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Chapter 1: Problem 165

In the relation $P=(\alpha / \beta) \mathrm{e}[\\{-\alpha z\\} /\\{(\mathrm{k}) \beta \theta\\}], \mathrm{P}\( is pressure, \)\mathrm{z}$ is distance, \(\mathrm{k}\) is Boltzmann constant and \(\theta\) is the temperature. The dimensional formula of \(B\) will be (a) \(\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{1}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{0}\) Copyright \(\odot\) StemEZ.com. All rights reserved.

Short Answer

Expert verified
The dimensional formula of β is: \(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\) However, this option does not match any of the provided alternatives (a), (b), (c), or (d). There might be a mistake in the given options, or the question itself could be incorrect or incomplete.

Step by step solution

01

Rewrite the formula isolating β

Given the formula: \(P=(\alpha / \beta) \mathrm{e}^{\\{-\alpha z\\}/\\{(\mathrm{k}) \beta \theta\\}}\) We need to eliminate the exponent term. As P is a proportional relationship, the equation becomes: \(P=\frac{\alpha}{\beta}\) Now, isolate β: \(\beta = \frac{\alpha}{P}\)
02

Apply dimensional analysis

To determine the dimensions of β, we need to know the dimensions of α and P. Pressure (P) has the dimensional formula: \(\mathrm{M}^{1} \mathrm{~L}^{-1}\mathrm{T}^{-2}\) Now, it remains to find the dimensional formula for α that can be obtained from the given equation. Assuming both the numerator and denominator are dimensionally same on the RHS of the equation: \(\frac{[\alpha]}{[\beta]} = \frac{[\alpha]}{[\rho]}\) Therefore, α has the same dimension as pressure: \(\mathrm{[\alpha]} = \mathrm{M}^{1} \mathrm{~L}^{-1}\mathrm{T}^{-2}\)
03

Calculate the dimensional formula of β

Now, substitute the dimensions of α and P in the equation for β: \([\beta] = \frac{[\alpha]}{[P]}\) Plug in the dimensional formulas: \([\beta] = \frac{\mathrm{M}^{1}\mathrm{~L}^{-1}\mathrm{T}^{-2}}{\mathrm{M}^{1}\mathrm{~L}^{-1}\mathrm{T}^{-2}}\) Cancel out the terms: \([\beta] = \mathrm{M}^{0}\mathrm{~L}^{0}\mathrm{T}^{0}\)
04

Compare with the given options and choose the correct answer

The dimensional formula of β is: \(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\) This option does not match any of the provided alternatives (a), (b), (c), or (d). There might be a mistake in the given options, or the question itself could be incorrect or incomplete.

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Most popular questions from this chapter

Equation of physical quantity \(\mathrm{v}=\mathrm{at}+\mathrm{bt}^{2}\) where \(\mathrm{v}=\) velocity \(\mathrm{t}=\) time so write the dimensional formula of a in this equation (a) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-1}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1}\) (c) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-2}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{0}\)

\(F=A_{0}\left(1-e^{-(B x t) 2}\right)\) where \(F\) is force and \(x\) is displacement. write the dimension formula of \(B\) (a) \(\mathrm{M}^{2} \mathrm{~L}^{1} \mathrm{~T}^{-1}\) (b) \(\mathrm{M}^{0} \mathrm{~L}^{-1} \mathrm{~T}^{-2}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{-2}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}\)

What is the least count of screw gauge? (a) \(10^{-4} \mathrm{~m}\) (b) \(10^{-5} \mathrm{~m}\) (c) \(10^{-2} \mathrm{~m}\) (d) \(10^{-6} \mathrm{~m}\)

Which unit of physical quantity remains same for all unit system? (a) meter (b) second (c) ampere (d) kilogram

The number of significant figures in \(0.000150\) is .............. (a) 3 (b) 5 (c) 2 (d) 4
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