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:

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