A standard tuning fork contains frequency ${n_0}$ another TF ${n_a}$ have $20\% $ more compared to ${n_0}$. ${n_b}$ $30\% $ more compared to ${n_0}$ . ${n_a}$ and ${n_b}$ have a beat relation of 6 beats per sec. Determine the values of ${n_0}$ , ${n_a}$.

Hint: In order to solve the question, we will first of all we will find the re${n_b}$lation between tuning fork one and tuning fork two after that we will find the relation between tuning fork one and tuning fork three and then we will use the relationship between tuning fork two and tuning fork three to find the frequency of tuning fork one after then using first fork we will find frequency of other forks.

Complete step by step answer:
In the question we are given;
standard tuning fork one (TF 1) contains frequency = ${n_0}$
standard tuning fork two (TF 2) contains frequency = ${n_a}$
standard tuning fork three (TF 3) contains frequency = ${n_b}$
relation between (TF 1) and (TF 2) according to the question
${n_a}$ have $20\% $ more compared to ${n_0}$
$\Rightarrow {n_a} = {n_0} + \dfrac{{20}}{{100}} \times {n_0}$
$\Rightarrow {n_a} = {n_0} + 0.2 \times {n_0}$
$\Rightarrow {n_a} = 1.2{\text{ }}{n_0}$
Now we will state ${n_a} = 1.2{\text{ }}{n_0}$ as equation 1 relation between (TF 1) and (TF 3) according to the question ${n_b}$ have $30\% $ more compared to ${n_0}$.
$\Rightarrow {n_b} = {n_0} + \dfrac{{30}}{{100}} \times {n_0}$
$\Rightarrow {n_b} = {n_0} + 0.3 \times {n_0}$
$\Rightarrow {n_b} = 1.3{\text{ }}{n_0}$

Now we will state ${n_b} = 1.3{\text{ }}{n_0}$ as equation 2.Beat is the difference between two frequencies in the question we are given: ${n_a}$ and ${n_b}$ have a beat relation of 6 beats per sec.Therefore, subtracting equation 2 from equation 1 we will get,
${n_b} - {n_a} = 1.3{\text{ }}{n_0} - 1.2{\text{ }}{n_0}$
substituting in the above equation we get
${n_b} - {n_a} = 6$
$\Rightarrow 6 = 0.1{\text{ }}{n_0}$
From above equation we will get the value of ${\text{ }}{n_0}$ as
${\text{ }}{n_0} = 60{\text{ }}beats{\text{ }}{s^{ - 1}}$
Now using equation 1 and the value of ${\text{ }}{n_0}$ we will find the ${\text{ }}{n_a}$
${n_a} = 1.2{\text{ }} \times 60{\text{ }}beats{\text{ }}{s^{ - 1}}$
$\Rightarrow {n_a} = 42{\text{ }}beats{\text{ }}{s^{ - 1}}$
Now using equation 2 and the value of ${\text{ }}{n_0}$ we will find the ${\text{ }}{n_b}$
${n_b} = 1.3{\text{ }} \times 60{\text{ }}beats{\text{ }}{s^{ - 1}}$
$\therefore {n_b} = 48{\text{ }}beats{\text{ }}{s^{ - 1}}$

Hence, the answers are ${\text{ }}{n_0} = 60{\text{ }}beats{\text{ }}{s^{ - 1}}$, ${n_a} = 42{\text{ }}beats{\text{ }}{s^{ - 1}}$ and ${n_b} = 48{\text{ }}beats{\text{ }}{s^{ - 1}}$.

Note: Many of the students will make the mistake by not solving for ${\text{ }}{n_0}$ instead of that solving for ${\text{ }}{n_a}$ and ${\text{ }}{n_b}$ but we should first find that variable which have relation with other two variable in that situation we only have to face two variable equation but if we do not use common variable it can result in three variable equation in answer which makes confusion.