The centre of mass of a non uniform rod of length L, whose mass per unit length varies as `rho=(k.x^2)/(L)` where k is a constant and x is the distance of any point from one end is (from the same end)

Options

A. (a) `(3/4)L`
B. (b) `(1/4)L`
C. (c) `(1/6)L`
D. (d) `(2/3)L`

Explanation

To find the center of mass of a non-uniform rod of length \( L \) with a mass per unit length varying as \( \rho = \frac{kx^2}{L} \), we can follow these steps:

### Step 1: Define the mass element
The mass element \( dm \) can be expressed in terms of the linear density \( \rho \) and an infinitesimal length \( dx \):
\[ dm = \rho \, dx = \frac{kx^2}{L} \, dx \]

### Step 2: Set up the integral for the center of mass
The center of mass \( x_{cm} \) is given by the formula:
\[ x_{cm} = \frac{\int_0^L x \, dm}{\int_0^L dm} \]
Substituting \( dm \) into the equation:
\[ x_{cm} = \frac{\int_0^L x \left(\frac{kx^2}{L}\right) dx}{\int_0^L \left(\frac{kx^2}{L}\right) dx} \]

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

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

### Step 5: Combine the results
Now we can substitute the results back into the expression for \( x_{cm} \):
\[ x_{cm} = \frac{\frac{kL^3}{4}}{\frac{kL^2}{3}} = \frac{L^3}{4} \cdot \frac{3}{kL^2} = \frac{3L}{4} \]

### Conclusion
Thus, the center of mass of the non-uniform rod is located at:
\[ x_{cm} = \frac{3L}{4} \]