About this IIT JEE Physics practice test

Mechanics, Electrostatics and Modern Physics carry the heaviest weight in JEE Physics, but the questions that actually cost students marks are the ones that quietly combine two chapters in a single problem — a banked curve with friction, or a photoelectric-effect question buried inside a de Broglie wavelength calculation. This set is built around exactly those combination questions, worked through with full unit tracking so a stray factor of ten never slips past unnoticed. Each answer shows the formula, the substitution and the final check, so you can see precisely where marks are usually lost. Once these stop feeling tricky, pull in a real JEE Mains paper and see how much faster you've gotten.

IIT JEE Physics Practice Test sample questions

These starter questions help you launch a physics mock test quickly. Swap them with your own worksheet, notebook, or textbook questions any time.

1. 1. In an experiment to determine the density of a small sphere, its mass is measured as 12.48 g ± 0.02 g and its radius as 1.50 cm ± 0.01 cm. Find the percentage error in the calculated value of density.

2. 2. A particle is projected from the ground with a speed of 40 m/s at an angle of 30° above the horizontal. Taking g = 10 m/s², find the speed and direction of the velocity vector 2 s after projection.

3. 3. A boatman wants to cross a 400 m wide river and land at a point directly opposite his starting point. The river flows at 4 km/h and the boat can move at 8 km/h relative to the water. Find the time taken to cross the river.

4. 4. Block A (4 kg) rests on a horizontal table (coefficient of kinetic friction μ = 0.2) and is connected by a light string over a frictionless pulley at the table's edge to block B (6 kg) hanging vertically. Taking g = 10 m/s², find the acceleration of the system and the tension in the string.

5. 5. A car moves on a banked circular track of radius 100 m banked at 30°. If the coefficient of friction between the tyres and the road is 0.2, find the maximum speed at which the car can travel without slipping. (g = 10 m/s²)

6. 6. A force F = (3x² + 2x) N acts on a 2 kg particle constrained to move along the x-axis. If the particle starts from rest at x = 0, find its speed when it reaches x = 3 m.

7. 7. A pump lifts water from a well of depth 20 m and delivers it through a pipe of cross-sectional area 0.01 m² with an exit speed of 5 m/s. Taking the density of water as 1000 kg/m³ and g = 10 m/s², find the power of the pump, assuming 100% efficiency.

8. 8. A solid sphere of mass 2 kg and radius 0.1 m starts from rest and rolls without slipping down an incline through a vertical height of 5 m. Find its speed at the bottom of the incline.

9. 9. A uniform rod of length 1.2 m and mass 3 kg, pivoted at one end, is released from rest in a horizontal position. Find its angular speed when it swings down to the vertical position. (g = 10 m/s²)

10. 10. A satellite orbits the Earth at a height of 3600 km above the surface. Taking the Earth's radius as 6400 km and GM(Earth) = 3.986 × 10^14 m³/s², find the satellite's orbital speed and time period.

11. 11. A rocket is fired vertically from the Earth's surface with a speed equal to half the escape velocity. Find the maximum height it reaches above the surface, in terms of the Earth's radius R.

12. 12. A steel ball of radius 2 mm and density 7800 kg/m³ falls through glycerine of density 1260 kg/m³ and coefficient of viscosity 0.83 Pa·s. Find its terminal velocity. (g = 9.8 m/s²)

13. 13. One mole of an ideal diatomic gas (γ = 7/5) at an initial temperature of 300 K is adiabatically compressed to half its original volume. Find the final temperature of the gas.

14. 14. A train approaches a stationary observer while sounding its whistle at a frequency of 500 Hz. If the train's speed is 30 m/s and the speed of sound in air is 340 m/s, find the frequency heard by the observer.

15. 15. Point charges of +4 μC and −4 μC are placed 6 cm apart. Find the magnitude of the electric field at a point on the perpendicular bisector of the line joining the charges, at a distance of 4 cm from the midpoint.

16. 16. A parallel plate capacitor has a plate area of 200 cm² and plate separation of 2 mm, and is connected to a 100 V battery. A dielectric slab of thickness 1 mm and dielectric constant K = 5 is inserted between the plates while the battery remains connected. Find the new capacitance and the charge stored. (ε0 = 8.85 × 10^-12 F/m)

17. 17. Three point charges of +2 μC each are placed at the vertices of an equilateral triangle of side 30 cm. Find the total electrostatic potential energy of the system.

18. 18. A battery of EMF 12 V and internal resistance 1 Ω is connected to an external circuit consisting of a 4 Ω resistor in series with a parallel combination of a 6 Ω and a 3 Ω resistor. Find the current drawn from the battery and the terminal voltage of the battery.

19. 19. A 10 Ω resistor and a 20 Ω resistor are connected in parallel across a 20 V battery of negligible internal resistance. Find the total power dissipated in the circuit and the power dissipated in the 20 Ω resistor.

20. 20. A 4 μF capacitor is charged through a 2 MΩ resistor by a 10 V battery. Find the time taken for the capacitor's voltage to reach 6.32 V, and the charge stored on the capacitor at that instant.

21. 21. A proton (mass 1.67 × 10^-27 kg, charge 1.6 × 10^-19 C) moves in a circular path of radius 0.5 m in a uniform magnetic field of 0.2 T acting perpendicular to its velocity. Find its speed and kinetic energy.

22. 22. Two long straight parallel wires carrying currents of 5 A and 8 A in the same direction are separated by 20 cm. Find the force per unit length between the wires and state whether the force is attractive or repulsive.

23. 23. A conducting rod of length 1 m rotates about an axis through one end, perpendicular to a uniform magnetic field of 0.5 T, with an angular speed of 20 rad/s. Find the EMF induced between the ends of the rod.

24. 24. A series LCR circuit has R = 30 Ω, L = 0.5 H, and C = 79.5 μF, connected to an AC source of frequency 50 Hz and rms voltage 200 V. Find the impedance of the circuit and the rms current flowing through it.

25. 25. A convex lens of focal length 20 cm is placed in contact with a concave lens of focal length 30 cm. Find the focal length and power of the combination, and state whether the combination is converging or diverging.

26. 26. In a Young's double-slit experiment, the slit separation is 0.5 mm, the screen is 1 m away, and the wavelength of light used is 500 nm. Find the fringe width and the distance of the 4th bright fringe from the central maximum.

27. 27. Light of wavelength 400 nm is incident on a metal surface with a work function of 2.0 eV. Find the maximum kinetic energy of the emitted photoelectrons and the stopping potential. (h = 6.63 × 10^-34 J·s, c = 3 × 10^8 m/s)

28. 28. Find the wavelength of the photon emitted when an electron in a hydrogen atom makes a transition from n = 4 to n = 2. (Rydberg constant R = 1.097 × 10^7 m^-1)

29. 29. A radioactive sample has a half-life of 20 minutes. Find the fraction of the original sample remaining after 1 hour, and the decay constant of the sample.

30. 30. In a half-wave rectifier circuit, the input AC voltage has a peak value of 20 V and the diode has a forward voltage drop of 0.7 V. If the load resistance is 1 kΩ, find the peak output voltage and the peak current through the load.

Answer key & quick explanations

Short answers for the sample questions above. Use this to self-check before generating a fresh AI-built mock test.

1. 1. In an experiment to determine the density of a small sphere, its mass is measured as 12.48 g ± 0.02 g and its radius as 1.50 cm ± 0.01 cm. Find the percentage error in the calculated value of density.

   ≈ 2.16%

   Density ρ = m/[(4/3)πr³], so the fractional error adds as Δρ/ρ = Δm/m + 3Δr/r. Here Δm/m = 0.02/12.48 ≈ 0.16% and Δr/r = 0.01/1.50 ≈ 0.67%, and the radius term gets tripled because ρ depends on r³. Adding 0.16% and 3×0.67% = 2.0% gives a total percentage error of about 2.16%.

2. 2. A particle is projected from the ground with a speed of 40 m/s at an angle of 30° above the horizontal. Taking g = 10 m/s², find the speed and direction of the velocity vector 2 s after projection.

   v ≈ 34.6 m/s, directed horizontally

   The horizontal component ux = 40cos30° = 34.6 m/s stays constant throughout the flight. The vertical component decays as vy = uy − gt = 40sin30° − 10(2) = 20 − 20 = 0, so at t = 2 s the particle is momentarily moving purely horizontally, exactly when it reaches its maximum height, since time to peak = uy/g = 20/10 = 2 s.

3. 3. A boatman wants to cross a 400 m wide river and land at a point directly opposite his starting point. The river flows at 4 km/h and the boat can move at 8 km/h relative to the water. Find the time taken to cross the river.

   t ≈ 3.46 min (≈208 s)

   Since the boat speed (8 km/h) exceeds the current (4 km/h), the boatman can angle upstream so the resultant velocity points straight across. The upstream angle satisfies sinθ = vr/vb = 0.5, so θ = 30°, and the effective crossing speed is vb cosθ = 8cos30° ≈ 6.93 km/h. Dividing the 0.4 km width by this speed gives t ≈ 0.0577 h, which is about 3.46 minutes.

4. 4. Block A (4 kg) rests on a horizontal table (coefficient of kinetic friction μ = 0.2) and is connected by a light string over a frictionless pulley at the table's edge to block B (6 kg) hanging vertically. Taking g = 10 m/s², find the acceleration of the system and the tension in the string.

   a = 5.2 m/s², T = 28.8 N

   For block B: 6g − T = 6a. For block A, friction opposes the tension: T − μ(4)(g) = 4a, i.e., T − 8 = 4a. Adding the two equations eliminates T: 60 − 8 = 10a, giving a = 5.2 m/s², and substituting back gives T = 4(5.2) + 8 = 28.8 N.

5. 5. A car moves on a banked circular track of radius 100 m banked at 30°. If the coefficient of friction between the tyres and the road is 0.2, find the maximum speed at which the car can travel without slipping. (g = 10 m/s²)

   v_max ≈ 29.6 m/s

   For a banked road with friction, the maximum speed before slipping is v_max = √[rg(tanθ + μ)/(1 − μtanθ)]. With tan30° ≈ 0.577, this becomes √[1000×0.777/0.885] = √878.9. Taking the square root gives v_max ≈ 29.6 m/s.

6. 6. A force F = (3x² + 2x) N acts on a 2 kg particle constrained to move along the x-axis. If the particle starts from rest at x = 0, find its speed when it reaches x = 3 m.

   v = 6 m/s

   By the work-energy theorem, the work done equals the gain in kinetic energy since the particle starts from rest. Integrating the force, W = ∫₀³(3x²+2x)dx = [x³+x²]₀³ = 27+9 = 36 J. Setting this equal to (1/2)(2)v² gives v² = 36, so v = 6 m/s.

7. 7. A pump lifts water from a well of depth 20 m and delivers it through a pipe of cross-sectional area 0.01 m² with an exit speed of 5 m/s. Taking the density of water as 1000 kg/m³ and g = 10 m/s², find the power of the pump, assuming 100% efficiency.

   P ≈ 10.6 kW (10625 W)

   The mass flow rate is ṁ = ρAv = 1000×0.01×5 = 50 kg/s. The pump must supply both gravitational PE per unit mass (gh) and kinetic energy (v²/2), so P = ṁ(gh + v²/2) = 50×(200 + 12.5) = 10625 W, or about 10.6 kW.

8. 8. A solid sphere of mass 2 kg and radius 0.1 m starts from rest and rolls without slipping down an incline through a vertical height of 5 m. Find its speed at the bottom of the incline.

   v ≈ 8.45 m/s

   For a solid sphere rolling without slipping, energy conservation gives mgh = (1/2)mv² + (1/2)Iω² with I = (2/5)mr² and ω = v/r, which simplifies to mgh = (7/10)mv². Solving, v² = (10/7)gh = (10/7)(10)(5) ≈ 71.4, so v ≈ 8.45 m/s.

9. 9. A uniform rod of length 1.2 m and mass 3 kg, pivoted at one end, is released from rest in a horizontal position. Find its angular speed when it swings down to the vertical position. (g = 10 m/s²)

   ω = 5 rad/s

   Using energy conservation, the loss in gravitational PE of the centre of mass equals the gain in rotational KE: mg(L/2) = (1/2)Iω², where I = (1/3)mL² for a rod pivoted at one end. The mass cancels, leaving ω² = 3g/L = 3(10)/1.2 = 25, so ω = 5 rad/s.

10. 10. A satellite orbits the Earth at a height of 3600 km above the surface. Taking the Earth's radius as 6400 km and GM(Earth) = 3.986 × 10^14 m³/s², find the satellite's orbital speed and time period.

    v ≈ 6.31 km/s, T ≈ 9950 s (≈2.76 hours)

    The orbital radius is r = R + h = 6400 + 3600 = 10,000 km = 1×10^7 m. Orbital speed follows from v = √(GM/r) = √(3.986×10^14/1×10^7) ≈ 6313 m/s. The period is T = 2πr/v = 2π(1×10^7)/6313 ≈ 9950 s, which is about 2.76 hours.

11. 11. A rocket is fired vertically from the Earth's surface with a speed equal to half the escape velocity. Find the maximum height it reaches above the surface, in terms of the Earth's radius R.

    h = R/3

    Using energy conservation with launch speed v = ve/2, (1/2)v² − GM/R = −GM/(R+h). Since GM/R = ve²/2, substituting v = ve/2 simplifies this to R+h = 4R/3, so h = R/3. With R = 6400 km, h ≈ 2133 km.

12. 12. A steel ball of radius 2 mm and density 7800 kg/m³ falls through glycerine of density 1260 kg/m³ and coefficient of viscosity 0.83 Pa·s. Find its terminal velocity. (g = 9.8 m/s²)

    v_t ≈ 0.0687 m/s (6.87 cm/s)

    At terminal velocity, the net force is zero, giving vt = 2r²g(ρ−σ)/(9η). Substituting r = 2×10^-3 m, ρ−σ = 7800−1260 = 6540 kg/m³, g = 9.8 m/s², and η = 0.83 Pa·s gives vt ≈ 0.0687 m/s.

13. 13. One mole of an ideal diatomic gas (γ = 7/5) at an initial temperature of 300 K is adiabatically compressed to half its original volume. Find the final temperature of the gas.

    T2 ≈ 396 K

    For an adiabatic process, TV^(γ−1) is constant, so T2 = T1(V1/V2)^(γ−1). With V1/V2 = 2 and γ−1 = 0.4, T2 = 300×2^0.4 ≈ 300×1.3195 ≈ 396 K. The temperature rises because compression does work on the gas.

14. 14. A train approaches a stationary observer while sounding its whistle at a frequency of 500 Hz. If the train's speed is 30 m/s and the speed of sound in air is 340 m/s, find the frequency heard by the observer.

    f' ≈ 548.4 Hz

    For a source approaching a stationary observer, the Doppler formula is f' = f·v/(v−vs). Substituting f = 500 Hz, v = 340 m/s, and vs = 30 m/s gives f' = 500×340/310 ≈ 548.4 Hz, higher than the source frequency as expected for an approaching source.

15. 15. Point charges of +4 μC and −4 μC are placed 6 cm apart. Find the magnitude of the electric field at a point on the perpendicular bisector of the line joining the charges, at a distance of 4 cm from the midpoint.

    E ≈ 1.73 × 10^7 N/C, directed from the +4 μC charge toward the −4 μC charge

    This pair forms a dipole with moment p = qd = (4×10^-6)(0.06) = 2.4×10^-7 C·m. On the equatorial line, the exact field is E = kp/(r² + (d/2)²)^(3/2). With r = 0.04 m and d/2 = 0.03 m, r²+(d/2)² = 0.0025, whose 3/2 power is 1.25×10^-4, giving E = (9×10^9)(2.4×10^-7)/1.25×10^-4 ≈ 1.73×10^7 N/C.

16. 16. A parallel plate capacitor has a plate area of 200 cm² and plate separation of 2 mm, and is connected to a 100 V battery. A dielectric slab of thickness 1 mm and dielectric constant K = 5 is inserted between the plates while the battery remains connected. Find the new capacitance and the charge stored. (ε0 = 8.85 × 10^-12 F/m)

    C ≈ 147.5 pF, Q ≈ 14.75 nC

    For a dielectric slab of thickness t partially filling the gap d, C = ε0A/(d − t + t/K). Here d−t = 0.001 m and t/K = 0.0002 m, so the effective gap is 0.0012 m, giving C = (8.85×10^-12×0.02)/0.0012 ≈ 147.5 pF. Since the battery stays connected, V remains 100 V, so Q = CV ≈ 14.75 nC.

17. 17. Three point charges of +2 μC each are placed at the vertices of an equilateral triangle of side 30 cm. Find the total electrostatic potential energy of the system.

    U = 0.36 J

    By symmetry, all three pairwise separations equal 30 cm and all charges are equal, so each pair contributes the same potential energy kq²/r = (9×10^9)(2×10^-6)²/0.3 = 0.12 J. Since there are three distinct pairs in the triangle, the total potential energy is 3×0.12 = 0.36 J.

18. 18. A battery of EMF 12 V and internal resistance 1 Ω is connected to an external circuit consisting of a 4 Ω resistor in series with a parallel combination of a 6 Ω and a 3 Ω resistor. Find the current drawn from the battery and the terminal voltage of the battery.

    I ≈ 1.71 A, terminal voltage ≈ 10.29 V

    The 6 Ω and 3 Ω resistors in parallel combine to (6×3)/(6+3) = 2 Ω, giving a total external resistance of 4+2 = 6 Ω. Including internal resistance, total circuit resistance is 7 Ω, so I = 12/7 ≈ 1.71 A. Terminal voltage is EMF minus internal drop: 12 − 1.71×1 ≈ 10.29 V.

19. 19. A 10 Ω resistor and a 20 Ω resistor are connected in parallel across a 20 V battery of negligible internal resistance. Find the total power dissipated in the circuit and the power dissipated in the 20 Ω resistor.

    Total power = 60 W, power in 20 Ω resistor = 20 W

    Since both resistors see the full 20 V across them, the branch currents are I1 = 20/10 = 2 A and I2 = 20/20 = 1 A, so the battery supplies a total current of 3 A and total power = 20×3 = 60 W. The power in the 20 Ω resistor alone is V²/R = 400/20 = 20 W.

20. 20. A 4 μF capacitor is charged through a 2 MΩ resistor by a 10 V battery. Find the time taken for the capacitor's voltage to reach 6.32 V, and the charge stored on the capacitor at that instant.

    t = 8 s, Q ≈ 25.28 μC

    The time constant of the RC circuit is τ = RC = (2×10^6)(4×10^-6) = 8 s. At t = τ, the capacitor voltage is V0(1−e^-1) = 10×0.632 = 6.32 V, which matches the target voltage exactly, so t = 8 s. The charge at that instant is Q = CV = (4×10^-6)(6.32) ≈ 25.28 μC.

21. 21. A proton (mass 1.67 × 10^-27 kg, charge 1.6 × 10^-19 C) moves in a circular path of radius 0.5 m in a uniform magnetic field of 0.2 T acting perpendicular to its velocity. Find its speed and kinetic energy.

    v ≈ 9.58 × 10^6 m/s, KE ≈ 7.67 × 10^-14 J (≈0.48 MeV)

    For circular motion in a magnetic field, r = mv/(qB), so v = qBr/m = (1.6×10^-19)(0.2)(0.5)/(1.67×10^-27) ≈ 9.58×10^6 m/s. The kinetic energy is (1/2)mv² = 0.5(1.67×10^-27)(9.58×10^6)² ≈ 7.67×10^-14 J, equivalent to roughly 0.48 MeV.

22. 22. Two long straight parallel wires carrying currents of 5 A and 8 A in the same direction are separated by 20 cm. Find the force per unit length between the wires and state whether the force is attractive or repulsive.

    F/L = 4 × 10^-5 N/m, attractive

    The force per unit length between two long parallel currents is F/L = μ0I1I2/(2πd). Substituting μ0 = 4π×10^-7, I1=5 A, I2=8 A and d=0.2 m, the π cancels to give F/L = (4×5×8×10^-7)/(2×0.2) = 4×10^-5 N/m. Since the currents flow in the same direction, the wires attract each other.

23. 23. A conducting rod of length 1 m rotates about an axis through one end, perpendicular to a uniform magnetic field of 0.5 T, with an angular speed of 20 rad/s. Find the EMF induced between the ends of the rod.

    EMF = 5 V

    For a rod rotating about one end in a uniform field, each element sweeps out an EMF that integrates to ε = (1/2)Bωl². Substituting B=0.5 T, ω=20 rad/s and l=1 m gives ε = 0.5×0.5×20×1 = 5 V.

24. 24. A series LCR circuit has R = 30 Ω, L = 0.5 H, and C = 79.5 μF, connected to an AC source of frequency 50 Hz and rms voltage 200 V. Find the impedance of the circuit and the rms current flowing through it.

    Z ≈ 120.8 Ω, I_rms ≈ 1.66 A

    The reactances are XL = 2πfL = 2π(50)(0.5) ≈ 157.1 Ω and XC = 1/(2πfC) = 1/(2π×50×79.5×10^-6) ≈ 40.0 Ω. The impedance is Z = √[R² + (XL−XC)²] = √[900 + 117.0²] ≈ 120.8 Ω, so I_rms = V/Z = 200/120.8 ≈ 1.66 A.

25. 25. A convex lens of focal length 20 cm is placed in contact with a concave lens of focal length 30 cm. Find the focal length and power of the combination, and state whether the combination is converging or diverging.

    f = 60 cm (converging), P ≈ 1.67 D

    For lenses in contact, 1/f = 1/f1 + 1/f2 = 1/20 + 1/(−30) = 1/60, giving f = 60 cm. Since f is positive, the combination behaves as a converging lens, and its power is P = 1/f (in metres) = 1/0.6 ≈ 1.67 D.

26. 26. In a Young's double-slit experiment, the slit separation is 0.5 mm, the screen is 1 m away, and the wavelength of light used is 500 nm. Find the fringe width and the distance of the 4th bright fringe from the central maximum.

    β = 1 mm, 4th bright fringe at 4 mm

    Fringe width in YDSE is β = λD/d = (500×10^-9)(1)/(0.5×10^-3) = 1×10^-3 m = 1 mm. The nth bright fringe lies at a distance nβ from the centre, so the 4th bright fringe is at 4×1 mm = 4 mm.

27. 27. Light of wavelength 400 nm is incident on a metal surface with a work function of 2.0 eV. Find the maximum kinetic energy of the emitted photoelectrons and the stopping potential. (h = 6.63 × 10^-34 J·s, c = 3 × 10^8 m/s)

    KE_max ≈ 1.11 eV, stopping potential ≈ 1.11 V

    The photon energy is E = hc/λ = (6.63×10^-34)(3×10^8)/(400×10^-9) ≈ 4.97×10^-19 J, which is about 3.11 eV. Subtracting the work function, KE_max = 3.11 − 2.0 ≈ 1.11 eV, and since eV0 = KE_max, the stopping potential is about 1.11 V.

28. 28. Find the wavelength of the photon emitted when an electron in a hydrogen atom makes a transition from n = 4 to n = 2. (Rydberg constant R = 1.097 × 10^7 m^-1)

    λ ≈ 486 nm

    Using the Rydberg formula, 1/λ = R(1/n1² − 1/n2²) with n1=2, n2=4, giving 1/λ = R(1/4 − 1/16) = R(3/16) ≈ 2.057×10^6 m^-1. Inverting gives λ ≈ 4.86×10^-7 m, or 486 nm — the well-known Hβ line of the Balmer series.

29. 29. A radioactive sample has a half-life of 20 minutes. Find the fraction of the original sample remaining after 1 hour, and the decay constant of the sample.

    Fraction remaining = 1/8 (12.5%); λ ≈ 5.78 × 10^-4 s^-1

    One hour is exactly 3 half-lives (60/20 = 3), so the remaining fraction is (1/2)³ = 1/8, or 12.5%. The decay constant follows from λ = ln2/T½ = 0.693/(20×60 s) ≈ 5.78×10^-4 s^-1.

30. 30. In a half-wave rectifier circuit, the input AC voltage has a peak value of 20 V and the diode has a forward voltage drop of 0.7 V. If the load resistance is 1 kΩ, find the peak output voltage and the peak current through the load.

    Peak output voltage ≈ 19.3 V, peak current ≈ 19.3 mA

    In a half-wave rectifier, the diode's forward drop subtracts directly from the peak input: V_out = 20 − 0.7 = 19.3 V. By Ohm's law, the peak current through the load is I = V_out/R = 19.3/1000 ≈ 19.3 mA.

Curriculum Mapping & Learning Guide

Use this breakdown to identify which skills each question tests and guide post-test review.

IIT JEE Physics units covered

1. Chapter 1: Units, Dimensions and Measurement
2. Chapter 2: Kinematics and Vectors
3. Chapter 3: Laws of Motion
4. Chapter 4: Work, Energy and Power
5. Chapter 5: Rotational Motion
6. Chapter 6: Gravitation
7. Chapter 7: Properties of Matter (Elasticity, Fluids)
8. Chapter 8: Thermodynamics and Kinetic Theory
9. Chapter 9: Oscillations and Waves
10. Chapter 10: Electrostatics
11. Chapter 11: Current Electricity
12. Chapter 12: Magnetic Effects of Current and Magnetism
13. Chapter 13: Electromagnetic Induction and AC
14. Chapter 14: Ray and Wave Optics
15. Chapter 15: Modern Physics (Atoms, Nuclei, Photoelectric)
16. Chapter 16: Semiconductor Devices

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