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

Step 1: Write the heat needed to melt the ice.
Both containers hold the same mass \( m \) of ice, and ice always needs the same latent heat \( L \) per gram to melt. So the total heat needed is the same in both:
\[ Q = mL \]

Step 2: Write the heat conducted through each container wall.
Heat flows into the ice through the container wall at a steady rate set by Fourier's law:
\[ Q = \frac{kA\,\Delta T}{d}\,t \]
Here \( k \) is the conductivity, \( A \) the wall area, \( d \) the wall thickness, \( \Delta T \) the temperature difference driving the heat flow, and \( t \) the time taken. Since P and Q are identical in size and shape, \( A \), \( d \), and \( \Delta T \) are the same for both.

Step 3: Apply this to P and Q separately.
For container P, melting takes time \( t_1 \):
\[ mL = \frac{k_P A\,\Delta T}{d}\,t_1 \]
For container Q, melting takes time \( t_2 \):
\[ mL = \frac{k_Q A\,\Delta T}{d}\,t_2 \]

Step 4: Divide the two equations.
Since the left side (\( mL \)) is the same in both equations, set the right sides equal to each other:
\[ k_P A\,\Delta T\,t_1 = k_Q A\,\Delta T\,t_2 \]
The area and temperature difference cancel out, leaving:
\[ k_P t_1 = k_Q t_2 \]

Step 5: Final Answer.
\[ \frac{k_P}{k_Q} = \frac{t_2}{t_1} \]
The ratio of conductivities of P and Q is \( \dfrac{t_2}{t_1} \).

Answer 2

A useful way to think about this problem is in terms of thermal resistance, the property that tells you how much a wall resists heat flow, similar to electrical resistance resisting current.

1. \( \frac{t_2}{t_1} \): The thermal resistance of a wall is \( R = \frac{d}{kA} \), so a lower conductivity \( k \) means a higher resistance and slower heat flow. A container with higher resistance takes longer to pass the same amount of heat through, so its melting time is longer. Since resistance is inversely proportional to conductivity, and melting time is directly proportional to resistance, melting time is inversely proportional to conductivity. That chain of proportionality lands exactly on \( \frac{k_P}{k_Q} = \frac{t_2}{t_1} \).
2. \( \sqrt{\frac{t_1}{t_2}} \): Thermal resistance and melting time both scale linearly with each other in this steady-flow setup, with nothing squared or rooted in the chain. A square root would need some other physical process, like heat spreading through a growing volume over time, which is not part of this problem.
3. \( \frac{t_2}{t_1^2} \): This would mean container P's melting time enters the ratio twice as strongly as container Q's, which has no basis here since both containers follow the exact same resistance-to-time relationship.
4. \( \frac{t_2}{t_2^2} \): Simplified, this is just \( \frac{1}{t_2} \) and drops \( t_1 \) out entirely, so it cannot represent a ratio that is supposed to compare P against Q.

Since the same heat has to pass through both walls, and the container with lower resistance (higher conductivity) lets that heat through fastest, the container with the shorter melting time is the better conductor. Working the resistance relationship through gives \( \frac{k_P}{k_Q} = \frac{t_2}{t_1} \).

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