Heat And Thermodynamics Ques 50
- Two identical conducting rods are first connected independently to two vessels, one containing water at $100^{\circ} \mathrm{C}$ and the other containing ice at $0^{\circ} \mathrm{C}$. In the second case, the rods are joined end to end and connected to the same vessels. Let $q_1$ and $q_2$ gram per second be the rate of melting of ice in the two cases respectively. The ratio $\frac{q_1}{q_2}$ is
(2004, 2M)
(a) $\frac{1}{2}$
(b) $\frac{2}{1}$
(c) $\frac{4}{1}$
(d) $\frac{1}{4}$
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Answer:
Correct Answer: 50.(c)
Solution:
Formula:
- $\frac{d Q}{d t}=L\left(\frac{d m}{d t}\right)$
or $\quad \frac{\text { Temperature difference }}{\text { Thermal resistance }}=L\left(\frac{d m}{d t}\right)$
$$ \begin{array}{rlrl} \text { or } & & \frac{d m}{d t} & \propto \frac{1}{\text { Thermal resistance }} \\ \Rightarrow & q & \propto \frac{1}{R} \end{array} $$
In the first case rods are in parallel and thermal resistance is $\frac{R}{2}$ while in second case rods are in series and thermal resistance is $2 R$.
$$ \frac{q _1}{q _2}=\frac{2 R}{R / 2}=\frac{4}{1} $$
BETA
