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Units and Measurements — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsUnits and Measurements3 Marks
Question
In the relation $\text{p}=\Big(\frac{\text{a}}{\text{b}}\Big)^{\text{e}^{-\Big(\frac{\text{az}}{\theta}\Big)}},\text{p}$ is the pressure, Z is the distance, and is the temperature. What is the dimensional formula of p?
✓

Answer

Since, ${\text{e}^{-\Big(\frac{\text{aZ}}{\theta}\Big)}}$is dimensionless, We have $\frac{\text{aZ}}{\theta}=1$ Or $\text{a}=\frac{\theta}{\text{Z}}=\frac{\text{K}}{\text{L}}=[\text{L}^{-1}\text{K}]$ We find that $\frac{\text{a}}{\text{b}}=$ dimensions of p and $\text{b} = [\text{ML}^{-1}\text{T}^{-2}].$ Therefore, dimensional formula of p is obtained as $\text{p}=\frac{\text{a}}{[\text{ML}^{-1}\text{T}^{-2}]}=\frac{[\text{L}^{-1}\text{K}]}{[\text{ML}^{-1}\text{T}^{-2}]}$ $=[​​\text{M}^{-1}\text{L}^0\text{T}^2\text{K}]$

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