Mass is non-uniformly distributed over the rod of length l, its linear mass density varies linearly with length as `lamda=kx^(2)`. The position of centre of mass (from lighter end) is given by-

Options

A. `2l//5`
B. `3//l`
C. `3l//4`
D. `2l//3`

Answer

To find the position of the center of mass of a rod with a non-uniform mass distribution, we can follow these steps:

### Step 1: Define the linear mass density
The linear mass density is given as:
\[ \lambda(x) = kx^2 \]
where \( k \) is a constant and \( x \) is the position along the rod.

### Step 2: Express the mass of a small segment
Consider a small segment of the rod of thickness \( dx \) at a distance \( x \) from the lighter end. The mass \( dm \) of this small segment can be expressed as:
\[ dm = \lambda(x) \cdot dx = kx^2 \cdot dx \]

### Step 3: Set up the integral for the center of mass
The center of mass \( x_{cm} \) of the rod can be calculated using the formula:
\[ x_{cm} = \frac{\int_0^L x \, dm}{\int_0^L dm} \]
Substituting \( dm \) into the equation:
\[ x_{cm} = \frac{\int_0^L x \cdot (kx^2) \, dx}{\int_0^L kx^2 \, dx} \]

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

### Step 5: Calculate the denominator
The denominator becomes:
\[ \int_0^L kx^2 \, dx = k \int_0^L x^2 \, dx \]
Calculating the integral:
\[ \int_0^L x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^L = \frac{L^3}{3} \]
Thus, the denominator is:
\[ k \cdot \frac{L^3}{3} \]

### Step 6: Combine the results
Now substituting the results back into the center of mass formula:
\[ x_{cm} = \frac{k \cdot \frac{L^4}{4}}{k \cdot \frac{L^3}{3}} = \frac{\frac{L^4}{4}}{\frac{L^3}{3}} = \frac{L^4 \cdot 3}{4 \cdot L^3} = \frac{3L}{4} \]

### Final Answer
The position of the center of mass from the lighter end is:
\[ x_{cm} = \frac{3L}{4} \]