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The resistance of a 10 m long wire is 10...

The resistance of a 10 m long wire is 10 `Omega`. Its length is increased by 25% by stretching the wire uniformly. The the resistance of the wire will be

A

12.5 `Omega`

B

14.5 `Omega`

C

15.6 `Omega`

D

16.6 `Omega`

Text Solution

Verified by Experts

The correct Answer is:
C
`R = (rho lambda)/(A)`
For given problem `R prop lambda^(2)`
Hence `(R_(1))/(R_(2)) = (lambda_(1)^(2))/(lambda_(2)^(2))`
`R_(2) = ((lambda_(2))/(lambda_(1)))^(2) xx R_(1) = 15.6 Omega`
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Knowledge Check

  • A wire of resistance 10Omega is elongated by 10% , by streching without changing its density. The resistance of the elongated wire is

    A
    `10.1Omega`
    B
    `11.1Omega`
    C
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  • The resistance of a wire is 5 ohm at 50°C and 6 ohm at 100°C. The resistance of the wire at 0°C will be:

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    1 ohm
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    4 ohm
    D
    3 ohm

  • The resistance of a wire at room temperature 30^(@)C is found to be 10Omega . Now to increase the resistance by 10%, the temperature of the wire must be [The temperature coefficient of resistance of the material of the wire is 0.002//""^(@)C ]

    A
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