Question

A hospital uses an ultrasonic scanner to locate tumours in a tissue. The operating frequency of the scanner is 3.2MHz. The speed of sound in the tissue is \[1.6\,{\text{km}}\,{{\text{s}}^{ - 1}}\] . The wavelength of sound in the tissue is:
(A) \[0.25\,{\text{mm}}\]
(B) \[0.5\,{\text{mm}}\]
(C) \[0.75\,{\text{mm}}\]
(D) \[1\,{\text{mm}}\]

Answer 1

Hint: First of all, we will find out the relation which relates velocity, wavelength and frequency. We will convert all the units into an S.I system of units and substitute the required values. After that we will manipulate accordingly to obtain the desired result.

Complete step by step solution:
In the given question, we are supplied with the following data:
There is an ultrasonic scanner in a hospital to locate the tumours in a tissue.The frequency at which the scanner operates is \[3.2\,{\text{MHz}}\] .The speed of the sound inside the tissue is \[1.6\,{\text{km}}\,{{\text{s}}^{ - 1}}\].We are asked to find out the wavelength of the sound which was used to locate the tumours.To begin with, we know sound is a type of wave, which propagates through a medium. We all know that sound waves cannot travel through vacuum. These are the types of mechanical waves where a medium is essential for its propagation. We know that for a constant frequency, the velocity of a sound wave is directly proportional to the wavelength of the wave. Higher the wavelength is, faster it travels through a medium.Before we proceed, let us convert the units of velocity and the frequency into its respective S.I system of units.
$3.2\,{\text{MHz}} \\
\Rightarrow 3.2 \times {10^6}\,{\text{Hz}}$
And, $1.6\,{\text{km}}\,{{\text{s}}^{ - 1}} \\
\Rightarrow 1.6 \times {10^3}\,{\text{m}}\,{{\text{s}}^{ - 1}}$
We all know a relation which relates velocity, frequency and wavelength of a sound wave, which is given by:
\[v = \lambda n\] …… (1)
Where,
\[v\] indicates the velocity of the sound wave.
\[\lambda \] indicates the wavelength of the sound wave.
\[n\] indicates the frequency at which it was operated.
Now, we substitute the required values in the equation (1) and we get:
$v = \lambda n \\
\Rightarrow \lambda = \dfrac{v}{n} \\
\Rightarrow \lambda = \dfrac{{1.6 \times {{10}^3}\,{\text{m}}\,{{\text{s}}^{ - 1}}}}{{3.2 \times {{10}^6}\,{{\text{s}}^{ - 1}}}} \\
\Rightarrow \lambda = 5 \times {10^{ - 4}}\,{\text{m}}$
Again, we can modify the result as:
$\Rightarrow \lambda = 0.5 \times {10^{ - 3}}\,{\text{m}} \\
\therefore \lambda = 0.5\,{\text{mm}}$
Hence, the wavelength of the sound which was used to locate the tumours is \[0.5\,{\text{mm}}\] .

The correct option is B.

Note: While solving this problem, it is better to convert all the units of the physical quantities involved, to the S.I system of units, before attempting to solve the problem. This substantially reduces the chance of getting wrong answers. Higher the wavelength is, longer distances the wave can travel.