Jitender Pal
Last Activity: 10 Years ago
Sol. A = 1 cm^2 = 1 × 10^–4 m^2, M = 0.03 g = 0.03 × 10^–3 kg,
P = 1 atm = 10^5 pascal, L= 40 cm = 0.4 m.
L base 1 = 80 cm = 0.8 m, P = 0.355 atm
The process is adiabati
P(V)^γ = P(V’)^γ = ⇒ 1 * (AL)^γ = 0.355 * (A2L)^γ ⇒ 1 1 = 0.355 2 ^γ ⇒ 1/0.355 = 2^γ
= γ log 2 = log (1/0.355) = 1.4941
V = √γP/f = √1.4941 * 10^5/m / v = √1.4941 * 10^5/(0.03 * 10^-3/10^-4 * 1 * 0.4) = √1.441 * 10^5 * 4 * 10^-5/3 * 10^-5 = 446.33 = 447 m/s