Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Concept Question How do we calculate instantaneous power when the force is time-dependent ? 🖼 Concept Image ✍️ Short Concept Power represents rate of doing work . For a moving particle: P = F ⃗ ⋅ v ⃗ P = \vec{F} \cdot \vec{v} If force varies with time, velocity must be found using Newton’s law and integration . 🔷 Step 1 — Instantaneous Power Definition 💯 Power is defined as: P = F ⃗ ⋅ v ⃗ P = \vec{F} \cdot \vec{v} This is a dot product . 👉 Only the component of force along velocity contributes to power. 🔷 Step 2 — When Force Depends on Time If: F ⃗ = F ( t ) \vec{F} = F(t) Then acceleration becomes: a ⃗ = F ⃗ m \vec{a} = \frac{\vec{F}}{m} Since velocity changes with time: v ⃗ = v ( t ) \vec{v} = v(t) 🔷 Step 3 — Finding Velocity Using Newton’s second law: F = m a F = ma and a = d v d t a = \frac{dv}{dt} So: d v d t = F ( t ) m \frac{dv}{dt} = \frac{F(t)}{m} Integrating: v = ∫ F ( t ) m   d t v = \int \frac{F(t)}{m} \, dt 👉 Integration gives vel...

  ❓ Concept Question What happens when two waves of slightly different frequencies superpose? Why do we hear beats ? 🖼 Concept Image ✍️ Short Concept When two waves with nearly equal frequencies interfere, the resultant wave shows periodic variation in amplitude . This phenomenon is called Beats . 🔷 Step 1 — Superposition of Two Waves 💯 When two harmonic waves move in same direction: x = x 1 + x 2 x = x_1 + x_2 If frequencies are close: x 1 = a cos ⁡ ( ω 1 t ) x_1 = a \cos(\omega_1 t) x 2 = a cos ⁡ ( ω 2 t ) x_2 = a \cos(\omega_2 t) Resultant motion becomes product form. 🔷 Step 2 — Product Form = Beat Signal After applying trigonometric identity: x = 2 a cos ⁡ ( ω 1 − ω 2 2 t ) cos ⁡ ( ω 1 + ω 2 2 t ) x = 2a \cos\left(\frac{\omega_1 - \omega_2}{2} t\right) \cos\left(\frac{\omega_1 + \omega_2}{2} t\right) Two parts appear: 👉 Slow oscillation 👉 Fast oscillation 🔷 Step 3 — Envelope Creates Beats Slow cosine term forms the envelope . Fast cosine term is...

  ❓ Question An aqueous solution of HCl has: p H = 1 pH = 1 The solution is diluted by adding equal volume of water . Find the new pH of the solution. (Ignore dissociation of water.) 🖼 Question Image ✍️ Short Concept pH is related to hydrogen ion concentration: p H = − log ⁡ [ H + ] pH = -\log[H^+] Dilution reduces concentration → pH increases. 🔷 Step 1 — Find Initial [ H + ] [H^+] [ H + ] 💯 Given: p H = 1 pH = 1 [ H + ] = 10 − 1   M [H^+] = 10^{-1} \, M Since HCl is strong acid , it dissociates completely. 🔷 Step 2 — Apply Dilution Rule Equal volume of water added ⇒ Total volume becomes double . So concentration becomes half . [ H + ] n e w = 10 − 1 2 [H^+]_{new} = \frac{10^{-1}}{2} = 5 × 10 − 2 = 5 \times 10^{-2} 🔷 Step 3 — Calculate New pH p H = − log ⁡ ( 5 × 10 − 2 ) pH = -\log(5 \times 10^{-2}) = − [ log ⁡ 5 + log ⁡ 10 − 2 ] = -[\log5 + \log10^{-2}] = − ( 0.699 − 2 ) = -(0.699 - 2) = 1.30 = 1.30 🔷 Step 4 — Concept Shortcut When concentration h...

  ❓ Question Group 14 elements A and B have first ionisation enthalpy values: A → 708 kJ mol⁻¹ B → 715 kJ mol⁻¹ These are the lowest values in the group . Find the nature of their ions: A 2 + and B 4 + A^{2+} \quad \text{and} \quad B^{4+} 🖼 Question Image ✍️ Short Concept Ionisation energy generally decreases down the group . Lowest values in Group 14 correspond to the heaviest elements . So these numbers help identify the elements. 🔷 Step 1 — Identify the Elements 💯 Group 14 elements: C → Si → Ge → Sn → Pb Ionisation energies: Sn ≈ 708 kJ/mol Pb ≈ 715 kJ/mol So: A = S n A = Sn B = P b B = Pb 🔷 Step 2 — Apply Inert Pair Effect In heavier p-block elements: ns² electrons become reluctant to participate in bonding. This is called: Inert   Pair   Effect \textbf{Inert Pair Effect} Because of this: Lower oxidation states become more stable . 🔷 Step 3 — Nature of A 2 + A^{2+}  (Sn²⁺) Tin: Stable oxidation states: + 2 and + 4 +2 \quad...

  ❓ Question Container contains: Liquid 1: Refractive index μ 1 = 1.2 \mu_1 = 1.2 Height = 60 cm Liquid 2: Refractive index μ 2 = 1.6 \mu_2 = 1.6 Height = H H Viewed from above, apparent shift of bottom = 40 cm Find H H . 🖼 Question Image ✍️ Short Concept For multiple layers: Apparent depth = ∑ real depth μ \text{Apparent depth} = \sum \frac{\text{real depth}}{\mu} Total apparent shift: Shift = Real depth − Apparent depth \text{Shift} = \text{Real depth} - \text{Apparent depth} This is the MOST IMPORTANT relation. 🔷 Step 1 — Write Real Depth 💯 Total real depth: = 60 + H = 60 + H 🔷 Step 2 — Write Apparent Depth Apparent depth = 60 1.2 + H 1.6 \text{Apparent depth} = \frac{60}{1.2} + \frac{H}{1.6} = 50 + H 1.6 = 50 + \frac{H}{1.6} 🔷 Step 3 — Use Shift Formula Given shift = 40 cm Shift = Real depth − Apparent depth \text{Shift} = \text{Real depth} - \text{Apparent depth} 40 = ( 60 + H ) − ( 50 + H 1.6 ) 40 = (60 + H) - ...

  ❓ 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 ×...