A piece of metal floats on mercury. The coefficient of volume expansion of metal and mercury are `gamma_1 and gamma_2`, respectively. if the temperature of both mercury and metal are increased by an amount `Delta T`, by what factor does the fraction of the volume of the metal submerged in mercury changes ?
A piece of metal floats on mercury. The coefficient of volume expansion of metal and mercury are `gamma_1 and gamma_2`, respectively. if the temperature of both mercury and metal are increased by an amount `Delta T`, by what factor does the fraction of the volume of the metal submerged in mercury changes ?
Text Solution
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Let the total volume of metal in air and mercury be V and `V_(s)` respectively, `rhoandsigma` be the the densities of metal and mercury respectively. when the metal floats in equilibrium, Fraction of the volume submerged,
`f_(s)=V_(s)/V=rho/sigma" "...(1)`
When the temperature changes, the fraction of volume submerged changes as densities change.
`(Deltaf_(s))/(f_(s))=(f_(s).)/(f_(s))-1=rho^(1)/sigma^(1)xxsigma/rho-1" "...(2)`
Densities of liquid and metal decrease as temperature increases `rho.=rho/(1+gamma_(1)DeltaT),sigma.=sigma/(1+gamma_(2)DeltaT)`
which on substitution in eqn. (2) yields
`(Deltaf_(s))/(f_(s))=(1+gamma_(2)DeltaT)/(1+gamma_(1)DeltaT)-1=((1+gamma_(2)DeltaT)-(1+gamma_(1)DeltaT))/((1+gamma_(1)DeltaT))`
= `((gamma_(2)-gamma_(1))DeltaT)/((1+gamma_(1)DeltaT))=(gamma_(2)-gamma_(1))DeltaT(1-gamma_(1)DeltaT)`
= `(gamma_(2)-gamma_(1))DeltaT`
[As `1/(1+gamma_(1)DeltaT)=(1+gamma_(1)DeltaT)^(-1)=(1-gamma_(1)DeltaT)andgamma_(1)gamma_(2)` small number, it has been neglected]
When the solid is completely submerged,
`1=f_(s)(1+gammaDeltaT)orDeltaT=(1-f_(s))/(gammaf_(s))`
b) `f_(s_(1))=f_(s)(1+gammaDeltaT_(1)),f_(s_(2))=f_(s)(1+gammaDeltaT_2)`
`(f_(s_(1)))/(f_(s_(2)))=((1+gammaDeltaT_(1))/(1+gammaDeltaT_(2)))`. On solving for `gamma`, we have
`gamma=(f_(s_(1))-f_(s_(2)))/(f_(s_(2))DeltaT_(1)-f_(s_(1))DeltaT_(2))`
c) As `f_(s)=rho/sigmaandf_(s).=(rho.)/(sigma.)`
`f_(s).=(rho/(1+gamma_(1)DeltaT))((1+gamma_(2)DeltaT))/sigmaf_(s).=f_(s)((1+gamma_(2)DeltaT)/(1+gamma_(1)DeltaT))`
If `gamma_(2)gtgamma_(1)`, the solid sinks.
If `gamma_(2)=gamma_(1)`, no effect on submergence.
If `gamma_(2)ltgamma_(1)`, the solid lifts up.
`w_(1)/w_(0)=(d_(0)(1+alphat))/d_(0)rho_(0)/((1+betat)rho_(0))=(1+alphat)/(1+betat)`
`w_(1)=w_(0)(1+alphat)(1-betat)=w_(0)[1+(alpha-beta)t]`
`f_(s)=V_(s)/V=rho/sigma" "...(1)`
When the temperature changes, the fraction of volume submerged changes as densities change.
`(Deltaf_(s))/(f_(s))=(f_(s).)/(f_(s))-1=rho^(1)/sigma^(1)xxsigma/rho-1" "...(2)`
Densities of liquid and metal decrease as temperature increases `rho.=rho/(1+gamma_(1)DeltaT),sigma.=sigma/(1+gamma_(2)DeltaT)`
which on substitution in eqn. (2) yields
`(Deltaf_(s))/(f_(s))=(1+gamma_(2)DeltaT)/(1+gamma_(1)DeltaT)-1=((1+gamma_(2)DeltaT)-(1+gamma_(1)DeltaT))/((1+gamma_(1)DeltaT))`
= `((gamma_(2)-gamma_(1))DeltaT)/((1+gamma_(1)DeltaT))=(gamma_(2)-gamma_(1))DeltaT(1-gamma_(1)DeltaT)`
= `(gamma_(2)-gamma_(1))DeltaT`
[As `1/(1+gamma_(1)DeltaT)=(1+gamma_(1)DeltaT)^(-1)=(1-gamma_(1)DeltaT)andgamma_(1)gamma_(2)` small number, it has been neglected]
When the solid is completely submerged,
`1=f_(s)(1+gammaDeltaT)orDeltaT=(1-f_(s))/(gammaf_(s))`
b) `f_(s_(1))=f_(s)(1+gammaDeltaT_(1)),f_(s_(2))=f_(s)(1+gammaDeltaT_2)`
`(f_(s_(1)))/(f_(s_(2)))=((1+gammaDeltaT_(1))/(1+gammaDeltaT_(2)))`. On solving for `gamma`, we have
`gamma=(f_(s_(1))-f_(s_(2)))/(f_(s_(2))DeltaT_(1)-f_(s_(1))DeltaT_(2))`
c) As `f_(s)=rho/sigmaandf_(s).=(rho.)/(sigma.)`
`f_(s).=(rho/(1+gamma_(1)DeltaT))((1+gamma_(2)DeltaT))/sigmaf_(s).=f_(s)((1+gamma_(2)DeltaT)/(1+gamma_(1)DeltaT))`
If `gamma_(2)gtgamma_(1)`, the solid sinks.
If `gamma_(2)=gamma_(1)`, no effect on submergence.
If `gamma_(2)ltgamma_(1)`, the solid lifts up.
`w_(1)/w_(0)=(d_(0)(1+alphat))/d_(0)rho_(0)/((1+betat)rho_(0))=(1+alphat)/(1+betat)`
`w_(1)=w_(0)(1+alphat)(1-betat)=w_(0)[1+(alpha-beta)t]`
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