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Question: \(1\) millimeter is _______ terameter. A) \({10^{ - 15}}\) B) \({10^{ - 14}}\) C) \({10^{ - 13...

11 millimeter is _______ terameter.
A) 10−15{10^{ - 15}}
B) 10−14{10^{ - 14}}
C) 10−13{10^{ - 13}}
D) 10−16{10^{ - 16}}

Explanation

Solution

In order to determine the size or magnitude of the physical quantity with some standard quantity of the same kind is known as measurement. If measurements of quantities are not done, then it will be difficult to conduct experiments or invent any theory.

Complete answer
Step I:
In the metric system, a millimeter is also defined as a standard unit of length. It is equal to one thousandth part of a meter, which is considered as the S.I. unit of length. Millimeter is represented by the symbol ‘mm’.
Step II:
Similarly, a terameter is also a unit of length. It is also related to the standard unit of length which is metres. One terameter is equal to a length of one trillion metres. Terameter is represented by the symbol ‘Tm’.
Step III:
Relation between millimeter and meter is:
1millim⁡eter=11000meter1mil\lim eter = \dfrac{1}{{1000}}meter
Or 1mm=10−3m1mm = {10^{ - 3}}m---(i)
Similarly, relation between terameter and meter can be written as:
1Tm=1012m1Tm = {10^{12}}m---(ii)
Step IV:
Relation between Millimeter and Terameter is
Therefore, equation (i) can be written as
1mm=10−3×10121012meter1mm = {10^{ - 3}} \times \dfrac{{{{10}^{12}}}}{{{{10}^{12}}}}meter
Or 1mm=10−31012×1Terameter1mm = \dfrac{{{{10}^{ - 3}}}}{{{{10}^{12}}}} \times 1Terameter
1mm=10−3−121mm = {10^{ - 3 - 12}}
1mm=10−15Tm1mm = {10^{ - 15}}Tm

Step V:
Therefore Option A is the right answer.
Note: It is important to remember that the term millimeter is used for measuring very small lengths as it is a very small metric unit of length. When the measurement is to be carried out more precisely, even than centimetres, then millimetres are used. But the term terameter is barely used to measure any length since it is even smaller than millimeter. So they are not used practically. But they are sometimes used to measure terrestrial lengths in astronomy.