Four point masses P, Q, R and S with respective masses 1 kg, 1 kg, 2 kg and 2 kg form the corners of a square of side a. The centre of mass of the system will be farthest from

Options

A. R and S
B. R only
C. P and Q
D. P and R

Explanation

To solve the problem of finding the center of mass of the four point masses arranged at the corners of a square, we will follow these steps:

### Step 1: Define the coordinates of the masses
Let's place the square in a Cartesian coordinate system. We can define the positions of the masses as follows:
- Mass P (1 kg) at (a/2, a/2)
- Mass Q (1 kg) at (a/2, -a/2)
- Mass R (2 kg) at (-a/2, a/2)
- Mass S (2 kg) at (-a/2, -a/2)

### Step 2: Calculate the total mass
The total mass \( M \) of the system is the sum of all the individual masses:
\[ M = m_P + m_Q + m_R + m_S = 1\, \text{kg} + 1\, \text{kg} + 2\, \text{kg} + 2\, \text{kg} = 6\, \text{kg} \]

### Step 3: Calculate the x-coordinate of the center of mass
The x-coordinate of the center of mass \( x_{cm} \) is given by the formula:
\[ x_{cm} = \frac{1}{M} \sum m_i x_i \]
Substituting the values:
\[ x_{cm} = \frac{1}{6} \left( 1 \cdot \frac{a}{2} + 1 \cdot \frac{a}{2} + 2 \cdot \left(-\frac{a}{2}\right) + 2 \cdot \left(-\frac{a}{2}\right) \right) \]
Calculating this:
\[ x_{cm} = \frac{1}{6} \left( \frac{a}{2} + \frac{a}{2} - a - a \right) = \frac{1}{6} \left( \frac{a}{2} + \frac{a}{2} - 2a \right) = \frac{1}{6} \left( a - 2a \right) = \frac{1}{6} \left( -a \right) = -\frac{a}{6} \]

### Step 4: Calculate the y-coordinate of the center of mass
The y-coordinate of the center of mass \( y_{cm} \) is given by:
\[ y_{cm} = \frac{1}{M} \sum m_i y_i \]
Substituting the values:
\[ y_{cm} = \frac{1}{6} \left( 1 \cdot \frac{a}{2} + 1 \cdot \left(-\frac{a}{2}\right) + 2 \cdot \frac{a}{2} + 2 \cdot \left(-\frac{a}{2}\right) \right) \]
Calculating this:
\[ y_{cm} = \frac{1}{6} \left( \frac{a}{2} - \frac{a}{2} + a - a \right) = \frac{1}{6} \left( 0 \right) = 0 \]

### Step 5: Determine the position of the center of mass
The center of mass is located at:
\[ \left( -\frac{a}{6}, 0 \right) \]
This means the center of mass is on the negative x-axis.

### Step 6: Identify which mass is farthest from the center of mass
Since the center of mass is located at \( \left( -\frac{a}{6}, 0 \right) \), we can analyze the distances from the center of mass to each mass:
- Distance to P: \( \sqrt{ \left( \frac{a}{2} + \frac{a}{6} \right)^2 + \left( \frac{a}{2} \right)^2 } \)
- Distance to Q: \( \sqrt{ \left( \frac{a}{2} + \frac{a}{6} \right)^2 + \left( -\frac{a}{2} \right)^2 } \)
- Distance to R: \( \sqrt{ \left( -\frac{a}{2} + \frac{a}{6} \right)^2 + \left( \frac{a}{2} \right)^2 } \)
- Distance to S: \( \sqrt{ \left( -\frac{a}{2} + \frac{a}{6} \right)^2 + \left( -\frac{a}{2} \right)^2 } \)

After calculating these distances, we find that the distances to P and Q are greater than those to R and S.

### Conclusion
The center of mass is farthest from points P and Q.