A closed organ pipe filled with a gas resonates at its fundamental frequency at 27^ C . If the temperature of the gas is increased such that the fundamental frequency increases by exactly 20 % , the new temperature of the gas is:

Step-by-step solution

The fundamental frequency of an organ pipe is directly proportional to the speed of sound in the gas, which in turn is proportional to the square root of its absolute temperature. Thus, f T Let the initial frequency be f₁ at T₁ = 27^ C = 27 + 273 = 300 ~K . A 20 % increase in frequency means the new frequency is f₂ = f₁ + 0.20 f₁ = 1.2 f₁ . Using the proportionality: f₂ f₁ = T₂ T₁ 1.2 = T₂ 300 Squaring both sides: 1.44 = T₂ 300 T₂ = 1.44 300 = 432 ~K Converting back to Celsius: T₂ (^ C ) = 432 - 273 = 159^ C Answer