Question

If the end correction of an open pipe is 0.8cm, then inner radius of that pipe will be:
a. \[\dfrac{1}{3}cm\]
b. \[\dfrac{2}{3}cm\]
c. \[\dfrac{3}{2}cm\]
d. \[0.2cm\]

Answer 1

Hint: When a strong blast of air is allowed to flow from the pipe then high frequency notes can be observed in the pipe known as overtones, the distance between the antinodes and the open end of the pipe is known as end correction.

Formula used: The formula of the open pipe is given by,
$ \Rightarrow \Delta l = 1 \cdot 2 \times r$
Where end correction is $\Delta l$ and inner radius is $r$.

Complete step by step answer:
Whenever there is a formation of wave in the open pipe then the wave is not actually starting from the pipe’s starting point but instead it is starting before the opening of the pipe and therefore we can have error in studying the properties of the wave theoretically and therefore we need end correction for better studying of the wave formation. The end correction depends only on the radius of the pipe.
It is given in the problem that the end correction of an open pipe is 0.8cm, then we need to find the inner radius of that pipe.
The formula of the open pipe is given by,
$ \Rightarrow \Delta l = 1 \cdot 2 \times r$
Where end correction is$\Delta l$ and inner radius is r.

The end correction is given as 0.8 cm therefore we get.
$ \Rightarrow \Delta l = 1 \cdot 2 \times r$
$ \Rightarrow 0 \cdot 8 = 1 \cdot 2 \times r$
$ \Rightarrow r = \dfrac{{0 \cdot 8}}{{1 \cdot 2}}$
$ \Rightarrow r = \dfrac{8}{{12}}$
$ \Rightarrow r = \dfrac{4}{6}$
$ \Rightarrow r = \dfrac{2}{3}cm$.
The inner radius of the pipe is $r = \dfrac{2}{3}cm$.

Hence, the correct answer is option (B).

Note: Remember the formula of the relation between the end correction and the inner relation. The antinodes are mid-way between the nodes and they are the points where the wave has maximum amplitude. When we design an organ or Boomwhacker then we must consider their diameter.