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A metal cylinder of length L is subjecte...


A metal cylinder of length `L` is subjected to a uniform compressive to a uniform compressive force `F` as shown in the figure. The material of the cylinder has Young's modulus `Y` and poisson's ratio `sigma` The change in volume of the cylinder is

A

`(sigamFL)/(Y)`

B

`((1-sigma)FL)/(Y)`

C

`((1+2sigma)FL)/(Y)`

D

`((1-2sigma)FL)/(Y)`

Text Solution

Verified by Experts

The correct Answer is:
D

Volume of the cylinder, `V = pir^(2)L`
Volumetric strain `=(DeltaV)/(V) = (Delta(pir^(2)L))/(pir^(2)L)`
`(DeltaV)/(V) = (pir^(2)DeltaL + 2pirlDeltar)/(pir^(2)L)`
`=(DeltaL)/(L) =(2Deltar)/(r)" "...(i)`
Poission's ratio,`sigma = - ((Deltar//r))/(DeltaL//L)`
or `(Deltar)/(r) = (sigma DeltaL)/(L)`
On substituting this value of `(Deltar)/(r)` in Eq. (i) , we get `(DeltaV)/(V) = (DeltaL)/(L) (1- 2sigma)" "....(ii)`
Young's modulus, `Y =((F//pir^(2)))/((DeltaL//L)) or (DeltaL)/(L) = (F)/(pir^(2)Y)`
On substituting this value of the value of `(DeltaL)/(L)` in Eq. (ii), we get
`(DeltaV)/(V) = (F)/(pir^(2)Y) (1-2sigma)`
`(DeltaV)/(pir^(2)L) =(F)/(pir^(2)Y) (1-2sigma)`
` Deltal V = (FL)/(Y) (1-sigma)`
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