The densities of wood and benzene at `0^(@)C` are `880 kg//m^(3)` and `900 kg//m^(3)` respectively. The coefficients of volume expansion are `1.2 xx 10^(-3)//'^(@)C` for wood and `1.5 xx 10^(-3)//'^(@)C` for benzene. At what temperature will a piece of wood just sink in benzene?

Options

A. `53^@C`
B. `63^@C`
C. `73^@C`
D. `83^@C`

Answer

D

Explanation

It is clear that at desired temperature `T^@C`, the dinsities of the wood and benzene must be equal for the wood to just sink.
i.e., `rho_W(T)=rho_B(T)`
If m is the mass of wood (which is assumed to be constant) then If (`V_0)_W` and `(V_B)_B` are the respective volumes at `0^@C` of mass m of wood and benzene,
`(rho_0)_W(V_0)_W=(rho_0)_B(B_0)_B=m`
`(rho_0)_W=880 kg//m^(3)` and `(rho_0)_B=900 kg//m^(3)`
Hence `(V_0)_W=(m)/(880)(m^3)`
End `(V_0)_B=(m)/(900)(m^3)`
we then have, `(V_T)_W=(V_0)_W(1+1.2xx10^-3T)`
`(V_T)_B=(V_0)_B(1+1.5xx10^-3T)`
thus `(V_T)_W/(V_T)_B=(rho_B)_T/((rho_W)_T)=1=((V_0)_W1+1.2xx10^-3T)/((V_0)_B1+1.5xx10^-3T)`
Thus `((V_0)_W)/((V_0)_B)=(900)/(880)=(90)/(88)=(1+1.5xx10^-3T)/(1+1.2xx10^-3T)`
soving for T, we have `T=83.2^@C`