Govt. Exams
For isolated capacitor, charge Q remains constant. U = Q²/(2C) = Q²d/(2ε₀A). Energy is proportional to d, so doubling d doubles energy... correction: U = CV²/2, but Q is constant so U = Q²/(2C) ∝ d, energy increases
By Gauss's law, in the region between spheres, only inner charge +Q contributes. E = kQ/r², directed radially outward
Using Gauss's law for an infinite plane sheet: E = σ/(2ε₀). The field is independent of distance and uniform on both sides
Distance from each vertex to centroid = a/√3. Total potential = 3 × kq/(a/√3) = 3√3kq/a. Correction: V = 3kq√3/a ≈ 3kq/a for approximation
Theoretical maximum COP = T_cold/(T_hot - T_cold) = 250/(350-250) = 250/100 = 2.5
For adiabatic process: TV^(γ-1) = constant. T₁V₁^0.4 = T₂(1.5V₁)^0.4. T₂/T₁ = (1/1.5)^0.4 = 0.73^0.4 ≈ 0.73
At equilibrium: T_f = (400 + 300)/2 = 350 K. ΔS_A = m·c·ln(T_f/T_A) = 1×400×ln(350/400) = -56 J/K. ΔS_B = 1×400×ln(350/300) = 56.4 J/K. ΔS_total ≈ 11.2 J/K
Internal energy is a state function, so for a cyclic process where the system returns to its initial state, ΔU = 0 always, regardless of the path.
From first law: ΔU = Q - W. If both Q and W are positive, ΔU = Q - W could be positive, negative, or zero depending on whether Q > W, Q < W, or Q = W respectively.
ΔS = Q/T = (2260 × 1000)/373 = 2260000/373 ≈ 6061 J/K ≈ 6.1 kJ/K
