Question

find conversion factor between S I and CGS units of work using dimensional analysis​

Answer 1

Answer:

The SI Unit of the Pressure = N/m².

The CGS unit of the Pressure = dyne/cm²

∴ N/m² = 10⁵ dyne/(100 cm)²

N/m²= 10⁵/10⁴ dyne/cm²

N/m²= 10 dyne/cm²

∴ Conversion Factor to convert SI Unit into the CGS unit is 10.

Answer 2

The conversion factor between S.I. and C.G.S. units of work using dimensional analysis​ is found as follows:

• The dimensional formula of work is ML²T⁻².
• Its unit in S.I. system is 'joule' and in C.G.S. system is 'erg'.
• Let M₁, L₁, T₁ represent kilogram (kg), metre (m), second (s) and M₂, L₂, T₂ represent gram (g), centimetre (cm), second (s) respectively.
• Then, the units of work in S.I. and C.G.S. systems will be (M₁L₁²T₁⁻²) and (M₂L₂²T₂⁻²) respectively.
• If the numerical values of work are n₁ and n₂ respectively, then

        $n_{1} (M_{1} L_{1}^{2} T_{1}^{-2}) = n_{2} (M_{2} L_{2}^{2} T_{2}^{-2})$

        $n_{2} = n_{1} (\frac{M_{1}}{M_{2}}) (\frac{L_{1}}{L_{2}})^{2} (\frac{T_{1}}{T_{2}})^{-2}$

• Here, n₁ = 1.

        $n_{2} = 1 (\frac{kg}{g}) (\frac{m}{cm})^{2} (\frac{s}{s})^{-2}$

             $= 1 (\frac{10^{3} g}{g}) (\frac{10^{2} cm}{cm})^{2} (\frac{s}{s})^{-2}$

             $= 1 * 10^{3} * 10^{4} * 1$

             $= 10^{7}$

• 1 joule = 10⁷ ergs.