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Question

(B) Let the original length of the wire be $l$ and its original resistance be $R = \frac{\rho l}{A}$.Let the fraction of the length stretched be $a$.

Vedclass · Accepted Answer

$\frac{1}{8}$

Vedclass · Answer

(B) Let the original length of the wire be $l$ and its original resistance be $R = \frac{\rho l}{A}$.Let the fraction of the length stretched be $a$. The length of the part to be stretched is $al$,and the remaining part is $l(1-a)$.The resistance of the unstretched part is $R_1 = \frac{\rho l(1-a)}{A} = R(1-a)$.When the part of length $al$ is stretched to a new length $l'$,the total length becomes $l' + l(1-a) = \frac{3}{2}l$,so $l' = \frac{3}{2}l - l + al = l(\frac{1}{2} + a)$.Since the volume of the stretched part remains constant,$A'l' = A(al)$,so $A' = \frac{A(al)}{l'} = \frac{A(al)}{l(\frac{1}{2} + a)} = \frac{Aa}{\frac{1}{2} + a}$.The resistance of the stretched part is $R' = \frac{\rho l'}{A'} = \frac{\rho l(\frac{1}{2} + a)}{\frac{Aa}{\frac{1}{2} + a}} = R \frac{(\frac{1}{2} + a)^2}{a}$.The total resistance is $R_1 + R' = 4R$.$R(1-a) + R \frac{(\frac{1}{2} + a)^2}{a} = 4R$.$(1-a) + \frac{\frac{1}{4} + a + a^2}{a} = 4$.$a - a^2 + \frac{1}{4} + a + a^2 = 4a$.$2a + \frac{1}{4} = 4a$.$2a = \frac{1}{4} \Rightarrow a = \frac{1}{8}$.

In order to increase the resistance of a given wire of uniform cross-section to four times its value,a fraction of its length is stretched uniformly until the full length of the wire becomes $\frac{3}{2}$ times the original length. What is the value of this fraction?

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