Two identical conducting rods are first connected parallel to two vessels, one containing water at 100 C o and the other containing ice at 0 C o. In the second case, the rods are joined end to end and connected to the same vessels. Let q1 and q2 be the rate of meting of ice in two cases respectively. The ratio of q1q2 is x. Find value of x.

Initially the rods are in parallelQt=(θ1−θ2)R⇒ (Qt)1=μAτ =q1L=(100−0)R/2 −−−(1)Finally when rods are inseries⇒ ( Qt)2=mLt=q2L=(100−0)2R −−−−(2)From Eqs. (1) and (2) q1q2=41

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Two identical conducting rods are first connected parallel to two vessels, one containing water at $100 {}^{o}C$ and the other containing ice at $0 {}^{o}C$ . In the second case, the rods are joined end to end and connected to the same vessels. Let $q_{1} a n d q_{2}$ be the rate of meting of ice in two cases respectively. The ratio of $\frac{q_{1}}{q_{2}}$ is x. Find value of x.

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Detailed Solution

Initially the rods are in parallel

$\begin{array}{l} \frac{Q}{t} = \frac{(θ_{1} - θ_{2})}{R} \\ \Rightarrow {(\frac{Q}{t})}_{1} = \frac{μ A}{τ} \\ = q_{1} L = \frac{(100 - 0)}{R / 2} - - - (1) \\ F i n a l l y w h e n r o d s a r e i n s e r i e s \\ \Rightarrow {(\frac{Q}{t})}_{2} = \frac{m L}{t} = q_{2} L = \frac{(100 - 0)}{2 R} - - - - (2) \\ F r o m E q s . (1) a n d (2) \\ \frac{q_{1}}{q_{2}} = \frac{4}{1} \end{array}$