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Units and Measurements — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsUnits and Measurements1 Mark
MCQ
If momentum $(P),$ area $(A)$ and time $(T)$ are taken to be fundamental quantities, then energy has the dimensional formula:
  • A
    $(\text{P}^1\text{A}^{-1}\text{T}^1).$
  • B
    $(\text{P}^2\text{A}^{1}\text{T}^1).$
  • C
    $(\text{P}^1\text{A}^\frac{-1}{2}\text{T}^1).$
  • ✓
    $(\text{P}^1\text{A}^\frac{1}{2}\text{T}^{-1}).$
✓

Answer

Correct option: D.
$(\text{P}^1\text{A}^\frac{1}{2}\text{T}^{-1}).$
According to the problem, fundamental quantities are momentum $(p),$ area $(A)$ and time $(T)$ and we have to express energy in these fundamental quantities.
Let energy $E$,
$\text{E}\propto\text{p}^\text{a}\text{A}^\text{A}\text{T}^\text{c}\Rightarrow\text{E}=\text{kp}^\text{a}\text{A}^\text{A}\text{T}^\text{c}$
where, $k$ is dimensionless constant of proportionality.
Dimensional formula of energy, $[\text{E}]=[\text{ML}^2\text{T}^{-2}]$ and $[\text{p}]=[\text{MLT}^{-1}]$
$[\text{A}]=[\text{L}^2],\ [\text{T}]=[\text{T}]$ and $[\text{E}]=[\text{K}][\text{p}]^\text{a}[\text{A}]^\text{b}[\text{T}]^\text{c}$
Putting all the dimensions, we get
$\text{ML}^2\text{T}^2=[\text{MLT}^{-1}]^\text{a}[\text{L}^2]^\text{b}[\text{T}]^\text{c}$
$\text{M}^\text{a}\text{L}^{\text{a}+2\text{b}}\text{T}^{\text{-a}+\text{c}}$
According to the principle of homogeneity of dimensions, we get,
$\text{a}=1\ \ \ ...(\text{i)}$
$\text{a}+2\text{b}=2\ \ \ ...(\text{ii)}$
$-\text{a}+\text{c}=-2\ \ \ ...(\text{iii)}$
By solving these equations $(i), (ii)$ and $(iii),$ we get
$\text{a}=1,\ \text{b}=\frac{1}{2},\ \text{c}=-1$
Dimensional formula for $E$ is $[\text{p}^1\text{A}^\frac{1}{2}\text{t}^{-1}].$

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