A RADAR sends a signal to an airplane at a distance 45km away with a speed of $3 \times {10^8}m{s^{ - 1}}$. After how much time is the signal received back from the airplane?
A) $3 \times {10^{ - 4}}s$
B) $3 \times {10^{ - 3}}s$
C) $2 \times {10^{ - 4}}s$
D) $5 \times {10^{ - 2}}s$

129k+ views
Hint: Sound bounces back from a solid or liquid like a rubber ball bounce off a wall. Like light, the sound gets reflected at the surface of a solid or liquid and follows the same laws of reflection obstacle of large size which may be polished or rough or curved is needed for the reflection of sound waves. It follows the laws of reflection:
(i) The reflected and incident sound waves make an equal angle.
(ii) The incident wave, normal, reflected sound waves lie in the same plane.

Formula used:
$v = \dfrac{{2d}}{t}$
Where $v$ is the speed of the sound wave, $d$ is the distance of the receiver is the time required for the signal.

Complete step by step solution:
We are provided with the data which is as given,
The speed of sound waves, $v = 3 \times {10^8}m{s^{ - 1}}$
Distance $d = 45km$
 We know that RADAR sends a signal to an airplane that is at a distance of 45km, then the signal reflects, and then it is received by the RADAR. Therefore the total distance traveled is two times the original distance.
Thus the total distance is equal to $2d = 2 \times 45 = 90km$
We know the formula of speed as shown here, ${\text{speed = }}\dfrac{{dis\tan ce}}{{time}}$
Which can be mathematically given as, $v = \dfrac{{2d}}{t}$
We are interested to find out the time a signal is received back from the airplane that is‘t’.
Therefore rearrange the equation for t we get, $t = \dfrac{{2d}}{v}$
Let's substitute the value in the rearranged equation we get,
 $t = \dfrac{{2 \times 45 \times {{10}^3}}}{{3 \times {{10}^8}}} = 3 \times {10^{ - 4}}s$ Or ${\text{300}}$ $\mu s$
$\therefore$ The time required to receive the signal back from the airplane is $3 \times {10^{ - 4}}s$. Hence, the correct option is (A).

Additional information:
If we shout or clap near suitable reflecting objects like a tall building or a mountain, we can hear the same sound, again and again, a little later. This sound which we hear is called echo. The sensation of sound persists in our brain for about $0.1s$. To hear a distinct echo the time interval between the original sound and reflected one must be at least $0.1s$. Echoes may be heard more than once due to successive or multiple reflections.

Two practical applications of sound waves:
1) Echo, which is used to find the distance between the two distant objects
2) Sound navigation and ranging (SONAR), which is used to find the depth of the ocean.
Sound can be reflected from the curved surface also and follows laws of reflection.