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**Hint:**In this question, we can solve the equation \[dB = 10\log {a_0}^2 - 10\log {a_1}^2\]. After then we can assume that the amplitude is changed by the factor $n$. Now, we can solve the above equation for $n$.

**Complete step by step solution: -**

We know that the loudness of sound is given by equation

$dB = 10\log \left( {\dfrac{{{I_0}}}{{{I_1}}}} \right) $

$\Rightarrow dB = 10\log {I_0} - 10\log {I_1} $

We know that intensity is directly proportional to the square of the amplitude i.e.

$I \propto {a^2}$

So,

\[dB = 10\log {a_0}^2 - 10\log {a_1}^2\]

According to the question, if amplitude of sound is multiplied by a factor of$\sqrt {10} $, the decibel level increases by \[10\] units. So,

$d{B^1} = 10\log {\left( {\sqrt {10} } \right)^2} - 10\log {a_1}^2 $

$\Rightarrow d{B^1} = 10\log 10 - 10\log {a_1}^2 $

$\Rightarrow d{B^1} = 10 - 10\log {a_1}^2 $

$\Rightarrow d{B^1} = 10 - dB $

Where \[dB = 10\log {a_1}^2$

Now, if \[70dB\] is being played at a function and if it is reduced to a level of\[30dB\] , then let the amplitude of the instrument playing music be reduced by a factor of $n$.

So,

$d{B^1} = 10\log \left( {\dfrac{{{a_1}^2}}{{{n^2}}}} \right) $

$\Rightarrow d{B^1} = 10\log {a_1}^2 - 10\log {n^2} $

$\Rightarrow d{B^1} = dB - 20\log n $

$\Rightarrow d{B^1} - dB = 20\log n$

$\Rightarrow 70 - 30 = 20\log n $

$\Rightarrow 40 = 20\log n $

$\Rightarrow 2 = \log n $

$\Rightarrow n = {10^2} $

$\Rightarrow n = 100 $

So, loud music of \[70dB\] is being played at a function. To reduce the loudness to a level of\[30dB\] , the amplitude of the instrument playing music to be reduced by a factor of $100$ .

**Hence, option C is correct.**

**Additional information: -**

Decibel is a logarithmic unit which is used to measure the loudness. It is used in electronics, signals and communications. Decibel is a logarithmic way of describing ratios of power, sound pressure, voltage, intensity, etc. Generally, it is used to measure the loudness of the sound. The level $0dB$ occurs when the intensity of the sound is equal to the reference level of the sound.

**Note:**

In this question, we have kept in mind that \[d{B^1}\] is the difference in decibel. We have to remember the calculations of logarithmic also such as $\log 10 = 1$ and $\log 1 = 0$.

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