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Two bottles $A$ and $B$ have radii $R_{A}$ and $R_{B}$ and heights $h_{A}$ and $h_{B}$ respectively, with $R_{B}=2 R_{A}$ and $h_{B}=2 h_{A}$. These are filled with hot water at $60^{\circ} C$. Consider that heat loss for the bottles takes place only from side surfaces. If the time, the water takes to cool down to $50^{\circ} C$ is $t_{A}$ and $t_{B}$ for bottles $A$ and $B$, respectively. Then, $t_{A}$ and $t_{B}$ are best related as
$t_{A}=t_{B}$
$t_{B}=2 t_{A}$
$t_{B}=4 t_{A}$
$t_{B}=t_{A} / 2$
Solution
$(b)$ Rate of heat loss for two bodies of same specific heat's and same density for which temperature difference (of bodies and surroundings) is same is given by
$\frac{\Delta T-h A}{\Delta t}$
where, $\Delta T=$ temperature change $\left(T_{f}-T_{i}\right)$, $\Delta t=$ time interval, $k=$ a constant
depending on shape of bodies, $A=$ surface area and $V=$ volume.
As in given case, bottles are identical and they cool down by same temperatures,
$\Delta t \propto \frac{V}{A}$
$\text { So, } \Delta t \propto \frac{A \cdot h}{A} \text { or } \Delta t \propto h$
$\therefore \quad \frac{t_{A}}{t_{B}}=\frac{h_{A}}{h_{B}} \Rightarrow \frac{t_{A}}{t_{B}}=\frac{h_{A}}{2 h_{A}} \quad\left[\therefore h_{B}=2 h_{A}\right]$
$\Rightarrow \quad t_{B}=2 t_{A}$
