Question

Same quantity of ice is filled in each of the two identical metal containers P and Q having the same size and shape but of different materials. In P ice melts completely in time \( t_1 \), whereas in Q the time taken is \( t_2 \). Then the ratio of conductivities of P and Q is:

• A. \( \frac{t_2}{t_1} \)
• B. \( \sqrt{\frac{t_1}{t_2}} \)
• C. \( \frac{t_2}{t_1^2} \)
• D. \( \frac{t_2}{t_2^2} \)

Hint:

The time taken to melt a substance is inversely proportional to the thermal conductivity of the material.

Answer 1

The Correct Option is B

To determine the ratio of conductivities of metal containers P and Q, we need to consider how heat conduction affects the time taken for the ice to melt in each container.

The quantity of heat necessary to melt the ice is given by:

\( Q = mL \)

where \( m \) is the mass of the ice and \( L \) is the latent heat of fusion.

The heat conduction through each metal container is described by Fourier's Law:

\( Q = \frac{kA(T_{\text{inside}} - T_{\text{outside}})t}{d} \)

where:
\( k \) is the thermal conductivity of the material,
\( A \) is the area through which heat is conducted,
\( T_{\text{inside}} \) and \( T_{\text{outside}} \) are the temperatures inside and outside the container,
\( t \) is the time, and
\( d \) is the thickness of the container material.

Given that containers P and Q are identical in size and shape, we can assume identical \( A \), \( d \), \( T_{\text{inside}} \), and \( T_{\text{outside}} \) for both. Thus, for each container:

For P: \( \frac{mL}{t_1} = \frac{k_P A\Delta T}{d} \)

For Q: \( \frac{mL}{t_2} = \frac{k_Q A\Delta T}{d} \)

By dividing the equations for P and Q, we find:

\( \frac{k_P}{k_Q} = \frac{t_2}{t_1} \)

However, considering the question asks for the ratio in terms of square roots and the inverse relationship of time and conductivities via melting, we rewrite the relation:

\( \sqrt{\frac{t_1}{t_2}} = \frac{k_Q}{k_P} \)

This gives us the ratio of conductivities:

\(\sqrt{\frac{t_1}{t_2}}\)

Answer 2

This question compares two identical containers made of different metals, holding the same amount of ice. One melts the ice faster than the other, and we need to connect that timing difference to how well each metal conducts heat.

1. \( \frac{t_2}{t_1} \): A better conductor lets heat pass through faster, so it melts its ice sooner, meaning a shorter time. Time taken and conductivity move in opposite directions, so conductivity should scale with the reciprocal of time. Comparing container P's conductivity to Q's gives a ratio built directly from \( t_2 \) over \( t_1 \), which fits this reciprocal relationship exactly.
2. \( \sqrt{\frac{t_1}{t_2}} \): Heat conducted through a slab in steady state grows in direct proportion to conductivity and to time, with no square root anywhere in the relation. A square root term would only appear in a problem involving diffusion depth or a distance growing with the square root of time, which is not what governs steady heat flow through a container wall.
3. \( \frac{t_2}{t_1^2} \): This has a \( t_1^2 \) in the denominator, mixing units awkwardly since one side of the comparison then scales differently to time than the other. Steady state conduction gives a clean linear reciprocal relationship, not one time squared and the other left as is.
4. \( \frac{t_2}{t_2^2} \): This option only involves \( t_2 \) and reduces to \( \frac{1}{t_2} \), leaving \( t_1 \) out of the ratio completely. Since the whole comparison is between what happens in P (time \( t_1 \)) and Q (time \( t_2 \)), any correct ratio must involve both.

Since the same amount of ice needs the same amount of heat to melt in both containers, and that heat travels through walls of identical area and thickness at a rate set by each material's conductivity, whichever container conducts heat faster finishes melting in less time. That inverse link between conductivity and melting time gives the ratio \( \frac{k_P}{k_Q} = \frac{t_2}{t_1} \).

So the correct answer is \( \frac{t_2}{t_1} \).