Question

Masses \(m\left(\dfrac{1}{3}\right)^N \dfrac{1}{N}\) are placed at \(x=N\), when \(N=2,3,4,\ldots,\infty\). If the total mass of the system is \(M\), then the centre of mass is

• A. \(\dfrac{1}{6}\dfrac{m}{M}\)
• B. \(\dfrac{1}{5}\dfrac{m}{M}\)
• C. \(\dfrac{1}{3}\dfrac{m}{M}\)
• D. \(\dfrac{1}{2}\dfrac{m}{M}\)

Hint:

For centre of mass problems involving infinite series, simplify the numerator carefully and use the geometric series formula: \[ \sum_{n=0}^{\infty} ar^n=\frac{a}{1-r}, \quad |r|\lt 1 \]

Answer 1

The Correct Option is A

Step 1: Write the expression for centre of mass.
The centre of mass of a discrete system is given by
\[ x_{cm}=\frac{\sum m_i x_i}{\sum m_i} \] Here,
\[ m_i=m\left(\frac13\right)^N\frac1N \] and the position is
\[ x_i=N \]

Step 2: Calculate the numerator \(\sum m_i x_i\).
Substituting the values,
\[ \sum m_i x_i = \sum_{N=2}^{\infty} m\left(\frac13\right)^N\frac1N \times N \] The \(N\) terms cancel, so
\[ \sum m_i x_i = m\sum_{N=2}^{\infty}\left(\frac13\right)^N \] This is a geometric series with first term
\[ a=\left(\frac13\right)^2=\frac19 \] and common ratio
\[ r=\frac13 \] Using the infinite geometric series formula,
\[ S=\frac{a}{1-r} \] \[ S= \frac{\frac19}{1-\frac13} \] \[ = \frac{\frac19}{\frac23} \] \[ = \frac19 \times \frac32 \] \[ = \frac16 \] Therefore,
\[ \sum m_i x_i = \frac{m}{6} \]

Step 3: Write the total mass.
The total mass of the system is given as
\[ M=\sum m_i \]

Step 4: Calculate the centre of mass.
Using the centre of mass formula,
\[ x_{cm} = \frac{\sum m_i x_i}{M} \] \[ x_{cm} = \frac{\frac{m}{6}}{M} \] \[ x_{cm} = \frac{1}{6}\frac{m}{M} \]

Step 5: Final conclusion.
Hence, the centre of mass is
\[ \boxed{\frac{1}{6}\frac{m}{M}} \]