Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Concept 🎬 Atomic Structure — Hydrogen-Like Ion Energy Trick in 59 Sec Hydrogen jaisa atom dikhe… energy difference diya ho… 👉 Samajh jao — game Z² ka hai! ⚡ Is concept ko pakad liya, toh Bohr-model wale questions 30–40 sec mein khatam 🔥 🖼️ Concept Image ✍️ Short Explanation Hydrogen-like ion problems mein calculation kam, structure samajhna zyada important hota hai. Bas 3 cheezein yaad rakho: One electron Bohr formula Z² scaling 🔹 Step 1 — Hydrogen-Like Ion Means… (FOUNDATION 💯)** Hydrogen-like ion ⇒ Sirf ek electron present hota hai. Examples: H He⁺ Li²⁺ Be³⁺ 👉 In sab par Bohr model directly apply hota hai. 🔹 Step 2 — Bohr Energy Formula (Most Important 🔥)** For hydrogen-like ion: E n = − 13.6   Z 2 n 2 ( in eV ) E_n = -\frac{13.6\,Z^2}{n^2}\quad (\text{in eV}) Where: Z Z  = atomic number n n  = principal quantum number ⚡ Z² factor sabse powerful clue hai. Energy hydrogen se compare karni ho, ...

  ❓ Question A cubic block of mass m m  is sliding down an inclined plane at 60 ∘ 60^\circ  with an acceleration of g / 2 g/2 . Find the coefficient of kinetic friction μ k \mu_k ​ . 🖼️ Question Image ✍️ Short Solution This is a direct Newton’s Second Law + friction problem. No tricks. Just resolve forces properly 🔥 🔹 Step 1 — Draw Free Body Diagram (MOST IMPORTANT 💯)** For block on incline at angle 60 ∘ 60^\circ : Forces acting: Weight m g mg  (vertical downward) Normal reaction N N Kinetic friction f k = μ k N f_k = \mu_k N  (up the plane) 🔹 Step 2 — Resolve Weight into Components Along incline: m g sin ⁡ 60 ∘ mg \sin 60^\circ Perpendicular to incline: m g cos ⁡ 60 ∘ mg \cos 60^\circ So normal reaction: N = m g cos ⁡ 60 ∘ N = mg \cos 60^\circ 🔹 Step 3 — Apply Newton’s Second Law (Along Incline) Net force down the plane: m g sin ⁡ 60 ∘ − f k = m a mg \sin 60^\circ - f_k = ma Since: f k = μ k N = μ k m g cos ⁡ 60 ∘ f_k =...

  ❓ 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 - ...