The frequency of the third overtone of a pipe of length ' L_c ', closed at one end is same as the frequency of the sixth overtone of a pipe of length ' L₀ ', open at both ends. Then the ratio L _ c : L ₀ is

Step-by-step solution

Third overtone of pipe closed at one end f = (2 n +1) V 4 l₁ = (2 3+1) V 4 l₁ = 7 ~V 4 l₁ Sixth overtone of pipe open at both ends aligned & f = nV 2 l₂ = 7 ~V 2 l₂ 7 ~V 4 l₁ = 7 ~V 2 l₂ & ~L _ c ~L _ o = 1 2 ( l ₁= L _ C and l ₂= L _ O ) aligned