Particle of masses `m, 2m,3m,…,nm` grams are placed on the same line at distance `l,2l,3l,…..,nl cm` from a fixed point. The distance of centre of mass of the particles from the fixed point in centimeters is :

Options

A. `((2n+1)L)/(4)`
B. `(L)/((2n+1))`
C. `(n(n^(2)+1)L)/(2)`
D. `((2n+1)L)/(3)`

Answer

To find the distance of the center of mass of particles with masses \( m, 2m, 3m, \ldots, nm \) placed at distances \( l, 2l, 3l, \ldots, nl \) from a fixed point, we can follow these steps:

### Step-by-step Solution:

1. **Identify the masses and their positions**:
- The masses are \( m_1 = m, m_2 = 2m, m_3 = 3m, \ldots, m_n = nm \).
- The corresponding positions are \( r_1 = l, r_2 = 2l, r_3 = 3l, \ldots, r_n = nl \).

2. **Use the formula for the center of mass (COM)**:
\[ x_{cm} = \frac{m_1 r_1 + m_2 r_2 + m_3 r_3 + \ldots + m_n r_n}{m_1 + m_2 + m_3 + \ldots + m_n} \]

3. **Calculate the numerator**:
\[ \text{Numerator} = m \cdot l + 2m \cdot 2l + 3m \cdot 3l + \ldots + nm \cdot nl \]
This can be simplified as:
\[ = ml + 4ml + 9ml + \ldots + n^2 ml = m l (1^2 + 2^2 + 3^2 + \ldots + n^2) \]

4. **Use the formula for the sum of squares**:
The sum of the squares of the first \( n \) natural numbers is given by:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]
Therefore, the numerator becomes:
\[ = m l \cdot \frac{n(n+1)(2n+1)}{6} \]

5. **Calculate the denominator**:
\[ \text{Denominator} = m + 2m + 3m + \ldots + nm = m(1 + 2 + 3 + \ldots + n) = m \cdot \frac{n(n+1)}{2} \]
(using the formula for the sum of the first \( n \) natural numbers).

6. **Substituting the numerator and denominator into the COM formula**:
\[ x_{cm} = \frac{m l \cdot \frac{n(n+1)(2n+1)}{6}}{m \cdot \frac{n(n+1)}{2}} \]

7. **Cancel out common terms**:
The \( m \) and \( n(n+1) \) terms cancel out:
\[ x_{cm} = \frac{l \cdot \frac{2n+1}{6}}{\frac{1}{2}} = \frac{l(2n+1)}{3} \]

### Final Answer:
Thus, the distance of the center of mass from the fixed point is:
\[ \boxed{\frac{l(2n+1)}{3}} \]