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A wire of length L,area of cross section...

A wire of length L,area of cross section A is hanging from a fixed support. The length of the wire changes to `L_1` when mass `M` is suspended from its free end. The expression for Young's modulus is:

A

`(Mg(L_(1)-L))/(AL)`

B

`(MgL)/(AL_(1))`

C

`(MgL)/(A(L_(1)-L))`

D

`(MgL_(1))/(AL)`

Text Solution

Verified by Experts

The correct Answer is:
C
Stress `=(Mg)/(A)`
Strain `=(Delta L)/(L) = (L_(1)- L)/(L)`
Young's modulus `=("Stress")/(Strain") = (MgL)/(A(L_(1)-L))`
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Knowledge Check

  • A wire of length L is hanging from a fixed support. The length changes to L_(1) and L_(2) when masses M_(1)and M_(2) are suspended respectively from its free end. Then L is equal to

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

  • 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

    A
    `([(Y_1Y_2)A])/([2(Y_1L_2+Y_2L_1)])`
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    D
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    C
    `k = YA//2L`
    D
    `k = YL//A`

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