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 Two wires of equal length and cross-sectional area are suspended as shown in figure. Their Young's modullare `Y_1` and`Y_2` respectively. The equivalent Young's modulil will bebr>

A

`Y_1+Y_2`

B

`(Y_1+Y_2)/(Y_1+Y_2)`

C

`(Y_1+Y_2)/(2)`

D

`sqrt(Y_1Y_2)`

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Knowledge Check

  • Two wires of equal length and cross-section area are suspended as shown in figure. Their Young's modulus are Y_1 and Y_2 , respectively. The equivalent Young's modulus will be

    A
    `Y_1 + Y_2`
    B
    `(Y_1 + Y_2)/2`
    C
    `(Y_1 Y_2)/(Y_1 + Y_2)`
    D
    `sqrt(Y_1 Y_2)`

  • Two wires of equal length and cross-section area suspended as shown in figure. Their Young's modulus are Y_(1) and Y_(2) respectively. The equavalent Young's modulus will be

    A
    `Y_(1) + Y_(2)`
    B
    `(Y_(1) - Y_(2))/(2)`
    C
    `(Y_(1)Y_(2))/(Y_(1) + Y_(2))`
    D
    `sqrt(Y_(1)Y_(2))`

  • Young's modules of material of a wire of length ' "L" ' and cross-sectional area "A" is "Y" .If the length of the wire is doubled and cross-sectional area is halved then Young's modules will be :

    A
    `Y/4`
    B
    Y
    C
    4Y
    D
    2Y

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