Question

It is the desire to increase the fundamental resonance frequency in a tube which is closed at one end. This can be achieved by
A. Replacing the air in the tube by the hydrogen gas
B. Increasing the length of the tube
C. Decreasing the length of the tube
D. Opening the closed end of the tube

Answer 1

This question has multiple correct options so we have to check all the options one by one and implement all four conditions that are given in the question. To do this we will use a formula of frequency which shows the relation between frequency, length and velocity. That will help us to find correct options amongst given options

Formula used:
$f=\dfrac{v}{4l}$

Complete step by step solution:
Now it is given that in the question that the tube is closed at one end so it has one node and one antinode.
We know that formula for resonance frequency in a tube which is closed at one end is given by
$f=\dfrac{v}{4l}....\left( 1 \right)$
Where f = frequency
l = length of the tube
v = velocity

Now let’s check the option (A)
A) Replacing the air in the tube by hydrogen gas
So we know that hydrogen is less dense and lighter than air so according to formula of density
$\rho =\dfrac{m}{v}$
From the equation (2) we can say that the velocity of hydrogen is more than velocity of air
So according to equation (1) when the velocity increases frequency will also increase hence the option (A) is correct.

Now option (B) Increasing the length of the tube:
From the equation (1) we can state that the frequency is inversely proportional to length l
$f\propto \dfrac{1}{l}.....\left( 3 \right)$
So if we increase length frequency will be decreased thus option (B) is incorrect.

Now option (C)
Decreasing length of the tube

From the equation (3) we can conclude that if we decrease the length of tube frequency it will get increased hence the option (C) is correct.

Option (D) opening the closed end of the tube.
When we open one closed end it will act as tube with open at both ends and formula of the frequency of tube which is open at both the end is
$f=\dfrac{v}{2l}......\left( 2 \right)$
As we can see in equation (2) length is decreased as compared to equation (1) hence the frequency is greater than frequency in the tube which is open at one end.
Hence the option (D) is correct.

Note:
This question can be altered or twisted as find that from below given options which option is wrong or in which the case frequency will decrease, so in that case the correct option will be option (B).