Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

❓ Question The electric field in a region is given by E = ( 2 i ^ + 4 j ^ + 6 k ^ ) × 10 3   N/C . The flux of the field through a rectangular surface parallel to the x–z plane is 6.0 N·m²·C⁻¹ . Find the area of the surface . 🖼️ Question Image ✍️ Short Solution We’ll use the electric flux formula : Φ = E ⃗ ⋅ A ⃗ = E A cos ⁡ θ Where: Φ \Phi  = Electric flux E ⃗ \vec{E}  = Electric field A ⃗ \vec{A}  = Area vector (perpendicular to the surface) θ \theta  = angle between E ⃗ \vec{E}  and A ⃗ \vec{A} 🔹 Step 1 — Determine direction of the area vector The surface is parallel to the x–z plane , ⇒ its normal vector is along the y-axis (positive or negative ĵ ). Thus, A ⃗ = A j ^​ 🔹 Step 2 — Find component of electric field perpendicular to surface Electric field: E ⃗ = ( 2 i ^ + 4 j ^ + 6 k ^ ) × 10 3 Only the ĵ-component (i.e., E y E_y E y ​ ) contributes to the flux through the x–z surface. So, E y = 4 × 10 3   N/C 🔹 Step 3...

  ❓ Question An inductor of reactance X L = 100   Ω X_L = 100~\Omega , a capacitor of reactance X C = 50   Ω X_C = 50~\Omega , and a resistor of resistance R = 50   Ω R = 50~\Omega  are connected in series with an AC source of 10 V , f = 50  Hz f = 50~\text{Hz} . The average power dissipated by the circuit is W W . Find the value of W W . 🖼️ Question Image ✍️ Step-by-Step Solution Step 1 — Given data R = 50   Ω , X L = 100   Ω , X C = 50   Ω , V rms = 10  V . Step 2 — Find net reactance The total reactance in a series RLC circuit is the difference between inductive and capacitive reactances: X = X L − X C = 100 − 50 = 50   Ω . Step 3 — Find impedance The total impedance Z Z  is given by Z = R 2 + X 2 = 50 2 + 50 2 = 5000 = 70.71   Ω . Step 4 — Find current (rms value) I rms = V rms Z = 10 70.71 = 0.1414  A . Step 5 — Find power factor (cos φ) Power factor for a series RLC circuit: cos ⁡ ϕ ...

  ❓ Question Two cylindrical rods A and B made of different materials are joined end-to-end in a straight line. The ratios are: L A L B = 1 2 , r A r B = 2 , K A K B = 1 2 . The free ends of rods A and B are held at 400 K and 200 K respectively. Find the interface temperature (temperature at the junction) when steady-state equilibrium is established. 🖼️ Question Image ✍️ Short Solution In steady state the heat current I I  (rate of heat flow) is the same through both rods. Thermal resistance of a rod of length L L , cross-sectional area A A , and thermal conductivity K K  is: R = L K A . For series rods, temperature drops are proportional to their resistances. If T i T_i ​ is the interface temperature then: 400 − T i R A    =    T i − 200 R B . So compute R A R_A  and R B R_B  up to a common factor using given ratios. Step 1 — Express areas via radii Area A = π r 2 A=\pi r^2 . Given r A / r B = 2 r_A/r_B=2  → A A / A B = ( r A / r B )...

  💡 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 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...