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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?A. increaseB. decreaseC. first increase and then decreaseD. first decrease and then increase |
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Answer» Correct Answer - C 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. |
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