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Question: A calorie is a unit of heat or energy and it equals about \[{\mathbf{4}}.{\mathbf{2}}\;J\] where \[1...

A calorie is a unit of heat or energy and it equals about 4.2  J{\mathbf{4}}.{\mathbf{2}}\;J where 1J=1  kgm2s−21J = 1\;kg{m^2}{s^{ - 2}}. Suppose we employ a system of units in which the units of mass equals α kg\alpha \,kg, the units of length equals β m\beta \,m, the unit of time is γ s\gamma \,s. Show that a calorie has a magnitude 4.2 α−1β−2γ2  {\mathbf{4}}.{\mathbf{2}}\,{\alpha ^{ - 1}}{\beta ^{ - 2}}{\gamma ^2}\; in terms of the new units.

Explanation

Solution

As we know the product of the numerical value (Say n) and its corresponding unit (say u) is a constant i.e.
n[u]=constantn[u] = {\text{constant}}
Or n1[u1]=n2[u2]{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 [M1L2T−2][{M^1}{L^2}{T^{ - 2}}]
As the unit of energy is kgm2s−2kg{m^2}{s^{ - 2}}
Now, we can use eqn (i) given in hint
n1[u1]=n2[u2]{n_1}[{u_1}] = {n_2}[{u_2}]
⇒n1[M11L12T1−2]\Rightarrow {n_1}[{M_1}^1{L_1}^2{T_1}^{ - 2}] =n2[M21L22T2−2] = {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
n1=4.2J{n_1} = 4.2J =1cal = 1cal
M1=1kg,{M_1} = 1kg, M2=α{M_2} = \alpha
L1=1m,{L_1} = 1m, L2=β{L_2} = \beta
T1=1sec⁡,{T_1} = 1\sec , T2=γ{T_2} = \gamma
we have to find n2{n_2}
putting the given value in eqn (ii)
⇒4.1[(1)2(1)2(T)−2]\Rightarrow 4.1[{(1)^2}{(1)^2}{(T)^{ - 2}}] =n2[α β2 γ−2] = {n_2}[\alpha {\text{ }}{\beta ^2}{\text{ }}{\gamma ^{ - 2}}]
⇒n2=4.2[α−1 β−2 γ2]\Rightarrow {n_2} = 4.2[{\alpha ^{ - 1}}{\text{ }}{\beta ^{ - 2}}{\text{ }}{\gamma ^2}]
Thus 1cal=4.2α−1β−2γ21cal = 4.2{\alpha ^{ - 1}}{\beta ^{ - 2}}{\gamma ^2}in new unit

Note: Always remember that Dimensional formula of heat (a form of energy) is [M1L2T−2][{M^1}{L^2}{T^{ - 2}}]
we can also convert by direct method as –
1cal=4.2J1cal = 4.2J
=4.2[kgm2s−2]= 4.2[kg{{\text{m}}^2}{{\text{s}}^{ - 2}}]
=4.2[(1kg)(1m)2(1s)−2]................(iii)= 4.2[(1kg){(1{\text{m)}}^2}{{\text{(1s)}}^{ - 2}}]................(iii)
As given α\alpha is equivalent to 1 kg
is equivalent to 1α\dfrac{1}{\alpha } in other unit
Similarly 1m is equivalent to 1β\dfrac{1}{\beta } in other unit
and 1s is equivalent to 1γ\dfrac{1}{\gamma }in other unit
Then putting the values in other unit system from the eqn (iii)
1cal=4.2[1α.(1β)2.(1γ)−2]1cal = 4.2\left[ {\dfrac{1}{\alpha }.{{\left( {\dfrac{1}{\beta }} \right)}^2}.{{\left( {\dfrac{1}{\gamma }} \right)}^{ - 2}}} \right]
1cal=4.2[α−1β−2γ2]1cal = 4.2\left[ {{\alpha ^{ - 1}}{\beta ^{ - 2}}{\gamma ^2}} \right]