Answer :
### Part (b): Dimensional Analysis
Escape Velocity Formula: [tex]\(\sqrt{\frac{R}{2GM}}\)[/tex]
1. Identify Dimensional Symbols:
- [tex]\(R\)[/tex]: Length (L)
- [tex]\(G\)[/tex]: Gravitational constant (L[tex]\(^3\)[/tex]M[tex]\(^-1\)[/tex]T[tex]\(^-2\)[/tex])
- [tex]\(M\)[/tex]: Mass (M)
2. Dimensional Formula Analysis:
- The formula for escape velocity is
[tex]\[ \sqrt{\frac{R}{2GM}} \][/tex]
- Dimensions of [tex]\(R\)[/tex] (L)
- Dimensions of [tex]\(2GM\)[/tex] (L[tex]\(^3\)[/tex]M[tex]\(^-1\)[/tex]T[tex]\(^-2\)[/tex]) * M = L[tex]\(^3\)[/tex]T[tex]\(^-2\)[/tex]
- Thus,
[tex]\[ \frac{R}{2GM} = \frac{[L]}{[L^3M^{-1}T^{-2}][M]} = \frac{[L]}{[L^3T^{-2}]} = [L^{-2}T^2] \][/tex]
- Taking the square root:
[tex]\[ \sqrt{\frac{R}{2GM}} = [L^{-2}T^2]^{1/2} = [L^{-1}T] \][/tex]
- But for velocity, dimensions should be:
[tex]\[ [LT^{-1}] \][/tex]
Therefore, the dimensional analysis shows that the given formula [tex]\(\sqrt{\frac{R}{2GM}}\)[/tex] is incorrect for representing velocity. The formula's derived dimensions [tex]\([L^{-1}T]\)[/tex] do not match the dimensions of velocity [tex]\([LT^{-1}]\)[/tex].
Time Period Formula: [tex]\(T= 2\pi \sqrt{\frac{M}{K}}\)[/tex]
1. Identify Dimensional Symbols:
- [tex]\(T\)[/tex]: Time (T)
- [tex]\(M\)[/tex]: Mass (M)
- [tex]\(K\)[/tex]: Force per unit displacement (ML[tex]\(^1\)[/tex]T[tex]\(^-2\)[/tex])
2. Dimensional Formula Analysis:
- The given formula for time period is
[tex]\[ 2\pi \sqrt{\frac{M}{K}} \][/tex]
- Dimensions of [tex]\(M\)[/tex] (M)
- Dimensions of [tex]\(K\)[/tex] (ML[tex]\(^1\)[/tex]T[tex]\(^-2\)[/tex])
- Thus,
[tex]\[ \frac{M}{K} = \frac{[M]}{[ML^{-1}T^{-2}]} = [L^1T^2] \][/tex]
- Taking the square root:
[tex]\[ \sqrt{\frac{M}{K}} = [L^1T^2]^{1/2} = [L^{1/2}T] \][/tex]
Hence, the formula
[tex]\[ 2\pi \sqrt{\frac{M}{K}} \][/tex]
has the correct dimension [tex]\([T]\)[/tex], confirming that the time period formula is dimensionally correct.
### Part (c): Measurement of the Rectangular Lamina
Given:
- Length [tex]\(L = 2.3 \, \text{cm} \pm 0.2 \, \text{cm}\)[/tex]
- Breadth [tex]\(B = 1.6 \, \text{cm} \pm 0.1 \, \text{cm}\)[/tex]
1. Area Calculation:
[tex]\[ \text{Area} = L \times B = 2.3 \, \text{cm} \times 1.6 \, \text{cm} = 3.68 \, \text{cm}^2 \][/tex]
2. Uncertainty in Area:
Using the formula for propagation of uncertainties for multiplication:
[tex]\[ \frac{\Delta A}{A} = \sqrt{\left(\frac{\Delta L}{L}\right)^2 + \left(\frac{\Delta B}{B}\right)^2} \][/tex]
Where:
- [tex]\(\Delta L = 0.2 \, \text{cm}\)[/tex]
- [tex]\(L = 2.3 \, \text{cm}\)[/tex]
- [tex]\(\Delta B = 0.1 \, \text{cm}\)[/tex]
- [tex]\(B = 1.6 \, \text{cm}\)[/tex]
Calculate:
[tex]\[ \frac{\Delta A}{3.68} = \sqrt{\left(\frac{0.2}{2.3}\right)^2 + \left(\frac{0.1}{1.6}\right)^2} \][/tex]
[tex]\[ \frac{\Delta A}{3.68} \approx \sqrt{(0.087)^2 + (0.0625)^2} \][/tex]
[tex]\[ \frac{\Delta A}{3.68} \approx \sqrt{0.007569 + 0.00390625} \approx \sqrt{0.01147525} \approx 0.107 \][/tex]
So,
[tex]\[ \Delta A \approx 3.68 \times 0.107 \approx 0.39476 \approx 0.39 \, \text{cm}^2 \][/tex]
Therefore, the area with uncertainty is:
[tex]\[ \text{Area} = 3.68 \, \text{cm}^2 \pm 0.39 \, \text{cm}^2 \][/tex]
To summarize:
- The formula for escape velocity [tex]\(\sqrt{\frac{R}{2GM}}\)[/tex] is dimensionally incorrect.
- The formula for time period [tex]\(T = 2\pi \sqrt{\frac{M}{K}}\)[/tex] is dimensionally correct.
- The area of the rectangular lamina is [tex]\(3.68 \, \text{cm}^2\)[/tex] with an uncertainty of [tex]\(\pm 0.39 \, \text{cm}^2\)[/tex].
Escape Velocity Formula: [tex]\(\sqrt{\frac{R}{2GM}}\)[/tex]
1. Identify Dimensional Symbols:
- [tex]\(R\)[/tex]: Length (L)
- [tex]\(G\)[/tex]: Gravitational constant (L[tex]\(^3\)[/tex]M[tex]\(^-1\)[/tex]T[tex]\(^-2\)[/tex])
- [tex]\(M\)[/tex]: Mass (M)
2. Dimensional Formula Analysis:
- The formula for escape velocity is
[tex]\[ \sqrt{\frac{R}{2GM}} \][/tex]
- Dimensions of [tex]\(R\)[/tex] (L)
- Dimensions of [tex]\(2GM\)[/tex] (L[tex]\(^3\)[/tex]M[tex]\(^-1\)[/tex]T[tex]\(^-2\)[/tex]) * M = L[tex]\(^3\)[/tex]T[tex]\(^-2\)[/tex]
- Thus,
[tex]\[ \frac{R}{2GM} = \frac{[L]}{[L^3M^{-1}T^{-2}][M]} = \frac{[L]}{[L^3T^{-2}]} = [L^{-2}T^2] \][/tex]
- Taking the square root:
[tex]\[ \sqrt{\frac{R}{2GM}} = [L^{-2}T^2]^{1/2} = [L^{-1}T] \][/tex]
- But for velocity, dimensions should be:
[tex]\[ [LT^{-1}] \][/tex]
Therefore, the dimensional analysis shows that the given formula [tex]\(\sqrt{\frac{R}{2GM}}\)[/tex] is incorrect for representing velocity. The formula's derived dimensions [tex]\([L^{-1}T]\)[/tex] do not match the dimensions of velocity [tex]\([LT^{-1}]\)[/tex].
Time Period Formula: [tex]\(T= 2\pi \sqrt{\frac{M}{K}}\)[/tex]
1. Identify Dimensional Symbols:
- [tex]\(T\)[/tex]: Time (T)
- [tex]\(M\)[/tex]: Mass (M)
- [tex]\(K\)[/tex]: Force per unit displacement (ML[tex]\(^1\)[/tex]T[tex]\(^-2\)[/tex])
2. Dimensional Formula Analysis:
- The given formula for time period is
[tex]\[ 2\pi \sqrt{\frac{M}{K}} \][/tex]
- Dimensions of [tex]\(M\)[/tex] (M)
- Dimensions of [tex]\(K\)[/tex] (ML[tex]\(^1\)[/tex]T[tex]\(^-2\)[/tex])
- Thus,
[tex]\[ \frac{M}{K} = \frac{[M]}{[ML^{-1}T^{-2}]} = [L^1T^2] \][/tex]
- Taking the square root:
[tex]\[ \sqrt{\frac{M}{K}} = [L^1T^2]^{1/2} = [L^{1/2}T] \][/tex]
Hence, the formula
[tex]\[ 2\pi \sqrt{\frac{M}{K}} \][/tex]
has the correct dimension [tex]\([T]\)[/tex], confirming that the time period formula is dimensionally correct.
### Part (c): Measurement of the Rectangular Lamina
Given:
- Length [tex]\(L = 2.3 \, \text{cm} \pm 0.2 \, \text{cm}\)[/tex]
- Breadth [tex]\(B = 1.6 \, \text{cm} \pm 0.1 \, \text{cm}\)[/tex]
1. Area Calculation:
[tex]\[ \text{Area} = L \times B = 2.3 \, \text{cm} \times 1.6 \, \text{cm} = 3.68 \, \text{cm}^2 \][/tex]
2. Uncertainty in Area:
Using the formula for propagation of uncertainties for multiplication:
[tex]\[ \frac{\Delta A}{A} = \sqrt{\left(\frac{\Delta L}{L}\right)^2 + \left(\frac{\Delta B}{B}\right)^2} \][/tex]
Where:
- [tex]\(\Delta L = 0.2 \, \text{cm}\)[/tex]
- [tex]\(L = 2.3 \, \text{cm}\)[/tex]
- [tex]\(\Delta B = 0.1 \, \text{cm}\)[/tex]
- [tex]\(B = 1.6 \, \text{cm}\)[/tex]
Calculate:
[tex]\[ \frac{\Delta A}{3.68} = \sqrt{\left(\frac{0.2}{2.3}\right)^2 + \left(\frac{0.1}{1.6}\right)^2} \][/tex]
[tex]\[ \frac{\Delta A}{3.68} \approx \sqrt{(0.087)^2 + (0.0625)^2} \][/tex]
[tex]\[ \frac{\Delta A}{3.68} \approx \sqrt{0.007569 + 0.00390625} \approx \sqrt{0.01147525} \approx 0.107 \][/tex]
So,
[tex]\[ \Delta A \approx 3.68 \times 0.107 \approx 0.39476 \approx 0.39 \, \text{cm}^2 \][/tex]
Therefore, the area with uncertainty is:
[tex]\[ \text{Area} = 3.68 \, \text{cm}^2 \pm 0.39 \, \text{cm}^2 \][/tex]
To summarize:
- The formula for escape velocity [tex]\(\sqrt{\frac{R}{2GM}}\)[/tex] is dimensionally incorrect.
- The formula for time period [tex]\(T = 2\pi \sqrt{\frac{M}{K}}\)[/tex] is dimensionally correct.
- The area of the rectangular lamina is [tex]\(3.68 \, \text{cm}^2\)[/tex] with an uncertainty of [tex]\(\pm 0.39 \, \text{cm}^2\)[/tex].
