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Question: A plastic tube 25m long and 4cm in diameter is dipped into a solution depositing a silver layer 0.1 ...

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

A. 4.14A B. 41.4A C. 414A D. 414mA  {\text{A}}{\text{. 4}}{\text{.14A}} \\\ {\text{B}}{\text{. 41}}{\text{.4A}} \\\ {\text{C}}{\text{. 414A}} \\\ {\text{D}}{\text{. 414mA}} \\\
Explanation

Solution

Hint : Fluorescent tube, is a low pressure mercury visible light. An electric current in the gas excites light that then causes a phosphor coating on the inside of the lamp to glow. A fluorescent lamp converts electrical incandescent lamps. The typical luminous efficiency of fluorescent lighting systems is 50-100 lumens per watt.

Complete step by step solution:
Given,
Length of the plastic tube is1 = 25m{\text{1 = 25m}}
Diameter of plastic tube is d = 4cm{\text{ = 4cm}}
Radius of plastic tube is r = 2cm{\text{ = 2cm}}
Thickness of silver layer is t = 0.1mm = 0.01cm{\text{ = 0}}{\text{.1mm = 0}}{\text{.01cm}}
New radius of plastic tube is r = 2 - t cm = 1.99 cm = 1.99×10−2m{\text{ = 2 - t cm = 1}}{\text{.99 cm = 1}}{\text{.99}} \times {\text{1}}{{\text{0}}^{ - 2}}{\text{m}}
Voltage of the battery is V = 12V
Density of silver isRAg=1.47×10−8Ω−m{{\text{R}}_{{\text{Ag}}}} = 1.47 \times {10^{ - 8}}\Omega - {\text{m}}
By ohm’s law the current through the circuit is
I = VR{\text{I = }}\dfrac{{\text{V}}}{{\text{R}}}
Whereas, R is the resistance given by
R = ℓAglπr2{\text{R = }}\dfrac{{{\ell _{{\text{Agl}}}}}}{{{\pi _{{r^2}}}}}
R = (1.47×10−8)π[(1(2)2−(199)2]×10−2)]2{\text{R = }}\dfrac{{(1.47 \times {{10}^{ - 8}})}}{{\pi [(1{{(2)}^2} - {{(199)}^2}] \times {{10}^{ - 2}}){]^2}}}

R = 36.75×10−412.5×10−2 R = 2.9×10−2Ω  {\text{R = }}\dfrac{{36.75 \times {{10}^{ - 4}}}}{{12.5 \times {{10}^{ - 2}}}} \\\ {\text{R = 2}}{\text{.9}} \times {10^{ - 2}}\Omega \\\

Using this value in the equation first we get
I = 122.9×10−2=4.14×102=414 A{\text{I = }}\dfrac{{12}}{{2.9 \times {{10}^{ - 2}}}} = 4.14 \times {10^2} = 414{\text{ A}}
Thus option C is the correct answer.

∗∗Additionalinformation:∗∗Electriccurrentflowsthroughthetubeinalenspressuredischarge.Electronscollidewithanionizednoblegasatominsidethebulbsurroundingthefilamenttoformaplasmabytheprocessofimpactionization,asaresultoftheavalancheionizationtheconductivityoftheionizedgasrapidlyrisesallowinghighercurrenttoflowthroughthelamp.∗∗Note:∗∗Currentinthecoatedtubeisabout414A.Checktheformulaaftersolvingtheanswerandputthevalueproperlyinthegivenformula.Sometimeswewritethewrongvalueduetothatanswer.Writethefinalanswerwiththestatement,don′twriteitdirectly.Thesearethepointsweneedtoremember. **Additional information:** Electric current flows through the tube in a lens pressure discharge. Electrons collide with an ionized noble gas atom inside the bulb surrounding the filament to form a plasma by the process of impact ionization, as a result of the avalanche ionization the conductivity of the ionized gas rapidly rises allowing higher current to flow through the lamp. **Note:** Current in the coated tube is about 414 A. Check the formula after solving the answer and put the value properly in the given formula. Sometimes we write the wrong value due to that answer. Write the final answer with the statement, don't write it directly. These are the points we need to remember.