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A metal wire of circular cross-section h...

A metal wire of circular cross-section has a resistance R. The wire is now stretched without breaking so that its length is doubled and the density is assumed to remain the same. If the resistance of the wire now becomes `R_2` then `R_2 : R_1` is

A

`1:1`

B

`1 :2`

C

`4 :1`

D

`1:4`

Text Solution

Verified by Experts

The correct Answer is:
C
As `R= ( rho l )/( A ) = ( rho l xx l )/(A xx l) =( rho l^2)/(V)`
here ` rho ` and V are constant `therefore R prop l^2 therefore (R_1)/( R_2) =(l_(1)^(2))/( l_(2)^(2))`
given `l_2 = 2l_1 `. Hence ` (R_1)/( R_2) =(l_1^(2) )/( 4l_(1)^(2)) = 1/4 implies R_2 : R_1 = 4 :1`
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Knowledge Check

  • A metal wire of circular cross-section has a resistance R_(1) . The wire is now stretched without breaking, so that its length is doubled and the density is assumed to remain the same. If the resistance of the wire now becomes R_(2) , then R_(2) : R_(1) is

    A
    `1 : 1`
    B
    `1 : 2`
    C
    `4 : 1`
    D
    `1 : 4`

  • The resistance of a wire is R. It is stretched uniformly so that its length is doubled. The resistance now becomes

    A
    2R
    B
    R/2
    C
    4R
    D
    R/4

  • If the area of cross-section of a resistance wire is halved, then its resistance becomes:

    A
    one-half
    B
    2 times
    C
    one-fourth
    D
    4 times

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