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)/(3)`
B. `(l)/(n+1)`
C. `(n(n^(2)+1)l)/(2)`
D. `(2l)/(n(n^(2)+1)`

Answer

To find the distance of the center of mass of the particles from the fixed point, we can follow these steps:

### Step 1: Identify the masses and their positions
The masses of the particles are given as \( m, 2m, 3m, \ldots, nm \) grams. The positions of these masses are at distances \( l, 2l, 3l, \ldots, nl \) cm from the fixed point.

### Step 2: Write down the formula for the center of mass
The formula for the center of mass \( x_{cm} \) of a system of particles is given by:

\[ x_{cm} = \frac{\sum m_i x_i}{\sum m_i} \]

where \( m_i \) is the mass of the \( i^{th} \) particle and \( x_i \) is its position.

### Step 3: Calculate the numerator \( \sum m_i x_i \)
Here, we have:
- Masses: \( m_1 = m, m_2 = 2m, m_3 = 3m, \ldots, m_n = nm \)
- Positions: \( x_1 = l, x_2 = 2l, x_3 = 3l, \ldots, x_n = nl \)

Thus, the numerator becomes:

\[ \sum m_i x_i = m \cdot l + 2m \cdot 2l + 3m \cdot 3l + \ldots + nm \cdot nl \]

This can be expressed as:

\[ = m \left( l + 4l + 9l + \ldots + n^2l \right) \]

Factoring out \( l \):

\[ = ml \left( 1^2 + 2^2 + 3^2 + \ldots + n^2 \right) \]

### Step 4: Use the formula for the sum of squares
The sum of the squares of the first \( n \) natural numbers is given by:

\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]

Thus, we have:

\[ 1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6} \]

Substituting this back into our expression for the numerator:

\[ \sum m_i x_i = ml \cdot \frac{n(n + 1)(2n + 1)}{6} \]

### Step 5: Calculate the denominator \( \sum m_i \)
The total mass \( \sum m_i \) is:

\[ m + 2m + 3m + \ldots + nm = m(1 + 2 + 3 + \ldots + n) = m \cdot \frac{n(n + 1)}{2} \]

### Step 6: Substitute into the center of mass formula
Now substituting the numerator and denominator into the center of mass formula:

\[ x_{cm} = \frac{ml \cdot \frac{n(n + 1)(2n + 1)}{6}}{m \cdot \frac{n(n + 1)}{2}} \]

### Step 7: Simplify the expression
Cancel \( m \) and \( n(n + 1) \):

\[ x_{cm} = \frac{l \cdot \frac{2n + 1}{6}}{\frac{1}{2}} = \frac{l(2n + 1)}{3} \]

### Final Answer
Thus, the distance of the center of mass of the particles from the fixed point is:

\[ \boxed{\frac{l(2n + 1)}{3}} \]