Question

Fill In the blanks for Resonance tube

(i) In resonance tube first resonating length is 25 cm then its second resonating length will be (if e = 0) ..........
(ii) In resonance tube first resonating length is 25 cm then its second resonating length will be (if e $\neq$ 0) ..........
(iii) Length of resonance tube is 140 cm. How many resonance are possible for wave have wave length 40 cm..........
(iv) Length of resonance tube is 150 cm. How many resonance are possible for wave having frequency 400 Hz and speed of sound 320 m/s. ..........
(v) Length of resonance tube ts 150 cm. Maximum level of liquid, in resonance condition for wave having wavelength 80cm. is........
(vi) Length of resonance tube is 150 cm Minimum level of liquid, in resonance condition for wave having wavelength 80 cm. is.........
(vii) ln resonance tube first resonating length is 17 cm, second resonating length is 55 cm then wavelength of wave is..........
(viii) In resonance tube first resonating length is 17 cm, second resonant length is 55 on then radius of tube is.........

Accepted Answer

$(i)$ Given, length of first resonating length $\neq$

Second resonating length $(for e = 0) l_{2} = \frac{3 λ}{4} = 3 \times 25 = 75 c m$

$(i i)$ Second resonating length, for $e \neq 0$ , $l_{2} = \frac{3 λ}{4} + e$

Hence, length slightly more than $75 c m$

$(i i i)$ Given, length $l = 140 c m$ , or $140 + 10 % = 154 c m$

$l_{7} = \frac{15 λ}{4} = 150 < 154 c m$ , Hence, $7$ resonance is possible.

$(i v)$ Given, length $l = 150 c m$ , Wavelength $λ = \frac{320}{400} = 0.8 m$

Hence, $\frac{(2 N - 1) (0.8 \times 100)}{4} \leq 150$

Integer value of $N = 4$ , Maximum number of resonance $= 4$

$(v)$ Given, length $l = 150 c m$ , wavelength $λ = 80 c m$

Hence, Maximum level of liquid $=$ Length of maximum resonance $= \frac{(2 N - 1) λ}{4}$

$\Rightarrow \frac{(2 \times 4 - 1) (80)}{4} - 10 = 130 c m$ , where $e = 10 c m$

$(v i)$ Given, length of resonance tube $l = 150 c m$ , wavelength $λ = 80 c m$

Minimum length of liquid $=$ Length of end correction $= 10 c m$

$(v i i)$ Length of first resonance $l_{1} = \frac{λ}{4} = (17 + e) c m$ , length of second resonance $l_{2} = \frac{3 λ}{4} = (55 + e) c m$

Hence, $λ = (17 + 2) \times 4 ≃ 76 c m$ , here, $e = 1.7 \times \frac{10}{100} ≃ 2$

$(v i i i)$ Length of first resonance $l_{1} = \frac{λ}{4} = (17 + e) c m$ , length of second resonance $l_{2} = \frac{3 λ}{4} = (55 + e) c m$

End correction $e = 0.3 D$ , where $D =$ diameter, $e = \frac{l_{2} - 3 l_{1}}{2} = \frac{55 - (3 \times 17)}{2} = 2$
Radius $r = \frac{D}{2} = \frac{e}{2 \times 0.3} = 3.33 c m$

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