Question

A plastic tube 25 m long and 4 cm in diameter is dipped into Ag solution deposting a silver layer 0.1 mm thick uniformly over its outer surface. Find the current if this coated tube is connected across a 12 V battery. ${\rho }_{Ag}=1.47\times {10}^{-8}\Omega -m$

• A. 4.14A
• B. 41.4A
• C. 414A
• D. 414mA

Answer 1

Answer: (C)

414A

Given
Length of plastic tube is, $l=25m$
Diameter of plastic tube is $d=4cm$
Radius of plastic tube is $r=2cm$
Thickness of silver layer is $t=0.1mm=0.01cm$
New radius of plastic tube is $r=2-tcm=1.99cm=1.99\times {10}^{-2}m$
Voltage of the battery is $V=12V$
Density of silver(solution) is ${\rho }_{Ag}=1.47\times {10}^{-8}\Omega m$
By Ohm's law, the current through the circuit is
$I=\frac{V}{R}$
where, R is the resistance given by
$R=\frac{{\rho }_{Ag}l}{A}$
where, $A$ is the area of cross section of tube.
$R=\frac{{\rho }_{Ag}l}{\pi r^{\prime 2}}$
$R=\frac{(1.47\times {10}^{-8})\left(25\right)}{\pi {(\left[\right(2{)}^2-\left(1.99{)}^2\right]\times {10}^{-2})}^2}$

$R=\frac{36.75\times {10}^{-4}}{\pi \left(0.0399\right)}$

$R=\frac{36.75\times {10}^{-4}}{12.5\times {10}^{-2}}$

$R=2.9\times {10}^{-2}\Omega$
Using this value in equation (1), we get,
$I=\frac{12}{2.9\times {10}^{-2}}$

$I=4.14\times {10}^2$
$I=414A$