Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Question A man of mass 75 kg stands in a lift. The lift starts moving upward with an acceleration of 2 m/s² . Find the force exerted on him by the floor . 🖼️ Question Image ✍️ Short Solution This is a Newton’s Laws + apparent weight question. When lift accelerates upward → normal force increases 💯 🔹 Step 1 — Understand Forces on the Man (MOST IMPORTANT 💯) Two forces act: Upward → Normal Reaction (N) Downward → Weight (mg) Since lift is accelerating upward , net force is upward. 🔹 Step 2 — Apply Newton’s Second Law N − m g = m a N - mg = ma N − m g = ma 👉 यही सबसे important equation है 🔹 Step 3 — Substitute Values Mass: m = 75  kg m = 75 \text{ kg} Acceleration: a = 2  m/s 2 a = 2 \text{ m/s}^2 Take: g = 10  m/s 2 g = 10 \text{ m/s}^2 N − ( 75 × 10 ) = 75 × 2 N - (75 \times 10) = 75 \times 2 N − 750 = 150 N - 750 = 150 N = 900  N N = 900 \text{ N} 🔹 Step 4 — Interpret the Result 👉 Normal force = apparent weight ...

❓ Question Among: Cr, Fe, Co, Ni Which metal has the lowest enthalpy of atomisation ? Find the number of valence electrons in that metal. 🖼 Question Image ✍️ Short Solution Enthalpy of atomisation depends on: 👉 Strength of metallic bonding 👉 Number of unpaired electrons More unpaired electrons ⇒ stronger bonding ⇒ higher enthalpy 🔷 Step 1 — Write Electronic Configurations 💯 Cr: [ A r ]   3 d 5 4 s 1 [Ar]\, 3d^5 4s^1 Fe: [ A r ]   3 d 6 4 s 2 [Ar]\, 3d^6 4s^2 Co: [ A r ]   3 d 7 4 s 2 [Ar]\, 3d^7 4s^2 Ni: [ A r ]   3 d 8 4 s 2 [Ar]\, 3d^8 4s^2 🔷 Step 2 — Count Unpaired Electrons Cr → 6 unpaired Fe → 4 unpaired Co → 3 unpaired Ni → 2 unpaired 🔷 Step 3 — Relation with Atomisation Enthalpy More unpaired electrons ⇒ stronger metallic bonding So: Cr > Fe > Co > Ni 👉 Lowest enthalpy = Ni 🔷 Step 4 — Valence Electrons in Ni For transition metals: Valence electrons = n s + ( n − 1 ) d ns + (n-1)d Ni: 3 d 8 4 s 2 3d^8 4s^2 Total valence elec...

  ❓ Question Reaction: A ( g ) → 2 B ( g ) + C ( g ) A(g) \rightarrow 2B(g) + C(g) First order reaction Given: t (min) Pressure (mm Hg) 10 160 ∞ 240 Find the incorrect statement . 🖼 Question Image ✍️ Short Solution In gas reactions: P ∝ n P \propto n So pressure can be used as a substitute for concentration. 🔷 Step 1 — Initial Pressure Idea 💯 At t = 0 t = 0 : Only A present → let pressure = P 0 P_0 ​ Reaction: A → 3  moles (2B + C) A \rightarrow 3 \text{ moles (2B + C)} So pressure increases during reaction. 🔷 Step 2 — Use Final Pressure At t = ∞ t = \infty : All A converted Final pressure: P ∞ = 3 P 0 P_\infty = 3P_0 Given: 240 = 3 P 0 240 = 3P_0 P 0 = 80 P_0 = 80 🔷 Step 3 — Pressure of A at Time t At time t: P t = 160 P_t = 160 Let pressure of A remaining = P A P_A ​ Total pressure: P t = P A + products P_t = P_A + \text{products} Using relation: P A = 3 P 0 − 2 P t P_A = 3P_0 - 2P_t P A = 240 − 2 ( 160 ) P_A = 240 - 2(160) ...

  ❓ Concept Question How do we calculate the apparent shift of bottom when light passes through multiple liquid layers ? 🖼 Concept Image ✍️ Short Concept In multi-layer systems, each layer affects light separately . Final apparent depth is obtained by adding contributions of each layer . 🔷 Step 1 — Apparent Depth Rule 💯 When viewed from air: Apparent depth = Real depth μ \text{Apparent depth} = \frac{\text{Real depth}}{\mu} 👉 Higher refractive index ⇒ object appears closer 🔷 Step 2 — Multiple Liquids = Layer-wise Effect For multiple layers: Total apparent depth = h 1 μ 1 + h 2 μ 2 + ⋯ \text{Total apparent depth} = \frac{h_1}{\mu_1} + \frac{h_2}{\mu_2} + \cdots ⚠️ Never combine into a single refractive index 👉 Each layer acts independently 🔷 Step 3 — Real Depth vs Apparent Depth Total real depth: h t o t a l = h 1 + h 2 h_{total} = h_1 + h_2 Shift: Shift = Real depth − Apparent depth \text{Shift} = \text{Real depth} - \text...

  ❓ Concept Question How do we calculate the moment of inertia of a system of different bodies about a given axis? 🖼 Concept Image ✍️ Short Concept MOI depends on: 👉 Mass distribution 👉 Shape of body 👉 Axis of rotation For multiple bodies → calculate individually and then add. 🔷 Step 1 — MOI Depends on Axis 💯 Moment of inertia is not fixed . It depends on: Shape Mass distribution Axis of rotation 👉 Axis change ⇒ MOI change This is the most important idea. 🔷 Step 2 — Different Bodies, Different MOI Even if mass and radius same , formulas differ: Disc → 1 4 M R 2 \frac{1}{4}MR^2  (about diameter) Solid sphere → 2 5 M R 2 \frac{2}{5}MR^2 Spherical shell → 2 3 M R 2 \frac{2}{3}MR^2 📌 JEE mixes these to create confusion. 🔷 Step 3 — Reference MOI Trick Often given: I = MOI of disc about its diameter I = \text{MOI of disc about its diameter} 👉 Convert all other MOIs into multiples of I This simplifies calculati...

  ❓ Question Two statements are given: Statement I: Mohr's salt is composed of only three types of ions — ferrous, ammonium and sulfate . Statement II: If molar conductance at infinite dilution of Fe²⁺, NH₄⁺ and SO₄²⁻ are x 1 , x 2 , x 3 x_1, x_2, x_3 x 1 ​ , x 2 ​ , x 3 ​ respectively, then molar conductance of Mohr’s salt at infinite dilution is: x 1 + x 2 + 2 x 3 x_1 + x_2 + 2x_3 Determine which statement is correct. 🖼 Question Image ✍️ Short Concept Mohr’s salt is a double salt . Formula: ( N H 4 ) 2 F e ( S O 4 ) 2 ⋅ 6 H 2 O (NH_4)_2Fe(SO_4)_2 \cdot 6H_2O When dissolved in water, it completely dissociates into simple ions . 🔷 Step 1 — Identify the Ions 💯 Mohr’s salt dissociation: ( N H 4 ) 2 F e ( S O 4 ) 2 → F e 2 + + 2 N H 4 + + 2 S O 4 2 − (NH_4)_2Fe(SO_4)_2 \rightarrow Fe^{2+} + 2NH_4^+ + 2SO_4^{2-} So ions present are: Ferrous ion Ammonium ion Sulfate ion Thus Statement I is correct . 🔷 Step 2 — Kohlrausch Law At infinite dilution: ...

❓ Question Find the total enthalpy change when 1 mol of water is converted from: 10 ∘ C → − 10 ∘ C  (ice) 10^\circ C \rightarrow -10^\circ C \text{ (ice)} 🖼 Question Image ✍️ Short Concept This process happens in three steps : 1️⃣ Cool water from 10°C → 0°C 2️⃣ Freeze water at 0°C 3️⃣ Cool ice from 0°C → −10°C Total enthalpy change = sum of all three. 🔷 Step 1 — Cooling Water (10°C → 0°C) 💯 Heat released: q 1 = n C w a t e r Δ T q_1 = n C_{water} \Delta T For water: C w a t e r = 75   J   m o l − 1 K − 1 C_{water} = 75 \, J\, mol^{-1} K^{-1} q 1 = 1 × 75 × ( 10 ) q_1 = 1 \times 75 \times (10) q 1 = 750   J q_1 = 750 \, J 🔷 Step 2 — Freezing at 0°C Latent heat of fusion: Δ H f = 6.01   k J / m o l \Delta H_f = 6.01 \, kJ/mol Heat released: q 2 = 6.01   k J q_2 = 6.01 \, kJ 🔷 Step 3 — Cooling Ice (0°C → −10°C) Heat released: q 3 = n C i c e Δ T q_3 = n C_{ice} \Delta T For ice: C i c e = 37.6   J   m o l − 1 K − 1 C_{ice} = 37.6 \, J\, mol^{-1}K^{-1} q ...

  ❓ Concept Question In an RC AC circuit , how do we determine the phase difference between voltages and how is it related to frequency? 🖼 Concept Image ✍️ Short Concept In AC circuits, resistors and capacitors behave differently with respect to phase . Understanding their phase relation helps determine frequency conditions . 🔷 Step 1 — RC Circuit Basics 💯 In AC circuits: Resistor: Current and voltage are in phase . Capacitor: Current leads voltage by 90° . So phase difference naturally appears in RC networks. 🔷 Step 2 — Voltage Across R and C Across resistor: V R V_R ​ is in phase with current . Across capacitor: V C V_C ​ lags current by 90° . 👉 That means V R V_R ​ and V C V_C  are perpendicular in phasor diagram. 🔷 Step 3 — Special Phase Difference Condition If question says: Phase difference = 90° That means resistive and capacitive effects are balanced . Mathematically this leads to a special condition involving frequency. 🔷 Step...