A tuning fork of frequency ' n ' is held near the open end of a tube which is dipped in water and length of the tube is adjusted until resonance occurs. If the two shortest lengths that produce resonance are l₁ and l₂ , the speed of sound in air is (neglect end correction)

Step-by-step solution

For a resonance tube, the water level acts as a closed end, forming a closed organ pipe. The first resonance occurs at length l₁ = 4 The second resonance occurs at length l₂ = 3 4 The difference between the two resonating lengths is: l₂ - l₁ = 3 4 - 4 = 2 = 2(l₂ - l₁) The speed of sound is given by v = n Substituting the value of , we get: v = 2n(l₂ - l₁) Answer: 2n(l₂ - l₁)