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Chapter 1: Problem 126

Dimensional formula for conductance is ........... (a) \(\mathrm{M}^{-1} \mathrm{~L}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{2}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{1}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{-2} \mathrm{~T}^{3} \mathrm{~A}^{2}\) (d) \(\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{3} \mathrm{~A}^{2}\)

Short Answer

Expert verified
The dimensional formula for conductance is (d) \(\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{3} \mathrm{~A}^{2}\).

Step by step solution

01

Define Conductance

Conductance is the measure of a material's ability to allow the flow of electric current. It is denoted by the symbol G and is the reciprocal of resistance (R). Mathematically, this can be represented as: \( G = \frac{1}{R} \)
02

Use the formula for resistance

The formula for resistance is given by Ohm's law: \( R = \rho \frac{l}{A} \) Where \(R\) is the resistance, \(\rho\) is the resistivity, \(l\) is the length of the material, and \(A\) is the cross-sectional area of the material. From the equation for conductance, we have: \( G = \frac{1}{\rho \frac{l}{A}} = \frac{A}{\rho l} \)
03

Derive the dimensional formula for conductance

We will now derive the dimensional formula for conductance by substituting the dimensions of the various terms in the equation for conductance. - Area (A) has dimensions of length squared (\(L^2\)) - Resistivity (\(\rho\)) has dimensions of \(\mathrm{M} \mathrm{~L}^{3} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\) - Length (l) has dimensions of length (\(L\)) Now, substitute these dimensions into the equation for conductance: \( G = \frac{([L^2])}{([\mathrm{M} \mathrm{~L}^{3} \mathrm{~T}^{-3} \mathrm{~A}^{-2} ] [L])} \) Simplify by canceling out like terms: \( G = [\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{3} \mathrm{~A}^{2}] \)
04

Match the derived formula with the given options

The derived dimensional formula for conductance is: \( [\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{3} \mathrm{~A}^{2}] \) Compare this to the given options: (a) \(\mathrm{M}^{-1} \mathrm{~L}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{2}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{1}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{-2} \mathrm{~T}^{3} \mathrm{~A}^{2}\) (d) \(\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{3} \mathrm{~A}^{2}\) Our derived dimensional formula for conductance matches option (d). Therefore, the correct answer is: (d) \(\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{3} \mathrm{~A}^{2}\)

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Most popular questions from this chapter

Find the dimensional formula for energy per unit surface area per unit time (a) \(\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{-2}\) (b) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-1}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{-3}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{-1} \mathrm{~T}^{1}\)

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