This one’s a little tricky. It hasn’t been mentioned anywhere that the resonances occur at the fundamental frequency. Let’s write lambda as Y Consider this. The end correction is usually small compared to the resonating length. So if we consider it to be resonating at the fundamental at 0.62 (since 0.62>0.6), then we should expect a wavelength lambda close to (4 x 0. 62) or __2.48__. We know that for a tube closed at one end, the allowed resonances correspond to Y/4, 3Y/4 , 5Y/4 i.e. (2n+1) Y/4 ; n = 0,1,2…… The difference between successive allowed lengths thus works out to (3Y/4) – Y/4 = Y/2 = 0.62 - 0.6 = 0.02 or __Y = 2 x 0.02 =0.04__ Compare this with the earlier value. Obviously something’s wrong. What’s actually happening is this. An allowed frequency higher than the fundamental is what causes the resonance, and the **wavelength** corresponding to this frequency (not the fundamental wavelength) is **equal to twice this difference in length**, Thus if L is the effective length and L’ the measured length, we may write L = L’ + 0.3d. Since the allowed Ys are given by, (2n+1)Y/4 = L, we may write for the two cases Y = 4 L / (2n+1) = 4(L’ + 0.3d)/(2n+1). Note that the two lengths **0.62 and 0.6 correspond to the same wavelength **__but different numbers of this wavelength.__ They do not represent resonance at different frequencies. Thus we may write, **4 (0.6+0.3d )/(2n+1) = 4(0.62+0.3d)/[ 2(n+1)+1] ……. (1)** which on simplification gives (0.6+0.3d )/(2n+1) = (0.62+0.3d)/(2n+3) Cross multiplication, componenedo dividendo (i think) etc can lead us to 2/(60+30d) = 2/(2n+1). Equating denominators, we get, 60+30d = 2n+1 or 30d = 2n+1 – 60. Now 30d is a +ve quantity . The lowest value of n for which this is true is **n=30** , which gives us 30d = 1 or d = 1/30 = 0.0333 m or __3.33 cm __ The required answer is thus **c** As a further check, if we plug n = 30 in eqn **(1)**, we get L.H.S. = R.H.S = 0.03999 |