In a new set of units, the unit of mass is $10kg$, unit of length is $1km$ and unit of time is $1\text{ minute}$. The value of $1\text{ joule}$ in this new hypothetical system is
a) $3.6\times {{10}^{-4}}\text{ new units}$
b) $6\times {{10}^{7}}\text{ new units}$
c) ${{10}^{11}}\text{ new units}$
d) $1.67\times {{10}^{4}}\text{ new units}$
Answer
674.4k+ views
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 $1\text{ joule}$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,
$1\text{ Joule = }1kg.{{m}^{2}}{{s}^{-2}}$ --(1)
Now in the new set of units, the unit of mass is $10kg$, the unit of length is $1km$ and the unit of time is $1\text{ minute}$. Therefore, the unit of Joule will be $10kg.k{{m}^{2}}\text{minut}{{\text{e}}^{-2}}$
Now, we know that
$1kg={{10}^{-1}}\times 10kg$ --(2)
$1km={{10}^{3}}m$
$\therefore 1m={{10}^{-3}}km$ --(3)
$1\text{ minute = 60s}$
$\therefore 1s=\dfrac{1}{60}\text{minute}$ ---(4)
Using (2), (3), (4) in (1), we get,
$1\text{ Joule = }1kg.{{m}^{2}}{{s}^{-2}}=\left( {{10}^{-1}}\times 10kg \right)\times {{\left( {{10}^{-3}}\times 1km \right)}^{2}}\times {{\left( \dfrac{1}{60}\times 1\text{minute} \right)}^{-2}}$
$\therefore 1\text{ Joule = }1kg.{{m}^{2}}{{s}^{-2}}=\left( {{10}^{-1}}\times 10kg \right)\times \left( {{10}^{-6}}\times 1k{{m}^{2}} \right)\times \left( \dfrac{1}{{{60}^{-2}}}\times 1\text{minut}{{\text{e}}^{-2}} \right)$
$\therefore 1\text{ Joule = }1kg.{{m}^{2}}{{s}^{-2}}=\left( {{10}^{-1}} \right)\times \left( {{10}^{-6}} \right)\times \left( 3600 \right)\times 10kg.1k{{m}^{2}}.1\text{minut}{{\text{e}}^{-2}}$
$\therefore 1\text{ Joule = }1kg.{{m}^{2}}{{s}^{-2}}=3.6\times {{10}^{-4}}\times 10kg.1k{{m}^{2}}.1\text{minut}{{\text{e}}^{-2}}=3.6\times {{10}^{-4}}\text{new units}$
Hence, the value of one Joule in the new system of units will be $3.6\times {{10}^{-4}}\text{new units}$.
Therefore, the correct option is A) $3.6\times {{10}^{-4}}\text{new units}$.
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.
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,
$1\text{ Joule = }1kg.{{m}^{2}}{{s}^{-2}}$ --(1)
Now in the new set of units, the unit of mass is $10kg$, the unit of length is $1km$ and the unit of time is $1\text{ minute}$. Therefore, the unit of Joule will be $10kg.k{{m}^{2}}\text{minut}{{\text{e}}^{-2}}$
Now, we know that
$1kg={{10}^{-1}}\times 10kg$ --(2)
$1km={{10}^{3}}m$
$\therefore 1m={{10}^{-3}}km$ --(3)
$1\text{ minute = 60s}$
$\therefore 1s=\dfrac{1}{60}\text{minute}$ ---(4)
Using (2), (3), (4) in (1), we get,
$1\text{ Joule = }1kg.{{m}^{2}}{{s}^{-2}}=\left( {{10}^{-1}}\times 10kg \right)\times {{\left( {{10}^{-3}}\times 1km \right)}^{2}}\times {{\left( \dfrac{1}{60}\times 1\text{minute} \right)}^{-2}}$
$\therefore 1\text{ Joule = }1kg.{{m}^{2}}{{s}^{-2}}=\left( {{10}^{-1}}\times 10kg \right)\times \left( {{10}^{-6}}\times 1k{{m}^{2}} \right)\times \left( \dfrac{1}{{{60}^{-2}}}\times 1\text{minut}{{\text{e}}^{-2}} \right)$
$\therefore 1\text{ Joule = }1kg.{{m}^{2}}{{s}^{-2}}=\left( {{10}^{-1}} \right)\times \left( {{10}^{-6}} \right)\times \left( 3600 \right)\times 10kg.1k{{m}^{2}}.1\text{minut}{{\text{e}}^{-2}}$
$\therefore 1\text{ Joule = }1kg.{{m}^{2}}{{s}^{-2}}=3.6\times {{10}^{-4}}\times 10kg.1k{{m}^{2}}.1\text{minut}{{\text{e}}^{-2}}=3.6\times {{10}^{-4}}\text{new units}$
Hence, the value of one Joule in the new system of units will be $3.6\times {{10}^{-4}}\text{new units}$.
Therefore, the correct option is A) $3.6\times {{10}^{-4}}\text{new units}$.
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.
Recently Updated Pages
Which will be the least stable resonating structure class 11 chemistry CBSE

How many 5 digit telephone numbers can be construc-class-11-maths-CBSE

How do you find the angle of the resultant vector class 11 physics CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

What is the need and importance of classification class 11 biology CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

10 examples of friction in our daily life

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

10 examples of diffusion in everyday life

