Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Question The output Y = 0 when: A = 1 and B = 1 A = 0 and B = 1 A = 1 and B = 0 A = 0 and B = 0 (Which logic gate does this condition represent?) 🖼️ Question Image ✍️ Short Solution To find the correct answer, we analyze when the output becomes 0 for different input combinations. Step 1 — Recall common two-input logic gates Gate Expression Output (Y) = 0 occurs when… AND Y = A·B When A = 0 or B = 0 OR Y = A + B When A = 0 and B = 0 NAND Y = ¬(A·B) When A = 1 and B = 1 NOR Y = ¬(A + B) When A = 1 or B = 1 XOR Y = A ⊕ B When A = B (both 0 or both 1) XNOR Y = ¬(A ⊕ B) When A ≠ B (0,1 or 1,0) Step 2 — Given condition The question states that Y = 0 when A = 1 and B = 1. Let’s check which gate satisfies this. For AND gate : A=1, B=1 → Y=1·1=1 → Y≠0 ❌ For OR gate : A=1, B=1 → Y=1+1=1 → Y≠0 ❌ For NAND gate : A=1, B=1 → Y=¬(1·1)=¬1=0 ✅ That’s the correct condition! Step 3 — Verify other inputs (optional cross-check) Let’s create the ...

  ❓ Question A mirror produces an image with a magnification of 1/4 . If the distance between the object and its image is 40 cm , find the focal length of the mirror. 🖼️ Question Image ✍️ Short Solution Step 1 — Recall the mirror and magnification formulas For any spherical mirror: 1 f = 1 u + 1 v​ and m = h i h o = − v u​ Given: m = 1 4 ⇒ v = − u 4​ ( Negative sign shows image is virtual and erect → convex mirror case. ) Step 2 — Use the separation between object and image Distance between object and image is given as 40 cm: ∣ v − u ∣ = 40 Since v = − u 4 v = -\frac{u}{4} ​ : ∣ − u 4 − u ∣ = 40 | -\frac{u}{4} - u | = 40 ∣ − 5 u 4 ∣ = 40 | -\frac{5u}{4} | = 40 5 ∣ u ∣ 4 = 40 ⇒ ∣ u ∣ = 32  cm \frac{5|u|}{4} = 40 \Rightarrow |u| = 32 \text{ cm} So, u = − 32  cm u = -32 \text{ cm}  (object in front of mirror). Then, v = − u 4 = − ( − 32 ) / 4 = + 8  cm v = -\frac{u}{4} = -(-32)/4 = +8 \text{ cm} ( Positive v → virtual image behind mirror → c...

  💡 Question A photoemissive substance is illuminated with radiation of wavelength λ i \lambda_i ​ so that it releases electrons with de-Broglie wavelength λ e \lambda_e ​ . The longest wavelength of radiation that can just emit photoelectrons (i.e., threshold wavelength) is λ 0 \lambda_0 . Find the expression for de-Broglie wavelength ( λ e \lambda_e ​ ) of the emitted electrons in terms of λ i \lambda_i  and λ 0 \lambda_0 ​ . 🖼️ Question Image ✍️ Short Solution Step 1 — Photoelectric effect equation When light of wavelength λ i \lambda_i  falls on a metal surface, the energy of the incident photon is: E = h c λ i​ The maximum kinetic energy of the emitted electron is: K max = h c λ i − h c λ 0​ where h c λ 0 \frac{hc}{\lambda_0} ​ is the work function ϕ \phi  (minimum energy required to eject an electron). Step 2 — Relate kinetic energy and de-Broglie wavelength For an electron having de-Broglie wavelength λ e \lambda_e ​ : K max = p 2 2...

  ⚡ Question Given below are two statements: Assertion (A): The outer body of an aircraft is made of metal which protects persons sitting inside from lightning strikes. Reason (R): The electric field inside the cavity enclosed by a conductor is zero. 🖼️ Question Image ✍️ Short Solution Step 1 — Understanding the concept When a conductor is placed in an external electric field (like a thundercloud’s field), charges on its surface rearrange themselves such that the net electric field inside becomes zero . This is a direct result of electrostatic shielding — a fundamental property of conductors. Step 2 — Applying to the aircraft The metal body of the aircraft acts as a closed conducting shell (essentially a Faraday cage ). Any external electric field (from lightning or thunderclouds) causes charge redistribution only on the outer surface . The field inside the cabin (the cavity) remains zero , keeping passengers completely safe from the electric discharge. ...

  ❓ Question A dipole made of two charges of magnitude 2   μ C 2\ \mu\text{C}  (equal and opposite) separated by a distance is placed between the plates of a capacitor that has a potential difference V = 5  V V=5\ \text{V} . The plate separation is 0.5  mm 0.5\ \text{mm} . The dipole axis is parallel to the electric field established between the plates. If the dipole is rotated by 30 ∘ 30^\circ  from the field axis, it experiences a torque tending to realign it. Find the value of the torque. 🖼️ Question Image ✍️ Short Solution Step 1 — Electric field between capacitor plates Uniform field magnitude: E = V d = 5  V 0.5 × 10 − 3  m = 5 5 × 10 − 4 = 1.0 × 10 4  V/m . Step 2 — Dipole moment Dipole moment magnitude p = q   .   With q = 2   μ C = 2 × 10 − 6  C q=2\ \mu\text{C}=2\times10^{-6}\ \text{C}  and assumed separation s = 0.5  mm = 5 × 10 − 4  m s=0.5\ \text{mm}=5\times10^{-4}\ \text{m} , p = 2 × 1...

  ❓ Question Match List - I with List - II: List - I List - II (A) Mass density (I) [ML²T⁻³] (B) Impulse (II) [MLT⁻¹] (C) Power (III) [ML²T⁰] (D) Moment of inertia (IV) [ML⁻³T⁰] 🖼️ Question Image ✍️ Short Solution To solve this, recall the dimensional formulas for each physical quantity: Step 1 — Mass density (ρ) Mass density is mass per unit volume: ρ = M V ⇒ [ ρ ] = M L 3 = [ M L − 3 T 0 ] ✅ Match → (A) → (IV) Step 2 — Impulse (J) Impulse is the product of force and time : J = F Δ t = ( M L T − 2 ) ( T ) = [ M L T − 1 ] ✅ Match → (B) → (II) Step 3 — Power (P) Power is work done per unit time : P = W t = F ⋅ L T = ( M L T − 2 ) L T = [ M L 2 T − 3 ] ✅ Match → (C) → (I) Step 4 — Moment of inertia (I) Moment of inertia about an axis: I = ∑ m r 2 ⇒ [ I ] = M L 2 ✅ Match → (D) → (III) 🧮 Image Solution ✅ Conclusion & Video Solution ✅ Final Matching: List - I List - II (A) Mass density (IV) [M L⁻³ T⁰] (B) Impulse (II) [M L T⁻¹] (C) Power (I) [M L² ...

  ❓ Question The dimension of μ 0 ε 0​ is equal to that of: Options: Velocity Charge Magnetic field Current (Given: μ 0 \mu_0  = vacuum permeability, ε 0 \varepsilon_0 = vacuum permittivity.) 🖼️ Question Image ✍️ Short Solution Let’s analyze the dimensions of μ 0 ε 0 . Step 1 — Recall known relations in electromagnetism From Maxwell’s equations, the speed of light c c c in vacuum is related to μ 0 \mu_0 μ 0 ​ and ε 0 \varepsilon_0 ε 0 ​ as: c = 1 μ 0 ε 0 . c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}. ​ 1 ​ . Now take the reciprocal and modify it: μ 0 ε 0 = 1 ε 0 c . \sqrt{\frac{\mu_0}{\varepsilon_0}} = \frac{1}{\varepsilon_0 c}. But let’s instead focus on the dimensional relation . We know that: μ 0 ε 0 = 1 c . So, μ 0 ε 0 = μ 0 μ 0 ε 0 = μ 0 c . Thus, the dimension of μ 0 ε 0 = (dimension of μ₀) × (dimension of velocity) . Step 2 — Dimensional formula of μ₀ From the definition of magnetic field B B : B = μ 0 I 2 π r ...

  ❓ Question The helium and argon gases are placed in a flask at the same room temperature (300 K) . What is the ratio of their average kinetic energies (per molecule) ? 🖼️ Question Image ✍️ Short Solution Step 1 — Recall the formula for average kinetic energy per molecule For any gas, according to the kinetic theory of gases , the average kinetic energy of a molecule is: E k ‾ = 3 2 k T where k k  = Boltzmann constant (1.38 × 10⁻²³ J/K) T T  = absolute temperature in kelvin Step 2 — Dependence on temperature Notice that the average kinetic energy per molecule depends only on temperature , not on the nature (mass or type) of the gas. That means — if two gases are at the same temperature, each molecule has the same average kinetic energy regardless of whether the gas is light (like helium) or heavy (like argon). Step 3 — Apply given data Given: T He = T Ar = 300  K T_{\text{He}} = T_{\text{Ar}} = 300~\text{K} So, E He = 3 2 k T He...