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**Hint:**The ratio of the difference of the boiling temperature and the freezing temperature in the mentioned scale in question to the difference of the boiling temperature and the freezing temperature in the degree Celsius scale will be equivalent to the ratio of difference of the temperature which is to be calculated and the corresponding temperature in degree Celsius scale to the corresponding temperature in degree Celsius scale. This will help you in answering this question.

**Complete step by step answer:**

it has been given in the question that the freezing point of the water is,

\[{{T}_{f}}=25{}^\circ P\]

The boiling point of the water has been given in the question as,

\[{{T}_{b}}=225{}^\circ P\]

As we all know, the ratio of the difference of the boiling temperature and the freezing temperature in the mentioned scale in question to the difference of the boiling temperature and the freezing temperature in the degree Celsius scale will be equivalent to the ratio of difference of the temperature which is to be calculated and the corresponding temperature in degree Celsius scale to the corresponding temperature in degree Celsius scale. That is we can write that,

\[\dfrac{{{T}_{b}}-{{T}_{f}}}{100{}^\circ C-0{}^\circ C}=\dfrac{T-25{}^\circ C}{25{}^\circ C}\]

Substituting the values in it will give,

\[\dfrac{225-25}{100{}^\circ C-0{}^\circ C}=\dfrac{T-25{}^\circ C}{25{}^\circ C}\]

Simplifying the equation can be written as,

\[\begin{align}

& \dfrac{200}{100}=\dfrac{T-25{}^\circ C}{25} \\

& \Rightarrow 2=\dfrac{T-25{}^\circ C}{25} \\

\end{align}\]

Rearranging this equation will give,

\[\begin{align}

& 50=T-25 \\

& \therefore T=75{}^\circ P \\

\end{align}\]

**So, the correct answer is “Option B”.**

**Note:**The degree Celsius is defined as a unit of temperature on the Celsius scale. This temperature scale is otherwise called the centigrade scale. The degree Celsius can be considered to a particular temperature on the Celsius scale or a unit to represent a difference between two temperatures.

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