 9.36 seconds.
2025 Syllabus Outcome (11.4.4): solve word problems using the relation f = 1/ T and v = f λ;
A wave generator produces 25 complete waves in 5 seconds with a wavelength of 0.4 m. Calculate the frequency, time period, and wave speed.
Key Formula: f = N/t, T = 1/f, v = fλ
f = 25 / 5 = 5 Hz; T = 1 / 5 = 0.2 s; v = 5 Hz × 0.4 m = 2.0 m/s.
2025 Syllabus Outcome (11.4.2): analyse key characteristics of waves by interpreting graphical representations of waveforms;
If the frequency of a wave is doubled, then the time period of the wave will
The time period (TT) and frequency (ff) are inversely proportional to each other, given by the relation T=1fT = \frac{1}{f}. If the frequency is doubled (2f2f), the time period becomes 12f\frac{1}{2f}, which is half of the original time period.
2025 Syllabus Outcome (11.1.3): solve word problems related to time period of a simple pendulum;
If the time period of a simple pendulum is 2 s on the surface of the Earth, then the time period of the same pendulum on the surface of the Moon will be
The time period of a simple pendulum is given by T=2πLgT = 2\pi\sqrt{\frac{L}{g}}. Since the acceleration due to gravity on the Moon (gmg_m) is less than that on Earth (geg_e) (gm≈16geg_m \approx \frac{1}{6}g_e), a smaller value of gg results in a larger time period TT. Therefore, the time period on the Moon will be greater than 2 s.
2025 Syllabus Outcome (11.1.2): discuss the factors on which the time period of a simple pendulum depends;
Which of the following pendulums has the maximum time period?
PendulumLengthMass
A50 cm50\text{ cm}50 gm50\text{ gm}
B60 cm60\text{ cm}40 gm40\text{ gm}
C70 cm70\text{ cm}30 gm30\text{ gm}
D80 cm80\text{ cm}10 gm10\text{ gm}
The time period of a simple pendulum is given by T=2πLgT = 2\pi\sqrt{\frac{L}{g}}. It is directly proportional to the square root of its length (L\sqrt{L}) and is completely independent of the mass of the bob. Since Pendulum D has the greatest length (80 cm80\text{ cm}), it has the maximum time period.
2025 Syllabus Outcome (11.4.4): solve word problems using the relation f = 1/ T and v = f λ;
If a periodic wave of wavelength 0.5 m0.5\text{ m} has a frequency of 2 Hz2\text{ Hz}, then the speed of the wave will be
Using the wave speed equation v=fλv = f\lambda, where frequency f=2 Hzf = 2\text{ Hz} and wavelength λ=0.5 m\lambda = 0.5\text{ m}:
v=2 Hz×0.5 m=1.0 m/sv = 2\text{ Hz} \times 0.5\text{ m} = 1.0\text{ m/s}.
UNIT 12 • 2 NUMERICAL PROBLEMS

Unit 12: Sound — Worked Numericals

2025 Syllabus Outcome (12.3.3): examine a method involving a measurement of distance and time for determining the speed of sound in air;
A ship sends an ultrasound pulse to the seabed and receives the echo after 1.6 s. If the speed of sound in seawater is 1500 m/s, calculate the depth of the sea.
Key Formula: d = (v × t) / 2
d = (1500 m/s × 1.6 s) / 2 = 1200 m.
2025 Syllabus Outcome (12.2.1): define the terms ‘pitch’, ‘loudness’ and ‘quality of sound’;
Four vibrating objects X1, X2, X3 and X4 are producing sounds. The object that produces the loudest sound is
Loudness of sound depends on both the surface area and the amplitude of vibration (specifically proportional to the square of the amplitude and the area). Object X4 has the maximum area (10 m²) combined with the largest amplitude (5 cm), making it produce the loudest sound.
UNIT 13 • 4 NUMERICAL PROBLEMS

Unit 13: Geometrical Optics — Worked Numericals

If an object is placed at a distance of 2 cm2\text{ cm} from a convex mirror of focal length of 4 cm4\text{ cm}, then find the position of the image formed.
📋 Official Checking Hints (Marking Scheme):
  • 1 mark for the correct mirror formula: 1/f = 1/p + 1/q.
  • 1 mark for the correct substitution with sign convention (for a convex mirror, f = -4 cm and p = +2 cm; or using standard positive focal length magnitude showing virtual image behind mirror).
  • 1 mark for the correct calculated image position with unit (q = -4/3 cm = -1.33 cm using f = -4 cm, or q = -4 cm as shown in E-Marking exemplar, indicating a virtual image behind the convex mirror).
Given Data:
•Object distance (pp) = +2 cm+2\text{ cm}
•Focal length of convex mirror (ff) = −4 cm-4\text{ cm} (virtual focus behind the mirror; note: in the AKU-EB E-Marking exemplar, f=4 cmf = 4\text{ cm} is substituted into 1f=1p+1q\frac{1}{f} = \frac{1}{p} + \frac{1}{q})
Formula:
1f=1p+1q  ⟹  1q=1f−1p\frac{1}{f} = \frac{1}{p} + \frac{1}{q} \implies \frac{1}{q} = \frac{1}{f} - \frac{1}{p}
Calculation (AKU-EB E-Marking Exemplar Method):
1q=14−12=1−24=−14\frac{1}{q} = \frac{1}{4} - \frac{1}{2} = \frac{1 - 2}{4} = -\frac{1}{4}
q=−4 cmq = -4\text{ cm}
*(Note: Using strict Cartesian sign convention f=−4 cmf = -4\text{ cm} gives 1q=−14−12=−34  ⟹  q=−1.33 cm\frac{1}{q} = -\frac{1}{4} - \frac{1}{2} = -\frac{3}{4} \implies q = -1.33\text{ cm}. In both conventions, the negative sign of qq indicates that a virtual, upright image is formed behind the convex mirror.)*
2025 Syllabus Outcome (13.2.4): illustrate the passage of light ray through parallel-sided transparent material (glass slab);
The floor of a filled water tank appears at an apparent depth of 3.0 m. If the refractive index of water is 1.33, calculate the real depth of the water.
Key Formula: n = Real Depth / Apparent Depth
Real Depth = n × Apparent Depth = 1.33 × 3.0 m = 3.99 m ≈ 4.0 m.
2025 Syllabus Outcome (13.5.3): differentiate between real and virtual images;
If an image is formed at a distance of 32 cm from a concave mirror with focal length of 12 cm, then the object’s distance is
Using the mirror formula: 1f=1p+1q\frac{1}{f} = \frac{1}{p} + \frac{1}{q}. For a concave mirror, focal length f=−12f = -12 cm, and assuming real image q=−32q = -32 cm. Substituting values: 1−12=1p+1−32  ⟹  1p=−112+132=−8+396=−596\frac{1}{-12} = \frac{1}{p} + \frac{1}{-32} \implies \frac{1}{p} = -\frac{1}{12} + \frac{1}{32} = \frac{-8 + 3}{96} = \frac{-5}{96}. Thus, p=−965=−19.2p = -\frac{96}{5} = -19.2 cm.
2025 Syllabus Outcome (13.2.4): illustrate the passage of light ray through parallel-sided transparent material (glass slab);
The floor of a filled water tank appears at a depth of 3 m. If the refractive index of water is 1.33, then the actual depth of the water is
Refractive index n=Real depthApparent depthn = \frac{\text{Real depth}}{\text{Apparent depth}}. Therefore, Real depth=n×Apparent depth=1.33×3 m=3.99 m\text{Real depth} = n \times \text{Apparent depth} = 1.33 \times 3\text{ m} = 3.99\text{ m}.
UNIT 14 • 5 NUMERICAL PROBLEMS

Unit 14: Electrostatics — Worked Numericals

2025 Syllabus Outcome (14.2.2): explain Coulomb’s law;
Calculate the electrostatic force of repulsion between two electrons separated by a distance of 1.0 m in air (k = 9 × 10⁹ N·m²/C², q_e = 1.6 × 10⁻¹⁹ C).
Key Formula: F = k q₁ q₂ / r²
F = (9 × 10⁹ × (1.6 × 10⁻¹⁹)²) / 1.0² = 2.304 × 10⁻²⁸ N.
Two capacitors of capacitance 6 µF and 3 µF are connected in a series across a cell of 18 V. The equivalent (effective) capacitance will be
For two capacitors connected in series, the equivalent capacitance CeqC_{eq} is given by 1Ceq=1C1+1C2=16+13=1+26=36=12\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{6} + \frac{1}{3} = \frac{1 + 2}{6} = \frac{3}{6} = \frac{1}{2}. Thus, Ceq=2 \muFC_{eq} = 2\text{ \mu F}.
2025 Syllabus Outcome (14.2.2): explain Coulomb’s law;
The electric force of repulsion between two electrons at a distance of 1 m is
(Note: The value of the constant of proportionality is 9×109 Nm2/C29 \times 10^9\text{ Nm}^2/\text{C}^2 and the charge on electron is 1.6×10−19 C1.6 \times 10^{-19}\text{ C}.)
Using Coulomb's law: F=k⋅q1⋅q2r2=9×109×(1.6×10−19)×(1.6×10−19)(1)2=9×109×2.56×10−381=23.04×10−29=2.304×10−28 NF = \frac{k \cdot q_1 \cdot q_2}{r^2} = \frac{9 \times 10^9 \times (1.6 \times 10^{-19}) \times (1.6 \times 10^{-19})}{(1)^2} = \frac{9 \times 10^9 \times 2.56 \times 10^{-38}}{1} = 23.04 \times 10^{-29} = 2.304 \times 10^{-28}\text{ N}.
The equivalent capacitance of the given combination of the capacitors will be
(Note: C1=5 \muFC_1 = 5\text{ \mu F}, C2=5 \muFC_2 = 5\text{ \mu F}, C3=5 \muFC_3 = 5\text{ \mu F} connected in parallel based on standard textbook diagram layouts)
Looking at standard parallel combinations for identical capacitors in AKU-EB problems where three equal capacitors are in parallel: Ceq=C1+C2+C3=5+5+5=15 \muFC_{eq} = C_1 + C_2 + C_3 = 5 + 5 + 5 = 15\text{ \mu F}.
2025 Syllabus Outcome (14.2.3): solve word problems related to the electrostatic charges by using Coulomb’s law;
If two capacitors C1=6 μFC_1 = 6\ \mu\text{F} and C2=12 μFC_2 = 12\ \mu\text{F} are connected in series, then the equivalent capacitance will be
For two capacitors connected in series:
1Ceq=1C1+1C2=16+112=2+112=312=14 μF−1  ⟹  Ceq=4 μF\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{6} + \frac{1}{12} = \frac{2 + 1}{12} = \frac{3}{12} = \frac{1}{4}\ \mu\text{F}^{-1} \implies C_{\text{eq}} = 4\ \mu\text{F}.
UNIT 15 • 3 NUMERICAL PROBLEMS

Unit 15: Electric Current — Worked Numericals

2025 Syllabus Outcome (15.6.4): calculate the total energy cost based on the cost price of kWh;
An electric heater rated at 2000 W operates on a 250 V mains supply for 3 hours daily. Calculate the current drawn and the energy consumed in kWh in 30 days.
Key Formula: I = P / V; E(kWh) = P(kW) × t(h)
I = 2000 W / 250 V = 8.0 A. Total hours = 3 × 30 = 90 h. Energy = 2.0 kW × 90 h = 180 kWh.
2025 Syllabus Outcome (15.3.3): describe the phenomenon of superconductivity in solids;
If the length and diametre of a solid cylindrical shape conductor is doubled, then the resistance will
Resistance R=ρLA=ρLπr2=ρ4Lπd2R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^2} = \rho \frac{4L}{\pi d^2}. If length LL is doubled (2L2L) and diameter dd is doubled (2d2d), then new resistance R′=ρ2Lπ(2d)2=ρ2L4πd2=12(ρLA)R' = \rho \frac{2L}{\pi (2d)^2} = \rho \frac{2L}{4\pi d^2} = \frac{1}{2} \left(\rho \frac{L}{A}\right), which is half of the original resistance.
2025 Syllabus Outcome (15.6.4): calculate the total energy cost based on the cost price of kWh;
If a 100 Watt100\text{ Watt} bulb glows for 200 hours200\text{ hours}, then the energy consumed by it will be
Electrical energy in kilowatt-hours (kWh\text{kWh}) is calculated by:
E=Power (in Watts)×Time (in hours)1000=100 W×200 h1000=20,0001000=20 kWhE = \frac{\text{Power (in Watts)} \times \text{Time (in hours)}}{1000} = \frac{100\text{ W} \times 200\text{ h}}{1000} = \frac{20,000}{1000} = 20\text{ kWh}.
UNIT 16 • 1 NUMERICAL PROBLEMS

Unit 16: D.C. Circuits — Worked Numericals

2025 Syllabus Outcome (16.1.3): calculate the effective/ equivalent resistance of resistors connected in series, parallel and both;
Two resistors of 6 Ω and 12 Ω are connected in parallel across a 12 V battery. Calculate the equivalent resistance and total current from the battery.
Key Formula: 1/R_eq = 1/R₁ + 1/R₂; I = V / R_eq
R_eq = (6 × 12) / (6 + 12) = 72 / 18 = 4.0 Ω. Total current I = 12 V / 4.0 Ω = 3.0 A.
UNIT 17 • 1 NUMERICAL PROBLEMS

Unit 17: Magnetism and Electromagnetism — Worked Numericals

2025 Syllabus Outcome (17.3.2): describe the effect of changing the magnitude and direction of the current on the magnetic field;
Why does a soft-iron core placed inside a current-carrying solenoid greatly increase its magnetic field strength, yet lose magnetism immediately when the current is switched off?
Key Formula: B_solenoid ∝ μ_r n I
Soft iron has high magnetic permeability so its magnetic domains align rapidly with the solenoid field, and low retentivity so domains randomise as soon as current stops.
UNIT 18 • 1 NUMERICAL PROBLEMS

Unit 18: Electromagnetic Induction — Worked Numericals

2025 Syllabus Outcome (18.5.2): state the two types of transformers and their uses;
A step-down transformer has 1000 turns on its primary coil and 50 turns on its secondary coil. If the primary is connected to 240 V AC, calculate the secondary output voltage.
Key Formula: V_s / V_p = N_s / N_p
V_s = 240 V × (50 / 1000) = 12 V.
UNIT 19 • 1 NUMERICAL PROBLEMS

Unit 19: Electromagnetic Waves and Electronics — Worked Numericals

2025 Syllabus Outcome (19.5.6): use universal gates (NAND and NOR) to create NOT, OR and AND;
For a two-input NAND gate with inputs A = 1 and B = 1, determine the output X, and state how a NAND gate can be wired to act as a NOT gate.
Key Formula: X = NOT(A AND B)
When A = 1 and B = 1, A AND B = 1, so NAND output X = 0. Joining both inputs together (A = B) converts a NAND gate into a NOT gate.
UNIT 20 • 4 NUMERICAL PROBLEMS

Unit 20: Atomic and Nuclear Physics — Worked Numericals

2025 Syllabus Outcome (20.6.2): calculate the half-life of radioactive elements;
Suppose a radioactive element has an initial number of atoms (XX) kept in a nuclear power station.
Calculate the number of radioactive atoms left after 1010 half-lives.
📋 Official Checking Hints (Marking Scheme):
  • 1 mark for writing the correct half-life decay formula: N = N₀ × (1/2)^n (or N = N₀ / 2^t).
  • 1 mark for substituting the correct values: N₀ = X and n = 10 -> N = X × (1/2)^10.
  • 1 mark for writing the correct final answer: N = X / 1024 (or (1/1024)X atoms remaining).
Given Data:
•Initial number of radioactive atoms (N0N_0) = XX
•Number of half-lives elapsed (nn) = 1010
Formula:
N=N0×(12)nN = N_0 \times \left(\frac{1}{2}\right)^n
Substitution & Calculation:
N=X×(12)10=X×1210=X1024N = X \times \left(\frac{1}{2}\right)^{10} = X \times \frac{1}{2^{10}} = \frac{X}{1024}
Answer:
After 1010 half-lives, X1024\frac{X}{1024} (or 11024th\frac{1}{1024}\text{th} of the original number of atoms XX) will remain undecayed.
2025 Syllabus Outcome (20.6.3): draw graphs showing decay curves of these elements;
If a radioactive isotope has a half-life of 60 seconds and an initial mass of 80 g, calculate the quantity of the undecayed isotope remaining after 120 seconds.
Key Formula: N = N₀ × (1/2)^n, where n = t / T_½
Number of half-lives n = 120 s / 60 s = 2. Remaining mass N = 80 g × (1/2)² = 80 / 4 = 20 g.
2025 Syllabus Outcome (20.6.3): draw graphs showing decay curves of these elements;
If a radioactive element has a half-life of 60 seconds, then the quantity of the isotope after 120 seconds will be
Number of half-lives = Total time / half-life = 120 s / 60 s = 2 half-lives. Remaining quantity = (1/2)2=1/4(1/2)^2 = 1/4 (one-fourth) of the initial quantity.
2025 Syllabus Outcome (20.2.3): show various nuclides by using the symbol of proton number (Z), nucleon number (A) and the nuclide notation (X);
The number of electrons in the isotope of uranium 92U238{}_{92}\text{U}^{238} is
In a neutral atom represented as ZXA_Z\text{X}^A, the atomic number Z=92Z = 92 gives both the number of protons in the nucleus and the number of orbital electrons (9292). (Note: A=238A = 238 is the mass/nucleon number and A−Z=238−92=146A - Z = 238 - 92 = 146 is the number of neutrons).