Question

Sunil is standing between two walls. The wall closest to him is at a distance of $660m$. If he shouts, he hears the first echo after $4s$ and after another $2$ seconds.
(A) What is the velocity of sound in air?
(B) What is the distance between the two walls?

Answer 1

The strategy of equivalent dividers relates particularly to the four-level headings. Its ability lies in its control over these headings, in unmistakable ways which can be used to make a doubt that everything is great and acceptable, course, and center interest.

The low-Reynolds-number development of a singular roundabout particle between equivalent dividers is settled from the particular impression of the speed field delivered by multiples of the force thickness on the atom's surface.
An incredible flexibility tensor is fabricated and couples these force multiples to depictions of the speed field in the fluid enveloping the particle.
Every segment of the incredible adaptability tensor is a restricted, mentioned measure of in reverse forces of the partition between the dividers.
These new explanations are used in a lot of Stoke Sian components multiplications to process the translational and rotational rates of a particle settling between equivalent dividers and the Brownian coast power on an atom diffusing between the dividers.

Formula used:
${\text{Velocity = }}\dfrac{{{\text{Distance}}}}{{{\text{Time}}}}$
${{Distance = Velocity \times Time}}$

Complete step by step solution:
Let us consider the given in the question,
Distance $= 660m$
First sound $= 4s$
Second sound $= 2s$
Now we have to calculate the velocity of sound in air:
${\text{Velocity = }}\dfrac{{{\text{Distance}}}}{{{\text{Time}}}}$
Velocity $= \dfrac{{660}}{2}m/s$
Velocity $= 330m/s$
Therefore, the velocity of sound in air is $330m/s$
Now we have to the distance between the two walls:
For the first wall,
Distance For the first wall,
Hence, we have already known that,
Velocity $= 330m/s$
Time $= 4{\text{s}}$
${{Distance = Velocity \times Time}}$
Distance $= 330 \times 4$
Distance $= 1320m \to \left( 1 \right)$
For the second wall,
Time $= 2{\text{s}}$
Velocity $= 330m/s$
Distance $= 330 \times 2$
Distance $= 330 \times 1$
${\text{Distance }} = 330{\text{ m}} \to \left( 2 \right)$
Adding equation $\left( 1 \right)$ and $\left( 2 \right)$,
Distance between ${1^{st}}$ wall $+$ distance between ${2^{nd}}$ wall = distance between two walls
Distance between two walls $= 1320 + 330$
Distance between two walls $= 1650m$

$\therefore$ The distance between $2$ the walls is $= 1650m$.

Note: The Einstein revision to the Newtonian consistency of a weakened suspension that represents the adjustment in stress conveyance because of the presence of the channel dividers is resolved. It is proposed how the technique and results can be reached out to calculations including numerous particles and intermittent reenactments of suspensions in kept calculations.

Separation is characterized as the measure of room between two things or the condition of being far separated. A case of separation is five feet between two tables. A case of separation is the contrast between different sides of an issue. Separation is a scalar amount that alludes to "how much ground an article has covered" during its movement. Dislodging is a vector amount that alludes to "how out of sight place an article is"; it is the item's general change in position.