Science M.C.Q. [1 M]

At room temperature, the minimum distance for echo is $17.2m.$

Loudness of sound is proportional to the intensity and amplitude of sound. So when we turn on the volume of $TV,$ the amplitude of sound wave increases hence, loudness of sound increases

Light waves, heat waves, $X-$rays are examples of electromagnetic waves, which do not require any medium for their propagation.
Sound waves cannot travel through a vacuum. It needs a material medium for its propagation. Sound waves can propagate only when the vibrational energy of the particles of the medium is travelling from one point to another.

The number of cycles per unit time is known as frequency. The SI unit of frequency is Hertz $(Hz).$

A vibrating body vibrates the surrounding air, and when these vibrations reach our ears we hear sound. But, some vibrations cannot be perceived as sound to humans as human ears can only detect sound in the frequency range of $20Hz$ to $20,000Hz,$ so depending upon the vibration frequency, can only we define a vibration as sound.

Frequency of $1\ Hz$ implies to $1$ complete vibration per second. This means, $60$ complete vibrations in $60$ second, or, in $1$ minute.

Audible sound has a frequency range of $20\ Hz$ to $20\ kHz.$
Infrasonic frequency is below audible range, i.e. below $20\ Hz.$
Ultrasonic frequency is above audible range, i.e. above $20\ KHz.$

An earthquake produces infra-sound before the main shock wave. Infra-sound is a low-frequency sound. It is sound that is lower in frequency than $20\ Hz$ (hertz) or cycles per second, the normal limit of human hearing. Hearing becomes gradually less sensitive as frequency decreases, so for humans to perceive infra-sound, the sound pressure must be sufficiently high.
Rhinos have extremely good ears, picking up infra-sound far deeper than the range of human hearing.
Rhinos can hear down to a frequency of four hertz, whereas even a human baby, with entirely undamaged ears, can normally only pick up sounds as low as $20\ hertz.$ Giraffes and elephants can also hear in this infra-sound range.

$d=?$
$\mathrm{v}=3 \times 10^8 \mathrm{~ms}^1$
$\mathrm{t}=2 \times 10^{-5} \mathrm{sec}$
$\text{d}=\frac{\text{v}\times\text{t}}{2}$
$\text{d}=\frac{3\times10^8\times2\times10^{-5}}{2}$
$=3\times10^3\text{m}$

Frequency is defined as the number of oscillations in one second , as the given object is oscillating, 50 times in one second therefore its frequency is $50\ Hz .$

$SONAR$ is a device that uses ultrasonic waves to measure the distance, direction and speed of under water objects.

The number of cycles an oscillator completes in each second is called its frequency.
The frequency $(f)$ of a wave is the number of full wave forms generated per second.
This is the same as the number of repetitions per second or the number of oscillations per second.

 $\text{v}=\frac{2\text{d}}{\text{t}}$
$\Rightarrow\text{v}=\frac{2\times640}{4}=320\text{m/ s}$
$\therefore\text{v}=\frac{2\text{d}}{\text{t}}$
$\Rightarrow320=\frac{2\text{x}}{7}$
$\text{x}=1120\text{m}$
Distance $=1120+640$
$=1760\text{m}$