Problem 65 - Master Resource Book in JEE Main Physics

Question

If \(E, m, J\) and \(G\) represent energy, mass, angular momentum and gravitational constant respectively, then the dimensional formula of \(E J^{2} / m^{5} G^{2}\) is (a) \(\left[\mathrm{MLT}^{-2}\right]\) (b) \(\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}\right]\) (c) \(\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]\) (d) dimensionless

Step-by-Step Solution

Answer

(d) dimensionless.

Step 1: Understand Dimensional Formula of Energy

Energy (E) has the dimensional formula \([ML^2T^{-2}]\). This comes from the basic formula for kinetic energy, which is \(\frac{1}{2}mv^2\), where \(m\) is mass and \(v\) is velocity (LT^{-1}). Therefore, the dimensional formula is \(ML^2T^{-2}\).

Step 2: Determine Dimensional Formula for Mass

Mass (m) is a fundamental quantity with a dimensional formula of \([M]\). There are no additional calculations needed for the mass dimension.

Step 3: Identify Dimensional Formula for Angular Momentum

Angular momentum (J) has the dimensional formula \([ML^2T^{-1}]\). This can be derived from the formula for angular momentum, which is momentum (\(mv\)) times radius (\(r\)), where \(v\) is velocity. Thus, \(J\) is expressed as \(mvr\), resulting in the dimensional formula \(ML^2T^{-1}\).

Step 4: Calculate Dimensional Formula for Gravitational Constant

The gravitational constant (G) has the dimensional formula \([M^{-1}L^3T^{-2}]\). This can be derived from Newton's law of universal gravitation which states \(F = G\frac{m_1m_2}{r^2}\), where \(F\) has the dimensions of force \([MLT^{-2}]\). Rearranging for \(G\), we find \([M^{-1}L^3T^{-2}]\).

Step 5: Combine Dimensions

The expression \(EJ^{2}/(m^5G^2)\) requires combining the dimensional formulas calculated in the previous steps:- \(E\) is \(ML^2T^{-2}\)- \(J^2\) is \((ML^2T^{-1})^2 = M^2L^4T^{-2}\)- \(m^5\) is \(M^5\)- \(G^2\) is \((M^{-1}L^3T^{-2})^2 = M^{-2}L^6T^{-4}\)Substituting these into the expression gives:\[(ML^2T^{-2})(M^2L^4T^{-2})/(M^5(M^{-2}L^6T^{-4}))\]

Step 6: Simplify Combined Dimensions

Simplify the combined dimensional expression:- Numerator: \(M^{1+2}L^{2+4}T^{-2-2} = M^3L^6T^{-4}\)- Denominator: \(M^{5-2}L^6T^{-4}\)- Combine numerator and denominator: \[\frac{M^3L^6T^{-4}}{M^3L^6T^{-4}} = M^0L^0T^0\]which is dimensionless.

Key Concepts

Dimensional Formula

The dimensional formula is a representation of physical quantities in terms of their fundamental dimensions: mass (M), length (L), and time (T). By expressing the dimensions of a quantity, we can ensure consistency in equations, no matter the unit system. This is useful when we need to verify equations or when dealing with dimensional analysis problems.

Key points for understanding dimensional formulas are:

• Each physical quantity has a unique dimensional formula, usually expressed as a combination of \(M, L,\) and \(T\).
• Dimensional analysis helps us to derive and check formulas by ensuring dimensional consistency.
• It simplifies the process of understanding the relationships between different physical quantities.

In our exercise, dimensional formulas were used to verify the dimensionless nature of the expression \( \frac{EJ^2}{m^5G^2} \). Understanding and applying these principles allows us to solve such problems confidently.

Energy

Energy, a fundamental concept in physics, is a measure of the ability of a system to perform work. Its dimensional formula is derived from the relationship found in kinetic energy, expressed mathematically as \( \frac{1}{2}mv^2 \). Here, \( m \) is mass with dimension \([M]\), and \( v \) is velocity, \([LT^{-1}]\). Thus, the dimensional formula for energy becomes \([ML^2T^{-2}]\).

Here's why understanding energy's dimensional formula is critical:

• It aids in analyzing various forms of energy, such as potential and kinetic.
• Ensures correct transformations between different energy types.
• Helps to maintain consistency when performing dimensional analysis in complex problems.

When energy plays a part in a formula, knowing its dimensional formula helps you check the balance and accuracy of physical equations.

Angular Momentum

Angular momentum, an extension of linear momentum for rotational systems, is a critical quantity in physics that describes the rotational state of an object. The dimensional formula for angular momentum is \([ML^2T^{-1}]\). It comes from the definition \( J = mvr \), where \( m \) is mass, \( v \) is velocity, and \( r \) is radius.

Understanding angular momentum through its dimensional formula is important because:

• It helps predict the transfer of momentum in closed systems.
• Allows the analysis of rotational movements in mechanical systems.
• Ensures balance in equations involving rotational forces and dynamics.

In the exercise, angular momentum squared \( J^2 = (ML^2T^{-1})^2 \) is used, emphasizing how dimensional analysis simplifies complex expressions by allowing us to easily handle powers of dimensional quantities.

Gravitational Constant

The gravitational constant, denoted as \( G \), is a crucial element in understanding gravitational interactions in physics. It appears in Newton's law of universal gravitation, an important formula that allows us to calculate the force of attraction between two masses. The dimensional formula for \( G \) is \([M^{-1}L^3T^{-2}]\), derived from rearranging the law equation \( F = G\frac{m_1m_2}{r^2} \), with force \( F \) having the dimensions \([MLT^{-2}]\).

Here's why knowing the dimensional formula of the gravitational constant is beneficial:

• It is vital for understanding the scale and strength of gravitational forces.
• Dimensional analysis with \( G \) helps in confirming the correctness of gravitational calculations.
• It plays an essential role in cosmology and astronomy for determining the motion of planets and stars.

In our given problem, \( G^2 \) appears, and understanding its dimensional formula is key in simplifying and resolving the expression \( \frac{EJ^2}{m^5G^2} \).