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