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Chapter 1: Problem 124

Dimensional formula for thermal conductivity (k) is.. (a) \(\mathrm{M}^{2} \mathrm{~L}^{1} \mathrm{~T}^{-2} \mathrm{~K}^{-1}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2} \mathrm{~K}^{1}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{-3} \mathrm{~K}^{-1}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-3} \mathrm{~K}^{-1}\)

Short Answer

Expert verified
The dimensional formula for thermal conductivity (k) is \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-3} \mathrm{~K}^{-1}\).

Step by step solution

01

Write down the formula

Write the formula for the Fourier's law of heat conduction: \(Q/t = -kA\frac{dT}{dx}\)
02

Rearrange the formula to solve for k

Rearrange the formula to isolate the thermal conductivity (k) on one side: \(k = \frac{Q}{t}\frac{dx}{A \, dT}\)
03

Find the dimensions of each term

Identify the dimensions of each physical quantity in the formula: - Rate of heat transfer (\(\frac{Q}{t}\)): \(\mathrm{kgs^{-3}}\) - Thickness (dx): \(\mathrm{L}\) - Cross-sectional area (A): \(\mathrm{L^2}\) - Change in temperature (dT): \(\mathrm{K}\)
04

Substitute the dimensions in the formula for k

Substitute the dimensions of all the physical quantities in the formula for k: \(k = \frac{\mathrm{kgs^{-3}}}{\mathrm{L^2} \, \mathrm{K}} \times \mathrm{L} = \mathrm{kgLs^{-3}K^{-1}}\)
05

Compare the dimensional formula with given options

Compare the obtained dimensional formula for the thermal conductivity with the given options: (a) \(\mathrm{M}^{2} \mathrm{~L}^{1} \mathrm{~T}^{-2} \mathrm{~K}^{-1}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2} \mathrm{~K}^{1}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{-3} \mathrm{~K}^{-1}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-3} \mathrm{~K}^{-1}\) We can see that our obtained dimensional formula for thermal conductivity (k) matches option (d): \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-3} \mathrm{~K}^{-1}\) Hence, the dimensional formula for thermal conductivity (k) is \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-3} \mathrm{~K}^{-1}\).

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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) \(3.1 \mathrm{gm} / \mathrm{cm}^{3}\)

Write the unit of angular acceleration in the SI system. (a) \(\mathrm{N} \cdot \mathrm{Kg}\) (b) \(\mathrm{rad} /(\mathrm{sec})^{2}\) (c) \(\mathrm{m} / \mathrm{sec}\) (d) \(\mathrm{N} / \mathrm{kg}\)

Dimensional formula of latent heat is ........ (a) \(\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{-2}\) (b) \(\mathrm{M}^{2} \mathrm{~L}^{0} \mathrm{~T}^{-2}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}\) (d) \(\mathrm{M}^{2} \mathrm{~L}^{2} \mathrm{~T}^{-1}\)

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

If the length of rod \(\mathrm{A}\) is \((2.35 \pm 0.01) \mathrm{cm}\) and that of \(\mathrm{B}\) is \((5.68 \pm 0.01) \mathrm{cm}\) then the rod \(\mathrm{B}\) is longer than \(\mathrm{rod} \mathrm{A}\) by \(\ldots\) (a) \((2.43 \pm 0.00) \mathrm{cm}\) (b) \((3.33 \pm 0.02) \mathrm{cm}\) (c) \((2.43 \pm 0.01) \mathrm{cm}\) (d) \((2.43 \pm 0.001) \mathrm{cm}\)
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