Answer
77.7k+ views
Hint: Both the microwave and ultrasonic sound waves are electromagnetic waves. Electromagnetic waves are the waves in which magnetic field vector and electric field vector are mutually perpendicular to each other and their propagation is also perpendicular to both the electric and magnetic field vector. Try to recall the formula in which the wavelength and frequencies of the waves are related to each other.
Complete step by step solution:
Velocity of the microwave is ${{{v}}_{{m}}}{{ = 3 \times 1}}{{{0}}^{{8}}}{{ m/s}}$ which is equal to the speed of light.
Velocity of the ultrasonic waves is ${{{v}}_{{u}}}{{ = 3 \times 1}}{{{0}}^{{2}}}{{ m/s}}$
The relationship between the wavelength and the frequency is given by the formula:
${{\upsilon = }}\dfrac{{{v}}}{{{\lambda }}}$
For microwave, the relation between wavelength and the frequency can be modified by using above formula as:
$\Rightarrow {{{\upsilon }}_{{m}}}{{ = }}\dfrac{{{{{v}}_{{m}}}}}{{{{{\lambda }}_{{m}}}}}$
Similarly, for ultrasonic waves we have
$\Rightarrow {{{\upsilon }}_{{u}}}{{ = }}\dfrac{{{{{v}}_{{u}}}}}{{{{{\lambda }}_{{u}}}}}$
It is given that microwave and an ultrasonic sound wave have the same wavelength i.e. $\Rightarrow {{{\lambda }}_{{m}}}{{ = }}{{{\lambda }}_{{u}}}{{ = \lambda }}$ so ${{{\upsilon }}_{{m}}}{{ = }}\dfrac{{{{{v}}_{{m}}}}}{{{\lambda }}}$ and ${{{\upsilon }}_{{u}}}{{ = }}\dfrac{{{{{v}}_{{u}}}}}{{{{{\lambda }}_{{u}}}}}$
Now, the ratio of the frequencies of a microwave and an ultrasonic sound wave is given by
$\Rightarrow \dfrac{{{{{\upsilon }}_{{m}}}}}{{{{{\upsilon }}_{{u}}}}}{{ = }}\dfrac{{\dfrac{{{{{v}}_{{m}}}}}{{{\lambda }}}}}{{\dfrac{{{{{v}}_{{u}}}}}{{{\lambda }}}}}$
Or $\dfrac{{{{{\upsilon }}_{{m}}}}}{{{{{\upsilon }}_{{u}}}}}{{ = }}\dfrac{{\dfrac{{{{{v}}_{{m}}}}}{{{1}}}}}{{\dfrac{{{{{v}}_{{u}}}}}{{{1}}}}}{{ }} \\
\therefore \dfrac{{{{{\upsilon }}_{{m}}}}}{{{{{\upsilon }}_{{u}}}}}{{ = }}\dfrac{{{{{v}}_{{m}}}}}{{{{{v}}_{{u}}}}} $
On substituting the values in above relation, we get
$\Rightarrow \dfrac{{{{{\upsilon }}_{{m}}}}}{{{{{\upsilon }}_{{u}}}}}{{ = }}\dfrac{{3 \times {{10}^8}}}{{3 \times {{10}^2}}} = {10^6}$
Thus, the ratio of the frequencies of a microwave and an ultrasonic sound wave is ${10^6}$.
Therefore, option (C) is the correct choice.
Note: The SI unit of velocity is ${{m/s}}$. The SI unit of frequency is hertz (${{Hz}}$). The SI unit of wavelength is meter (${{m}}$). As the SI unit of frequency is hertz (${{Hz}}$) but in the answer no SI unit is mentioned. It is so because the ratio of frequencies of both the electromagnetic waves are given so it is a unit less quantity.
Complete step by step solution:
Velocity of the microwave is ${{{v}}_{{m}}}{{ = 3 \times 1}}{{{0}}^{{8}}}{{ m/s}}$ which is equal to the speed of light.
Velocity of the ultrasonic waves is ${{{v}}_{{u}}}{{ = 3 \times 1}}{{{0}}^{{2}}}{{ m/s}}$
The relationship between the wavelength and the frequency is given by the formula:
${{\upsilon = }}\dfrac{{{v}}}{{{\lambda }}}$
For microwave, the relation between wavelength and the frequency can be modified by using above formula as:
$\Rightarrow {{{\upsilon }}_{{m}}}{{ = }}\dfrac{{{{{v}}_{{m}}}}}{{{{{\lambda }}_{{m}}}}}$
Similarly, for ultrasonic waves we have
$\Rightarrow {{{\upsilon }}_{{u}}}{{ = }}\dfrac{{{{{v}}_{{u}}}}}{{{{{\lambda }}_{{u}}}}}$
It is given that microwave and an ultrasonic sound wave have the same wavelength i.e. $\Rightarrow {{{\lambda }}_{{m}}}{{ = }}{{{\lambda }}_{{u}}}{{ = \lambda }}$ so ${{{\upsilon }}_{{m}}}{{ = }}\dfrac{{{{{v}}_{{m}}}}}{{{\lambda }}}$ and ${{{\upsilon }}_{{u}}}{{ = }}\dfrac{{{{{v}}_{{u}}}}}{{{{{\lambda }}_{{u}}}}}$
Now, the ratio of the frequencies of a microwave and an ultrasonic sound wave is given by
$\Rightarrow \dfrac{{{{{\upsilon }}_{{m}}}}}{{{{{\upsilon }}_{{u}}}}}{{ = }}\dfrac{{\dfrac{{{{{v}}_{{m}}}}}{{{\lambda }}}}}{{\dfrac{{{{{v}}_{{u}}}}}{{{\lambda }}}}}$
Or $\dfrac{{{{{\upsilon }}_{{m}}}}}{{{{{\upsilon }}_{{u}}}}}{{ = }}\dfrac{{\dfrac{{{{{v}}_{{m}}}}}{{{1}}}}}{{\dfrac{{{{{v}}_{{u}}}}}{{{1}}}}}{{ }} \\
\therefore \dfrac{{{{{\upsilon }}_{{m}}}}}{{{{{\upsilon }}_{{u}}}}}{{ = }}\dfrac{{{{{v}}_{{m}}}}}{{{{{v}}_{{u}}}}} $
On substituting the values in above relation, we get
$\Rightarrow \dfrac{{{{{\upsilon }}_{{m}}}}}{{{{{\upsilon }}_{{u}}}}}{{ = }}\dfrac{{3 \times {{10}^8}}}{{3 \times {{10}^2}}} = {10^6}$
Thus, the ratio of the frequencies of a microwave and an ultrasonic sound wave is ${10^6}$.
Therefore, option (C) is the correct choice.
Note: The SI unit of velocity is ${{m/s}}$. The SI unit of frequency is hertz (${{Hz}}$). The SI unit of wavelength is meter (${{m}}$). As the SI unit of frequency is hertz (${{Hz}}$) but in the answer no SI unit is mentioned. It is so because the ratio of frequencies of both the electromagnetic waves are given so it is a unit less quantity.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write an article on the need and importance of sports class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Choose the one which best expresses the meaning of class 9 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)