Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Question A, B and C are: Disc Solid sphere Spherical shell All have same mass (M) and radius (R) . Moment of inertia of system about PQ axis is: x 15 I \frac{x}{15} I Where: I = MOI of disc about its diameter I = \text{MOI of disc about its diameter} Find x . 🖼 Question Image ✍️ Short Concept This is a standard MOI formula + addition question. Golden idea: 👉 Write MOI of each body about its own center 👉 Convert to required axis (if needed) 👉 Add them 👉 Compare with given I 🔷 Step 1 — Write Standard MOI Formulas 💯 For same mass M and radius R: 🔹 Disc (about diameter) I d = 1 4 M R 2 I_d = \frac{1}{4} MR^2 (This is given as I) 🔹 Solid Sphere (about diameter) I s = 2 5 M R 2 I_s = \frac{2}{5} MR^2 🔹 Spherical Shell (about diameter) I s h = 2 3 M R 2 I_{sh} = \frac{2}{3} MR^2 🔷 Step 2 — Express Everything in Terms of I Given: I = 1 4 M R 2 I = \frac{1}{4} MR^2 So, Solid sphere: 2 5 M R 2 = 2 5 ÷ 1 4   ...

  ❓ Question Given: R = 100 kΩ C = 100 pF Phase difference between: V i n and ( V B − V A ) V_{in} \quad \text{and} \quad (V_B - V_A) is 90°. Input signal frequency: ω = 10 x  rad/s \omega = 10^x \text{ rad/s} Find x . 🖼 Question Image ✍️ Short Concept In an RC circuit: Capacitive reactance: X C = 1 ω C X_C = \frac{1}{\omega C} 90° phase condition usually occurs when: X C = R X_C = R 👉 That means: ω = 1 R C \omega = \frac{1}{RC} 🔷 Step 1 — Write Key Relation 💯 For 90° phase condition: ω = 1 R C \boxed{\omega = \frac{1}{RC}} This is the MOST IMPORTANT step. 🔷 Step 2 — Convert Values R = 100 kΩ = 10 5   Ω = 10^5 \, \Omega C = 100 pF = 10 − 10   F = 10^{-10} \, F 🔷 Step 3 — Calculate RC R C = 10 5 × 10 − 10 RC = 10^5 \times 10^{-10} R C = 10 − 5 RC = 10^{-5} 🔷 Step 4 — Find Angular Frequency ω = 1 R C \omega = \frac{1}{RC} = 1 10 − 5 = \frac{1}{10^{-5}} = 10 5  rad/s = 10^5 \text{ rad/s} So, x = 5 x = 5 ✅ Final Answer x = 5 \boxed{x = 5...

  ❓ Question How do we calculate the equivalent power when two thin lenses are placed at a distance from each other? 🖼 Concept Image ✍️ Short Concept This is a lens combination formula-based question. Golden idea: 👉 Touching lenses → simple addition 👉 Separated lenses → correction term aata hai 🔷 Step 1 — Power of a Lens 💯 P = 1 f P = \frac{1}{f} (f in metres ) Convex lens → positive power Concave lens → negative power 🔷 Step 2 — When Lenses Are Touching If lenses are in contact: P t o t a l = P 1 + P 2 \boxed{P_{total} = P_1 + P_2} 👉 Bas simple addition. No extra term. 🔷 Step 3 — When Lenses Are Separated (MOST IMPORTANT) If separation distance = d: P t o t a l = P 1 + P 2 − d P 1 P 2 \boxed{P_{total} = P_1 + P_2 - d P_1 P_2} ⚠️ Distance term always minus ⚠️ d must be in metres Yahin JEE trap lagata hai. 🔷 Step 4 — Why Distance Matters? First lens image banata hai. Wahi image second lens ke liye object ban jata hai. Isliye system simple ad...

  ❓ Question A wire of: Length = 10 cm Diameter = 0.5 mm Temperature = 1727°C Power radiated = 94.2 W Find the emissivity (e) of the wire. 🖼 Question Image ✍️ Short Solution This is a Stefan–Boltzmann Law question. Golden formula: P = e σ A T 4 P = e \sigma A T^4 👉 Convert temperature to Kelvin 👉 Find curved surface area 👉 Substitute and solve for e 🔷 Step 1 — Convert All Units (MOST IMPORTANT 💯) Length: L = 10   cm = 0.1   m L = 10\,\text{cm} = 0.1\,\text{m} Diameter: d = 0.5   mm = 0.0005   m d = 0.5\,\text{mm} = 0.0005\,\text{m} Radius: r = 0.00025   m r = 0.00025\,\text{m} Temperature: T = 1727 + 273 = 2000   K T = 1727 + 273 = 2000\,K 🔷 Step 2 — Surface Area of Wire Wire is cylindrical → Use curved surface area: A = 2 π r L A = 2\pi r L A = 2 π ( 0.00025 ) ( 0.1 ) A = 2\pi (0.00025)(0.1) A = 1.57 × 10 − 4   m 2 A = 1.57 \times 10^{-4} \, m^2 🔷 Step 3 — Apply Stefan’s Law P = e σ A T 4 P = e \sigma A T^4 Given: P = 94.2 P = 94.2 σ = 5.67 ×...

  ❓ 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...

  ❓ 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 =...

  ❓ 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} ...

  ❓ 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 ^...