Step 1

We are required to find the $\overline y $ of the centroid C of the beam.

 
Step 2

The diagram can be represented as below:

 
Step 3

To locate the centroid $\overline y $ we will use the relation,

\[\overline y = \frac{{{a_1}{y_1} + {a_2}{y_2} + {a_3}{y_3}}}{{{a_1} + {a_2} + {a_3}}}\]….. (1)

Here, ${a_1}$ represents the area of the rectangle 1, ${a_2}$ represents the area of the rectangle 2,${a_3}$ represents the area of the rectangle 3, ${y_1}$ represents the distance of center of gravity of rectangle 1 from the top, ${y_2}$ represents the distance of center of gravity of rectangle 2 from the bottom and ${y_3}$ represents the distance of center of gravity of rectangle 3 from the bottom.

 
Step 4

To find the center of gravity of rectangle 1, we will use the relation,

\[\begin{array}{c} {y_1} = \frac{{{b_1}}}{2}\\ = \frac{{15\;{\rm{mm}}}}{2}\\ = {\rm{7}}{\rm{.5 mm}} \end{array}\]
 
Step 5

To find the center of gravity of rectangle 2, we will use the relation,

\[\begin{array}{c} {y_2} = {b_1} + \frac{{{l_2}}}{2}\\ = 15{\rm{ mm}} + \frac{{150\;{\rm{mm}}}}{2}\\ = 15{\rm{ mm}} + 75\,{\rm{mm}}\\ = {\rm{90 mm}} \end{array}\]
 
Step 6

To find the center of gravity of rectangle 3, we will use the relation,

\[\begin{array}{c} {y_3} = {b_1} + {l_2} + \frac{{{b_3}}}{2}\\ = 15{\rm{ mm}} + 150{\rm{ mm}} + \frac{{15{\rm{ mm}}}}{2}\\ = 15{\rm{ mm}} + 150{\rm{ mm}} + 7.5{\rm{ mm}}\\ = 172.5{\rm{ mm}} \end{array}\]
 
Step 7

To find the area of rectangle 1, we will use the relation.

\[{a_1} = {l_1} \times {b_1}\]

Here, ${l_1}$ and ${b_1}$ represents the length and breadth of the rectangle 1.

 
Step 8

On plugging the values in the above relation, we get,

\[\begin{array}{c} {a_1} = 150{\rm{ mm}} \times 15{\rm{ mm}}\\ = {\rm{2250}}\;{\rm{m}}{{\rm{m}}^2} \end{array}\]
 
Step 9

To find the area of rectangle 2, we will use the relation.

\[{a_2} = {l_2} \times {b_2}\]

Here, ${l_2}$ and ${b_2}$ represents the length and breadth of the rectangle 2.

 
Step 10

On plugging the values in the above relation, we get,

\[\begin{array}{c} {a_2} = 150{\rm{ mm}} \times 15{\rm{ mm}}\\ = {\rm{2250}}\;{\rm{m}}{{\rm{m}}^2} \end{array}\]
 
Step 11

To find the area of rectangle 3, we will use the relation.

\[{a_3} = {l_3} \times {b_3}\]

Here, ${l_3}$ and ${b_3}$ represents the length and breadth of the rectangle 3.

 
Step 12

On plugging the values in the above relation, we get,

\[\begin{array}{c} {a_3} = 100{\rm{ mm}} \times 15{\rm{ mm}}\\ = {\rm{1500}}\;{\rm{m}}{{\rm{m}}^2} \end{array}\]
 
Step 13

On plugging the values in the equation (1), we get,

\[\begin{array}{c} \overline y = \frac{{\left( {2250{\rm{ m}}{{\rm{m}}^2} \times 7.5{\rm{ mm}}} \right) + \left( {2250\,{\rm{m}}{{\rm{m}}^2} \times 90{\rm{ mm}}} \right) + \left( {1500\,{\rm{m}}{{\rm{m}}^2} \times 172.5{\rm{ mm}}} \right)}}{{2250{\rm{ m}}{{\rm{m}}^2} + 2250{\rm{ m}}{{\rm{m}}^2} + 1500{\rm{ m}}{{\rm{m}}^2}}}\\ = \frac{{16875{\rm{ m}}{{\rm{m}}^3} + 202500{\rm{ m}}{{\rm{m}}^3} + 258750{\rm{ m}}{{\rm{m}}^3}}}{{6000\,{\rm{m}}{{\rm{m}}^2}}}\\ = \frac{{{\rm{478125 m}}{{\rm{m}}^3}}}{{100800\,{\rm{m}}{{\rm{m}}^2}}}\\ = 79.687{\rm{ mm}} \end{array}\]