Question

The dimensions of the area \( A \) of a black hole can be written in terms of the universal constant \( G \), its mass \( M \), and the speed of light \( c \) as \( A = G^\alpha M^\beta c^\gamma \). Here

• A. \( \alpha = -2, \beta = -2, \gamma = 4 \)
• B. \( \alpha = 2, \beta = 2, \gamma = -4 \)
• C. \( \alpha = 3, \beta = 3, \gamma = -2 \)
• D. \( \alpha = -3, \beta = -3, \gamma = 2 \)

Hint:

In dimensional analysis, always express each physical quantity in terms of fundamental dimensions and solve for unknowns by equating powers of \( M \), \( L \), and \( T \).

Answer 1

The Correct Option is B

Step 1: Dimensional analysis.
In dimensional analysis, we express each physical quantity in terms of its fundamental dimensions: mass \( [M] \), length \( [L] \), and time \( [T] \). The dimensions of the variables are: - \( [G] = \frac{M^{-1}L^3}{T^2} \) (gravitational constant) - \( [M] = M \) (mass) - \( [c] = \frac{L}{T} \) (speed of light) We need to find the dimensions of \( A \). Let the dimensions of \( A \) be \( [A] = L^2 \) (since area is in terms of length squared).
Step 2: Set up the equation.
We are given the equation \( A = G^\alpha M^\beta c^\gamma \), and we need to equate the dimensions of both sides: \[ [L^2] = \left( \frac{M^{-1}L^3}{T^2} \right)^\alpha \times (M^\beta) \times \left( \frac{L}{T} \right)^\gamma \] Now, expanding and simplifying the dimensions: \[ [L^2] = M^{-\alpha} L^{3\alpha} T^{-2\alpha} \times M^\beta \times L^\gamma T^{-\gamma} \] Simplifying the powers of \( M \), \( L \), and \( T \): \[ [L^2] = M^{-\alpha + \beta} L^{3\alpha + \gamma} T^{-2\alpha - \gamma} \]
Step 3: Compare powers.
Equating the powers of \( M \), \( L \), and \( T \) on both sides: - For \( M \), \( -\alpha + \beta = 0 \) - For \( L \), \( 3\alpha + \gamma = 2 \) - For \( T \), \( -2\alpha - \gamma = 0 \)
Step 4: Solve the system of equations.
From the equations: - \( -\alpha + \beta = 0 \Rightarrow \beta = \alpha \) - \( -2\alpha - \gamma = 0 \Rightarrow \gamma = -2\alpha \) - \( 3\alpha + \gamma = 2 \Rightarrow 3\alpha - 2\alpha = 2 \Rightarrow \alpha = 2 \) Thus, \( \alpha = 2 \), \( \beta = 2 \), and \( \gamma = -4 \).
Step 5: Conclusion.
Therefore, the correct answer is (2) \( \alpha = 2, \beta = 2, \gamma = -4 \).

Answer 2

We need to find which combination of exponents \( \alpha, \beta, \gamma \) makes the equation \( A = G^{\alpha} M^{\beta} c^{\gamma} \) dimensionally consistent, where \( A \) is an area. Instead of solving the exponent equations directly, we can substitute each option into the dimensional equation and check which one balances. The dimensions involved are \( [G] = M^{-1}L^{3}T^{-2} \), \( [M] = M \), \( [c] = LT^{-1} \), and \( [A] = L^{2} \), so we need

\[ M^{-\alpha+\beta} L^{3\alpha+\gamma} T^{-2\alpha-\gamma} = M^{0}L^{2}T^{0} \]

1. Option (1) \( \alpha=-2,\beta=-2,\gamma=4 \): The mass exponent is \( -(-2)+(-2) = 2-2 = 0 \), which matches. But the length exponent is \( 3(-2)+4 = -6+4 = -2 \), which should be \( 2 \). Since \( -2 \neq 2 \), this option fails.
2. Option (2) \( \alpha=2,\beta=2,\gamma=-4 \): The mass exponent is \( -2+2 = 0 \), correct. The length exponent is \( 3(2)+(-4) = 6-4 = 2 \), correct. The time exponent is \( -2(2)-(-4) = -4+4 = 0 \), correct. All three exponents match, so this option is dimensionally consistent.
3. Option (3) \( \alpha=3,\beta=3,\gamma=-2 \): The mass exponent is \( -3+3 = 0 \), which matches, but the length exponent is \( 3(3)+(-2) = 9-2 = 7 \), which does not equal \( 2 \). This option fails.
4. Option (4) \( \alpha=-3,\beta=-3,\gamma=2 \): The mass exponent is \( -(-3)+(-3) = 3-3 = 0 \), which matches, but the length exponent is \( 3(-3)+2 = -9+2 = -7 \), which does not equal \( 2 \). This option fails.

Only option (2) satisfies all three dimensional equations (mass, length and time) simultaneously.

Therefore, the correct answer is \( \alpha = 2, \beta = 2, \gamma = -4 \).