Question

A calorie is a unit of heat or energy and it equals about $$\mathbf{4}.\mathbf{2}J$$ where $$1J=1kg{m}^2{s}^{-2}$$. 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)
$\Rightarrow 4.1[(1{)}^2(1{)}^2(T{)}^{-2}]$ $=n_2[\alpha {\beta }^2{\gamma }^{-2}]$
$\Rightarrow n_2=4.2[{\alpha }^{-1}{\beta }^{-2}{\gamma }^2]$
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 ${\displaystyle \frac{1}{\alpha }}$ in other unit
Similarly 1m is equivalent to ${\displaystyle \frac{1}{\beta }}$ in other unit
and 1s is equivalent to ${\displaystyle \frac{1}{\gamma }}$in other unit
Then putting the values in other unit system from the eqn (iii)
$1cal=4.2\left[{\displaystyle \frac{1}{\alpha }}.{\left({\displaystyle \frac{1}{\beta }}\right)}^2.{\left({\displaystyle \frac{1}{\gamma }}\right)}^{-2}\right]$
$1cal=4.2\left[{\alpha }^{-1}{\beta }^{-2}{\gamma }^2\right]$