Problem 41 - Master Resource Book in JEE Main Physics

Question

Two rods of same length and material transfer a given amount of heat in \(12 \mathrm{~s}\), when they are joined end to end (i.e., in series). But when they are joined in parallel, they will transfer same heat under same conditions in (a) 245 (b) \(3 \underline{5}\) (c) \(48 \mathrm{~s}\) (d) \(1.5 \mathrm{~s}\)

Step-by-Step Solution

Answer

Correct duration in parallel arrangement is approximately 3 seconds.

Step 1: Understanding Series and Parallel Arrangements

When rods are in series, the effective length doubles while the cross-sectional area remains the same. When rods are in parallel, the effective length is the same as one rod, but the effective cross-sectional area is doubled. This affects the rate of heat transfer.

Step 2: Applying Heat Transfer Formula for Series

The formula for the rate of heat transfer through a rod is given by \( Q/t = kA\Delta T/L \), where \( k \) is the thermal conductivity, \( A \) is the cross-sectional area, \( \Delta T \) is the temperature difference, and \( L \) is the length. For series configuration, \( L = 2L_0 \) and \( A = A_0 \).

Step 3: Rate of Heat Transfer in Series Configuration

For series, the equation becomes \( Q/t = kA_0\Delta T / 2L_0 \). Given that it transfers in 12 seconds, we have \( Q/12 = kA_0\Delta T / 2L_0 \).

Step 4: Applying Heat Transfer Formula for Parallel

For parallel configuration, the length \( L = L_0 \) remains the same, and the area doubles \( A = 2A_0 \). Substitute these into the heat transfer formula: \( Q/t' = k(2A_0)\Delta T / L_0 \).

Step 5: Rate of Heat Transfer in Parallel Configuration

The equation becomes \( Q/t' = 2kA_0\Delta T / L_0 \), which is four times the rate in series, so \( t' = 12/4 = 3 \).

Step 6: Comparing Results with Options

Upon comparing the calculated time for parallel configuration, 3 seconds, with the given options, none of them exactly match due to a misprint. Based on reasoning, option \( \underline{35} \) could be chosen as a potential misrepresentation of the correct answer such as something close to 3 seconds.

Key Concepts

Heat Transfer

When dealing with heat transfer, one must understand the factors influencing it. Heat transfer is the process by which thermal energy moves from a region of high temperature to a region of lower temperature. This transfer can occur through mechanisms like conduction, convection, and radiation. In this context, conduction is involved, which is the transfer of heat through a material without the movement of the material itself.

The formula used for conduction is critical: \[\frac{Q}{t} = \frac{kA\Delta T}{L}\]where:

• \(Q\) is the amount of heat transferred,
• \(t\) is the time taken for the transfer,
• \(k\) is the thermal conductivity of the material,
• \(A\) is the cross-sectional area perpendicular to the heat flow direction,
• \(\Delta T\) is the temperature difference across the material, and
• \(L\) is the length of the path through which heat flows.

In series and parallel configurations, the arrangement of materials affects exactly how heat moves through them, as will be discussed in-depth for the circuit analogies.

Series and Parallel Circuits

Understanding the configuration of series and parallel circuits provides a useful analogy for comprehending heat transfer through rods. In electrical terms:

• Series circuits have components arranged along a single path, so all current flows through each component sequentially. Here, the resistance increases with the length that the current must travel.
• Parallel circuits have components arranged in multiple paths, allowing currents to travel through multiple branches simultaneously. The total resistance decreases because the current has multiple pathways.

Similarly, when heat flows through rods arranged in series, it must travel a greater distance — effectively doubling the length, which reduces the rate of heat transfer as it deals with greater equivalent resistance. In contrast, arranging rods in parallel creates a scenario where heat has multiple paths to travel through (larger cross-sectional area), thus reducing time for the same amount of heat to be transferred.

This explanation aligns heat conduction with the familiar rules of electrical circuits but applies them to thermal resistance, which varies based on physical configuration.

Physics Problem Solving

Physics problem-solving often involves breaking down complex scenarios into manageable parts. In the exercise at hand, this involves understanding how different configurations (series vs. parallel) affect the rate of thermal conductivity.

To methodically tackle such problems:

• Carefully examine the problem statement and identify which physical laws or principles apply. Here, the principle of thermal conduction is key.
• Apply relevant equations systematically, substituting known values and isolating unknowns. Remember, the configuration changes the denominator of the heat transfer equation due to alterations in effective length and area.
• Evaluate your results with respect to expected patterns. For instance, intuitively a parallel setup should provide a quicker heat transfer due to reduced resistance, analogous to reduced electrical resistance in parallel circuits.
• When results seemingly don’t match provided options, considering possible misprints or logical reasoning based on close approximations might be needed, which was considered in resolving the given exercise.

By practicing this structured approach, complex physics problems can be broken down into simpler ones, making them more approachable and solvable with confidence.