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# A 3.6 m long vertical pipe resonates with a source of frequency 212.5 Hz when water level is at certain height in the pipe. Find the height of water level (from the bottom of the pipe) at which resonance occurs. Neglect end correction. Now, the pipe is filled to a height H (≈ 3.6 m). A small hole is drilled very close to its bottom and water is allowed to leak. Obtain an expression for the rate of fall of water level in the pipe as a function of H. If the radii of the pipe and the hole are 2 x 10-2 m and 1 x 10-3 m respectively, calculate the time interval between the occurrences of first two resonances. Speed of sound in air is 340 m/s and g = 10 m/s2.

7 years ago
Hello Student,
Speed of sound, v = 340 m/s
Let ℓ0 be the length of air column corresponding to the fundamental frequency. Then
v/4 ℓ0 = 212.5
or ℓ0 = v/4 (212.5) = 340/4 (212.5) = 0.4 m.
NOTE : In closed pipe only odd harmonic s are obtained. Now, let ℓ1, ℓ2, ℓ3, ℓ4, etc. be the lengths corresponding to the 3rd harmonic, 5th harmonic, 7th harmonic etc. Then
3(v/4 ℓ1) = 212.5 ⇒ ℓ1 = 1.2 m;
5 (4 ℓ2 = 212.5 ⇒ ℓ2 = 2.0 m;
7 (v/ 4 ℓ3) = 212.5 ⇒ ℓ3 = 2.8 m;
9 (v/4 ℓ4) = 212.5 ⇒ ℓ4 = 3.6 m;
Or heights of water of water level are (3.6 – 0.4) m, (3.6 – 1.2) m, (3.6 – 2.0) m and (3.6 – 2.8) m.
Therefore heights of water level are 3.2 m, 2.4 m, 1.6 m and 0.8 m.
Let A and a be the area of cross – sections of the pipe and hole respectively. Then
A = π ( 2 x 10-2)2 = 1.26 x 10-3 m-2
And a = π (10-3)2 = 3.14 x 10-6 m2
Velocity of efflux, v = √2 g H
Continuity equation at 1 and 2 gives,
a √2 g H = A (-d H/dt)
Therefore, rate of fall of water level in the pipe,
(-d H/dt) = a/A √2 g H
Substituting the values, we got
-d H/dt = 3.14 x 10-6/1.26 x 10-3 √2 x 10 x H
⇒ -d H/dt = (1.11 x 10-2) √H
Between first two resonances, the water level falls from 3.2 m to 2.4 m.
∴ d H/√H = -1.11 x 10-2 dt
$\int _{3.2}^{2.4}\frac{dH}{\sqrt{H}}$= -(1.11 x 10-2) $\int _{0}^{t} dt$
⇒ 2[ √2.4 - √3.2 ] = -(1.1 x 10-2) t
⇒ t = 43 second

Thanks