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Question

If the speed of light $c$, Planck's constant $h$ and gravitational constant $G$ are chosen as fundamental units, then the dimensions of time in terms of these units are:

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [Option A]
## CONCEPT FRAMEWORK
**Primary Concept**: Dimensional Analysis
**Related Topics**: Fundamental Units, Derived Units
**Prerequisites**: Understanding of dimensional formulas for physical quantities, basic algebra
## SOLUTION PATHWAY
**Given Information**:
    * Fundamental units: Speed of light (c), Planck's constant (h), Gravitational constant (G)
    * Target dimension: Time (T)
**Method & Steps**:
    1.  Assume that time (T) can be expressed as a combination of the fundamental units with unknown powers:  $T = c^x h^y G^z$
        Reasoning: This is the fundamental idea of dimensional analysis, where a derived quantity is expressed as a product of fundamental quantities raised to some powers.
        Formula/Rule: None
    2.  Write down the dimensions of each quantity involved:
        * [c] = $LT^{-1}$
        * [h] = $ML^2T^{-1}$
        * [G] = $M^{-1}L^3T^{-2}$
        * [T] = $T$
        Working: These are the standard dimensional formulas for speed of light, Planck's constant, and gravitational constant.
        Key Point: Correct dimensions are crucial for successful dimensional analysis.
    3.  Substitute the dimensions into the equation from step 1:
        $T = (LT^{-1})^x (ML^2T^{-1})^y (M^{-1}L^3T^{-2})^z$
        $T = L^{x+2y+3z} M^{y-z} T^{-x-y-2z}$
         Working: Expanding and grouping the terms with the same dimensions.
    4.  Equate the powers of M, L, and T on both sides of the equation:
        * For M: $y - z = 0$  =>  $y = z$  (1)
        * For L: $x + 2y + 3z = 0$  (2)
        * For T: $-x - y - 2z = 1$  (3)
        Working: Because the LHS has dimensions of time only, the powers of L and M must be zero.
    5.  Solve the system of linear equations:
        From (1), $y = z$. Substituting in (2): $x + 2z + 3z = 0 => x + 5z = 0 => x = -5z$.  Substituting in (3): $-(-5z) -z -2z = 1 => 5z - 3z = 1 => 2z = 1 => z = 1/2$. Then $y = 1/2$ and $x = -5/2$
        Working: Standard algebraic manipulation to solve for the unknown exponents.
        Key Point: Careful algebra is critical.
    6.  Substitute the values of x, y, and z back into the original equation:
        $T = c^{-5/2} h^{1/2} G^{1/2}$
        Result: The dimensions of time are $c^{-5/2}G^{1/2}h^{1/2}$
        Verification: Substituting these values back into the dimensional equation confirms the equality of dimensions.
 ## OPTION ANALYSIS
[A] ✓: The correct dimensional combination for time is $c^{-5/2}G^{1/2}h^{1/2}$, as derived through dimensional analysis.
[B] :x:  The power of c is incorrect, it should be -5/2, not 5/2. 
[C] :x: The power of c is incorrect, it should be -5/2, not -3/2. 
[D] :x: The power of c is incorrect, it should be -5/2, not 1/2.
 ## LEARNING AIDS
**Common Mistakes**:
    1. [Error 1]: Incorrect dimensional formulas for c, h, or G. [How to avoid]: Memorize or correctly derive these dimensions.
    2. [Error 2]: Mistakes in solving the system of linear equations. [How to avoid]: Double-check your algebra and substitutions.
**Quick Tips**:
    1. [Helpful shortcut]: Always check if your final dimensional formula matches the expected dimensions.
    2. [Memory aid]: Remember the basic dimensional formulas for force, energy, and momentum as they are frequently used.

VENKATESWARARAO · Answer

The correct Option is "$c^{-5/2}G^{1/2}h^{1/2}$"

VENKATESWARARAO · Answer

$c^{5/2}G^{1/2}h^{1/2}$

VENKATESWARARAO · Answer

$c^{-3/2}G^{1/2}h^{1/2}$

VENKATESWARARAO · Answer

$c^{1/2}G^{1/2}h^{1/2}$

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