Question

A cylindrical glass tube is partially filled with water to create an air column of length $0.1m$. The air column is in resonance, in its fundamental mode, with a tuning fork. The level of the water is now lowered and the next resonance is observed at $0.35m$. The end correction is

• A. $0.025m$
• B. $0.020m$
• C. $0.015m$
• D. $0.010m$

Answer 1

Answer: (A)

$0.025m$

$\text{Given}:$ $l_1=0.1m$, $l_2=0.35m$

$\text{Solution}:$

The wave velocity, v is given as:

$v=\lambda \nu$

Where,$\lambda$ =wavelength $\nu$ = frequency of wave.

In first case, length of air column =$l_1+e$

This length is equal to one-fourth of the wavelength in the fundamental mode. $⟹\frac{\lambda }{4}=l_1+e$

$\therefore \lambda =4(l_1+e)$

Therefore, velocity of wave is given as:

$v=4(l_1+e)\nu$…………. Equation (1)

In the second case, length of air column $=l_2+e$

This length is equal to three-fourth of the wavelength in the first harmonic mode.

$⟹\frac{3\lambda }{4}=l_2+e$

$⟹\lambda =\frac{4(l_2+e)}{3}$

Therefore, velocity of wave is given as:

$v=\frac{4(l_2+e)\nu }{3}$ …………. Equation (2)

Since the wave velocity remains constant, thus equating equations (1) and (2),

$⟹4(l_1+e)\nu =\frac{4(l_2+e)\nu }{3}$

Cancelling 4 and $\nu$ from both sides, we get

$⟹(l_1+e)=\frac{(l_2+e)}{3}$

$⟹3l_1+3e=l_2+e$

$⟹2e=l_2-3l_1$

Now we have $l_1=0.1m$ and $l_2=0.35m$,

$⟹2e=0.35-3\left(0.1\right)$

$⟹2e=0.35-0.30$

$⟹2e=0.05$

$⟹e=\frac{0.05}{2}$

$\therefore e=0.025m$

$\text{Hence A is the correct opton}$