Govt. Exams
For isobaric process: Q = nCₚΔT. For monoatomic gas, Cₚ = (5/2)R. Q = 5 × (5/2) × 8.314 × 100 = 62.355 kJ
In Carnot cycle, isothermal processes involve heat exchange with reservoirs (Q₁ at hot reservoir, Q₂ at cold reservoir), while adiabatic processes have no heat exchange (Q=0)
For adiabatic process: TVγ⁻¹ = constant. T₁V₁^(γ-1) = T₂V₂^(γ-1). 300 × 1^0.4 = T₂ × 2^0.4. T₂ = 300/2^0.4 = 300/1.6818 = 178.9 K
For isothermal process: W = nRT ln(Vf/Vi) = 1 × 8.314 × 300 × ln(5/2) = 2494.2 × 1.609 = 3.04 kJ
Acceleration a = dv/dt = 6t + 2. At t=2s: a = 6(2) + 2 = 14 m/s². Force F = ma = 2 × 14 = 28 N
For Atwood-like system with one block horizontal: a = (m_B × g)/(m_A + m_B) = (2 × 10)/(3 + 2) = 20/5 = 4 m/s²
Using s = ut + ½gt² for complete journey (s=0): 0 = 30t - 5t². Solving: t(30 - 5t) = 0, giving t = 6 s (non-zero solution)
Resultant force = √(8² + 6²) = √(64 + 36) = √100 = 10 N. Acceleration = F/m = 10/6 = 1.67 m/s²
Net force = 20 - 12 = 8 N. Work done by net force = 8 × 5 = 40 J. This equals KE gained = 40 J
Stability increases with larger base area and lower center of gravity. The 20×30 cm face down provides the largest base area, ensuring maximum stability.
