The persistence of sound in a room after the source of sound is t… - Vidyadip

MCQ

The persistence of sound in a room after the source of sound is turned off is called reverberation. The measure of reverberation time is the time required for sound intensity to decrease by $60 \,dB$. It is given that the intensity of sound falls off as $I_0 \exp \left(-c_1 \alpha\right)$ where $I_0$ is the initial intensity, $c_1$ is a dimensionless constant with value $1 / 4$. Here, $\alpha$ is a positive constant which depends on the speed of sound, volume of the room, reverberation time, and the effective absorbing area $A_e$. The value of $A_e$ is the product of absorbing coefficient (with value between $0$ and $1,1$ being a perfect absorber) and the area of the room. For a concert hall of volume $600 \,m ^3$, the value of $A_e$ (in $m ^2$ ) required to give a reverberation time of $1 s$ is closest to (speed of sound in air $=340 \,m / s$ )

A
$50$
✓
$100$
C
$110$
D
$67$

✓

Answer

Correct option: B.

$100$

b
(B)

$\beta_1-\beta_2=10 \log \frac{I_1}{I_2}$

$6=\log \frac{I_1}{I_2}$

$I_2=10^{-6} I_1$

$I = I _0 e ^{-\alpha / 4}$

$10^{-6} I _0= I _0 e ^{- a / 4} \Rightarrow \alpha=24 \ln 10$

Using dimensional analysis we get

$\alpha=\frac{ A _{ e } V _{ s } t }{ V }$

[Although it cannto be calculated technically, as we have less equations and more variables but this is done just by observation]

$24 \ell \operatorname{n} 10=\frac{ A _{ e } v _{ s } t }{ v }$

$A _{ e }=\frac{24 \ell n 10 \times v }{ v _{ s } t }$

$A _{ e }=\frac{24 \times 6400 \times 2.303}{340 \times 1}$

$A _{ e }=97.6 \approx 100 \,m ^2$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from 14. waves and sound→14. waves and sound — all question types→Physics STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

A coconut fruit hanging high in a palm tree has $.........$ owing to its location.

The respective speeds of five molecules are $2, 1.5, 1.6, 1.6$ and $1.2 \,km/sec.$ The most probable speed in $km/sec$ will be

Given below are two statements :

Statement$-I:$ Acceleration due to gravity is different at different places on the surface of earth.

Statement$-II:$ Acceleration due to gravity increases as we go down below the earth's surface.

In the light of the above statements, choose the correct answer from the options given below

Thief's car is moving with a speed of $10\,m/s$. A police van chasing this car with a speed of $5\,m/s$ fires a bullet at the thief's car with muzzle velocity $72\,km/h$. Find the speed with which the bullet with hit the car.............$m/s$

Ratio between maximum range and square of time of flight in projectile motion is

A satellite moves round the earth in a circular orbit of radius $R$ making one revolution per day. A second satellite moving in a circular orbit, moves round the earth once in $8$ days. The radius of the orbit of the second satellite is

Four identical spheres each of mass $m$ are placed at the corners of square of side $2\,metre$. Taking the point of intersection of the diagonals as the origin, the co-ordinates of the centre of mass are

Two satellites $A$ and $B$ of masses $200\, kg$ and $400\, kg$ are revolving round the earth at height of $600\, km$ and $1600\, km$ respectively. If $T _{ A }$ and $T _{ B }$ are the time periods of $A$ and $B$ respectively then the value of $T _{ B }- T _{ A }$

[Given : radius of earth $=6400\, km$, mass of earth $=6 \times 10^{24}\, kg$ ]

A satellite is orbiting the earth close to its surface. A particle is to be projected from the satellite to just escape from the earth. The escape speed from the earth is νe. Its speed with respect to the satellite:

In the 5th overtone of an open organ pipe, these are $(N-$ stands for nodes and $A-$ for antinodes)