If the linear density (mass per unit length) of a rod of length `3 m` is proportional to `x`, where `x`, where `x` is the distance from one end of the rod, the distance of the centre of gravity of the rod from this end is.

Options

A. 2.5 m
B. 1 m
C. 1.5 m
D. 2 m

Explanation

To find the distance of the center of gravity of a rod of length 3 m, where the linear density is proportional to the distance from one end, we can follow these steps:

### Step 1: Define the linear density
We know that the linear density \( \lambda \) (mass per unit length) is proportional to the distance \( x \) from one end of the rod. We can express this as:
\[ \lambda = kx \]
where \( k \) is a constant of proportionality.

### Step 2: Determine the mass of an infinitesimal element
Consider an infinitesimal element of the rod of length \( dx \) at a distance \( x \) from the end. The mass \( dm \) of this element can be expressed as:
\[ dm = \lambda \cdot dx = kx \cdot dx \]

### Step 3: Set up the integrals for the center of mass
The center of mass \( x_{cm} \) of the rod can be calculated using the formula:
\[ x_{cm} = \frac{\int_0^L x \, dm}{\int_0^L dm} \]
where \( L \) is the length of the rod (3 m in this case).

### Step 4: Calculate the numerator
Substituting \( dm \) into the integral for the numerator:
\[ \int_0^L x \, dm = \int_0^L x \cdot (kx \, dx) = k \int_0^L x^2 \, dx \]
Calculating the integral:
\[ \int_0^L x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^L = \frac{L^3}{3} \]
Thus, the numerator becomes:
\[ k \cdot \frac{L^3}{3} \]

### Step 5: Calculate the denominator
Now, we calculate the denominator:
\[ \int_0^L dm = \int_0^L kx \, dx = k \int_0^L x \, dx \]
Calculating this integral:
\[ \int_0^L x \, dx = \left[ \frac{x^2}{2} \right]_0^L = \frac{L^2}{2} \]
Thus, the denominator becomes:
\[ k \cdot \frac{L^2}{2} \]

### Step 6: Substitute into the center of mass formula
Now we can substitute the results from steps 4 and 5 into the center of mass formula:
\[ x_{cm} = \frac{k \cdot \frac{L^3}{3}}{k \cdot \frac{L^2}{2}} = \frac{\frac{L^3}{3}}{\frac{L^2}{2}} = \frac{2L}{3} \]

### Step 7: Calculate the distance from the end
Substituting \( L = 3 \, \text{m} \):
\[ x_{cm} = \frac{2 \cdot 3}{3} = 2 \, \text{m} \]

Thus, the distance of the center of gravity from one end of the rod is **2 meters**.

### Final Answer
The distance of the center of gravity of the rod from the end is **2 meters**.

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