Problem 51 - Master Resource Book in JEE Main Physics

Question

The ratio of thermal conductivity of two rods is \(5: 4\). The ratio of their cross-sectional areas is \(1: 1\) and they have the same thermal resistances. The ratio of their lengths, must will be (a) \(4: 5\) (b) \(9: 1\) (c) \(1: 9\) (d) \(5: 4\)

Step-by-Step Solution

Answer

The ratio of their lengths is \( 5:4 \) (option d).

Step 1: Understanding Thermal Resistance Formula

Thermal resistance \( R \) is given by the formula \( R = \frac{L}{kA} \), where \( L \) is the length of the rod, \( k \) is thermal conductivity, and \( A \) is the cross-sectional area. The problem states that the rods have the same thermal resistance.

Step 2: Setting Up the Equation for Equal Resistance

Since the thermal resistances are the same for both rods, we equate their resistances: \( \frac{L_1}{k_1 A_1} = \frac{L_2}{k_2 A_2} \). Given that the cross-sectional areas are equal \((A_1 = A_2)\), the equation simplifies to \( \frac{L_1}{k_1} = \frac{L_2}{k_2} \).

Step 3: Substituting Given Ratios

We know the ratio of thermal conductivities is \( k_1: k_2 = 5:4 \). Substitute this into the equation: \( \frac{L_1}{5} = \frac{L_2}{4} \).

Step 4: Solving for Length Ratio

Cross multiply to solve for the ratio: \( 4L_1 = 5L_2 \). Therefore, the ratio of the lengths \( L_1: L_2 \) is \( 5:4 \).

Step 5: Identifying the Correct Answer

The ratio of the lengths \( 5:4 \) matches option (d) in the given choices.

Key Concepts

Thermal Conductivity

Thermal conductivity is a material's ability to conduct heat. The higher the thermal conductivity, the better the material is at transferring heat. In the context of the exercise, we are given a ratio of the thermal conductivities of two rods as \(5:4\). This means that one rod is better at conducting heat than the other by this proportion.

Thermal conductivity is a crucial material property when analyzing heat transfer problems, as it directly influences how quickly heat can move through a material. The symbol \(k\) is often used to represent thermal conductivity in equations and formulas such as

• \( R = \frac{L}{kA} \) where \(R\) represents thermal resistance,
• \(L\) is the length of the material, and
• \(A\) is the cross-sectional area.

Understanding this concept is vital when dealing with problems involving multiple rods or materials, as different materials will conduct heat differently depending on their thermal conductive properties.

Cross-Sectional Area

The cross-sectional area of a material is essentially the size of the surface that is perpendicular to the direction of heat flow. It plays a significant role in determining the rate of heat transfer. In the context of this exercise, the cross-sectional area ratio is given as \(1:1\), which means both rods have the same cross-sectional area.

The symbol \(A\) typically represents the cross-sectional area in equations. It appears in the formula for thermal resistance, such as

• \( R = \frac{L}{kA} \),

where a larger cross-sectional area will result in a lower thermal resistance and thus a greater heat transfer rate.

Students must grasp the role of cross-sectional area in thermal scenarios because it can affect how quickly or slowly heat is conducted through a rod. A larger area means more space for heat to pass through, thus increasing the efficiency of heat transfer.

Lengths of Rods

The length of a rod is the distance over which heat must travel to get from one end to the other. In this exercise, the lengths of the rods determine the thermal resistance, along with thermal conductivity and cross-sectional area. The formula for thermal resistance \( R = \frac{L}{kA} \) shows how critical the length \(L\) is in calculating resistance.

From the problem, we're asked to find the ratio of the lengths of the rods assuming equal thermal resistances and given thermal conductivities. The step-by-step solution provides a clear path:

• With thermal resistance equal, \( \frac{L_1}{k_1} = \frac{L_2}{k_2} \).
• Using the ratio \( k_1:k_2 = 5:4 \), we derive \( \frac{L_1}{5} = \frac{L_2}{4} \).

From here, we solve to find that the length ratio \( L_1:L_2 \) is \(5:4\). Understanding these calculations is pivotal for solving practical problems involving thermal resistance and length determinations.