Convert 76cm hg to Nm2 using dimensions

Answers (1)

Using nu=constant

\begin{aligned} &P_{1}\left[M_{1}^{a} L_{1}^{b} T_{1}^{c}\right]=P_{2}\left[M_{2}^{a} L_{2}^{b} T_{2}^{c}\right]\\ &\text { We have } P_{2}=P_{1}\left[\frac{M_{1}}{M_{2}}\right]^{a}\left[\frac{L_{1}}{L_{2}}\right]^{b}\left[\frac{T_{1}}{T_{2}}\right]^{c} \end{aligned}

The dimensional formula of pressure P is [ML^{-1}T^{-2}]

\text { So } a=1, b=-1, \text { and } c=-2

\begin{array}{l} \mathrm{M}_{1}=1 \mathrm{g}, \mathrm{M}_{2}=1 \mathrm{kg} \\ \mathrm{L}_{1}=1 \mathrm{cm}, \mathrm{L}_{2}=1 \mathrm{m} \\ \mathrm{T}_{1}=1 \mathrm{s}, \mathrm{T}_{2}=1 \mathrm{s} \end{array}

In \ cgs \ s ystem $76 \mathrm{cm}$\ of \ mercury \ \ pressure $=76 \times 13.6 \times 980$ dyne \ \ $\mathrm{cm}^{-2}$

Then 

\begin{aligned} P_{2} &=76 \times 13.6 \times 980\left[\frac{1 g}{1 \mathrm{kg}}\right]^{1}\left[\frac{1 \mathrm{cm}}{1 \mathrm{m}}\right]^{-1}\left[\frac{1 \mathrm{s}}{1 \mathrm{s}}\right]^{-2} \\ &=76 \times 13.6 \times 980\left[\frac{10^{-3} \mathrm{kg}}{1 \mathrm{kg}}\right]\left[\frac{10^{-2} \mathrm{m}}{1 \mathrm{m}}\right]^{-1}\left[\frac{1 \mathrm{s}}{1 \mathrm{s}}\right]^{-2} \\ &=76 \times 13.6 \times 980 \times\left[10^{-3}\right] \times 10^{2} \\ P_{2} &=1.01 \times 10^{5} \mathrm{Nm}^{-2} \end{aligned}

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