If mass is written as \( m = k c^P G^{-1/2} h^{1/2} \), then the value of \( P \) will be:
If mass is written as \( m = k c^P G^{-1/2} h^{1/2} \), then the value of \( P \) will be:
\( \frac{1}{3} \)
\( \frac{1}{2} \)
2
\( -\frac{1}{3} \)
The Correct Option is B
Approach Solution - 1
The given equation is:
\[ m = k c^P G^{-1/2} h^{1/2}, \]
where \( k \) is a dimensionless constant, \( c \) is the speed of light (\([c] = [L][T]^{-1}\)), \( G \) is the gravitational constant (\([G] = [M]^{-1}[L]^3[T]^{-2}\)), \( h \) is Planck's constant (\([h] = [M][L]^2[T]^{-1}\)).
The dimensions of mass are:
\[ [m] = [M]. \]
The dimensions of each term in the equation are:
\[ [c^P] = ([L][T]^{-1})^P = [L]^P[T]^{-P}, \]
\[ [G^{-1/2}] = ([M]^{-1}[L]^3[T]^{-2})^{-1/2} = [M]^{1/2}[L]^{-3/2}[T], \]
\[ [h^{1/2}] = ([M][L]^2[T]^{-1})^{1/2} = [M]^{1/2}[L][T]^{-1/2}. \]
Substituting the dimensions into the equation:
\[ [M] = k \cdot [L]^P[T]^{-P} \cdot [M]^{1/2}[L]^{-3/2}[T]^{-1} \cdot [M]^{1/2}[L][T]^{-1/2}. \]
Combine the dimensions of each term:
\[ [M] = [M]^{1/2 + 1/2}[L]^{P - 3/2 + 1}[T]^{-P + 1 - 1/2}. \]
Equating powers of each dimension:
For mass \([M]\):
\[ 1 = \frac{1}{2} + \frac{1}{2}. \]
For length \([L]\):
\[ 0 = P - \frac{3}{2} + 1. \]
Simplify:
\[ P = \frac{1}{2}. \]
For time \([T]\):
\[ 0 = -P + 1 - \frac{1}{2}. \]
Simplify:
\[ P = \frac{1}{2}. \]
Therefore, the value of \( P \) is:
\[ \frac{1}{2} \]
Approach Solution -2
This problem requires us to find the value of the exponent \(P\) in a given physical equation by using the principle of dimensional analysis. The equation relates mass to fundamental constants: the speed of light (\(c\)), the gravitational constant (\(G\)), and Planck's constant (\(h\)).
Concept Used:
The principle of homogeneity of dimensions states that for a physical equation to be dimensionally correct, the dimensions of all the terms on both sides of the equation must be the same. We will write down the dimensional formula for each quantity and then equate the powers of the fundamental dimensions (Mass [M], Length [L], Time [T]) on both sides of the equation to solve for the unknown exponent \(P\).
Step-by-Step Solution:
Step 1: List the dimensional formulas for all the physical quantities involved.
- Mass (\(m\)): \( [M^1 L^0 T^0] \)
- Speed of light (\(c\)): This is a velocity, so its dimension is \( [L T^{-1}] \).
- Gravitational constant (\(G\)): From Newton's law of gravitation, \( F = \frac{G m_1 m_2}{r^2} \), we get \( G = \frac{F r^2}{m_1 m_2} \). The dimensional formula is: \[ [G] = \frac{[M L T^{-2}] [L^2]}{[M][M]} = [M^{-1} L^3 T^{-2}] \]
- Planck's constant (\(h\)): From the energy of a photon, \( E = hf \), we get \( h = \frac{E}{f} \). The dimensional formula is: \[ [h] = \frac{[M L^2 T^{-2}]}{[T^{-1}]} = [M L^2 T^{-1}] \]
- \(k\) is a dimensionless constant, so it has no dimensions.
Step 2: Write the dimensional equation for the given formula \( m = k c^P G^{-1/2} h^{1/2} \).
According to the principle of homogeneity:
\[ [m] = [k] [c]^P [G]^{-1/2} [h]^{1/2} \]Step 3: Substitute the dimensional formulas of each quantity into the equation.
\[ [M^1 L^0 T^0] = (1) \cdot [L T^{-1}]^P \cdot [M^{-1} L^3 T^{-2}]^{-1/2} \cdot [M L^2 T^{-1}]^{1/2} \]Step 4: Simplify the right-hand side of the equation by applying the exponents to each dimension.
\[ [M^1 L^0 T^0] = [L^P T^{-P}] \cdot [M^{(-1)(-1/2)} L^{(3)(-1/2)} T^{(-2)(-1/2)}] \cdot [M^{1/2} L^{(2)(1/2)} T^{(-1)(1/2)}] \] \[ [M^1 L^0 T^0] = [L^P T^{-P}] \cdot [M^{1/2} L^{-3/2} T^{1}] \cdot [M^{1/2} L^{1} T^{-1/2}] \]Step 5: Combine the powers of M, L, and T on the right-hand side.
\[ [M^1 L^0 T^0] = [M^{(1/2 + 1/2)}] \cdot [L^{(P - 3/2 + 1)}] \cdot [T^{(-P + 1 - 1/2)}] \] \[ [M^1 L^0 T^0] = [M^1] \cdot [L^{(P - 1/2)}] \cdot [T^{(-P + 1/2)}] \]Final Computation & Result:
Step 6: Equate the powers of the corresponding dimensions (M, L, T) on both sides of the equation.
- For M: \( 1 = 1 \) (This confirms the consistency of the equation for mass).
- For L: \( 0 = P - \frac{1}{2} \)
- For T: \( 0 = -P + \frac{1}{2} \)
Solving the equation for L:
\[ P = \frac{1}{2} \]Solving the equation for T also gives:
\[ P = \frac{1}{2} \]Both results are consistent. Thus, the value of \(P\) is \(\frac{1}{2}\).
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Concepts Used:
Gravitational Potential Energy
The work which a body needs to do, against the force of gravity, in order to bring that body into a particular space is called Gravitational potential energy. The stored is the result of the gravitational attraction of the Earth for the object. The GPE of the massive ball of a demolition machine depends on two variables - the mass of the ball and the height to which it is raised. There is a direct relation between GPE and the mass of an object. More massive objects have greater GPE. Also, there is a direct relation between GPE and the height of an object. The higher that an object is elevated, the greater the GPE. The relationship is expressed in the following manner:
PEgrav = mass x g x height
PEgrav = m x g x h
Where,
m is the mass of the object,
h is the height of the object
g is the gravitational field strength (9.8 N/kg on Earth) - sometimes referred to as the acceleration of gravity.