Q. A copper wire is stretched such that it's cross section area decrea

1.	Q. A copper wire is stretched such that it's cross section area decreases by 0.25 % . The Percentage Change in its electric resistance is ?o 0.5%o 2.5%o 1%o 2.0%
Answer» ★ Concept :- Here the concept of Percentage increase in electric resistance has been used. We see that we are given a wire which is stretched. This means the volume of the wire won't be changing. From here we can take a relationship between the length, area and volume of wire. Then we can apply it in the formula of the ELECTRICAL RESISTIVITY and from there we can find the initial resistance. Then we can compute the change in Resistance by taking logarithm and find the answer. Let's do it !! _____________________________________ ★ Formula Used :- $\;\boxed{\sf{R\;=\;\rho\:\dfrac{l}{A}}}$ _____________________________________ ★ Solution :- Given, » Resistivity of the material = ρ This will be same for both cases even when the area of cross section has been increased. » Initial length of the wire = l » Initial area of cross section of the wire = A » Diamter of the wire = d » RADIUS of the wire = r = ½ × d » Volume of the wire = V » Length = l = V/A (since volume remains same on stretching) Then, » Area of cross-section = πr² = π × ¼ × r² Let the Intial Resistance of wire be R. $\;\sf{\rightarrow\;\;R\;=\;\rho\:\dfrac{l}{A}}$ By applying values here, we get $\;\tt{\rightarrow\;\;R\;=\;\rho\:\dfrac{\dfrac{V}{A}}{A}}$ $\;\tt{\rightarrow\;\;R\;=\;\rho\:\dfrac{V}{A\:\times\:A}}$ $\;\tt{\rightarrow\;\;R\;=\;\rho\:\dfrac{V}{\pi\:\times\:\dfrac{D^{2}}{4}\:\times\:\pi\:\times\:\dfrac{D^{2}}{4}}}$ $\;\tt{\rightarrow\;\;R\;=\;\rho\:\dfrac{V}{\pi^{2}\:\times\:\dfrac{D^{4}}{16}}}$ $\;\tt{\rightarrow\;\;R\;=\;\rho\:\dfrac{16V}{\pi^{2}\:\times\:D^{4}}}$ Now here we see that the change is small or LARGE, we shall apply differentiation and logarithm first. ∆R is the change in electrical resistance. ∆D is the change in cross sectional area From this we get, $\;\tt{\mapsto\;\;\dfrac{\Delta R}{R}\;=\;-4\dfrac{\Delta D}{D}}$ Here the factor is achieved from logarithmic function. Multiplying both sides by 100, we get $\;\tt{\mapsto\;\;\dfrac{\Delta R}{R}\:\times\:100\;=\;-4\dfrac{\Delta D}{D}\:\times\:100}$ This will give ∆R % that is percentage increase in electrical resistance . Now applying the value, we get $\;\tt{\mapsto\;\;\dfrac{\Delta R}{R}\:\times\:100\;=\;4\:\times\:0.25\%}$ (since we don't INCLUDE -ve sign in calculation of percentage) $\;\tt{\mapsto\;\;\Delta R\%\;=\;1\%}$ This is the required answer. $\;\:\underline{\boxed{\tt{Required\;\:\%\;\:increase\;\:in\;\;Resistance\;=\;\bf{\purple{1\%}}}}}$ _______________________________ ★ More to know :- $\;\sf{\leadsto\;\: Resistance\;=\;\dfrac{Voltage}{Current}}$ $\;\sf{\leadsto\;\;Power\;=\;Voltage\;\times\;Current}$ $\;\sf{\leadsto\;\:Current\;=\;\dfrac{Charge}{Time}}$ $\;\sf{\leadsto\;\:Voltage\;=\;\dfrac{Workdone}{Charge}}$ $\;\sf{\leadsto\;\;F\;=\;\dfrac{1}{4\pi \epsilon_{0}}\:\times\:\dfrac{q_{1}q_{2}}{r^{2}}}$ where, ϵ｡ is the permittivity of free space. q and it's denotions are two charges. r is the distance between two charges. F is the force between two charges. This can be either attractive or repulsive on the basis of charges.

Q. A copper wire is stretched such that it's cross section area decreases by 0.25 % . The Percentage Change in its electric resistance is ?o 0.5%o 2.5%o 1%o 2.0%

Answer»

★ Concept :-

Here the concept of Percentage increase in electric resistance has been used. We see that we are given a wire which is stretched. This means the volume of the wire won't be changing. From here we can take a relationship between the length, area and volume of wire. Then we can apply it in the formula of the ELECTRICAL RESISTIVITY and from there we can find the initial resistance. Then we can compute the change in Resistance by taking logarithm and find the answer.

Let's do it !!

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