Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Question Two thin convex lenses of focal lengths f 1 = 30  cm , f 2 = 10  cm f_1 = 30\text{ cm}, \quad f_2 = 10\text{ cm} are placed coaxially, 10 cm apart . Find the power of the combination . 🖼️ Question Image ✍️ Short Solution This is NOT the simple “lenses in contact” case. Because they are separated by distance d d d , we must use the correct combination formula 🔥 🔹 Step 1 — Formula for Two Lenses Separated by Distance Equivalent focal length: 1 F = 1 f 1 + 1 f 2 − d f 1 f 2 \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} Where: f 1 , f 2 f_1, f_2 ​ in same units d d  = separation 🔹 Step 2 — Convert Everything to Same Unit Given: f 1 = 30  cm , f 2 = 10  cm , d = 10  cm f_1 = 30\text{ cm},\quad f_2 = 10\text{ cm},\quad d = 10\text{ cm} All already in cm — good 👍 🔹 Step 3 — Substitute Values 1 F = 1 30 + 1 10 − 10 30 × 10 \frac{1}{F} = \frac{1}{30} + \frac{1}{10} - \frac{10}{30\times 10} Ca...

  ❓ Concept 🎬 Rotational Motion — COM of Bent Rod in 59 Sec Rod seedhi nahi hai… right angle pe bent hai? 👉 Samajh jao — COM ko todna padega! 🧠⚙️ Is type ke questions JEE mein frequently aate hain, aur solution hamesha break–combine method se hota hai 🔥 🖼️ Concept Image ✍️ Short Explanation Centre of mass shape se directly nahi milta — mass distribution se milta hai. Isliye bent rod dekhte hi: 👉 use parts mein tod do. 🔹 Step 1 — Golden Rule of COM (FOUNDATION 💯)** Centre of mass depends on: Mass distribution NOT directly on bending or fancy shape Uniform rod ⇒ Mass ∝ Length \text{Mass} \propto \text{Length} 📌 Density same ⇒ longer part = heavier part. 🔹 Step 2 — Bent Rod ⇒ Break Into Parts Right-angle bent rod ko hamesha: 👉 2 straight rods mein tod do Har part: Uniform Known length Known orientation (x-axis ya y-axis) Yahin se problem simple ho jaati hai 😎 🔹 Step 3 — COM of Each Straight Rod (MOST IMPORTANT 🔥)** Unifo...

  ❓ 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 — ...