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/m3. The length of the wire increases by 1mm. If the diameter of wire is 0.6mm. Find te initial length of wire.

(X= 2 * 1011 N/m2 and Vof the metal piece is 800cm3)


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)


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



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)


e = (mg - ρgV)(1/AY). dx


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


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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