Answer

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**Hint:-**

- Fundamental frequency is given by the equation\[f = \dfrac{1}{{2L}}\sqrt {\dfrac{T}{M}} \].

- The immersed object will get an upthrust from the water.

**Complete step by step solution:-**

The fundamental frequency is \[f = \dfrac{1}{{2L}}\sqrt {\dfrac{T}{M}} \]

Where \[L\] is the length of the string.

\[M\] is the linear mass density of the string.

\[T\] is the tension of the string.

The fundamental frequency for transverse standing waves in the wire at air is given is \[300Hz\].

Consider the tension \[T\] on the string in air. If an object is immersed in a liquid the tension is changed to\[T'\].

Consider \[V\] is the volume of liquid, \[p\]is the specific gravity, \[g\] is the acceleration due to gravity.

\[T' = T - \](Upthrust force).

Here only the half of the volume is displaced, so that

Mass of the volume displace is half of the product of volume and density

\[ = \dfrac{1}{2}V\rho \]

\[\rho \]is the density of liquid.

The upthrust force \[ = \dfrac{1}{2}V\rho g\]

\[T' = T - \dfrac{1}{2}V\rho g\]

Tension in the air \[T = Vpg\]

Density of water \[\rho = 1g/cc\]

\[T' = Vpg - \dfrac{1}{2}Vg\]

\[T' = Vg(p - \dfrac{1}{2})\]

New fundamental frequency, after the object is immersed in water,

\[f' = \dfrac{1}{{2L}}\sqrt {\dfrac{{T'}}{M}} \]

Substitute new tension value in this equation we get

\[f' = \dfrac{1}{{2L}}\sqrt {\dfrac{{Vg(p - \dfrac{1}{2})}}{M}} \]

Compare this with the fundamental frequency before the object immersed in water,

\[

f'/300 = \sqrt {\dfrac{{p - \dfrac{1}{2}}}{P}} \dfrac{{\sqrt {(p - \dfrac{1}{2})} }}{{\sqrt p }} \\

\\

\]

Simplify the equation,

\[f'/f = \dfrac{{\sqrt {Vg(p - \dfrac{1}{2})} }}{{\sqrt {Vpg} }}\]

The fundamental frequency (\[f\]) is already given, \[f = 300Hz\]

\[f'/300 = \dfrac{{\sqrt {Vg(p - \dfrac{1}{2})} }}{{\sqrt {Vpg} }}\]

We cancel the terms inside the root also,

\[f'/300 = \dfrac{{\sqrt {(p - \dfrac{1}{2})} }}{{\sqrt p }}\]

\[f'/300 = \sqrt {\dfrac{{p - \dfrac{1}{2}}}{P}} \]

Now rearrange the equation, like the given options,

\[f'/300 = {(\dfrac{{p - \dfrac{1}{2}}}{P})^{1/2}}\]

\[f' = 300{(\dfrac{{2p - 1}}{{2P}})^{1/2}}\]

**So the answer is (A) \[f' = 300{(\dfrac{{2p - 1}}{{2P}})^{1/2}}\]**

**Note:-**

- The value of acceleration due to gravity is \[g = 9.8m/s\].

- The unit of frequency is Hertz (\[Hz\]).

- The sound in the water is faster than sound in air.

- The immersed object weight is equivalent to the volume displaced times the density and acceleration due to gravity.

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