A rod is of length `3` m and its mass acting per unit length is directly proportional to distance x from its one end. The centre of gravity of the rod from that end will be at

Options

A. 1.5 m
B. 2 m
C. 2.5 m
D. 3.0 m

Explanation

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

### Step-by-Step Solution:

1. **Define the Problem**:
We have a rod of length \( L = 3 \) m. The mass per unit length \( \rho \) is directly proportional to the distance \( x \) from one end. This can be expressed as:
\[ \rho(x) = kx \]
where \( k \) is a constant of proportionality.

2. **Determine the Mass Element**:
The mass of a small element \( dx \) of the rod at position \( x \) can be expressed as:
\[ dm = \rho(x) \, dx = kx \, dx \]

3. **Set Up the Integral for Center of Mass**:
The center of mass \( x_{cm} \) can be calculated using the formula:
\[ x_{cm} = \frac{\int_0^L x \, dm}{\int_0^L dm} \]
Substituting \( dm \):
\[ x_{cm} = \frac{\int_0^3 x \cdot (kx \, dx)}{\int_0^3 (kx \, dx)} \]

4. **Calculate the Numerator**:
The numerator becomes:
\[ \int_0^3 x \cdot (kx) \, dx = k \int_0^3 x^2 \, dx = k \left[ \frac{x^3}{3} \right]_0^3 = k \cdot \frac{27}{3} = 9k \]

5. **Calculate the Denominator**:
The denominator is:
\[ \int_0^3 (kx) \, dx = k \left[ \frac{x^2}{2} \right]_0^3 = k \cdot \frac{9}{2} = \frac{9k}{2} \]

6. **Substitute Back into the Center of Mass Formula**:
Now substituting the numerator and denominator into the center of mass formula:
\[ x_{cm} = \frac{9k}{\frac{9k}{2}} = 9k \cdot \frac{2}{9k} = 2 \]

7. **Final Result**:
Therefore, the center of gravity of the rod from the end where \( x = 0 \) is:
\[ x_{cm} = 2 \text{ m} \]