Question

The frequency of Radar is $780MHz$ . The frequency of the reflected aeroplane is increased by $2.6KHz$. The frequency of the aeroplane is:
a. $2km/s$
b. $1km/s$
c. $0.5km/s$
d. $0.25km/s$

Answer 1

Hint: We can approach the given problem with the help of the Doppler Effect. While solving the given problem remember that the frequency of the aeroplane is lesser than the velocity of the light.

Formula used:
To calculate the frequency of the aeroplane:
$\Rightarrow {\nu }_a={\displaystyle \frac{2{\nu }_{a}f}{c-{\nu }_a}}$
Remember: $c-{\nu }_a\simeq c$
Where ${\upsilon }_a$ frequency of the aeroplane, $f$frequency and $c$ is the velocity of light and it is equal to the value of $3.8\times {10}^{8}m/s$

Complete step by step answer:
Consider the given values in the question. It is given that,
Frequency of the radar is $780MHz$.
Frequency from reflected waves from an aeroplane is increased by$2.6KHz$.
Solve the given problem with the help of the Doppler Effect. In physics, the Doppler effect is defined as the decrease or increase in the frequency of light, sound, or other waves, as the observer and the source move away or towards each other.
We can derive the formula with the help of the apparent frequency $f^{\prime }$. That can be calculated as,
$\Rightarrow f^{\prime }=f\left({\displaystyle \frac{c+{\nu }_a}{c}}\right)$
We can consider the frequency of the aeroplane. We have,
$\Rightarrow f_a=f\left({\displaystyle \frac{c+{\nu }_a}{c}}\right)\left({\displaystyle \frac{c}{c-{\nu }_a}}\right)$
On cancelling out the common terms we get,
$\Rightarrow f_a=f\left({\displaystyle \frac{c+{\nu }_a}{c-{\nu }_a}}\right)$

The difference of frequency of the aeroplane and the frequency is
$$\Rightarrow f_a-f=f\left({\displaystyle \frac{c+{\nu }_a}{c-{\nu }_a}}\right)-1$$
From this formula we can derive the formula to find the frequency of the aeroplane. That is,
To calculate the frequency of the aeroplane:
$\Rightarrow {\nu }_a={\displaystyle \frac{2{\nu }_{a}f}{c-{\nu }_a}}$
Remember: $c-{\nu }_a\simeq c$
Where ${\upsilon }_a$ frequency of the aeroplane, $f$ frequency and $c$ is the velocity of light and it is equal to the value of $3.8\times {10}^{8}m/s$.
Remember that the frequency of the aeroplane is lesser than the velocity of the light.
On substitute the values in the equation we get,
$\Rightarrow 2.6\times {10}^3={\displaystyle \frac{2\times 780\times {10}^6\times {\nu }_a}{3\times {10}^8}}$
Simplify the given equation we get,
$\Rightarrow {\displaystyle \frac{2.6\times {10}^3}{2\times 7.8}}\times 3$
$\Rightarrow {\nu }_a=500m/s$
On convert the meter into kilometre,
$\Rightarrow {\nu }_a=0.5km/s$
$\therefore {\nu }_a=0.5km/s$

Hence, the correct answer is option (C).

Note: Doppler Effect can be calculated if the values of the source, observer values are given. If the source that is moving towards the observer at rest, we have the formula as,
$f^{\prime }={\displaystyle \frac{V}{\left(V-V_S\right)}}f$
Where, $f^{\prime }$is the apparent frequency, $f$ is the actual frequency, $V_S$ is the velocity of the sound waves and $V$is the actual velocity of source.