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. `(2 l)/(n(n^2 + 1))`

Explanation

To find the center of mass of the system 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 1: Identify the masses and their positions
The masses of the particles are:
- Mass of first particle = \( m \)
- Mass of second particle = \( 2m \)
- Mass of third particle = \( 3m \)
- ...
- Mass of nth particle = \( nm \)

The positions of the particles are:
- Position of first particle = \( l \)
- Position of second particle = \( 2l \)
- Position of third particle = \( 3l \)
- ...
- Position of nth particle = \( nl \)

### Step 2: Write the formula for the center of mass
The center of mass \( x_{cm} \) of a system of particles is given by the formula:
\[ x_{cm} = \frac{\sum m_i x_i}{\sum m_i} \]
where \( m_i \) is the mass of the ith particle and \( x_i \) is its position.

### Step 3: Calculate the numerator \( \sum m_i x_i \)
We need to calculate:
\[ \sum m_i x_i = m \cdot l + 2m \cdot (2l) + 3m \cdot (3l) + \ldots + nm \cdot (nl) \]
This can be factored out as:
\[ = m \cdot l \left( 1 \cdot 1 + 2 \cdot 2 + 3 \cdot 3 + \ldots + n \cdot n \right) \]
This simplifies to:
\[ = m \cdot l \sum_{i=1}^{n} i^2 \]

### Step 4: Use the formula for the sum of squares
The formula for the sum of the squares of the first n natural numbers is:
\[ \sum_{i=1}^{n} i^2 = \frac{n(n + 1)(2n + 1)}{6} \]
Thus, we can substitute this into our equation:
\[ \sum m_i x_i = m \cdot l \cdot \frac{n(n + 1)(2n + 1)}{6} \]

### Step 5: Calculate the denominator \( \sum m_i \)
Now we calculate the total mass:
\[ \sum m_i = m + 2m + 3m + \ldots + nm = m(1 + 2 + 3 + \ldots + n) \]
Using the formula for the sum of the first n natural numbers:
\[ 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2} \]
Thus, the total mass becomes:
\[ \sum m_i = m \cdot \frac{n(n + 1)}{2} \]

### Step 6: Substitute into the center of mass formula
Now we can substitute the values into the center of mass formula:
\[ x_{cm} = \frac{m \cdot l \cdot \frac{n(n + 1)(2n + 1)}{6}}{m \cdot \frac{n(n + 1)}{2}} \]
The \( m \) cancels out:
\[ x_{cm} = \frac{l \cdot \frac{n(n + 1)(2n + 1)}{6}}{\frac{n(n + 1)}{2}} \]
This simplifies to:
\[ x_{cm} = l \cdot \frac{2n + 1}{3} \]

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