The densities of wood and benzene at `0^(@)`C are 880 kg `m^(-3)` and 900 kg `m^(-3)`, respectively. The coefficient of volume expansion is
`12 xx 10^(-3) C^(-10)` for wood and `1.5 xx 10^(-30) C^(-1)` for
benzene. Then the temperature at which a piece of wood just sinks in benzene is

Options

A. `88^(@)`C
B. `90^(@)`C
C. 83.3 C
D. `90.3^(@)`C

Answer

C

Explanation

Given , density of wood at `0^(@) C, rho_(w) = 880 " kg " m^(-3)`
density of benzene at `0^(@)C, rho_(b) = 900 "kg " m^(-3)`
coefficient of volume expansion of wood,
` gamma_(w) = 1.2 xx 10^(-3)""^(@)C^(-1)`
Coefficient of vulume expansion of benzene,
`gamma_(b) = 1.5 xx 10^(-3)""^(@)C^(-1)`
and initial termperature, `T_(1) = 0^(@)` C
Let `T_(2)` be the temperature at which pieces of wood will just sink in benzene and `Delta T = T_(2) - T_(1)`
The piece of wood begins to sink when it weight is equal to the weight of benzene displaced.
Mass = volume `xx` Density
Therefore, `V rho_(w) g = V rho_(b) g`
`therefore " " (rho_(w))/(1 + gamma_(w) Delta T) = ( rho_(b))/(1 + gamma_(b) Delta T)`
` (880)/( 1 + 1.2 xx 10^(-3) Delta T) = (900)/(1 + 1.5 xx 10^(-3) Delta T) `
800 + 880 `xx 1.5 xx 10^(-3) Delta T = 900 + 900 xx 1.2 xx 10^(-3) Delta T `
( 1320 - 1080 ) `xx 10^(-3) Delta T = 20 `
`Delta T = (20)/(240 xx 10^(-3))`
`Delta T = 83.33""^(@)C`
`T_(2) - T_(1) = 83.33""^(@)C`
Hence, `T_(2) = 83.3""^(@)C`