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Answer 1

Question:

Some cubic meters of earth is dug out to sink a well which is 16 m deep and which has a radius of 3.5 m. If that amount of earth when out is spread over a rectangular plot of dimensions 25 m × 16 m, what is the height of the platform so formed?

(a) 1.54 m (b) 1.50 m (c) 1.52 m (d) 1.53 m

Answer:

a) 1.54 m

Step-by-step explanation:

Shape of well is cylindrical in shape. So, we have to find the volume of cylinder.

Volume of cylinder = πr²h

Where r is 3.5 m and h is 16 m.

Substitute the values,

→ Volume of well = 22/7 × 3.5 × 3.5 × 16

→ Volume of well = 616.00

→ Volume of well = 616 m³

The amount of earth when out is spread over a rectangular plot of dimensions 25 m × 16 m. So,

Volume of well = Volume of rectangular plot

(Volume of rectangle = length × breadth × height)

Assume that the height of the rectangular plot is x m.

→ 616 = 25 × 16 × x

→ 616 = 400x

→ 1.54 = x

Hence, the height of the rectangular plot is 1.54 m.

Answer 2

$answer$

Radius of well= 3.5 m

Depth to sink well= 16m

Volume of the earth dug out =

$= \pi {r}^{2} h = \frac{22}{7} \times {(3.5)}^{2} \times 16$

$= \frac{4312}{7} = 616 {m}^{3}$

Volume of the earth piled upon the given plot

=25 ×16×h m^3

Volume of the earth piled upon the given plot=Volume of the earth dug out

So

25×16×h=616

h=616/400

=1.54m.

Therefore Height of the platform formed =1.54m

Option A.