A 16 m deep well with diameter 3.5 m is dug up and the earth from

Assume the well as a solid right circular cylinder. Then, the radius of the solid right circular cylinder is

$r=\frac{3.5}{2}=1.75 \mathrm{~m}$

The well is 16m deep. Thus, the height of the solid right circular cylinder ism.

Therefore, the volume of the solid right circular cylinder is

$V_{1}=\pi r^{2} h$

$=\frac{22}{7} \times(1.75)^{2} \times 16$ cubic meters

Let the height of the platform formed be x m. The length and the breadth of the platform are l=27.5m and b=7m respectively. Therefore, the volume of the platform is

$V_{2}=l b x=27.5 \times 7 \times x=192.5 x$ cubic meters

Since, the well is spread to form the platform; the volume of the well is equal to the volume of the platform. Hence, we have

$V_{1}=V_{2}$

$\Rightarrow \frac{22}{7} \times(1.75)^{2} \times 16=192.5 x$

$\Rightarrow \quad x=\frac{22}{7 \times 192.5} \times(1.75)^{2} \times 16$

$\Rightarrow \quad=\frac{22 \times 3.0625 \times 16}{7 \times 192.5}$

$\Rightarrow \quad=0.8$

Hence, the height of the platform is 0.8 m = 80 cm.