Question

CASE STUDY (Rain water harvesting)
A farmer has a rectangular field of length 30 m and breadth 15 m. By the farmer a pit of diameter 7 m is dug 12 m deep for rain water harvesting. The earth taken out is spread in the field. Based on the above information, answer the following questions. (31) Find the volume of the earth taken out.

(A) 460 m3 (B) 462 m3 (C) 465 m3 (D) 468 m3.
(32) The area of the rectangular ficld is

(A) 420 m2 (B) 430 m2 (C) 440 m2 (D) 450 m2
(33) Find the area of the top of the pit.

(A) 38.5 m2 (B) 40.5 m2 (C) 41.5 m2 (D) None of these
(34) The area of the remaining field is

(A) 402.3 m2 (B) 405 m2 (C) 410 m2 (D) 411.5 m2
(35) Find the level rise in the field.

(A) 0.5 m (B) 3 m (C) 1.12 m (D) 2.12 m


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Answer 1

Step-by-step explanation:

Given : CASE STUDY (Rain water harvesting)

A farmer has a rectangular field of length 30 m and breadth 15 m. By the farmer a pit of diameter 7 m is dug 12 m deep for rain water harvesting. The earth taken out is spread in the field.

To find: Based on the above information, answer the following questions.

Solution:

(31) Find the volume of the earth taken out.

(A) 460 m³ (B) 462 m³ (C) 465 m³ (D) 468 m³.

Ans: Find volume of pit.

Pit is having a cylindrical shape.

Diameter: 7 m

Depth : 12 m

Volume of cylinder(V) = $\pi {r}^{2} h$

place the values

$V = \frac{22}{7} \times ( { \frac{7}{2} )}^{2} \times 12$

$V = \frac{22}{7} \times { \frac{7}{2} } \times \frac{7}{2} \times 12$

after cancel common terms

$V = 11 \times 7 \times 6$

or

$\bf V = 462 \: {m}^{3}$

Option b is correct.

(32) The area of the rectangular field is

(A) 420 m² (B) 430 m² (C) 440 m² (D) 450 m²

Ans: Area of Rectangle= length×breadth

Length of field= 30 m

Breadth of field =15 m

$area = 30 \times 15$

$\bf Area = 450 \: {m}^{2}$

Option d is correct.

(33) Find the area of the top of the pit.

(A) 38.5 m² (B) 40.5 m² (C) 41.5 m² (D) None of these

Ans: Area of top of cylinder is a circle.

Thus,

Area of top of pit is a circle of radius 3.5 m

$Area= \pi {r}^{2}$

$area = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$

$\bf Area = 38.5 \: {m}^{2}$

Option A is correct.

(34) The area of the remaining field is

(A) 402.3 m² (B) 405 m² (C) 410 m² (D) 411.5 m²

Sol: To find Area of remaining field; Subtract area of top of pit from area of cylinder.

Area of the remaining field=

$450 - 38.5$

$\bf \text{Area of the remaining field}= 411.5 \: {m}^{2}$

Option d is correct.

(35) Find the level rise in the field.

(A) 0.5 m (B) 3 m (C) 1.12 m (D) 2.12 m

Sol: Volume of earth taken out from dig= Volume of remaining field.

Since, after raising the height of field it is converted into a cuboid.

So,

$462 = 411.5 \times h$

or

$h = \frac{462}{411.5}$

or

$\bf h = 1.12 \: m$

Option C is correct.

Final answer:

31) Option b is correct.

32) Option d is correct.

33) Option a is correct.

34) Option d is correct.

35) Option c is correct.

Hope it helps you.

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