Question

A big box at $100{}^\circ C$is connected to an ice cube through identical metal rods as shown. Ice is melting at the rate of ${{Q}_{1}}$gram/sec. Rods are rearranged so that they are connected in series between box and ice cube. The new rate of melting is ${{Q}_{2}}$gram/sec. Then$\dfrac{{{Q}_{2}}}{{{Q}_{1}}}$?

A. 4
B. 2
C. $\dfrac{1}{2}$
D. $\dfrac{1}{4}$

Answer 1

As a first step, one could read the question and thus divide the question into two parts one where the rods are connected as in the given figure, other where the rods are connected in series. Then you could recall the formula accordingly and thus find the answer.

Formula used:
Rate of melting,
$Q=\dfrac{KA\Delta T}{l}$

Complete step-by-step solution:
In the question we are given a big box and an ice cube of temperatures $100{}^\circ C$and $0{}^\circ C$respectively. When the identical rods are connected as shown in the above figure, the rate of melting of ice is given to be ${{Q}_{1}}g/s$and when these rods are connected in series with each other the rate of melting would be ${{Q}_{2}}g/s$. We are supposed to find the ratio of these rates of melting.
For the arrangement as shown in the figure the rate of melting could be given by,
${{Q}_{1}}=\dfrac{K\times 2A\times 100}{l}=\dfrac{200KA}{l}$…………………………………………………. (1)
When the rods are connected in series with each other, the rate of melting would be,
${{Q}_{2}}=\dfrac{K\times A\times 100}{2l}=\dfrac{50KA}{l}$……………………………………………………. (2)
Now on dividing equation (2) by equation (1), we get,
$\dfrac{{{Q}_{2}}}{{{Q}_{1}}}=\dfrac{\dfrac{50KA}{l}}{\dfrac{200KA}{l}}$
$\therefore \dfrac{{{Q}_{2}}}{{{Q}_{1}}}=\dfrac{1}{4}$
Therefore, the required ratio is found to be $\dfrac{1}{4}$. Hence, option D is correct.

Note: In the questions, where we are dealing with the ratios of quantities, one should take care to note which quantity is in the numerator and which is in the denominator. For the given question, if we have taken the ratio wrong, we would have got 4 as the ratio, which also exists in the options.