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A cylindrical conductor has length l and...

A cylindrical conductor has length `l` and area of cross section A. Its conductivity changes with distance `(x)` from one of its ends as `sigma = sigma_(0) (l)/(x). [sigma_(0) "is a constant"]`. Calculate electric field inside the conductor as a function of `x`, when a cell of emf V is connected across the ends.

Text Solution

Verified by Experts

The correct Answer is:
`(2 V_(x))/(l^(2))`
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Knowledge Check

  • A conductor of length l and area of cross-section A has a resistance R then its specific resistance is

    A
    `rho=(RA)/(l)`
    B
    `rho=RA l`
    C
    `rho=(l)/(RA)`
    D
    `rho=(R^(2)A)/(l)`

  • A cylindrical conductor AB of length l and area of cross-section a is connected to a battery having emf E and negligible internal resistance. The specific conductivity of cylindrical conductor varies as sigma = sigma_(0) (1)/(sqrt(x)) , where sigma_(0) is constant and x is distance from end A. What is the electric field just near the end B of cylinder?

    A
    `(E)/(l)`
    B
    `(3E)/(2l)`
    C
    `(2E)/(3l)`
    D
    `(E)/(2l)`

  • The resistance of wire of length l and area of cross-section a is x ohm. If the wire is stretched to double its length, its resistance would become:

    A
    2 x ohm
    B
    0.5 x ohm
    C
    4 x ohm
    D
    6 x ohm

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