Answer
396.9k+ views
Hint: To solve the given question, take the help of the known values of the temperatures of melting ice and boiling water at standard atmospheric pressure and find the error in the readings of the thermometer. Then take any two temperatures with their errors and calculate the difference.
Complete step by step answer:
It is given that a thermometer gave the readings of $-1{}^{\circ }C$ and $99{}^{\circ }C$ when inserted into melting ice and boiling water respectively and both the observations were done at standard atmospheric pressure. However, by other accurate experiments we know that the temperature of a melting ice at standard atmospheric pressure is equal to $0{}^{\circ }C$ and the temperature of boiling water at standard atmospheric pressure is equal to $100{}^{\circ }C$.
This means that the given thermometer gives a reading that is less than the actual reading by $1{}^{\circ }C$. Therefore, we can say that the thermometer has an error of $-1{}^{\circ }C$.Now, suppose the same thermometer is used to measure the temperatures of two bodies whose actual temperatures are ‘x’ ${}^{\circ }C$ and ‘y’ ${}^{\circ }C$ respectively.
The actual difference between the two temperatures is equal to $(x-y){}^{\circ }C$
Then the thermometer will show the readings of the temperatures as $(x-1){}^{\circ }C$ and $(y-1){}^{\circ }C$.
Now, the difference between the two temperatures according to the given thermometer is $\left[ (x-1)-(y-1) \right]{}^{\circ }C=(x-y){}^{\circ }C$. With this we get that the difference in the reading is as same as the actual difference in the readings. Therefore, the error in the difference in the readings in zero.
Hence, the correct option is C.
Note: Some may say that an accurate thermometer can show a reading of $-1{}^{\circ }C$ when it is used to measure the temperature of a melting ice. Since the ice is melting, it will have a lower temperature. However, note that a substance changes its phases at a constant temperature. Therefore, until an ice cube melts and becomes water it will be at temperature of $0{}^{\circ }C$.
Complete step by step answer:
It is given that a thermometer gave the readings of $-1{}^{\circ }C$ and $99{}^{\circ }C$ when inserted into melting ice and boiling water respectively and both the observations were done at standard atmospheric pressure. However, by other accurate experiments we know that the temperature of a melting ice at standard atmospheric pressure is equal to $0{}^{\circ }C$ and the temperature of boiling water at standard atmospheric pressure is equal to $100{}^{\circ }C$.
This means that the given thermometer gives a reading that is less than the actual reading by $1{}^{\circ }C$. Therefore, we can say that the thermometer has an error of $-1{}^{\circ }C$.Now, suppose the same thermometer is used to measure the temperatures of two bodies whose actual temperatures are ‘x’ ${}^{\circ }C$ and ‘y’ ${}^{\circ }C$ respectively.
The actual difference between the two temperatures is equal to $(x-y){}^{\circ }C$
Then the thermometer will show the readings of the temperatures as $(x-1){}^{\circ }C$ and $(y-1){}^{\circ }C$.
Now, the difference between the two temperatures according to the given thermometer is $\left[ (x-1)-(y-1) \right]{}^{\circ }C=(x-y){}^{\circ }C$. With this we get that the difference in the reading is as same as the actual difference in the readings. Therefore, the error in the difference in the readings in zero.
Hence, the correct option is C.
Note: Some may say that an accurate thermometer can show a reading of $-1{}^{\circ }C$ when it is used to measure the temperature of a melting ice. Since the ice is melting, it will have a lower temperature. However, note that a substance changes its phases at a constant temperature. Therefore, until an ice cube melts and becomes water it will be at temperature of $0{}^{\circ }C$.
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