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Modern Physics – JEE Main PYQs (2026) Quiz Important Constants: $h = 6.63 \times 10^{-34}\ \text{J·s}$ $c = 3 \times 10^8\ \text{m/s}$ $e = 1.6 \times 10^{-19}\ \text{C}$ $k = 1.38 \times 10^{-23}\ \text{J/K}$ $1\text{ eV} = 1.6 \times 10^{-19}\ \text{J}$ Score: 0 Q1. Two electrons are moving in orbits of two hydrogen-like atoms with speeds $3 \times 10^5$ m/s and $2.5 \times 10^5$ m/s respectively. If the radii of these orbits are nearly same then the possible order of energy states are respectively: 9 and 8 8 and 10 10 and 12 6 and 5 Check Answer For hydrogen-like atom: $v \propto \frac{1}{n}$ $\frac{v_1}{v_2} = \frac{n_2}{n_1}$ $\frac{3}{2.5} = 1.2$ Closest integer ratio → 9 and 8. Q2. Number of photons of equal energy emitted per second by a 6 mW laser source operating at 663 nm is: $1\times10^{16}$ $5\times10^{16}$ $2\times10^{16}$ $5\times10^{15}$ Check Answer $E = \frac{hc}{\lambda}$ $E = \frac{6.63\times10^{-34...

Waves – Complete Formula Revision          (JEE Mains) Every Formula | Every Trap | PYQ Mapping 1️⃣ Basic Wave Parameters Wave equation: \[ y(x,t) = A \sin(kx - \omega t + \phi) \] Amplitude = \( A \) Angular frequency = \( \omega = 2\pi f \) Wave number = \( k = \frac{2\pi}{\lambda} \) Wave speed = \( v = \frac{\omega}{k} \) Also: \( v = f\lambda \) ⚠ Trap: Sign in wave equation decides direction. 2️⃣ Speed of Transverse Wave on String \[ v = \sqrt{\frac{T}{\mu}} \] T = tension \( \mu = \frac{m}{L} \) (linear density) ⚠ Increasing tension increases speed, but frequency remains constant. 3️⃣ Longitudinal Sound Waves Speed in medium: \[ v = \sqrt{\frac{B}{\rho}} \] B = Bulk modulus \( \rho \) = density For gas: \[ v = \sqrt{\frac{\gamma P}{\rho}} \] ⚠ Trap: Speed independent of frequency. 4️⃣ Principle of Superposition Resultant displacement: \[ y = y_1 + y_2 \] Resultant amplitude: ...

Current Electricity – Complete Formula, Laws & PYQ Profiling (JEE Main Only) 1. Electric Current & Drift Velocity Electric current: I = Q / t Current density: J = nqv d Drift velocity: v d = eEτ / m Relation: I = nAev d Mobility: μ = v d / E Conductivity: σ = nqμ Resistivity: ρ = 1 / σ JEE Trap: Drift velocity is extremely small, but electrical signal propagates almost at speed of light. 2. Ohm’s Law & Resistance Ohm’s law: V = IR Resistance: R = ρL / A Temperature dependence: R = R₀(1 + αΔT) Combination of resistors: Series: R eq = R₁ + R₂ + … Parallel: 1/R eq = 1/R₁ + 1/R₂ + … PYQ Trap: For same material & volume, R ∝ L² (since A ∝ 1/L). 3. Electric Power & Heating Effect P = VI = I²R = V² / R Electrical energy: E = Pt Joule’s law of heating: H = I²Rt 4. EMF & Internal Resistance Terminal voltage (discharging): V = E − Ir Terminal voltage (charging): V = E + Ir Current: I = E / (R + r) M...

Capacitors – Complete Formula, Traps & Question Profiling (JEE Main Only) 1. Basic Concept of Capacitor Capacitor: Device to store electric charge Consists of two conductors separated by dielectric C = Q / V SI unit → Farad (F) 1 μF = 10⁻⁶ F 2. Parallel Plate Capacitor C = ε₀ A / d A → area of plates d → separation With dielectric (k): C = k ε₀ A / d 3. Effect of Dielectric Capacitance increases k times Electric field reduces: E = E₀ / k Potential reduces: V = V₀ / k Dielectric constant: k = ε / ε₀ 4. Series & Parallel Combination Series: 1/C eq = 1/C₁ + 1/C₂ + ... Charge same on each capacitor V distributes Parallel: C eq = C₁ + C₂ + ... Potential same Charge distributes 5. Energy Stored in Capacitor U = ½ C V² U = Q² / (2C) U = ½ QV Energy density: u = ½ ε E² 6. Capacitor with Battery Connected / Disconnected Battery connected: V constant Q increases with dielectric Energy increas...

Thermal Properties of Matter – Complete Formula Sheet (JEE Main Only) 1. Temperature & Heat Temperature: Measure of degree of hotness Heat: Energy transferred due to temperature difference Heat always flows: Higher T → Lower T SI unit of heat: Joule (J) 2. Temperature Scales K = °C + 273 °C = (5/9)(°F − 32) Zero points: 0 K → Absolute zero Triple point of water = 273.16 K 3. Thermal Expansion (a) Linear Expansion ΔL = α L ΔT (b) Areal Expansion ΔA = β A ΔT    (β = 2α) (c) Volume Expansion ΔV = γ V ΔT    (γ = 3α) 4. Expansion of Solids – PYQ Traps Hole expands as if material absent α is same in all directions (isotropic solid) If expansion prevented → thermal stress develops Thermal stress: σ = Y α ΔT 5. Calorimetry Heat absorbed/released: Q = m c ΔT c → specific heat capacity Water has maximum specific heat Principle: Heat lost = Heat gained 6. Latent Heat Q = m L L → latent heat Temper...

⚛️ Physics Quiz – Refraction, Electricity, Friction & Rolling Motion Challenge your understanding of optics, electromagnetism, and mechanics with these NTA-style numerical problems. Q1. The image of an object placed in air formed by a convex refracting surface is at a distance of 10 m behind the surface. The image is real and at 2/3 of the distance of the object from the surface. The wavelength of light inside the surface is 2/3 of its wavelength in air. The radius of curvature of the surface is \( R = x \, \text{m} \). Find the value of x . Options: (1) 10 (2) 15 (3) 30 (4) 45 Show Solution Official Answer: (3) Explanation: Using the refraction formula for a spherical surface: \[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \] Given: \( \frac{v}{u} = \frac{2}{3}, \, v = 10 \text{ m},...

  📘 Physics Quiz: Orbital Motion, Waves & Communication Sharpen your problem-solving skills with these NTA-style conceptual physics questions! Q1. A geostationary satellite is orbiting around an arbitrary planet 'P' at a height of 11R above the surface of 'P' , where R is the radius of the planet. The time period of another satellite at a height of 2R from the surface of 'P' is ____ hours, given that the time period of the geostationary satellite is 24 hours . Options: 6√2 6 3 5 Show Solution Official Answer: (3) Explanation: The time period of a satellite is proportional to \( r^{3/2} \). For the geostationary satellite: \[ T_1 \propto (12R)^{3/2} \] For the second satellite: \[ T_2 \propto (3R)^{3/2} \...

  💡 Physics Quiz: Test Your Concepts! Are you ready to check your physics preparation? Tackle these 5 challenging questions! Loading quiz... Quiz Results You scored 0 / 5. Try Again Challenge your friends and see who scores higher!

15 practice questions of KTG and Thermodynamics for JEE Advanced 15 Practice questions on KTG and Thermodynamics  for JEE Advanced Are you ready to challenge yourself with some of the toughest Kinetic Theory of Gases (KTG) and Thermodynamics problems? Here are 15 carefully selected advanced-level questions to help you gear up for JEE Advanced. Make sure you understand the concepts from our KTG and Thermodynamics notes before attempting these questions. Question 1: Calculate the number of degrees of freedom for a diatomic gas at room temperature and find the total internal energy of 1 mole of the gas at 300 K. Assume ideal gas behavior. Solution: A diatomic gas has 5 degrees of freedom (3 translational and 2 rotational) at room temperature. Total Internal Energy = \( U = \frac{5}{2} nRT \) Substituting the values: \( U = \frac{5}{2} \times 1 \times 8.314 \times 300 = 6235....

 Q. Four particles each of mass 1 kg are placed at four corners of a square of side 2 m. Moment of inertia of system about an axis perpendicular to its plane and passing through one of its vertex is ______ kgm2.                                                         ( JEE mains 2024) Show Answer Moment of Inertia of the System The moment of inertia (I) of four particles, each of mass 1 kg, placed at the four corners of a square of side 2 m, about an axis perpendicular to its plane and passing through one of its vertices is given by: \[ I = 16 \, \text{kg} \cdot \text{m}^2 \]

 Q. An alternating voltage v(t) = 220 sin 100tr volt is applied to a purely resistive load of 50 2. The time taken for the current to rise from half of the peak value to the peak value is                                                            ( JEE mains 2029) A. 2.2 ms B. 5 ms C. 3.3 ms 4.7.2 ms Show Answer Ans:- (C) Phase Analysis Answer cos θ = (V 0 / 2) / V 0 = 1/2 ∴ θ = π / 3 Time taken for phase change by π / 3 is t 1 = θ / ω = (π / 3) / 100π = 1/300 = 3.33 ms More Questions

 Q.Α10 Ω resistance is connected across 220 V - 50 Hz AC supply. The time taken by the current to change from its maximum value to the rms value is:                                                           ( JEE mains 2021) A. 2.5 ms B. 4.5 ms C. 3.3 ms D. 7.2 msl Show Answer (a) \(I = I_0 \cos \omega t\) Since the current is changing from its maximum value to RMS value, therefore, \(I = I_0 / \sqrt{2}\) \(\therefore \frac{I_0}{\sqrt{2}} = I_0 \cos \omega t\) \(\cos \omega t = \frac{1}{\sqrt{2}}\) \(\cos(2\pi ft) = \cos \frac{\pi}{4} \implies 2\pi \times 50t = \frac{\pi}{4}\) \(t = 2.5 \, \text{ms}\) More Questions

 Q. A solid circular disc of mass 50kg rolls along a horizontal floor so that its center of mass has a speed of 0.4 m/s. The absolute value of work done on the disc to stop it is _______ J.                                                         (jee mains 2024) Show Answer Ans:-  (6) Here, we will use the Work-Energy Theorem: \[ W = \Delta KE = 0 - \left( \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \right) \] Simplifying, \[ W = - \left( \frac{1}{2} \times 50 \times 0.4^2 \times \left( 1 + \frac{1}{2} \right) \right) \] Absolute work = +6 J // // //

 Q. In a linear simple harmonic motion (SHM) (A) Restoring force is directly proportional to the displacement.  (B) The acceleration and displacement are opposite in direction.  (C) The velocity is maximum at mean position. (D) The acceleration is minimum at extreme points.  Choose the correct answer from the options given below:                                                         (JEE mains 2023) 1. (A), (B) and (C) only   2. (C) and (D) only 3. (A), (B) and (D) only 4. (A), (C) and (D) only Show Answer Ans. (1)   F=-kx,          A true  a=-w²x.      B true  Velocity is maximum at mean position, C true  Acceleration is maximum at extreme point, D false

 Q. The height of transmitting antenna is 180 m and the height of the receiving antenna is 245 m. The maximum distance between them for satisfactory communication in line of sight will be :                                                         ( JEE mains 2023) 1. 48 km 2. 56 km 3. 96 km 4. 104 km Ans. (4)   dmax = √(2Rh, +2Rh) = √2x64×105×180+2×64×105×245  = {(8× 6 × 10³) + (8 × 7 × 10³)} m  = (48 +56) km  = 104 km

 Q. The half-life of a radioactive nucleus is 5 years, The fraction of the original sample that would decay in 15 years is :                                                         ( JEE mains 2023) 1. 1/8 2. 1/4 3. 7/8 4. 3/4 Show Answer Ans. (3)  15 year = 3 half lives  Number of active nuclei = N/8  Number of decay = 7/8N

 Q. The position of a particle related to time is given by x = (5t² - 4t + 5)m. The magnitude of velocity of the particle at t = 2s will be :                                                         ( JEE mains 2023) 1. 10 m/s 2. 14 m/s 3. 16 m/s 4. 06 m/s Show Answer Ans. (3)  x=5t²-4t+ 5  v = 10t-4  At t = 2s.  v = 16m/s

  Q. The position vector of a particle related to time t is given by r =(10ti+15t²j+7k)m The direction of net force experienced by the particle is :                                                         (JEE mains 2023) 1. Positive y-axis   2. Positive x-axis 3. Positive z-axis 4. In x-y plane Show Answer Ans:- 1

  Q. A block of mass 10 kg is moving along x-axis under the action of force F = 5x N. The work done by the force in moving the block from x = 2m to 4m will be__________ J.                                                         (JEE mains 2023) Show Answer Work done = ∫Fdx  ∫ 5x.dx = 5[x²/2] (limit 2=>4) =5/2[16-4] = 30 J.     Understand the Concept

 Q. An object is placed in a medium of refractive index 3. An electromagnetic wave of intensity 6 × 108 W/m² falls normally on the object and it is absorbed completely. The radiation pressure on the object would be (speed of light in free space = 3 × 108 m/s) :                                                         (jee mains 2024) 1. 36 Nm-² 2. 18 Nm-² 3. 6 Nm-² 4.  2 Nm-² Show Answer (3) -  Radiation pressure = I /V = Ιμ/C  =6×108×3/3×108  = 6 N/m²