Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/460

The weight of mercury which fills the whole tube is

$w=\rho \int 's\;dx=\rho C\Sigma \left({\frac {1}{\lambda }}\right){\frac {l}{n}}$,

where $n$ is the number of points, at equal distances along the tube, where $\lambda$ has been measured, and $\rho$ is the mass of unit of volume.

The resistance of the whole tube is

$R=\int {\frac {r}{s}}\;ds={\frac {r}{C}}\Sigma (\lambda ){\frac {l}{n}}$,

where $r$ is the specific resistance per unit of volume.

Hence

$wR=r\rho \Sigma (\lambda )\Sigma \left({\frac {1}{\lambda }}\right){\frac {l^{2}}{n^{2}}}$,

and

$r={\frac {wR}{\rho l^{2}}}{\frac {n^{2}}{\Sigma (\lambda )\Sigma \left({\frac {1}{\lambda }}\right)}}$

gives the specific resistance of unit of volume.

To find the resistance of unit of length and unit of mass we must multiply this by the density.

It appears from the experiments of Matthiessen and Hockin that the resistance of a uniform column of mercury of one metre in length, and weighing one gramme at 0°C, is 13.071 Ohms, whence it follows that if the specific gravity of mercury is 13.595, the resistance of a column of one metre in length and one square millimetre in section is 0.96146 Ohms.

362.] In the following table $R$ is the resistance in Ohms of a column one metre long and one gramme weight at 0°C, and $r$ is the resistance in centimetres per second of a cube of one centimetre, according to the experiments of Matthiessen.^[1]

	Specific gravity			$R$		$r$	Percentage increment of resistance for 1°C at 20°C.
Silver	10	.50	hard drawn	0	.1689	1609	0	.377
Copper	8	.95	hard drawn	0	.1469	1642	0	.388
Gold	19	.27	hard drawn	0	.4150	2154	0	.365
Lead	11	.391	pressed	2	.257	19847	0	.387
Mercury	13	.595	liquid	13	.071	96146	0	.072
Gold 2, Silver 1	15	.218	hard or annealed	1	.668	10988	0	.065
Selenium at 100°C			Crystalline form			6 × 10¹³	1	.00

1. Phil. Mag., May, 1865.