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Chapter 1: Problem 145

Dimensional formula of \(\mathrm{CV}\) ? where C - capacitance and \(\mathrm{V}\) - potential difference (a) \(\mathrm{M}^{1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{1}\) (c) \(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{1} \mathrm{~A}^{-1}\) (d) \(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{1} \mathrm{~A}^{1}\)

Short Answer

Expert verified
The dimensional formula of CV is \(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{1} \mathrm{~A}^{1}\).

Step by step solution

01

Obtain the Dimensional formulas for Capacitance (C) and Potential Difference (V)

First, we need to know the dimensional formulas for capacitance and potential difference. The base dimensions are Mass (M), Length (L), Time (T), and Current (A). The dimensional formulas are as follows: Capacitance(C): [C] = \([M^{-1}L^{-2}T^4A^2]\) Potential Difference (V): [V] = \([M^1L^2T^{-3}A^{-1}]\)
02

Multiply the Dimensional Formulas of Capacitance and Voltage

To get the dimensional formula of CV, we need to multiply the dimensional formulas of capacitance and potential difference. \([CV] = [C] \times [V]\) \([CV] = [M^{-1}L^{-2}T^4A^2] \times [M^1L^2T^{-3}A^{-1}]\)
03

Simplify the Dimensional Formula for CV

Simplify by adding the exponents of the same base dimensions: \([CV] = [M^{-1+1}L^{-2+2}T^{4-3}A^{2-1}]\) \([CV] = [M^0L^0T^1A^1]\) Now let's check which of the given options matches the derived dimensional formula: (a) \(\mathrm{M}^{1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{1}\) (c) \(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{1} \mathrm{~A}^{-1}\) (d) \(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{1} \mathrm{~A}^{1}\) The correct dimensional formula for CV is option (d), which is \(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{1} \mathrm{~A}^{1}\).

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Most popular questions from this chapter

The mass of substance is \(75.5 \mathrm{gm}\) and its volume is $25 \mathrm{~cm}^{3}$. It's density up to the correct significant figure is \(\ldots \ldots \ldots \ldots \ldots\) (a) \(3.02 \mathrm{gm} / \mathrm{cm}^{3}\) (b) \(3.200 \mathrm{gm} / \mathrm{cm}^{3}\) (c) \(3.02 \mathrm{gm} / \mathrm{cm}^{3}\) (d) \(3.1 \mathrm{gm} / \mathrm{cm}^{3}\)

Equation of \(\ell_{1}=\ell_{0}\left[1+\alpha\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)\right]\) find out the dimensions of the coefficient of linear expansion \(\alpha\) suffix. (a) \(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{1} \mathrm{~K}^{1}\) (b) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{1} \mathrm{~K}^{1}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{0} \mathrm{~K}^{1}\) (d) \(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0} \mathrm{~K}^{-1}\)

\([(1\) femtometer \() /(100\) nanometer \()]=\ldots \ldots \ldots \ldots \ldots\) (a) \(10^{-6}\) (b) \(10^{-8}\) (c) \(10^{24}\) (d) \(10^{-24}\)

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.

Acceleration due to gravity is given by $\mathrm{g}=\left(\mathrm{GM} / \mathrm{R}^{2}\right)\( what is the equation of the fractional error \)\Delta \mathrm{g} / \mathrm{g}\( in measurement of gravity \)\mathrm{g}$ ? [G \& M constant] (a) \(-(\Delta \mathrm{R} / \mathrm{R})\) (b) \(2(\Delta \mathrm{R} / \mathrm{R})\) (c) \(-2(\Delta \mathrm{R} / \mathrm{R})\) (d) \((1 / 2)(\Delta \mathrm{R} / \mathrm{R})\)
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