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To determine the dimensions of time [tex]\(T\)[/tex] in terms of [tex]\(G\)[/tex] (gravitational constant), [tex]\(h\)[/tex] (Planck's constant), and [tex]\(c\)[/tex] (speed of light), we need to conduct a dimensional analysis using these fundamental units.
Let's examine the dimensions of each unit:
1. Gravitational constant [tex]\(G\)[/tex]:
[tex]\[ [G] = \frac{L^3}{M T^2} \][/tex]
2. Planck's constant [tex]\(h\)[/tex]:
[tex]\[ [h] = M L^2 T^{-1} \][/tex]
3. Speed of light [tex]\(c\)[/tex]:
[tex]\[ [c] = \frac{L}{T} \][/tex]
We want to express the dimension of time [tex]\([T]\)[/tex] in the form:
[tex]\[ [T] = G^a \cdot h^b \cdot c^c \][/tex]
### Step-by-Step Dimensional Analysis:
1. Expressing Dimensions:
Using the given forms:
[tex]\[ [G^a] = \left( \frac{L^3}{M T^2} \right)^a = \frac{L^{3a}}{M^a T^{2a}} \][/tex]
[tex]\[ [h^b] = (M L^2 T^{-1})^b = M^b L^{2b} T^{-b} \][/tex]
[tex]\[ [c^c] = \left( \frac{L}{T} \right)^c = L^c T^{-c} \][/tex]
2. Combining Dimensions:
Multiply these expressions together to form:
[tex]\[ [G^a h^b c^c] = \frac{L^{3a} M^b L^{2b} L^c}{M^a T^{2a} T^b T^c} = \frac{L^{3a + 2b + c} M^b}{M^a T^{2a + b + c}} \][/tex]
Simplifying the combined expressions,
[tex]\[ [G^a h^b c^c] = \frac{L^{3a + 2b + c} M^{b - a}}{T^{2a + b + c}} \][/tex]
3. Equating with Time (on both sides):
Since we want this to represent the dimension of time [tex]\([T]\)[/tex], equate powers of [tex]\(L\)[/tex], [tex]\(M\)[/tex], and [tex]\(T\)[/tex] to match [tex]\(T\)[/tex] which is simply:
[tex]\[ [T] = T^1 \][/tex]
Thus, we establish the following equations by matching the exponents of [tex]\(L\)[/tex], [tex]\(M\)[/tex], and [tex]\(T\)[/tex]:
- For [tex]\(L\)[/tex] (length):
[tex]\[ 3a + 2b + c = 0 \][/tex]
- For [tex]\(M\)[/tex] (mass):
[tex]\[ b - a = 0 \quad \text{or}\quad b = a \][/tex]
- For [tex]\(T\)[/tex] (time):
[tex]\[ -2a - b - c = 1 \][/tex]
4. Solving the System of Equations:
Substitute [tex]\(b = a\)[/tex] into the first and third equations:
[tex]\[ 3a + 2a + c = 0 \Rightarrow 5a + c = 0 \Rightarrow c = -5a \][/tex]
[tex]\[ -2a - a - (-5a) = 1 \Rightarrow -3a + 5a = 1 \Rightarrow 2a = 1 \Rightarrow a = \frac{1}{2} \][/tex]
Since [tex]\(b = a\)[/tex]:
[tex]\[ b = \frac{1}{2} \][/tex]
And from [tex]\(c = -5a\)[/tex]:
[tex]\[ c = -5 \left(\frac{1}{2}\right) = -\frac{5}{2} \][/tex]
Thus, the dimensions of time [tex]\( [T] \)[/tex] in terms of [tex]\(G\)[/tex], [tex]\(h\)[/tex], and [tex]\(c\)[/tex] are:
[tex]\[ [T] = G^{\frac{1}{2}} \cdot h^{\frac{1}{2}} \cdot c^{-\frac{5}{2}} \][/tex]
The correct answer is:
[tex]\[ \boxed{a} \][/tex]
Let's examine the dimensions of each unit:
1. Gravitational constant [tex]\(G\)[/tex]:
[tex]\[ [G] = \frac{L^3}{M T^2} \][/tex]
2. Planck's constant [tex]\(h\)[/tex]:
[tex]\[ [h] = M L^2 T^{-1} \][/tex]
3. Speed of light [tex]\(c\)[/tex]:
[tex]\[ [c] = \frac{L}{T} \][/tex]
We want to express the dimension of time [tex]\([T]\)[/tex] in the form:
[tex]\[ [T] = G^a \cdot h^b \cdot c^c \][/tex]
### Step-by-Step Dimensional Analysis:
1. Expressing Dimensions:
Using the given forms:
[tex]\[ [G^a] = \left( \frac{L^3}{M T^2} \right)^a = \frac{L^{3a}}{M^a T^{2a}} \][/tex]
[tex]\[ [h^b] = (M L^2 T^{-1})^b = M^b L^{2b} T^{-b} \][/tex]
[tex]\[ [c^c] = \left( \frac{L}{T} \right)^c = L^c T^{-c} \][/tex]
2. Combining Dimensions:
Multiply these expressions together to form:
[tex]\[ [G^a h^b c^c] = \frac{L^{3a} M^b L^{2b} L^c}{M^a T^{2a} T^b T^c} = \frac{L^{3a + 2b + c} M^b}{M^a T^{2a + b + c}} \][/tex]
Simplifying the combined expressions,
[tex]\[ [G^a h^b c^c] = \frac{L^{3a + 2b + c} M^{b - a}}{T^{2a + b + c}} \][/tex]
3. Equating with Time (on both sides):
Since we want this to represent the dimension of time [tex]\([T]\)[/tex], equate powers of [tex]\(L\)[/tex], [tex]\(M\)[/tex], and [tex]\(T\)[/tex] to match [tex]\(T\)[/tex] which is simply:
[tex]\[ [T] = T^1 \][/tex]
Thus, we establish the following equations by matching the exponents of [tex]\(L\)[/tex], [tex]\(M\)[/tex], and [tex]\(T\)[/tex]:
- For [tex]\(L\)[/tex] (length):
[tex]\[ 3a + 2b + c = 0 \][/tex]
- For [tex]\(M\)[/tex] (mass):
[tex]\[ b - a = 0 \quad \text{or}\quad b = a \][/tex]
- For [tex]\(T\)[/tex] (time):
[tex]\[ -2a - b - c = 1 \][/tex]
4. Solving the System of Equations:
Substitute [tex]\(b = a\)[/tex] into the first and third equations:
[tex]\[ 3a + 2a + c = 0 \Rightarrow 5a + c = 0 \Rightarrow c = -5a \][/tex]
[tex]\[ -2a - a - (-5a) = 1 \Rightarrow -3a + 5a = 1 \Rightarrow 2a = 1 \Rightarrow a = \frac{1}{2} \][/tex]
Since [tex]\(b = a\)[/tex]:
[tex]\[ b = \frac{1}{2} \][/tex]
And from [tex]\(c = -5a\)[/tex]:
[tex]\[ c = -5 \left(\frac{1}{2}\right) = -\frac{5}{2} \][/tex]
Thus, the dimensions of time [tex]\( [T] \)[/tex] in terms of [tex]\(G\)[/tex], [tex]\(h\)[/tex], and [tex]\(c\)[/tex] are:
[tex]\[ [T] = G^{\frac{1}{2}} \cdot h^{\frac{1}{2}} \cdot c^{-\frac{5}{2}} \][/tex]
The correct answer is:
[tex]\[ \boxed{a} \][/tex]
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