A conducting wire of length 1 has resistance R. If its length is increased to nl by stretching it uniformly, what would be the new resistance of wire ? (Assume that there is no change in the volume of the wire when it is stretched.)

A conducting wire of length 1 has resistance R. If its length is incre

1.	A conducting wire of length 1 has resistance R. If its length is increased to nl by stretching it uniformly, what would be the new resistance of wire ? (Assume that there is no change in the volume of the wire when it is stretched.)
Answer» Solution :Original length of the wire = l and AREA of cross-section=A. New length of the wire l.= NL New area of the cross-section of wire = A. Here, Al-A.l. = A. (nl) ( `:.` It is GIVEN that the volume of the wire remains constant when it is stretched.) `:.` New area of cross-section of the wire, `A. = A/n` Oiginal resistance of the wire, `R = rho . l/A` Newresistance of the wire. `R.= rho. (l.)/(A.)` `= rho . (nl)/(A/n) = n^2 (rho . l/A)` `:. R. = n^2 R` (Note: If the length of the wire is doubled (n=2), the new resistance of the wire will be `R. = (2)^2 R = 4R`.] * You need not memorise these values.You can use these values for solving NUMERICAL problems.