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Inclined Plane + Friction Trick 🔥 | JEE in 60 Sec

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  ❓ Question On an inclined plane of angle θ \theta , a block slides down with given acceleration. How do we find the coefficient of kinetic friction μ k \mu_k ​ ? 🖼 Concept Image ✍️ Short Solution This is a force resolution + Newton’s 2nd Law question. Golden idea: 👉 Resolve gravity 👉 Apply F = m a F = ma  along the plane 👉 Extract μ k \mu_k ​ 🔷 Step 1 — Forces on an Inclined Plane 💯 Inclined angle = θ \theta Gravity splits into: Along plane → m g sin ⁡ θ mg \sin\theta Perpendicular → m g cos ⁡ θ mg \cos\theta Normal reaction: N = m g cos ⁡ θ N = mg \cos\theta 🔷 Step 2 — Direction of Kinetic Friction If block slides downwards : Friction always acts upwards (opposes motion). Friction force: f k = μ k N f_k = \mu_k N f k = μ k m g cos ⁡ θ f_k = \mu_k mg \cos\theta ⚠️ Direction mistake = direct negative marking in JEE. 🔷 Step 3 — Apply Newton’s 2nd Law Along Plane Net force along plane: m g sin ⁡ θ − μ k m g cos ⁡ θ mg \sin\theta - \mu_k mg...
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Cp/Cv Ratio Trick — Degrees of Freedom in 60 Sec 🔥

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  ❓ Question In Kinetic Theory of Gases, how do we quickly determine the value of γ = C p C v \gamma = \frac{C_p}{C_v} for different types of gases? 🖼 Concept Image ✍️ Short Solution This is a direct Degrees of Freedom → γ mapping concept . Golden relation: γ = C p C v = f + 2 f \gamma = \frac{C_p}{C_v} = \frac{f+2}{f} 👉 If you know f , you know everything. No derivation needed in exam. 🔷 Step 1 — KTG ka Golden Rule 💯 For any ideal gas: γ = f + 2 f \boxed{\gamma = \frac{f+2}{f}} Where: f = degrees of freedom f = \text{degrees of freedom} 👉 Bas f pata ho, answer automatic. 🔷 Step 2 — Degrees of Freedom (Core Memory Table) 🔹 Monatomic Gas f = 3 f = 3 (Only translational motion) γ = 5 3 \gamma = \frac{5}{3} 🔹 Diatomic Rigid Gas f = 5 f = 5 (3 translational + 2 rotational) γ = 7 5 \gamma = \frac{7}{5} 🔹 Diatomic Non-Rigid Gas f = 7 f = 7 (Vibrational mode ON) γ = 9 7 \gamma = \frac{9}{7} 🔹 Triatomic Rigid Gas f = 6 f = 6 γ = 4 3 \gamma =...
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Work Done in Cyclic Process — PV Diagram Trick 🔥

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  ❓ Question An ideal gas undergoes a cyclic process as shown in the figure. Work done by the gas in the entire cycle is: 🖼 Question Image ✍️ Short Solution This is a PV diagram cycle question . Key Idea: Work   done   in   one   complete   cycle   =   Area   enclosed   in   PV   graph \textbf{Work done in one complete cycle = Area enclosed in PV graph} Direction matters: Clockwise → Work positive Anticlockwise → Work negative 🔷 Step 1 — MOST IMPORTANT RULE 💯 For cyclic process: W = ∮ P   d V W = \oint P\, dV Graphically: W = Area enclosed by the cycle \boxed{W = \text{Area enclosed by the cycle}} No need to calculate individual paths unless asked separately. 🔷 Step 2 — Identify Shape of Cycle Look carefully at the diagram: If rectangle → Area = Δ P × Δ V \Delta P \times \Delta V If triangle → Area = 1 2 × base × height \frac{1}{2} \times \text{base} \times \text{height} ...
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JEE Main: Find Power Using F · v Concept 💡

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  ❓ Question An object of mass 1000 g experiences a time dependent force F ⃗ = ( 2 t   i ^ + 3 t 2   j ^ )   N \vec F = (2t \,\hat i + 3t^2 \,\hat j)\, \text{N} The power generated by the force at time t is: 🖼 Question Image ✍️ Short Solution This is a Power = F · v question. 👉 Time dependent force 👉 Velocity changes with time 👉 First find velocity using Newton’s 2nd Law Then apply: P = F ⃗ ⋅ v ⃗ P = \vec F \cdot \vec v 🔷 Step 1 — Convert Mass & Use Newton’s Law (MOST IMPORTANT 💯) Given mass: 1000  g = 1  kg 1000 \text{ g} = 1 \text{ kg} Using: F ⃗ = m a ⃗ \vec F = m\vec a So acceleration: a ⃗ = F ⃗ \vec a = \vec F a ⃗ = ( 2 t i ^ + 3 t 2 j ^ ) \vec a = (2t \hat i + 3t^2 \hat j) 🔷 Step 2 — Find Velocity by Integration Since: a ⃗ = d v ⃗ d t \vec a = \frac{d\vec v}{dt} Integrate: x-component: v x = ∫ 2 t   d t = t 2 v_x = \int 2t\, dt = t^2 y-component: v y = ∫ 3 t 2   d t = t 3 v_y = \int 3t^2\, dt = t^3 So velocity: v ⃗ = ( t 2 i ^...
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JEE Main: ΔU Depends Only on Start & End 💡

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  ❓ Concept 🎬 Electrostatics — Potential Energy is Path Independent (59 Sec) Charge circular path pe move kare, curve dikhe, arc dikhe… 👉 Par JEE poochta hai sirf start aur end ka khel! ⚡🧠 Electrostatics ka yeh golden rule pakad liya, toh half chapter khud simplify ho jaata hai 🔥 🖼️ Concept Image ✍️ Short Explanation Electrostatic force conservative hoti hai. Matlab energy change path pe depend nahi karta . Isliye circular path sirf distraction hai 😎 🔹 Step 1 — Key Rule: Electrostatic Force is Conservative (FOUNDATION 💯)** Electrostatic field ka sabse powerful rule: Δ U = U f − U i \Delta U = U_f - U_i 👉 Depends only on: Initial position Final position ❌ Path does NOT matter Circular, zig-zag, straight — sab same. 🔹 Step 2 — Potential Energy Change Formula For a charge q 3 q_3 ​ placed in electric field: Δ U = q 3 ( V f − V i ) \Delta U = q_3 (V_f - V_i) ⚡ Sirf potential difference important hai. Force ka work = potential energy change...
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JEE Main Wave Optics: Superposition Amplitude Formula 💡

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  ❓ Concept 🎬 Wave Optics – Resultant Amplitude in 60 Sec Do light waves mil rahi hain… par amplitude seedha add hoga ya cancel? 👉 Answer hamesha phase difference batata hai! 🌊✨ Wave optics mein calculation kam, vector understanding zyada important hota hai 🔥 🖼️ Concept Image ✍️ Short Explanation Superposition ka matlab simple addition nahi hota. Light waves → electric field vectors hote hain. Isliye: 👉 Vector addition apply hota hai. 🔹 Step 1 — Superposition Principle (FOUNDATION 💯)** When two waves meet at a point: E resultant = E 1 + E 2 But important: ⚠️ Vector sum , not simple arithmetic sum. 🔹 Step 2 — Same Frequency, Same Direction Given waves: E 1 = E 0 sin ⁡ ( ω t ) E_1 = E_0 \sin(\omega t) E 2 = E 0 sin ⁡ ( ω t + ϕ ) E_2 = E_0 \sin(\omega t + \phi) ✔ Same ω \omega  ⇒ stable interference ✔ Same direction ⇒ direct vector addition 🔹 Step 3 — Phase Difference Controls Everything Different cases: ϕ = 0 \phi = 0 → Maximum const...
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JEE Main Physics: Beats from cos A cos B Form 💡

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  ❓ Question Two harmonic waves moving in the same direction superimpose to form a wave: x = a cos ⁡ ( 1.5 t ) cos ⁡ ( 50.5 t ) x = a \cos(1.5t)\cos(50.5t) (where t t  is in seconds) Find the period with which they beat . (Answer close to nearest integer) 🖼️ Question Image ✍️ Short Solution This is a standard beats question hidden inside a product form. Key idea: 👉 Product of cosines = amplitude modulation 🔥 🎯 HOOK (Before Reading) “Wave ke andar wave dikhe? Samajh jao beats chal rahe hain!” 🎵 🔹 Step 1 — Identify the Structure (MOST IMPORTANT 💯)** Given: x = a cos ⁡ ( 1.5 t ) cos ⁡ ( 50.5 t ) x = a \cos(1.5t)\cos(50.5t) x = a cos ( 1.5 t ) cos ( 50.5 t ) This is of the form: x = ( slow term ) × ( fast term ) x = (\text{slow term}) \times (\text{fast term}) x = ( slow term ) × ( fast term ) Here: cos ⁡ ( 50.5 t ) \cos(50.5t) cos ( 50.5 t ) → rapid oscillation cos ⁡ ( 1.5 t ) \cos(1.5t) cos ( 1.5 t ) → slowly varying...
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