Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Question A wire of total resistance R is bent to form a triangular pyramid (tetrahedron) as shown in the figure, with each segment having the same length (and hence same resistance). The equivalent resistance between points A and B is given as R n . Find the value of n n . 🖼️ Question Image ✍️ Short Solution This is a classic JEE symmetry-based resistance problem . No Kirchhoff, no long equations — sirf symmetry + equivalent paths 🔥 🔹 Step 1 — Count number of equal segments A triangular pyramid (tetrahedron) has: 4 vertices 6 edges The wire of total resistance R R  is bent uniformly into these 6 equal segments . So resistance of each edge : r = R 6​ 🔹 Step 2 — Identify symmetry in the network We are asked resistance between A and B . Important symmetry observation 👇 Points C and D are symmetrically placed with respect to A and B So their potentials will be equal 📌 Hence: No current flows in the wire CD We can safely r...

❓ Concept 🎬 Solenoid + Magnetic Material = New Magnetic Field | Concept in 59 Sec Solenoid ke andar vacuum/air ki jagah agar magnesium ya koi magnetic material bhar diya… 👉 Magnetic field badal jaati hai 🧲 Par kyun , aur kitni ? Yeh concept JEE mein baar-baar aata hai 🔥 🖼️ Concept Image ✍️ Short Explanation Is type ke questions mein calculation kam aur definition zyada important hoti hai. Agar tum μ, μr aur χ ka relation pakad loge, toh answer seconds mein nikal jaata hai 😎 🎯 HOOK (Write & Read First) “Agar solenoid ke andar vacuum ki jagah magnesium bhar diya… toh magnetic field badh jaati hai — par kyun aur kitni?” 🧲🤔 🔹 Step 1 — Magnetic Field of a Solenoid (in Air / Vacuum)** Long solenoid ke liye: B 0 = μ 0 n I Yahaan: n n  = turns per unit length I I  = current μ 0 \mu_0 ​ = permeability of free space 📌 Yeh reference field hota hai (without material). 🔹 Step 2 — When a Magnetic Material is Inserted Material daaln...

❓ Question For a hydrogen atom , the ratio of the largest wavelength of the Lyman series to that of the Balmer series is equal to ? 🖼️ Question Image ✍️ Short Solution This is a pure concept + formula-based JEE question . The key is to remember: Largest wavelength ⇔ smallest energy difference So we must identify the closest transition in each series. 🔹 Step 1 — Rydberg formula (foundation 💯)** For hydrogen spectrum: 1 λ = R ( 1 n 1 2 − 1 n 2 2 ) , n 2 > n 1 Where: n 1 n_1 ​ = lower energy level (series identifier) n 2 n_2 ​ = higher energy level 📌 Larger wavelength ⇒ smaller value of 1 λ \frac{1}{\lambda} ​ . 🔹 Step 2 — Largest wavelength of Lyman series Lyman series: n 1 = 1 Largest wavelength occurs for smallest jump : n 2 = 2 → 1 So: 1 λ L = R ( 1 − 1 4 ) = 3 R 4 Thus: λ L = 4 3 R 🔹 Step 3 — Largest wavelength of Balmer series Balmer series: n 1 = 2 Largest wavelength ⇒ nearest upper level : n 2 = 3 → 2 So: 1 λ B = R ( 1 4 − 1 9 ...

  ❓ Concept 🎬 Motion of Charge at Boundary of Two Magnetic Fields — Concept in 59 Sec Ek charged particle uniform magnetic field mein gaya — 👉 circular motion start 🔄 Par jaise hi particle boundary cross karta hai aur magnetic field change hoti hai, 👉 radius bhi change ho jaata hai! Ab sawal yeh hota hai: Net displacement kaise aur kyun aata hai? 🔥 🖼️ Concept Image ✍️ Short Explanation Yeh question formula-based nahi , pure concept + geometry ka test hota hai. Agar tum radius–bending logic samajh gaye, toh answer automatically build ho jaata hai 😎 🔹 Step 1 — Magnetic Force = Circular Motion (FOUNDATION 💯)** Jab charge q q , speed v v  ke saath magnetic field B B  ke perpendicular move kare: q v B = m v 2 r qvB = \frac{mv^2}{r} So radius: r = m v q B r=\frac{mv}{qB} 📌 Direct observations : B B  ↑ ⇒ r r  ↓ (particle zyada bend karega) B B  ↓ ⇒ r r  ↑ (particle kam bend karega) 🔹 Step 2 — Two Regions = ...

❓ Question If ε 0 \varepsilon_0 ​ denotes the permittivity of free space and ϕ E \phi_E  is the electric flux through the area bounded by a closed surface, then the dimensions of ε 0   d ϕ E d t are the same as that of what physical quantity? 🖼️ Question Image ✍️ Short Solution This is a pure dimensional-analysis + concept question from Electrostatics + Maxwell’s equations . No numbers, no calculus — just definitions and dimensions . 🔹 Step 1 — Meaning of electric flux Electric flux through a closed surface is: ϕ E = ∮ E ⃗ ⋅ d A ⃗ So dimensionally: [ ϕ E ] = [ E ]   [ A ] 🔹 Step 2 — Dimensions of electric field Electric field: E = F q Where: Force F F  has dimensions [ M L T − 2 ] [MLT^{-2}] Charge q q q has dimensions [ I T ] [IT] So: [ E ] = M L T − 2 I T = M L T − 3 I − 1 🔹 Step 3 — Dimensions of electric flux Area: [ A ] = L 2 Hence: [ ϕ E ] = ( M L T − 3 I − 1 ) ( L 2 ) = M L 3 T − 3 I − 1 🔹 Step 4 — Take time derivative of flux ...

❓ Concept 🎬 Focal Length of Lens in Water — Concept in 59 Sec Lens ko air se nikaal kar paani mein daal diya… aur suddenly focal length change ho gaya? 🤔 👉 Isme koi magic nahi — sirf relative refractive index ka game hai 🔥 🖼️ Concept Image ✍️ Short Explanation JEE mein “lens in water” wale questions numerical kam, conceptual zyada hote hain. Agar tum lens maker + relative refractive index samajh gaye, toh answer seconds mein aa jaata hai 😎 🔹 Step 1 — Lens Maker’s Formula (in Air) For a thin lens in air: 1 f air = ( μ lens − 1 ) ( 1 R 1 − 1 R 2 ) Yahaan: μ lens \mu_{\text{lens}} ​ = refractive index of lens w.r.t air R 1 , R 2 R_1, R_2 ​ = radii of curvature 📌 Lens ka shape fixed hota hai — curvatures change nahi hote. 🔹 Step 2 — Lens Placed in Another Medium Agar lens ko kisi aur medium (jaise water) mein rakh diya: 👉 Refractive index relative ho jaata hai: μ relative = μ lens μ medium Modified lens maker formula: 1 f medium = ( ...

  ❓ Concept 🎬 Singular Matrix Trick in 60 Sec | JEE Matrices 2×2 matrix dikhi? Aur question bole “singular” ? 👉 Bas ek chhoti si identity yaad rakh lo — baaki saara game automatic ho jaata hai 🔥 🖼️ Concept Image ✍️ Short Explanation JEE mein “singular matrix” ka matlab definition-based hota hai. No heavy algebra, no tricks — sirf determinant logic . 🔹 Step 1 — Singular Matrix Meaning (CORE RULE 💯)** A matrix is singular ⟺ its determinant is zero . 📌 Yeh rule har order ke matrix ke liye true hai. 🔹 Step 2 — 2×2 Determinant Formula (MOST IMPORTANT 🔥)** For a 2×2 matrix: ( a b c d ) \begin{pmatrix} a & b\\ c & d \end{pmatrix} Determinant: ∣ A ∣ = a d − b c |A| = ad - bc So, Matrix is singular  ⟺ a d = b c \boxed{\text{Matrix is singular } \Longleftrightarrow ad = bc} 👉 Bas yahi identity poore chapter ko control karti hai. 🔹 Step 3 — What These JEE Questions Are REALLY Doing Typical JEE statement: “Elements are chosen f...

  ❓ Question Two projectiles are fired from the ground with the same initial speed from the same point , at angles ( 45 ∘ + α ) and ( 45 ∘ − α ) with the horizontal. Find the ratio of their times of flight . 🖼️ Question Image ✍️ Short Solution This is a classic JEE symmetry question from projectile motion. The trick is to remember what depends on sine and what depends on sine double-angle . 🔹 Step 1 — Time of flight formula For a projectile fired from ground with speed u u u at angle θ \theta : T = 2 u sin ⁡ θ g 📌 Time of flight depends on sin ⁡ θ \sin\theta sin θ (not on sin ⁡ 2 θ \sin 2\theta sin 2 θ ). 🔹 Step 2 — Write times for both projectiles Let: T 1 T_1 ​ = time of flight at angle ( 45 ∘ + α ) (45^\circ+\alpha) T 2 T_2 ​ = time of flight at angle ( 45 ∘ − α ) Using the formula: T 1 = 2 u sin ⁡ ( 45 ∘ + α ) g T_1=\frac{2u\sin(45^\circ+\alpha)}{g} T 2 = 2 u sin ⁡ ( 45 ∘ − α ) g T_2=\frac{2u\sin(45^\circ-\alpha)}{g} 🔹 Step 3 — Take the r...