A block of wood is floating in water at `0^@ C`. The temperature of water is slowly raised from `0^@ C` to `10^@ C`. How will the precentage of volume of block above water level change with rise in temperature?

Options

A. increase
B. decrease
C. first increase and then decrease
D. first decrease and then increase

Answer

C

Explanation

Fraction of wooden block immersed at `0^@C`
`(V_1)/(V_0)=((rho_("wood"))_(0^@C))/((rho_(H)_(2)O)_(0^@C))`
`f_1=(V_0-V_1)/(V_0)=((rho_(H2O))_(0^@C)-(rho_("wood"))_(0^@C))/((rho_(H_2O))_(0^@C))`
`V_1`- volume of wood immersed in water at `0^@C`
`V_0`- Volume of wood
`(rho_("wood"))_(0^@C)`- Density of wood at `0^@C`
When the temperature is raised to `10^@C` the volume of wood immersed in water changes to `V_2`.
`(V_2)/(V_0)=((rho_("wood"))_(0^@C))/((rho_(H_2O))_(0^@C))`
`f_2=(V_0-V_2)/(V_0)=((rho_(H_2O))_(10^@C)-(rho_("wood"))_(10^@C))/((rho_(H_2O))_(10^@C))`
From `0^@C` to `4^@C`, the density of water increases, and from `4^@C` to `10^@C` the density of water decreases. But for wood density decreases as temperature increases. The volume of block above water level will first increase and then decrease.