Question

The temperature of two bodies measured by a thermometer are ${{t}_{1}}={{20}^{0}}C\pm {{0.5}^{0}}C$ and ${{t}_{2}}={{50}^{0}}C\pm {{0.5}^{0}}C$. The temperature difference and the error there is
A. ${{30}^{0}}C\pm {{1}^{0}}C$
B. ${{70}^{0}}C\pm {{0.5}^{0}}C$
C. ${{30}^{0}}C\pm {{0.5}^{0}}C$
D. ${{70}^{0}}C\pm {{1}^{0}}C$

Answer 1

Hint: This problem can be solved by finding out the absolute temperature difference and then finding the maximum absolute error using the respective formulas. The maximum absolute error will be the sum of the individual absolute errors of the measurements.

Formula used:
For a quantity $Z$ which is defined as
$Z=\left| X-Y \right|$
where $X,Y$ are two other physical quantities (defined as $x+\Delta x,y+\Delta y$ respectively), the absolute error in $Z$, that is $\left| \Delta z \right|$ is defined as
$\left| \Delta z \right|=\left| \Delta x \right|+\left| \Delta y \right|$
where $\left| \Delta x \right|,\left| \Delta y \right|$ are the absolute errors in $X,Y$ respectively.
Also, $z=\left| x-y \right|$
$\therefore Z=z+\Delta z$

Complete step-by-step answer:
First we will find the absolute temperature difference between the two measured temperatures and then find the maximum absolute error in the temperature difference. The required answer is the combination of the absolute difference and the maximum absolute error.
To follow this process, we will use the following formulas.
For a quantity $Z$ which is defined as
$Z=\left| X-Y \right|$
where $X,Y$ are two other physical quantities (defined as $x+\Delta x,y+\Delta y$ respectively), the absolute error in $Z$, that is $\left| \Delta z \right|$ is defined as
$\left| \Delta z \right|=\left| \Delta x \right|+\left| \Delta y \right|$ --(1)
where $\left| \Delta x \right|,\left| \Delta y \right|$ are the absolute errors in $X,Y$ respectively.
Also, $z=\left| x-y \right|$ --(2)
$\therefore Z=z+\Delta z$ --(3)

Now, let us analyze the question.
We are given ${{t}_{1}}={{20}^{0}}C\pm {{0.5}^{0}}C$.
Also, ${{t}_{2}}={{50}^{0}}C\pm {{0.5}^{0}}C$.
Therefore, let ${{T}_{1}}={{20}^{0}}C,\text{ }\Delta {{T}_{1}}=\pm {{0.5}^{0}}C$.
Similarly, ${{T}_{2}}={{50}^{0}}C,\text{ }\Delta {{T}_{2}}=\pm {{0.5}^{0}}C$
Let our required answer be $t=T+\Delta T$ --(4) [Using (3)]
where $t=\left| {{t}_{1}}-{{t}_{2}} \right|$
Now, using (1), we get,
$\left| \Delta T \right|=\left| \Delta {{T}_{1}} \right|+\left| \Delta {{T}_{2}} \right|=\left| \pm {{0.5}^{0}}C \right|+\left| \pm {{0.5}^{0}}C \right|={{0.5}^{0}}C+{{0.5}^{0}}C={{1}^{0}}C$
$\therefore \left| \Delta T \right|={{1}^{0}}C$
$\therefore \Delta T=\pm {{1}^{0}}C$ --(5)
Using (2), we get,
$T=\left| {{T}_{1}}-{{T}_{2}} \right|=\left| {{20}^{0}}C-{{50}^{0}}C \right|=\left| -{{30}^{0}}C \right|={{30}^{0}}C$ --(6)
Hence, using (4), (5) and (6), we get our required answer as ${{30}^{0}}C\pm {{1}^{0}}C$.
Hence, the correct option is A) ${{30}^{0}}C\pm {{1}^{0}}C$.

Note: Students often get confused about what to do with the error, whether to subtract them or add them. The options are also set in such a way so as to confuse them even more. However, students should remember that the error of the required quantity will always be the maximum of the absolute errors of the individual quantities. Proceeding in the mathematical way shown above is the best way to avoid these types of confusion and mistakes.