Answer

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**Hint**To solve this question, we need to perform the dimensional analysis for the given relation between the quantities in the question. To do this, we have to consider the dimensions of the quantities on the LHS and RHS and equate separately the powers of the dimensions to get the equations corresponding to the unknown variables.

**Complete step by step answer**

According to the question, the frequency $(n)$ is proportional to the length $L$ raised to the power$a$, density $d$ raised to the power $b$ and the Young’s modulus $Y$ raised to the power $c$, that is,

$\Rightarrow n \propto {L^a}{d^b}{Y^c}$

Removing the proportionality sign with the constant$c$, we get

$\Rightarrow n = c\left( {{L^a}{d^b}{Y^c}} \right)$ (1)

For the above equation to be correct, the dimensions of the quantity in the LHS should be equal to the dimensions of the quantities in the RHS.

Replacing the quantities of the above equation with their dimensions, we get

$\Rightarrow \left[ n \right] = \left[ {{M^0}{L^0}{T^{ - 1}}} \right]$, $\left[ L \right] = \left[ {{M^0}{L^1}{T^0}} \right]$, $\left[ d \right] = \left[ {{M^1}{L^{ - 3}}{T^0}} \right]$ and $\left[ Y \right] = \left[ {{M^1}{L^{ - 1}}{T^{ - 2}}} \right]$

$\because c$is a constant, so it has no dimensions.

Substituting these in (1) we get

$\Rightarrow \left[ {{M^0}{L^0}{T^{ - 1}}} \right] = {\left[ {{M^0}{L^1}{T^0}} \right]^a}{\left[ {{M^1}{L^{ - 3}}{T^0}} \right]^b}{\left[ {{M^1}{L^{ - 1}}{T^{ - 2}}} \right]^c}$

$\Rightarrow \left[ {{M^0}{L^0}{T^{ - 1}}} \right] = \left[ {{M^0}{L^a}{T^0}} \right]\left[ {{M^b}{L^{ - 3b}}{T^0}} \right]\left[ {{M^c}{L^{ - c}}{T^{ - 2c}}} \right]$

On simplifying, we get

$\Rightarrow \left[ {{M^0}{L^0}{T^{ - 1}}} \right] = \left[ {{M^{b + c}}{L^{a - 3b - c}}{T^{ - 2c}}} \right]$

Comparing the exponents of similar dimensions, we get

$\Rightarrow b + c = 0$ (2)

$\Rightarrow a - 3b - c = 0$ (3)

And

$\Rightarrow - 2c = - 1$ (4)

From (4), we get $c = \dfrac{1}{2}$

Putting this in (2)

$\Rightarrow b + \dfrac{1}{2} = 0$

$\Rightarrow b = - \dfrac{1}{2}$

Putting the values of$b,c$in (3)

$\Rightarrow a + \dfrac{3}{2} - \dfrac{1}{2} = 0$

$\Rightarrow a + 1 = 0$

Finally, $a = - 1$

$\therefore a = - 1,b = - \dfrac{1}{2},c = \dfrac{1}{2}$

Putting these values in (1)

$\Rightarrow n = c\left( {{L^{ - 1}}{d^{ - \dfrac{1}{2}}}{Y^{\dfrac{1}{2}}}} \right)$

Or, $f = \dfrac{c}{L}\sqrt {\dfrac{Y}{d}} $

Therefore, the formula for the frequency of a tuning fork is

$\Rightarrow f = \dfrac{c}{L}\sqrt {\dfrac{Y}{d}} $, where $c$is a constant.

So, the values of $a$, $b$ and $c$ are $ - 1, - \dfrac{1}{2},\dfrac{1}{2}$ respectively.

**Hence, the correct answer is option B.**

**Note**

The dimensions of all the quantities of the question have been deduced using their respective fundamental formulae. We can use any formula in physics corresponding to a quantity to find out the dimension of that quantity.

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