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In the formula \(X=3 Y Z^{2}, X\) and \(Z\) have the dimensions of capacitance and magnetic induction, respectively. The dimensions of \(Y\) in M.K.S. system are a. \(M^{-3} L^{-2} T^{-2} A^{-4}\) b. \(M L^{-2} \dot{A}\) c. \(M^{-3} L^{-2} T^{8} A^{4}\) d. \(M^{-3} L^{-2} T^{4} A^{4}\)

Short Answer

Expert verified
The dimensions of \(Y\) in the M.K.S system are \(M^{-3} L^{-2} T^{8} A^{4}\) (Option c).

Step by step solution

01

Identify dimensions of given quantities

Capacitance, denoted by \(X\), has dimensions \([C] = M^{-1} L^{-2} T^{4} A^{2}\). Magnetic induction, denoted by \(Z\), has dimensions \([B] = M T^{-2} A^{-1}\).
02

Express given formula in dimension terms

The formula is \(X = 3 Y Z^2\). Substitute \(X\) and \(Z\) with their respective dimensions:\([C] = [Y][B]^2\).
03

Substitute dimensions into equation

Substitute the dimensions for \([C]\) and \([B]\) into \([C] = [Y][B]^2\):\(M^{-1} L^{-2} T^{4} A^{2} = [Y] \cdot (M T^{-2} A^{-1})^{2}\).
04

Expand the power of Z dimensions

Calculating \([B]^2\):\((M T^{-2} A^{-1})^{2} = M^2 T^{-4} A^{-2}\).
05

Solve for dimensions of Y

Substitute \((M^2 T^{-4} A^{-2})\) into the equation:\(M^{-1} L^{-2} T^{4} A^{2} = [Y] M^{2} T^{-4} A^{-2}\).Solve for \([Y]\) dimensions:\([Y] = M^{-1-2} L^{-2} T^{4+4} A^{2+2}\) which simplifies to \(M^{-3} L^{-2} T^{8} A^{4}\).
06

Compare with options given

Among the options: - a. \(M^{-3} L^{-2} T^{-2} A^{-4}\)- b. \(M L^{-2} \dot{A}\)- c. \(M^{-3} L^{-2} T^{8} A^{4}\)- d. \(M^{-3} L^{-2} T^{4} A^{4}\)Option c, \(M^{-3} L^{-2} T^{8} A^{4}\) matches the calculated dimensions for \(Y\).

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Key Concepts

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

Capacitance
Capacitance is an important concept in the field of electromagnetism as it measures a system's ability to store an electric charge. This is often denoted by the letter "C" and mathematically defined by the ratio of the electric charge "Q" stored on a conductor to the potential difference "V" across its plates. The formula for capacitance is given by:\[ C = \frac{Q}{V} \],where:
  • \(C\) is the capacitance, measured in farads (F).
  • \(Q\) is the charge in coulombs (C).
  • \(V\) is the voltage in volts (V).
When dealing with dimensional analysis, the dimensions of capacitance can be expressed using the fundamental M.K.S. units (meter, kilogram, second). Therefore, in the M.K.S. system, the dimensions of capacitance are represented as \[ [C] = M^{-1} L^{-2} T^{4} A^{2} \].This representation helps connect the concept of storing charges to physical dimensions, as it shows the interplay with mass \(M\), length \(L\), time \(T\), and electric current \(A\). Understanding these relations is crucial for solving problems involving electrical components in physics.
Magnetic Induction
Magnetic induction, often symbolized as "B," is closely related to the magnetic field, another fundamental concept in physics. It represents the magnetic force per unit length on a moving charge or current-carrying wire in the presence of a magnetic field. A practical way to visualize this is to think about the magnetic field lines around a magnet, where magnetic induction is a measure of how dense these lines are.The dimensional formula for magnetic induction, when evaluated through the lens of the M.K.S. system, is given by:\[ [B] = M T^{-2} A^{-1} \].This highlights how magnetic induction is inherently connected to the forces acting on moving charges, as it involves mass \(M\) and time \(T\) dimensions similar to those in mechanical forces. The inverse relation with the ampere \(A\) demonstrates the connection to the electrical current, outlining how magnetism and electricity are interlinked.
M.K.S. System
The M.K.S. system, which stands for Meter-Kilogram-Second, is one of the oldest and most unified systems used for expressing physical quantities. It was developed as a pathway to create a standardized measurement system.This system lays the foundation for the modern International System of Units (SI) by focusing on three fundamental physical quantities:
  • Meter (M) for measuring length.
  • Kilogram (K) for measuring mass.
  • Second (S) for measuring time.
It facilitates consistency between disciplines and simplifies the process of dimensional analysis. In dimensional analysis, it allows you to express other physical properties by relating them to length, mass, time, and other derived units.
Understanding these foundational units is crucial for accurately solving problems in physics, such as determining the correct dimensions of a given physical quantity. In our original exercise, we calculated the dimensions of the quantity \(Y\) using these M.K.S. units, leading us to options in the answer choices that relate directly to these fundamental measurements.

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