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Re: Lead needed to coat a cable
Radius of uncoated cable = 55/2 = 27.5 mm = 0.0275 m
Radius of coated cable = 55/2 + 4 = 31.5 mm = 0.0315 m
Length of cable = 5000 m
Required volume of lead
$V = \pi (0.0315^2 - 0.0275^2)(5000) = 1.18\pi ~ \text{ m}^3$
Required mass of lead
$m = \rho V = (11.4 \times 10^3)(1.18\pi) = 42,260.70 ~ \text{ kg}$
Re: Lead needed to coat a cable
Thank you, sir!! :)