Question

A calorie is a unit of heat or energy and it equals about 4.2J where $1J = 1kg{m^2}{s^{ - 2}}$. Suppose we employ a system or units in which the unit of mass equals $\alpha $ kg, the unit of length equals $\beta $ m, the unit of time is $\gamma $s. Show that a calorie has a magnitude 4.2${\alpha ^{ - 1}}{\beta ^{ - 2}}{\gamma ^2}$ in terms of the new units.

Answer 1

Hint – In order to solve such problems we have to consider the dimensional formulas of the quantities given and use the information provided in the problem.

Step-By-Step answer:
We know that 1 calorie is 4.2 joules
And $1J = 1kg{m^2}{s^{ - 2}}$
The dimensional formula of $1J = 1kg{m^2}{s^{ - 2}}$ can be written as $\left[ {M{L^2}{T^{ - 2}}} \right]$ and $\left[ {(kg){{(m)}^2}{{(s)}^{ - 2}}} \right]$.
It is said that the unit of mass, length and time is $\alpha $ kg, $\beta $m and $\gamma $s.
Let the new unit of calorie be P the old one is 4.2J = 4.2 $\left[ {(kg){{(m)}^2}{{(s)}^{ - 2}}} \right]$.
As the units will be same so, we can do,
4.2 $\left[ {(kg){{(m)}^2}{{(s)}^{ - 2}}} \right]$=P$\left[ {(\alpha kg){{(\beta m)}^2}{{(\gamma s)}^{ - 2}}} \right]$

On solving it further we get the value of the P as,
$P = 4.2\dfrac{{\left[ {(kg){{(m)}^2}{{(s)}^{ - 2}}} \right]}}{{\left[ {(\alpha kg){{(\beta m)}^2}{{(\gamma s)}^{ - 2}}} \right]}} = 4.2\left[ {(kg){{(m)}^2}{{(s)}^{ - 2}}} \right]$

Hence it is proved.

Note – Here you just need to know the dimensional formulas and its usage. It is already given that $1J = 1kg{m^2}{s^{ - 2}}$ and 1 calorie is 4.2 joules but you should also remember these values since it is very useful in competitive exams if the problem arises from this concept. We also recommend you to study the chapter of units and measurement thoroughly so that you can do those problems correct if it arises. The problems arising from this topic are very simple but only those who have read this concept can. Students also need to know that units will always be the same as a term whatever may the symbol be so we can equate those units as done in the problem above.