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

Answer: 64.60

Let volume of water be $V$ and density of water at $4^{\circ }C$ be $d$
So initially,
For floatation (or equilibrium)
$W=$ Upthrust
$W=0.98Vdg$ ...(1)
Now let the temperature at which entire body gets immersed in water be $T$ and let density of water at that temperature be $dT$
So, for floatation (or equilibrium)
$W=VdTg$ ...(2)
By (1) & (2),
$0.98Vdg=VdTg$
$d_T=0.98d$
Now for water.
$d_T=\frac{d_0}{1+{\gamma }_W\Delta T}$
$0.98d=\frac{d}{1+3.3\times 10^{-4}(T-4)}$
$0.98[1+3.3\times 10^{-4}(T-4)=1$
$0.98+3.234\times 10^{-4}(T-4)=1$
$3.234\times 10^{-4}(T-4)=0.02$
$T-4=\frac{0.02}{3.234\times 10^{-4}}$
$T-4=61.842$
$T=65.842^{\circ }C$