The maximum height of a mountain on earth is limited by the rock flowing under the enormous weight above it. Studies show that maximum height depends on young’s modulus (Y) of the rod, acceleration due to gravity (g) and the density of the rock (d).
(a) Write an equation showing the dependence of maximum height (h) of mountain on Y, g and d. It is given that unit of Y is `Nm^(-2)`.
(b) Take `d = 3 xx 10^(3) kg m^(-3), Y = 1 xx 10^(10) Nm^(-2) and g = 10 ms^(-2)` and assume that maximum height of a mountain on the surface of earth is limited to 10 km [height of mount Everest is nearly 8 km]. Write the formula for h.

Answer

To solve the problem, we will follow these steps:

### Part (a): Deriving the Equation for Maximum Height (h)

1. **Identify the Variables**: We know that the maximum height \( h \) depends on Young's modulus \( Y \), acceleration due to gravity \( g \), and the density of the rock \( d \).

2. **Assume a Functional Form**: We can express the relationship as:
\[ h = K \cdot Y^A \cdot g^B \cdot d^C \]
where \( K \) is a dimensionless constant and \( A \), \( B \), and \( C \) are the powers to which each variable is raised.

3. **Determine Dimensions**:
- The dimensions of \( h \) (height) are \( [L] \).
- The dimensions of \( Y \) (Young's modulus) are \( [M L^{-1} T^{-2}] \).
- The dimensions of \( g \) (acceleration due to gravity) are \( [L T^{-2}] \).
- The dimensions of \( d \) (density) are \( [M L^{-3}] \).

4. **Set Up the Dimensional Equation**:
The left-hand side (LHS) has dimensions:
\[ [L] = [M^0 L^1 T^0] \]
The right-hand side (RHS) becomes:
\[ [M^A L^{-A} T^{-2A}] \cdot [L^B T^{-2B}] \cdot [M^C L^{-3C}] = [M^{A+C} L^{-A+B-3C} T^{-2A-2B}] \]

5. **Equate Dimensions**:
- For mass: \( A + C = 0 \)
- For length: \( -A + B - 3C = 1 \)
- For time: \( -2A - 2B = 0 \)

6. **Solve the Equations**:
From \( -2A - 2B = 0 \), we get \( B = -A \).
Substituting \( B = -A \) into the second equation:
\[ -A - A - 3C = 1 \implies -2A - 3C = 1 \]
From \( A + C = 0 \), we have \( C = -A \).
Substituting \( C = -A \) into \( -2A - 3(-A) = 1 \):
\[ -2A + 3A = 1 \implies A = 1 \]
Therefore, \( C = -1 \) and \( B = -1 \).

7. **Final Equation**:
Substituting \( A \), \( B \), and \( C \) back into the equation for \( h \):
\[ h = K \cdot \frac{Y}{g \cdot d} \]

### Part (b): Writing the Formula for h with Given Values

1. **Substituting the Given Values**:
Given:
- \( d = 3 \times 10^3 \, \text{kg/m}^3 \)
- \( Y = 1 \times 10^{10} \, \text{Nm}^{-2} \)
- \( g = 10 \, \text{m/s}^2 \)

2. **Calculating h**:
We can express the maximum height \( h \) as:
\[ h = K \cdot \frac{Y}{g \cdot d} \]
We need to find \( K \) using the maximum height of a mountain, which is given as \( 10 \, \text{km} = 10 \times 10^3 \, \text{m} \).

3. **Finding K**:
Rearranging the equation:
\[ K = h \cdot \frac{g \cdot d}{Y} \]
Substituting the values:
\[ K = (10 \times 10^3) \cdot \frac{10 \cdot (3 \times 10^3)}{1 \times 10^{10}} \]
\[ K = 10 \times 10^3 \cdot \frac{30 \times 10^3}{10^{10}} = 10 \cdot 30 \cdot 10^{-4} = 300 \cdot 10^{-4} = 3 \times 10^{-2} \]

4. **Final Formula for h**:
Thus, the formula for \( h \) becomes:
\[ h = 0.03 \cdot \frac{Y}{g \cdot d} \]