Question

a cylindrical metallic wire is stretched to increase its length by 5% calculate the percentage change in its resistance​

Answer 1

Answer:

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Answer 2

The percent change in the resistance of the cylindrical metallic wire is equal to 10.25%.

Given:

Percent increase in length of the cylindrical metallic wire = 5%

To Find:

The percent change in the resistance of the cylindrical metallic wire.

Solution:

• The resistance of a cylindrical metallic wire is directly proportional to its length (L) and inversely proportional to its cross-sectional area (A). So the resistance (R) of a cylindrical wire is given as:

                $R=\rho(\frac{L}{A})\:\:\:where\:\rho=Resistivity\:of\:the\:Material$

→ Let the initial length, cross-sectional area, and resistivity of the cylindrical metallic wire be L₁, A₁, and ρ respectively.

 ∴ The initial resistance of the metallic wire (R₁) = ρ(L₁/A₁)

∵ The percent increase in the length of the metallic wire is 5%.

∴ The new length of the wire (L₂) = L₁ (1 +0.05)

                        →  L₂= 1.05L₁ = (105L₁ )/100 = (21L₁ )/20

→ Let the new cross-sectional of the metallic wire be A₂.

∵ The volume of the metallic wire will remain constant.: A₁L₁ = A₂L₂

                                          $Volume = Constant\\\\A_{1} L_{1} =A_{2} L_{2} \\\\A_{1} L_{1} =A_{2} (\frac{21L_{1} }{20} )\\\\A_{2}=\frac{20}{21}A_{1}$

→ Final resistance of the metallic wire (R₂) = ρ(L₂/A₂)    

                                         $R_{2} =p(\frac{L_{2} }{A_{2} } )\\\\=p(\frac{21L_{1} }{20} )(\frac{21}{20A_{1} } )\\\\=(p\frac{L_{1} }{A_{1} } )(\frac{21}{20} )^{2} \\\\R_{2} =R_{1} (\frac{21}{20} )^{2}$                      

→ We can now calculate the % change in resistance of the metallic wire:

                                         $=\frac{R_{2} -R_{1} }{R_{1} } (100)\\\\=\frac{(R_{1} (\frac{21}{20} )^{2}-R_{1}) }{R_{1} }(100)\\\\=((\frac{21}{20} )^{2}-1) (100)\\\\=(\frac{441}{400}-1)(100) \\\\=\frac{41}{4} \\\\= 10.25$

Therefore the percent change in the resistance of the cylindrical metallic wire is equal to 10.25%.

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