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A mild steel wire of length 1.0m and cro...

A mild steel wire of length `1.0m` and cross sectional area `2L` is strethched, within its elastic limit horizontally between two pillars(figure). A mass of `m` is suspended form the midpont of the wire. Strain in the wire is

A

`(y^2)/(l)`

B

`(y^2)/(3l)`

C

`(y^2)/(4l)`

D

`(2y^2)/(l)`

Text Solution

Verified by Experts

The correct Answer is:
A

`Delta l = 2(y^(2) + l^(2))^(1//2) - 2l = 2l (1 + (y^(2))/(l^(2)))^(1//2) - 2l`
`= 2l (1 + (1)/(2) (y^(2))/(l^(2)))-2l = (y^(2))/(2l^(2)) xx 2l = (y^(2))/(l)`
energy `(1)/(2) xx (Y A Delta l^(2))/(2l) - 2l = (1)/(2) "stress" xx "strain" xx "volime"`
`= (1)/(2)y ("strain")^(2) A(2l) = (1)/(2) y((Delta l)/(2l))^(2) A.2l = (yA(Delta l)^(2))/(4l)`
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Knowledge Check

  • A mild steel wire of length 2L and cross-sectional area A is stretched, well with in the elastic limit, horizontally between two pillars as shown in figure. A mass m is suspended from the mid-point of the wire strain in the wire is

    A
    `(x^(2))/(2L^(2))`
    B
    `(x)/(L)`
    C
    `(x^(2))/(L)`
    D
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  • A mild steel wire of length 2L and cross-sectional area A is stretched, well within elastic limit, horizontally between two pillars, A mass m is suspended from the midpoint of the wire. Strain in the wire is

    A
    `x^2/(2L^2)`
    B
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  • A metal wire of length L_1 and area of cross section A is ttached to a rigid support. Another metal wire of length L_2 and of the same cross sectional area is attached to the free end of the first wire. A body of mass M is then suspended from the free end of the second wire, if Y_1 and Y_2 are the Young's moduli of the wires respectively the effective force constant of the system of two wires is

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    C
    `([(Y_1Y_2)A])/((Y_1L_2+Y_2L_1))`
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