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Question: In a new set of units, the unit of mass is \(10kg\), unit of length is \(1km\) and unit of time is \...

In a new set of units, the unit of mass is 10kg10kg, unit of length is 1km1km and unit of time is 1 minute1\text{ minute}. The value of 1 joule1\text{ joule} in this new hypothetical system is
a) 3.6×10−4 new units3.6\times {{10}^{-4}}\text{ new units}
b) 6×107 new units6\times {{10}^{7}}\text{ new units}
c) 1011 new units{{10}^{11}}\text{ new units}
d) 1.67×104 new units1.67\times {{10}^{4}}\text{ new units}

Explanation

Solution

Hint: This problem can be solved by breaking down Joule into the fundamental units of length, time and mass in the SI system and applying the required conversions individually to these units as given in the question. In this way we can get a numerical value equal to 1 joule1\text{ joule}in the new units.

Complete step by step answer:
As explained in the hint, we will write the unit Joule as a combination of the fundamental SI units of mass, length and time and then apply the required conversions individually.
In the fundamental system of units,
1 Joule = 1kg.m2s−21\text{ Joule = }1kg.{{m}^{2}}{{s}^{-2}} --(1)
Now in the new set of units, the unit of mass is 10kg10kg, the unit of length is 1km1km and the unit of time is 1 minute1\text{ minute}. Therefore, the unit of Joule will be 10kg.km2minute−210kg.k{{m}^{2}}\text{minut}{{\text{e}}^{-2}}
Now, we know that
1kg=10−1×10kg1kg={{10}^{-1}}\times 10kg --(2)
1km=103m1km={{10}^{3}}m
∴1m=10−3km\therefore 1m={{10}^{-3}}km --(3)
1 minute = 60s1\text{ minute = 60s}
∴1s=160minute\therefore 1s=\dfrac{1}{60}\text{minute} ---(4)
Using (2), (3), (4) in (1), we get,
1 Joule = 1kg.m2s−2=(10−1×10kg)×(10−3×1km)2×(160×1minute)−21\text{ Joule = }1kg.{{m}^{2}}{{s}^{-2}}=\left( {{10}^{-1}}\times 10kg \right)\times {{\left( {{10}^{-3}}\times 1km \right)}^{2}}\times {{\left( \dfrac{1}{60}\times 1\text{minute} \right)}^{-2}}
∴1 Joule = 1kg.m2s−2=(10−1×10kg)×(10−6×1km2)×(160−2×1minute−2)\therefore 1\text{ Joule = }1kg.{{m}^{2}}{{s}^{-2}}=\left( {{10}^{-1}}\times 10kg \right)\times \left( {{10}^{-6}}\times 1k{{m}^{2}} \right)\times \left( \dfrac{1}{{{60}^{-2}}}\times 1\text{minut}{{\text{e}}^{-2}} \right)
∴1 Joule = 1kg.m2s−2=(10−1)×(10−6)×(3600)×10kg.1km2.1minute−2\therefore 1\text{ Joule = }1kg.{{m}^{2}}{{s}^{-2}}=\left( {{10}^{-1}} \right)\times \left( {{10}^{-6}} \right)\times \left( 3600 \right)\times 10kg.1k{{m}^{2}}.1\text{minut}{{\text{e}}^{-2}}
∴1 Joule = 1kg.m2s−2=3.6×10−4×10kg.1km2.1minute−2=3.6×10−4new units\therefore 1\text{ Joule = }1kg.{{m}^{2}}{{s}^{-2}}=3.6\times {{10}^{-4}}\times 10kg.1k{{m}^{2}}.1\text{minut}{{\text{e}}^{-2}}=3.6\times {{10}^{-4}}\text{new units}
Hence, the value of one Joule in the new system of units will be 3.6×10−4new units3.6\times {{10}^{-4}}\text{new units}.
Therefore, the correct option is A) 3.6×10−4new units3.6\times {{10}^{-4}}\text{new units}.

Note: This is the easiest and shortest method to convert a unit into the new set of units. In fact this principle can be employed even when converting between different standard systems of units such as the CGS system of units and the FPS system of units. Students should properly understand this approach because many questions purposefully set their information in the CGS system of units but require the answer in the SI units. Hence, students should be adept at converting between different systems of units using this approach.