Question

At $4^{\circ} C , 0.98$ of the volume of a body is immersed in water. The temperature (in ${ }^{\circ} C$ ) at which the entire body gets immersed in water $\left(\gamma_{w}=3.3 \times 10^{-4} K^{-1}\right)$ is (neglect the expansion of the body) -

Answer 1

Let volume of water be $V$ and density of water at $4^{\circ} C$ be $d$
So initially,
For floatation (or equilibrium)
$W =$ Upthrust
$W =0.98\, V\, dg$ ...(1)
Now let the temperature at which entire body gets immersed in water be $T$ and let density of water at that temperature be $d T$
So, for floatation (or equilibrium)
$W = Vd Tg$ ...(2)
By (1) & (2),
$0.98\, Vdg = Vd Tg$
$d _{ T } =0.98\, d$
Now for water.
$d _{ T } =\frac{ d _{0}}{1+\gamma_{ W } \Delta T }$
$0.98 d =\frac{ d }{1+3.3 \times 10^{-4}( T -4)}$
$0.98\left[1+3.3 \times 10^{-4}(T-4)=1\right.$
$0.98+3.234 \times 10^{-4}(T-4)=1$
$3.234 \times 10^{-4}(T-4)=0.02$
$T -4 =\frac{0.02}{3.234 \times 10^{-4}}$
$T -4 =61.842$
$T =65.842^{\circ} C$