Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Question Match the following: LIST–I A. Triatomic rigid gas B. Diatomic non-rigid gas C. Monoatomic gas D. Diatomic rigid gas LIST–II C p / C v = 5 / 3 C_p/C_v = 5/3 C p / C v = 7 / 5 C_p/C_v = 7/5 C p / C v = 4 / 3 C_p/C_v = 4/3 C p / C v = 9 / 7 C_p/C_v = 9/7 🖼️ Question Image ✍️ Short Solution This is a pure degrees of freedom question . No formulas to memorize separately — just remember: γ = C p C v = f + 2 f Where f f = degrees of freedom 🔥 🔹 Step 1 — Formula to Remember (MOST IMPORTANT 💯)** For ideal gas: C v = f 2 R C_v = \frac{f}{2}R C p = C v + R = f + 2 2 R C_p = C_v + R = \frac{f+2}{2}R So: γ = f + 2 f \boxed{\gamma = \frac{f+2}{f}} 🔹 Step 2 — Monoatomic Gas Monoatomic gas has: f = 3 f = 3 So: γ = 3 + 2 3 = 5 3 \gamma = \frac{3+2}{3} = \frac{5}{3} ✔️ Matches option 1 So: C → 1 \boxed{C \rightarrow 1} 🔹 Step 3 — Diatomic Rigid Gas Rigid diatomic gas: 3 translational 2 rotational f = 5 f = 5 γ = 5 + 2 5 = 7 5 \g...

  ❓ Question Two plane-polarized light waves combine at a point. Their electric field components are E 1 = E 0 sin ⁡ ( ω t ) , E 2 = E 0 sin ⁡  ⁣ ( ω t + π 2 ) . E_1 = E_0 \sin(\omega t), \qquad E_2 = E_0 \sin\!\left(\omega t + \frac{\pi}{2}\right). Find the amplitude of the resultant wave . 🖼️ Question Image ✍️ Short Solution This is a classic JEE superposition + phase difference problem. No trigonometric expansion marathon needed — phasor (vector) method makes it instant 🔥 🔹 Step 1 — Identify Phase Difference (MOST IMPORTANT 💯)** Given: E 1 = E 0 sin ⁡ ( ω t ) , E 2 = E 0 sin ⁡ ( ω t + π 2 ) E_1 = E_0 \sin(\omega t), \quad E_2 = E_0 \sin(\omega t + \tfrac{\pi}{2}) So the phase difference : Δ ϕ = π 2 = 90 ∘ \Delta\phi = \frac{\pi}{2} = 90^\circ 📌 This means the two waves are in quadrature . 🔹 Step 2 — Use Resultant Amplitude Formula For two waves of amplitudes E 0 E_0 ​ and E 0 E_0 ​ with phase difference Δ ϕ \Delta\phi : E res = E 0 2 + E 0 2 + 2 E 0...

  ❓ Concept 🎬 Charged Particle in Magnetic Field — r vs E Graph in 59 Sec Charged particle magnetic field mein gaya… energy badhi … par sawal yeh hai 👇 👉 radius linearly badhega ya curve banayega? ⚡🤔 Is concept ko pakad liya, toh graph-based magnetism questions JEE mein free ho jaate hain 🔥 🖼️ Concept Image ✍️ Short Explanation Yeh question calculation ka nahi , relation samajhne ka test hai. Bas teen steps yaad rakho — force → speed → energy . 🔹 Step 1 — Magnetic Force Causes Circular Motion (FOUNDATION 💯)** Jab charged particle ki velocity magnetic field ke perpendicular hoti hai: q v B = m v 2 r qvB=\frac{mv^2}{r} 👉 Magnetic force centripetal force ka kaam karta hai aur particle circular path follow karta hai. 🔹 Step 2 — Radius Depends on Speed Upar wale relation se: r = m v q B r=\frac{mv}{qB} So: r ∝ v \boxed{r\propto v} 📌 Intuition: Zyada speed ⇒ magnetic field ke liye bend karna mushkil Result ⇒ larger radius 🔹 Step 3 — ...

  ❓ Question Two charges q 1 q_1 ​ and q 2 q_2 ​ are separated by a distance of 30 cm . A third charge q 3 q_3 ​ , initially at C (as shown), is moved along a circular path of radius 40 cm from C to D . If the change in potential energy of q 3 q_3 ​ due to this movement is q 3 K 4 π ε 0 , \frac{q_3 K}{4\pi\varepsilon_0}, find the value of K K . 🖼️ Question Image ✍️ Short Solution This is a pure potential-energy difference problem . 👉 Path does NOT matter — only initial and final positions matter. We compare the electric potential at C and D due to fixed charges q 1 , q 2 q_1, q_2 . 🔹 Step 1 — Understand the Geometry (MOST IMPORTANT 💯)** From the figure: q 1 q_1 ​ at point A q 2 q_2 ​ at point B A B = 30  cm AB = 30\text{ cm} A C = 40  cm AC = 40\text{ cm}  (vertical) Arc CD is a quarter circle of radius 40 cm , centred at A So: A D = 40  cm AD = 40\text{ cm} B D = A D − A B = 40 − 30 = 10  cm BD = AD - ...

  ❓ Concept 🎬 Mechanical Properties of Solids — Wire Elongation Concept in 59 Sec Same material… same force … phir bhi ek wire zyada stretch karegi! 👉 Reason length mein hai ya thickness mein? 🤯 Is concept ko pakad liya, toh ratio-based elasticity questions turant ho jaate hain 🔥 🖼️ Concept Image ✍️ Short Explanation Elasticity ke questions mein formula se zyada dependence matter karta hai. Yahan ek hi idea kaam karta hai: 👉 Elongation kis cheez par depend karti hai? 🔹 Step 1 — Elongation Formula (FOUNDATION 💯)** For a stretched wire: Δ L = F   L A   Y \Delta L=\frac{F\,L}{A\,Y} Where: F F  = applied force L L  = length A A  = cross-sectional area Y Y  = Young’s modulus 📌 Core idea: elongation depends on L and A . 🔹 Step 2 — Same Material ⇒ Same Y Y ** Given: both wires are of the same material . So: Y A = Y B ⇒ Y  cancels in ratio Y_A = Y_B \quad \Rightarrow \quad Y \text{ cancels in ratio} ...

  ❓ Question A uniform rod of total length 5L is bent at a right angle , such that one arm has length 2L and the other arm has length 3L . Find the position of the centre of mass of the system. 🖼️ Question Image ✍️ Short Solution This is a standard JEE Centre of Mass problem . No integration needed — treat each arm separately and use weighted average 🔥 🔹 Step 1 — Split the system into two rods Since the rod is uniform : Mass ∝ length Let: Rod 1: length = 2L , along x-axis Rod 2: length = 3L , along y-axis Assume: Total mass = M Mass per unit length = λ \lambda λ Then: Mass of rod 1 = 2 λ L 2\lambda L Mass of rod 2 = 3 λ L 3\lambda L 🔹 Step 2 — Locate centre of mass of each part For a uniform straight rod , COM lies at its midpoint. COM of rod 1 (along x-axis): ( x 1 , y 1 ) = ( L , 0 ) (x_1, y_1) = (L, 0) COM of rod 2 (along y-axis): ( x 2 , y 2 ) = ( 0 , 3 L 2 ) (x_2, y_2) = (0, \tfrac{3L}{2}) 🔹 Step 3 — Use Centre...

  ❓ Concept 🎬 Zener Diode in Circuit — Ammeter Reading Concept in 59 Sec Circuit mein Zener diode laga ho… aur ammeter ka reading poocha jaaye — toh sabse pehla sawal yeh hai 👇 👉 Zener ON hai ya OFF? ⚡🤔 Is concept ko clear rakhoge, toh numbers khud line mein aa jaate hain 😎 🖼️ Concept Image    ✍️ Short Explanation Zener-based questions calculation se pehle logic maangte hain. Ek baar ON/OFF decision ho gaya, baaki steps straightforward hote hain 🔥 🔹 Step 1 — Zener Diode: Basic Role (FOUNDATION 💯)** Zener diode ka kaam: Voltage regulation Reverse bias mein kaam karta hai Rules: If V applied ≥ V Z V_{\text{applied}} \ge V_Z ​ → Zener ON (breakdown) If V applied < V Z V_{\text{applied}} < V_Z ​ → Zener OFF 📌 First check = breakdown condition 🔹 Step 2 — Voltage Across Parallel Branch (MOST IMPORTANT 🔥)** Agar Zener ON ho jaata hai: Zener node voltage clamp kar deta hai Parallel branch (load + ammeter) ...