It has been found that in a new scale of temperature known as the ${}^\circ P$ scale which is linear, the freezing and boiling points of water are $25{}^\circ P$ and $225{}^\circ P$respectively. Find out the temperature on this fresh scale, which will be corresponding to a temperature of \[25{}^\circ C\] on the Celsius scale?
\[\begin{align}
& A.27{}^\circ P \\
& B.75{}^\circ P \\
& C.50{}^\circ P \\
& D.275{}^\circ P \\
\end{align}\]
Answer
623.7k+ views
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.
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Trending doubts
Difference Between Prokaryotic Cells and Eukaryotic Cells

Two of the body parts which do not appear in MRI are class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

What will happen if the mucus is not secreted by the class 11 biology CBSE

