The new resistance of a wire becomes [[1]]times the original resistance if its diameter is doubled and its length remains unchanged.

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Detailed Solution

The new resistance of a wire becomes

\frac{1}{4}

times the original resistance if its diameter is doubled and its length remains unchanged. We know that the resistance of a wire is directly proportional to the length and inversely proportional to the area of the cross section of wire. So, the resistance of wire can be written as-

R = \frac{ρl}{A}

[where

R = resistance of wire

;

l = length of wire

;

ρ = resistivity

;

A = area of cross section of wire

]

= > R = \frac{ρl}{π r^{2}}

…………………………… (i)
we know [

A = π r^{2}]

and [

radius (r) = \frac{diameter}{2}]

Put the above values in equation (i) we get,

= > R = \frac{ρl}{π {(\frac{d}{2})}^{2}}

= > R = \frac{ρl}{\frac{π d^{2}}{4}}

= > R = \frac{4 ρl}{π d^{2}}

…………………………. (ii)
If the diameter is doubled and length remains unchangedthenput the values in equation (ii) . As a result we get new resistance of the wire (R’) is: