A `6Omega` resistance wire is doubled up by folding. Calculate the new resistance of the wire.

Options

A. `1.6 Omega`
B. `1.8 Omega`
C. `1.5 Omega`
D. `1.9 Omega`

Answer

C

Explanation

Suppose the length of `6 Omega` resistance wire is l, its area of cross-section is A and its resistivity is `rho`. Then:
`R = (rho xx l)/(A)`
or `6 = (rho xx l)/(A)` ...(1)
Now, when this wire is doubled up by folding, then its length will become half, that is, the length will become `(l)/(2)`. But on doubling the wire by folding, its area of cross-section will becomes double, that its, the area of cross-section will become 2A. Suppose the new resistance of the doubled up wire (or folded wire) is R. So,
`R = (rho xx l)/(2 xx 2A)`
or `R = (rho xx l)/(4A)` ...(2)
Now, dividing equation (2) by equation (1), we get:
`(R)/(6) = (rho xx l xx A)/(4A xx rho xx l)`
or `(R)/(6) = (1)/(4)`
`4R = 6`
`R = (6)/(4)`
`R = 1.5 Omega`
Thus, the new resistance of the doubled up wire is `1.5 Omega`.