Question

A calorie is a unit of heat or energy and it equals about \[{\mathbf{4}}.{\mathbf{2}}\;J\] where \[1J = 1\;kg{m^2}{s^{ - 2}}\]. Suppose we employ a system of units in which the units of mass equals \[\alpha \,kg\], the units of length equals \[\beta \,m\], the unit of time is \[\gamma \,s\]. Show that a calorie has a magnitude \[{\mathbf{4}}.{\mathbf{2}}\,{\alpha ^{ - 1}}{\beta ^{ - 2}}{\gamma ^2}\;\] in terms of the new units.

Answer 1

Hint: As we know the product of the numerical value (Say n) and its corresponding unit (say u) is a constant i.e.
$n[u] = {\text{constant}}$
Or ${n_1}[{u_1}] = {n_2}[{u_2}]$ ……………………(i)

Complete step by step solution:
As we know that the dimensional formula of heat is the dimensional formula of energy because heat is the form of energy.
Hence,
Dimensional formula of heat is $[{M^1}{L^2}{T^{ - 2}}]$
As the unit of energy is \[kg{m^2}{s^{ - 2}}\]
Now, we can use eqn (i) given in hint
${n_1}[{u_1}] = {n_2}[{u_2}]$
$ \Rightarrow {n_1}[{M_1}^1{L_1}^2{T_1}^{ - 2}]$$ = {n_2}[{M_2}^1{L_2}^2{T_2}^{ - 2}]$ …………….(ii)
Where M1 , L1, T1 are the fundamental units in one system and M2 , L2, T2 are the fundamental units in other system.
Here
${n_1} = 4.2J$ $ = 1cal$
${M_1} = 1kg,$ ${M_2} = \alpha $
${L_1} = 1m,$ ${L_2} = \beta $
${T_1} = 1\sec ,$ ${T_2} = \gamma $
we have to find ${n_2}$
putting the given value in eqn (ii)
$ \Rightarrow 4.1[{(1)^2}{(1)^2}{(T)^{ - 2}}]$ $ = {n_2}[\alpha {\text{ }}{\beta ^2}{\text{ }}{\gamma ^{ - 2}}]$
$ \Rightarrow {n_2} = 4.2[{\alpha ^{ - 1}}{\text{ }}{\beta ^{ - 2}}{\text{ }}{\gamma ^2}]$
Thus $1cal = 4.2{\alpha ^{ - 1}}{\beta ^{ - 2}}{\gamma ^2}$in new unit

Note: Always remember that Dimensional formula of heat (a form of energy) is $[{M^1}{L^2}{T^{ - 2}}]$
we can also convert by direct method as –
\[1cal = 4.2J\]
\[ = 4.2[kg{{\text{m}}^2}{{\text{s}}^{ - 2}}]\]
\[ = 4.2[(1kg){(1{\text{m)}}^2}{{\text{(1s)}}^{ - 2}}]................(iii)\]
As given $\alpha $ is equivalent to 1 kg
is equivalent to $\dfrac{1}{\alpha }$ in other unit
Similarly 1m is equivalent to $\dfrac{1}{\beta }$ in other unit
and 1s is equivalent to $\dfrac{1}{\gamma }$in other unit
Then putting the values in other unit system from the eqn (iii)
$1cal = 4.2\left[ {\dfrac{1}{\alpha }.{{\left( {\dfrac{1}{\beta }} \right)}^2}.{{\left( {\dfrac{1}{\gamma }} \right)}^{ - 2}}} \right]$
$1cal = 4.2\left[ {{\alpha ^{ - 1}}{\beta ^{ - 2}}{\gamma ^2}} \right]$