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A wire of 1Omega resistance is stretched...

A wire of `1Omega` resistance is stretched and its length becomes twice, then the new resistance of the wire

A

`(1)/(4) Omega`

B

`2Omega`

C

`(1)/(2)Omega`

D

`4Omega`

Text Solution

Verified by Experts

The correct Answer is:
D
According to the question, Resistance, `R= (rho l)/(A)`
Here,l= initial length of wire and A= cross-section area of wire. In the first case.
`1= (rho l_(1))/(A_(1))` ...(i)
Length of new wire, `l_(2)= 2l_(1)`. The volume of wire remains constant. Hence, `V_(1)= V_(2)`
`A_(1) l_(1)= A_(2)l_(2)`
`A_(1)l_(1)= A_(2) (2l_(1))`
`A_(2) = (A_(1))/(2)`
Resistance of new wire, `R.= rho (l_(2))/(A_(2))= (rho (2l_(1)))/((A_(1))/(2)) =4 xx (rho l_(1))/(A_(1))`
From Eq (i ) we get `=4 xx 1= 4 Omega` Hence, the new resistance of wire `=4Omega`
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