Units & Measurement

We know,

$R = {V \over I}$

$ \therefore $ $\left[ R \right] = {{\left[ V \right]} \over {\left[ I \right]}}$ = ${{\left[ {M{L^2}{T^{ - 3}}{I^{ - 1}}} \right]} \over {\left[ I \right]}}$ = ${\left[ {M{L^2}{T^{ - 3}}{I^{ - 2}}} \right]}$

Given, v = at + ${b \over {t + c}}$

Dimension of at = Dimension of velocity v

$ \therefore $ [at] = [LT^-1]

$ \Rightarrow $ [a] = ${{\left[ {L{T^{ - 1}}} \right]} \over {\left[ T \right]}}$ = [LT^-2]

As c is added to t,

$ \therefore $ [c] = [T]

${{\left[ b \right]} \over {\left[ T \right]}}$ = [LT^-1]

$ \therefore $ [b] = [L]

Plank constant (h) = ${{E\lambda } \over c}$ = ${{\left[ {M{L^2}{T^{ - 2}}} \right]\left[ L \right]} \over {\left[ {L{T^{ - 1}}} \right]}}$ = [ML²T^-1]

Moment of inertia (I) = mr² = [ML²]

$ \therefore $ ${h \over I} = $ ${{\left[ {M{L^2}{T^{ - 1}}} \right]} \over {\left[ {M{L^2}} \right]}}$ = [T^-1] = frequency

F = ${{G{m_1}{m_2}} \over {{r^2}}}$

$ \Rightarrow $ G = ${{F \times {r^2}} \over {{m_1}{m_2}}}$

$ \therefore $ Dimension of G = ${{\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]} \over {\left[ M \right]\left[ M \right]}}$ = $\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]$

F = ${1 \over {4\pi {\varepsilon _0}}}{{{q^2}} \over {{r^2}}}$

$ \Rightarrow $ ${\varepsilon _0} = {{{q^2}} \over {4\pi F{r^2}}}$

$ \therefore $ Unit of ${\varepsilon _0}$ = coulomb²/newton-metre²

Plank constant (h) = ${{E\lambda } \over c}$ = ${{\left[ {M{L^2}{T^{ - 2}}} \right]\left[ L \right]} \over {\left[ {L{T^{ - 1}}} \right]}}$ = [ML²T^-1]

Angular momentum (L) = I$ \times $$\omega $ = [ML²][T^-1] = [ML²T^-1]

Dimension of force = [MLT^-2]

Dimension of impulse = [MLT^-1]

So both of them have different dimensions.