Question

Resistance of a given wire of length '/' is 4n. The wire is stretched uniformly such that
its length becomes 3 /. Find the new resistance of the stretched wire.

Answer 1

Answer:

1/8

Explanation:

Solution:

Let R,ρ,L and A be the original wire’s resistance, resistivity, length and area of cross-section.

Suppose, a part of wire’s length equal to l1 is stretched uniformly to make the resistance of the new wire 4R. Consequently, the area of cross section of the stretched portion will also change such that the volume of the stretched portion remains unchanged in the process. Therefore, if l1’ and A’ are the length and area of cross section of the stretched out part of the wire, then 

l1’×A’=l1×A…………..(1)

Also, (L−l1)+l1’=1.5L……………..(2)

Now, in the new configuration of the wire, we have,

Resistance due to the unstretched part, Runstretched=ρA(L−l1)

Resistance due to the stretched part, Rstretched=ρA’l1’=ρl1A(l1’)2 (from (1))

By the problem, 4R=Runstretched+Rstretched

which implies

4ρAL=ρA(L−l1)+ρl1A(