Problem 37 - Master Resource Book in JEE Main Physics

Question

The dimensions of electrical conductivity are (a) \(\left[\mathrm{ML}^{3} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]\) (b) \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{3} \mathrm{~A}^{2}\right]\) (c) \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{3} \mathrm{~A}^{2}\right]\) (d) \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{-3} \mathrm{~A}^{2}\right]\)

Step-by-Step Solution

Answer

Option (b) and (c) are both correct, having dimensions \([M^{-1}L^{-3}T^{3}A^{2}]\).

Step 1: Understand Electrical Conductivity

Electrical conductivity (\(\sigma\)) is a measure of a material's ability to conduct electric current. It's generally represented as \(\sigma = \frac{1}{\rho}\), where \(\rho\) is resistivity.

Step 2: Determine Dimensions of Resistivity

The dimensions of resistivity \(\rho\) derive from the formula \[\rho = \frac{E}{J}\], where \(E\) is the electric field with dimensions \([MLT^{-3}A^{-1}]\) and \(J\) is the current density with dimensions \([L^{-2}A]\).

Step 3: Calculate the Dimensions of Conductivity

Considering conductivity as the inverse of resistivity, \[\sigma = \frac{1}{\rho} = \frac{J}{E}\]. Replace \(J\) with its dimension \([L^{-2}A]\) and \(E\) with \([MLT^{-3}A^{-1}]\).

Step 4: Simplify the Dimensions

Plug in the dimensions: \[\sigma = \frac{[L^{-2}A]}{[MLT^{-3}A^{-1}]} = [M^{-1}L^{-3}T^{3}A^{2}]\].

Step 5: Match the Final Result

Compare the result \([M^{-1}L^{-3}T^{3}A^{2}]\) with the given options. It matches option (b) and (c).

Key Concepts

Electrical Conductivity

Electrical conductivity is a fundamental property that indicates how well a material can allow the flow of electric current. When we think of conductivity, we often consider metals, as they generally have high conductivity values. This property is essential in applications ranging from electrical wiring to electronics. The greater the conductivity, the better the material is at transmitting electricity. The symbol for electrical conductivity is \( \sigma \) and it is inversely related to resistivity, \( \rho \). Thus, \( \sigma = \frac{1}{\rho} \). The dimensional formula derived for electrical conductivity is \([M^{-1}L^{-3}T^{3}A^{2}]\). This detail is crucial, because it helps us understand the underlying physical principles behind how materials transmit electricity.

Resistivity

Resistivity is a property of a material that indicates how much it opposes the flow of electric current. It is the inverse of electrical conductivity and is represented by the Greek letter \( \rho \). Mathematically, resistivity is expressed in terms of the electric field \( E \) and current density \( J \), where \( \rho = \frac{E}{J} \).

• The dimensions of electric field \( E \) are \([MLT^{-3}A^{-1}]\).
• The dimensions of current density \( J \) are \([L^{-2}A]\).

Resistivity helps us understand how materials resist the passage of electric current, making it a crucial concept in physics and engineering. Low resistivity materials like metals have high conductivity, while insulators like rubber have high resistivity, making them poor conductors.

Electric Field

The electric field is a vector field around charged particles that represents the force experienced by other charges in its vicinity. It is a fundamental concept in electromagnetism and plays a crucial role in electric circuits, fields, and forces.

• Electric fields are measured in newtons per coulomb (N/C) or volts per meter (V/m).
• The dimensional formula for electric field \( E \) is \([MLT^{-3}A^{-1}]\).

Electric fields are responsible for the force exerted between charges, and they influence how current flows in conductive materials. Understanding electric fields helps us harness electricity in various technologies, from capacitors to AC and DC circuits.

Current Density

Current density is a measure of electric current per unit area of cross-section. It is a vector quantity and provides insight into how current flows through a conductor. This concept is particularly useful in understanding how efficiently electricity moves through different materials.

• Represented by \( J \), it is calculated as the current \( I \) over the area \( A \) as \( J = \frac{I}{A} \).
• The dimensional formula for current density is \([L^{-2}A]\).

Current density is a key factor in electrical applications, showing where and how electricity is distributed across conductive bodies. It helps in designing circuits and improving the performance of electronic devices by ensuring the efficient flow of electrons.

Physics JEE Main

The JEE Main physics examination covers a wide range of topics, with dimensional analysis being one of them. Dimensional analysis is a technique used to check the correctness of physical equations and to derive relationships between different physical quantities. This method is invaluable for solving complex problems and confirming the consistency of units used in equations.

• Dimensional analysis simplifies understanding and solving physics problems by ensuring units align correctly.
• It is particularly useful in examinations like JEE Main, where quick thinking and verification of solutions can lead to time efficiency and accuracy.

Mastering this technique enables JEE candidates to confidently tackle questions related to various physics concepts, including those involving electrical conductivity, resistivity, electric fields, and current density.