2022 A/L Physics Past Paper · Question 26 (English Medium)

Answer: (4) $100$

Given information

What we know:

All the instruments are identical and stand on the circle, so each one is the same distance from the centre. Each gives the same sound intensity there.

With $10$ instruments the sound intensity level at the centre is $50\ \text{dB}$.

We want the number of instruments that makes the level $60\ \text{dB}$.

What the question asks

We want: the number $N_2$ of identical instruments that give $60\ \text{dB}$ at the centre.

What each letter means:

$\beta$ ($\text{dB}$): the sound intensity level.

$I$ ($\text{W m}^{-2}$): the sound intensity.

$I_0$ ($\text{W m}^{-2}$): the intensity at the threshold of hearing.

$N_1$, $N_2$: the number of instruments at the start and at the end.

$I_1$, $I_2$ ($\text{W m}^{-2}$): the intensity at the centre for $N_1$ and for $N_2$ instruments.

Relevant theory

Therefore, a quality called intensity level has been introduced to represent the above mentioned range. It is defined as follows. It's unit is decibel (dB).

Intensity level of sound (β) = 10 log₁₀ ( I/I₀ )

If I is the sound intensity at the given point due to one operating electric drill, then the sound intensity at the same point due to four operating drills = 4I

For example, one child singing in a room is quiet. Ten children singing together are much louder, because their sound energy adds up.

The same happens here. The intensity from all the instruments together is the intensity of one, taken as many times as there are instruments.

$$I = N\,I_{\text{one}}$$

In words: the total intensity is the number of instruments times the intensity of one.

The level uses the intensity in the formula. If we take the level before and after, the threshold intensity $I_0$ cancels in the difference.

$$\beta_2 - \beta_1 = 10 \log_{10}\left(\dfrac{I_2}{I_1}\right)$$

In words: the rise in level is ten times the $\log_{10}$ of the ratio of the two intensities.

Required equations

$I = N\,I_{\text{one}}$

$\beta_2 - \beta_1 = 10 \log_{10}\left(\dfrac{I_2}{I_1}\right)$

$N_2 = N_1 \times \dfrac{I_2}{I_1}$

Applying it to this question

The level is in $\text{dB}$ and the ratio $\dfrac{I_2}{I_1}$ has no unit. The number of instruments is a pure number.

$10$

$\beta_2 - \beta_1 = 60 - 50 = 10$

$10 \log_{10}\left(\dfrac{I_2}{I_1}\right) = 10 \Rightarrow \log_{10}\left(\dfrac{I_2}{I_1}\right) = 1 \Rightarrow \dfrac{I_2}{I_1} = 10$

$100$

$N_2 = N_1 \times \dfrac{I_2}{I_1} = 10 \times 10 = 100$

Reasoning and calculation

A rise of $10\ \text{dB}$ means the intensity is ten times larger.

The intensity is in proportion to the number of identical instruments. So the number must also be ten times larger: $10 \times 10 = 100$. That is option (4).

Checks

$n2 / n1$

Does it make sense?

Every $10\ \text{dB}$ more means ten times the intensity. Going from $50\ \text{dB}$ to $60\ \text{dB}$ is one such step.

The number of instruments is only $10$ now, so a ten-fold increase gives $100$. A small number like $20$ or $50$ could not give a ten-fold intensity.

Conclusion

$N_2 = 100$

The intensity must be $10$ times larger, so the number of instruments must be $10$ times larger. We need $N_2 = 100$ instruments.

The answer is (4).

Every $10\ \text{dB}$ means ten times the intensity. So $+10\ \text{dB}$ is $\times 10$, $+20\ \text{dB}$ is $\times 100$ and $+30\ \text{dB}$ is $\times 1000$.

Do not add decibels the way you add numbers of instruments.

Exam tip