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**Hint:**To solve this problem, we need to use a formula for the effective centre of mass of the system at a given point. In this question we have a system of 3 particles present on 3 corners of a right triangle, so the centre of mass will lie somewhere between them.

**Formula used:**

The centre of mass is given as,

Along X coordinate \[{X_{COM}} = \dfrac{{\sum {{m_i}{x_i}} }}{{\sum {{m_i}} }}\]

Along Y coordinate \[{Y_{COM}} = \dfrac{{\sum {{m_i}{y_i}} }}{{\sum {{m_i}} }}\]

**Complete answer:**

Given (0,0), (3,0) and (0,4) to be the coordinates of mass 1kg, 1.5kg and 2.5kg by assuming 1kg at origin.

Image: Three point particles are placed at three corners of a right triangle.

\[{X_{COM}}\] can be given as,

\[{X_{COM}} = \dfrac{{\sum {{m_i}{x_i}} }}{{\sum {{m_i}} }}\]

\[{X_{COM}} = \dfrac{{1 \times 0 + 1.5 \times 3 + 2.5 \times 0}}{{1 + 1.5 + 2.5}}\]

\[ = \dfrac{{4.5}}{5} = 0.9cm\]

\[{Y_{COM}}\] can be given as,

\[{Y_{COM}} = \dfrac{{\sum {{m_i}{y_i}} }}{{\sum {{m_i}} }}\]

\[{Y_{COM}} = \dfrac{{1 \times 0 + 1.5 \times 0 + 2.5 \times 4}}{{1 + 1.5 + 2.5}}\]

\[ = \dfrac{4}{2} = 2cm\]

Therefore, the centre of mass is 0.9 cm to the right and 2 cm above the mass of 1kg.

**Hence option A is the correct answer**

**Note:**We use the formula for centre of mass for individual coordinates in the x-y-z plane which is only applicable for point objects, for an extended object or non-uniform object like a rod, we need to consider differential mass and its position and integrate over its entire length.

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