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Two rods of equal length and diameler but o thermal conductivitics 2 unit and 3 unit respeclivcly are joined in parallel. The tharmal conductivity of a rod of geomelrical dimension cqual to cither of the rod, which can conduct the same heat is (a) 1 unit (b) \(1.5\) unit (c) \(2.5\) unit (d) 5 unil

Short Answer

Expert verified
(c) 2.5 unit

Step by step solution

01

Identify the Given Information

We are given two rods with equal length and diameter, having thermal conductivities of 2 units and 3 units respectively. These rods are joined in parallel.
02

Understanding Parallel Conductivity

When two rods are placed in parallel, the effective thermal conductivity can be found by calculating the average of their conductivities. This is analogous to the parallel configuration of resistances in electrical circuits.
03

Calculate Effective Thermal Conductivity

For rods in parallel configuration, the effective thermal conductivity, \( k_{eff} \), is the average of the two given conductivities: \[ k_{eff} = \frac{k_1 + k_2}{2} = \frac{2 + 3}{2} \]
04

Perform the Calculation

Substitute the values into the formula to get \[ k_{eff} = \frac{2 + 3}{2} = \frac{5}{2} = 2.5 \]
05

Conclude the Solution

The thermal conductivity of a rod, geometrically similar to either of the original rods that can conduct the same heat, is 2.5 units.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Conduction
Heat conduction is the process of heat transfer through a material without any movement of the material itself. It occurs at a molecular level when energetic molecules transfer their energy to neighboring less energetic molecules. In solid rods, heat conduction is the primary mode of heat transfer. Think of it as a relay race where runners pass the baton of energy from one to another.
The rate of this energy transfer depends on the material's thermal conductivity, which measures how easily heat can pass through the material. Materials with high thermal conductivity, like metals, are excellent conductors of heat, while materials with low thermal conductivity, like wood or rubber, are insulators.
Understanding heat conduction helps in determining how materials will behave in different thermal environments, offering insights into their suitability for various applications such as insulation, electronics, and construction.
Parallel Thermal Conductivity
When two or more materials with different thermal conductivities are placed together, understanding how they conduct heat when configured in parallel is crucial. In a parallel configuration, each material has the same temperature difference across its ends, and this setup affects the overall heat conduction characteristics.
The effective thermal conductivity of a parallel arrangement of two rods can be thought of like resistors in parallel in electrical circuits. The total conductivity is not a simple sum; instead, it's the average of their individual conductivities. This is because both rods share the same temperature gradient and can balance each other's heat flow capacity.
  • Use the formula for average thermal conductivity: \[ k_{eff} = \frac{k_1 + k_2}{2} \]
For example, if the rods have conductivities of 2 and 3 units, their effective thermal conductivity becomes \(2.5\) units, facilitating the same heat conduction as a single rod with these properties.
Physics Problem Solving
Solving physics problems like parallel thermal conductivity requires a systematic approach that allows students to dissect the problem effectively.
Begin by identifying what is given in the problem, such as dimensions, material properties, and any specific conditions (like parallel configuration). Next, understand the relationships between these variables using the appropriate physics concepts.
In this case, employ the concept of averaging in a parallel setup to find effective conductivity. Always keep the bigger picture in mind. Verify whether your solution logically fits within the provided parameters. This strategic approach not only simplifies problem-solving but also enhances the understanding of core concepts.
  • Break down the problem into smaller parts.
  • Utilize formulas effectively.
  • Double-check calculations to ensure accuracy.
JEE Main and Advanced Physics Concepts
The Joint Entrance Examination (JEE) Main and Advanced exams for engineering in India include many concepts from physics that require a deep understanding and practical application.
Understanding principles like thermal conductivity is essential, as questions often involve applying them in complex scenarios, such as composite rods or parallel thermal setups. Success in these exams depends on not just rote memorization but a thorough understanding of underlying physics concepts that can be applied flexibly.
Being familiar with different configurations and the mathematics behind them, such as the formula for effective thermal conductivity, provides a strong foundation. Such knowledge is crucial for solving multi-step physics problems under time constraints. Reviewing past papers and practicing problem-solving enhances both speed and accuracy, which are key to performing well in these exams.

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Most popular questions from this chapter

Statement-1 : If \(a, r\) and \(t\) are absorptive, rellective and transmitting powers of body then \(a+r+t=1\) Statement-2: For \(t \rightarrow 1\) the body will be diathermanous.

\(\Lambda\) student takes a calorimeter of mass \(50 \mathrm{~g}\), made of copper of specific heat \(0.1\) cal \(\mathrm{g}^{1} \mathrm{C}^{\circ} \mathrm{}^{\circ}\) IIc puts \(100 \mathrm{~g}\) of water al \(40^{\circ} \mathrm{C}\) into il. IIc drops \(100 \mathrm{~g}\) of ice at \(-20^{\circ} \mathrm{C}\) into it. Specific heat of ice is \(0.5 \mathrm{cal} \mathrm{g}^{1} \mathrm{C}^{\circ} 1 .\) IIc then passes steam so that water starts boiling. Latent heat of ice \(80 \mathrm{cal} / \mathrm{g}\) and of water \(540 \mathrm{cal} / \mathrm{g}\). Temperature of water after adding ice into water: (a) \(0^{\circ} \mathrm{C}\) (b) \(5^{\circ} \mathrm{C}\) (c) \(10^{\circ} \mathrm{C}\). (d) \(20^{\circ} \mathrm{C}\)

Three rods cach of same length and cross section are joined in scrics. The thermal conductivity of the materials are \(k, 2 k\) and \(3 k\) respectively. If one end is kepl al \(200^{\circ} \mathrm{C}\) and the other at \(100^{\circ} \mathrm{C}\). What would be the temperature of the junctions in the steady state? \Lambdassume that no heat is lost duc to radiation from the sides of the rods.

Ice at \(0^{\circ} \mathrm{C}\) is added to \(200 \mathrm{~g}\) of water initially at \(70^{\circ} \mathrm{C}\) in a vacuum flask. When \(50 \mathrm{~g}\) of icc has been added and has all melted the tempcrature of the flask and contents is \(40{ }^{\circ} \mathrm{C}\). When a further \(80 \mathrm{~g}\) of ice has been added and has all melted the temperature of the whole becomes \(10^{\circ} \mathrm{C}\). Find the latent heat of fusion of ice.

Statement-1: For steady state during heat conduction across a rod of uniform cross-section temperature gradient is same for all points. Statement-2: Thermal current decreases as we move across rod from higher temperature to lower temperature.

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