Apiece of metal of 2kg weight is suspended from 1end of a vertical wire whose other end is fixed and the piece is fully emerged in an oil of density 700kg/m^{3}. The length of the wire increases by 1mm. If the diameter of wire is 0.6mm. Find te initial length of wire.

(X= 2 * 10^{11} N/m^{2} and Vof the metal piece is 800cm^{3})

Here,

the final length (L) = initial length (l) + total elongation

The tension in the wire would be the difference of the weight of the body and the upthrust exerted by the liquid (oil).

T = mg - ρgV (1)

here

m is the mass of the metal, ρ is teh density of oil and V is the volume of the metal

the elongation of a small length segment dx would be given as

e = Tdx / AY

where

dx is the small length segment, A is the area of cross-section and Y is the Young's Modulus.

so, from (1)

e = T(dx / AY) = (mg - ρgV)(dx / AY)

or

e = (mg - ρgV)(1/AY). dx

thus, total elongation can be calculate by integrating over the length of the whole wire

so,

**E = (mg - ρgV).(1/AY). l**

we can calculate the elongation by substituting the appropriate values and then the final length by

L = E + l

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