Question

Fill in the blanks. $1\text{ micrometer = }\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\text{ decameter}$.
$A)\text{ 1}{\text{0}}^{-6}$
$B)\text{ 1}{\text{0}}^{-7}$
$C)\text{ 1}{\text{0}}^{-9}$
$D)\text{ 1}{\text{0}}^{-8}$

Answer 1

Hint: This problem can be solved by using the mathematical relations of the metric unit ‘micro’ and the metric unit ‘deca’. Using these definitions, we can get the relation between micrometer and meter, and decameter and meter separately. Combining these two, we can get the required relation between the micrometer and the decameter.

Complete step by step answer:
To solve this problem, we will first recall the metric system and find out the mathematical relation between the micrometer and the meter, and the decameter and the meter and use these two mathematical relations to get the required answer.
Firstly, let $1\text{ micrometer = }K\text{ decameter}$. --(1)
We have to find out $K$.
Now, $1\text{ micrometer = }{10}^{-6}\text{meter}$ --(2)
Since, the relation of the metric unit ‘micro’ with the base unit is $1\text{ micro = }{10}^{-6}\text{base}$.
Also, $1\text{ decameter = }10\text{meter}$ --(3)
Since, the relation of the metric unit ‘deca’ with the base unit is $1\text{ deca = }10\text{ base}$.
Dividing (1) by (2), we get,
${\displaystyle \frac{1\text{ micrometer}}{1\text{ decameter}}}={\displaystyle \frac{{10}^{-6}\text{ meter}}{10\text{ meter}}}={10}^{-7}$
$\therefore 1\text{ micrometer = }{10}^{-7}\times 1\text{ decameter = }{10}^{-7}\text{decameter}$ --(4)
Comparing (4) with (1), we get,
$K={10}^{-7}$.
Hence, our required value is ${10}^{-7}$.
Therefore, the correct option is $B)\text{ 1}{\text{0}}^{-7}$.

Additional information:
There are also other smaller units such as the femtometer, picometer and nanometer. Femtometer and picometer are commonly used in the study of atoms, molecular physics and molecular chemistry.

Note: Students must properly memorize the full metric system of units and their mathematical relations with the base. This is because many questions purposefully provide the values of the physical quantities and other information in some uncommon metric unit which has to be converted to the base for ease of calculation. These questions typically test the knowledge of the student about the metric system and his or her ability to work with and convert the different metric units. If the student does not know the conversion of a unit to the base, he or she may not be able to attempt the problem at all.