Question

In a new system of units called star units, \[1k{g^*} = 10kg\] , \[1{m^*} = 1km\] and \[1{s^*} = 1\min \] , what will be the value of \[1J\] in the new system?
A.\[2.4 \times {10^{ - 5}}{J^*}\]
B. \[3.6 \times {10^{ - 4}}{J^*}\]
C. \[4.2 \times {10^{ - 3}}{J^*}\]
D. \[4.2 \times {10^{ - 2}}{J^*}\]

Answer 1

Hint:First write the values of old systems converted into new systems. Then write the unit of 1 joule and convert them into the new system and calculate to obtain the required answer.

Formula used:
The unit of 1 joule is,
\[1J = \dfrac{{(kg).{m^2}}}{{{s^2}}}\]

Complete step by step solution:
It is given that \[1k{g^*} = 10kg\] , \[1{m^*} = 1km\] and \[1{s^*} = 1\min \].
Therefore,
\[kg = \dfrac{1}{{10}}k{g^*}\]
\[\Rightarrow 1000\,m = 1{m^*}\\
\Rightarrow m = \dfrac{1}{{1000}}{m^*}\]
And
\[60s = 1{s^*}\\
\Rightarrow s = \dfrac{1}{{60}}{s^*}\]

Now, we know that,
\[1J = \dfrac{{(kg).{m^2}}}{{{s^2}}}\]
Substitute the obtained values of kg, m and s in the equation \[1J = \dfrac{{(kg).{m^2}}}{{{s^2}}}\] to obtain the required solution.
\[1\,J = \dfrac{{(\dfrac{1}{{10}}k{g^*}).{{\left( {\dfrac{1}{{1000}}{m^*}} \right)}^2}}}{{{{\left( {\dfrac{1}{{60}}{s^*}} \right)}^2}}} \\ \]
\[\Rightarrow 1\,J = \dfrac{{3600}}{{10000000}}k{g^*}{\left( {{m^*}} \right)^2}{\left( {{s^*}} \right)^{ - 2}} \\ \]
\[\Rightarrow 1\,J = 3.6 \times {10^{ - 4}}k{g^*}{\left( {{m^*}} \right)^2}{\left( {{s^*}} \right)^{ - 2}}\]
\[\therefore 1\,J = 3.6 \times {10^{ - 4}}{J^*}\]

Hence, the correct option is B.

Additional information: We are shifting the units of Joule to a new unit system, as we already know the old system so just express the new system in terms of the old system.

Note: To solve this type of problems one should have a strong knowledge about unit convergence, as we are converting km to m so we need to know that 1km is equal to 1000 m also we have to know the unit of 1 joule to solve this question otherwise we cannot be able to solve this kind of questions.