RS AGGARWAL CLASS 9 CHAPTER 15 VOLUMES AND SURFACE AREA OF SOLIDS EXERCISE 15A

 EXERCISE 15A

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Question 1:

Find the volume, the lateral surface are and the total surface area of the cuboid whose dimensions are:
(i) length = 12 cm, breadth = 8 cm and height = 4.5 cm
(ii) length = 26 m, breadth = 14 m and height = 6.5 m
(iii) length = 15 m, breadth = 6 m and height = 5 dm
(iv) length = 24 m, breadth = 25 cm and height = 6 m

Answer 1:

(i)
Here, l = 12 cm, b = 8 cm, h = 4.5 cm
Volume of the cuboid = $l\times b\times h$
                                  $$\begin{gathered} =(12\times 8\times 4.5) {\mathrm{cm}}^3 \\ = 432 {\mathrm{cm}}^3 \end{gathered}$$

Total Surface area = 2(lb + lh+ bh)
                              $$\begin{gathered} =2(12\times 8 + 12\times 4.5 +8\times 4.5) {\mathrm{cm}}^2 \\ =2(96 +54 + 36) {\mathrm{cm}}^2 \\ =2\times 186 {\mathrm{cm}}^2 \\ =372 \mathrm{cm}{ }^2 \end{gathered}$$

Lateral surface area = $2\left(l+b\right)\times h$
                              $$\begin{gathered} =\left[2\left(12+8\right)\times 4.5\right] {\mathrm{cm}}^2 \\ =\left[2\left(20\right)\times 4.5\right] {\mathrm{cm}}^2 \\ =40\times 4.5 {\mathrm{cm}}^2 \\ =180 {\mathrm{cm}}^2 \end{gathered}$$

(ii)
Here, l = 26 m; b = 14 m; h =6.5 m
Volume of the cuboid =  $l\times b\times h$
                                $$\begin{gathered} =\left(26\times 14\times 6.5\right){\mathrm{m}}^3 \\ =2366 {\mathrm{m}}^3 \end{gathered}$$

Total surface area = 2(lb + lh+ bh)
                             $$\begin{gathered} =2(26\times 14+26\times 6.5+6.5\times 14) {\mathrm{m}}^2 \\ =2(364+169+91) {\mathrm{m}}^2 \\ =2\times 624 {\mathrm{m}}^2 \\ =1248 {\mathrm{m}}^2 \end{gathered}$$

Lateral surface area = $2\left(l+b\right)\times h$
                              $$\begin{gathered} =\left[2\left(26+14\right)\times 6.5\right] {\mathrm{m}}^2 \\ =\left[2\times 40\times 6.5\right] {\mathrm{m}}^2 \\ =520 {\mathrm{m}}^2 \end{gathered}$$

(iii)
Here, l = 15 m; b = 6 m; h = 5 dm = 0.5 m
Volume of the cuboid =  $l\times b\times h$
                                $$\begin{gathered} =(15\times 6\times 0.5) {\mathrm{m}}^3 \\ =45 {\mathrm{m}}^3 \end{gathered}$$


Total surface area = 2(lb + lh+ bh) 
                          $$\begin{gathered} =2(15\times 6+15\times 0.5+6\times 0.5) {\mathrm{m}}^2 \\ =2(90+7.5 +3) {\mathrm{m}}^2 \\ =2\times 100.5 {\mathrm{m}}^2 \\ = 201{\mathrm{m}}^2 \end{gathered}$$

Lateral surface area = $2\left(l+b\right)\times h$
                               $$\begin{gathered} =\left[2\left(15+6\right)\times 0.5\right] {\mathrm{m}}^2 \\ =\left[2\times 21\times 0.5\right] {\mathrm{m}}^2 \\ =21 {\mathrm{m}}^2 \end{gathered}$$

(iv)
Here, l = 24 m; b = 25 cm = 0.25 m; h =6 m
Volume of the cuboid =  $l\times b\times h$
                                $$\begin{gathered} =\left(24\times 0.25\times 6\right) {\mathrm{m}}^3 \\ =36 {\mathrm{m}}^3 \end{gathered}$$

Total Surface area = 2(lb + lh+ bh)
                            $$\begin{gathered} =2(24\times 0.25 +24\times 6 +0.25\times 6) {\mathrm{m}}^2 \\ =2(6+144 + 1.5) {\mathrm{m}}^2 \\ =2\times 151.5 {\mathrm{m}}^2 \\ =303 {\mathrm{m}}^2 \end{gathered}$$

Lateral surface area = $2\left(l+b\right)\times h$
                             $$\begin{gathered} =\left[2\left(24+0.25\right)\times 6\right] {\mathrm{m}}^2 \\ =\left[2\times 24.25\times 6\right] {\mathrm{m}}^2 \\ =291 {\mathrm{m}}^2 \end{gathered}$$

Question 2:

A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What is the volume of a packet containing 12 such matchboxes?

Answer 2:

Volume of each matchbox = 4 × 2.5 × 1.5 = 15 cm³

∴ Volume of 12 matchboxes = 12 × 15 = 180 cm³

Thus, the volume of a packet containing 12 such matchboxes is 180 cm³.

Question 3:

A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (Given, 1 m³ = 1000 litres.)

Answer 3:

Volume of water in the tank = Length × Breadth × Height = 6 × 5 × 4.5 = 135 m³

∴ Volume of water in litres = 135 × 1000 = 135000 L        (1 m³ = 1000 L)

Thus, the water tank can hold 135000 L of water.

Question 4:

The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank if its length and depth are respectively 10 m and 2.5 m. (Given, 1000 litres = 1 m³.)

Answer 4:

Capacity of the tank = 50000 L = $\frac{50000}{1000}$ = 50 m³           (1000 L = 1 m³)

Length of the tank = 10 m

Height (or depth) of the tank = 2.5 m

Now,

Volume of the cuboidal tank = Length × Breadth × Height

∴ Breadth of the tank = $\frac{\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{tank}}{\mathrm{Length} \times \mathrm{Height}}=\frac{50}{10\times 2.5}=\frac{50}{25}$ = 2 m

Thus, the breadth of the tank is 2 m.

Question 5:

A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates, each measuring 1.5 m × 1.25 m × 0.5 m, that can be stored in the godown.

Answer 5:

Volume of the godown = 40 m × 25 m × 15 m = 15000 m³

Volume of each wooden crate = 1.5 m × 1.25 m × 0.5 m = 0.9375 m³

∴ Maximum number of wooden crates that can be stored in the godown

$$\begin{gathered} =\frac{\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{godown}}{\mathrm{Volume} \mathrm{of} \mathrm{each} \mathrm{wooden} \mathrm{crate}} \\ =\frac{15000}{0.9375} \\ =16000 \end{gathered}$$

Question 6:

How many planks of dimensions (5 m × 25 cm × 10 cm) can be stored in a pit which is 20 m long, 6 m wide and 80 cm deep?

Answer 6:

Number of planks = $\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{pit} \mathrm{in} {\mathrm{cm}}^3}{\mathrm{volume} \mathrm{of} 1 \mathrm{plank} \mathrm{in} {\mathrm{cm}}^3}$

Volume of one plank = $(l\times b\times h) {\mathrm{cm}}^3$
                                  $$\begin{gathered} =500\times 25\times 10 {\mathrm{cm}}^3 \\ = 125000 {\mathrm{cm}}^3 \end{gathered}$$
$$\begin{gathered} \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{pit}=(l\times b\times h) {\mathrm{cm}}^3 \\ \mathrm{Here}, l=20 \mathrm{m}=2000 \mathrm{cm}; b=6 \mathrm{m}=600 \mathrm{cm}; h=80 \mathrm{cm} \\ \mathrm{i}.\mathrm{e}., \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{pit} =2000\times 600\times 80 {\mathrm{cm}}^3 \\ =96000000 {\mathrm{cm}}^3 \end{gathered}$$

∴ Number of planks = $\frac{96000000}{125000} = \frac{96000}{125}=768$

Question 7:

How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick if each brick measures (25 cm × 11.25 cm × 6 cm)?

Answer 7:

Length of the wall = 8 m = 800 cm
Breadth of the wall = 22.5 cm
Height of the wall = 6 m = 600 cm
i.e., volume of wall$=800\times 22.5\times 600 {\mathrm{cm}}^3=10800000 {\mathrm{cm}}^3$

Length of the brick = 25 cm
Breadth of the brick = 11.25 cm
Height of the brick = 6 cm
i.e., volume of one brick $=25\times 11.25\times 6 =1687.5 {\mathrm{cm}}^3$

∴ Number of bricks = $\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{wall}}{\mathrm{volume} \mathrm{of} \mathrm{one} \mathrm{brick}}$ = $\frac{10800000}{1687.5}=6400$

Question 8:

Find the capacity of a closed rectangular cistern whose length is 8 m, breadth 6 m and depth 2.5 m. Also, find the area of the iron sheet required to make the cistern.

Answer 8:

Length of the cistern, l = 8 m

Breadth of the cistern, b = 6 m

Height (or depth) of the cistern, h = 2.5 m

∴ Capacity of the cistern

= Volume of the cistern

= l × b × h

= 8 × 6 × 2.5

= 120 m³

Also,

Area of the iron sheet required to make the cistern

= Total surface area of the cistern

= 2(lb + bh + hl)

= 2(8 × 6 + 6 × 2.5 + 2.5 × 8)

= 2 × 83

= 166 m²

Question 9:

The dimensions of a room are (9 m × 8 m × 6.5 m). It has one door of dimensions (2 m × 1.5 m) and two windows, each of dimensions (1.5 m × 1 m). Find the cost of whitewashing the walls at Rs 25 per square metre.

Answer 9:

Length of the room, l = 9 m

Breadth of the room, b = 8 m

Height of the room, h = 6.5 m

Now,

Area of the walls to be whitewashed

= Curved surface area of the room − Area of the door − 2 × Area of each window

= 2h(l + b) − 2 m × 1.5 m − 2 × 1.5 m × 1 m

= 2 × 6.5 × (9 + 8) − 3 − 3

= 221 − 6

= 215 m²

∴ Cost of whitewashing the walls at Rs 25 per square metre

= Area of the walls to be whitewashed × Rs 25 per square metre

= 215 × 25

= Rs 5,375

Question 10:

A wall 15 m long, 30 cm wide and 4 m high is made of bricks, each measuring (22 cm × 12.5 cm × 7.5 cm). If $\frac{1}{12}$ of the total volume of the wall consists of mortar, how many bricks are there in the wall?

Answer 10:

Length of the wall = 15 m = 1500 cm
Breadth of the wall = 30 cm
Height of the wall = 4 m = 400 cm
Volume of wall =  $=1500\times 30\times 400 {\mathrm{cm}}^3= 18000000 {\mathrm{cm}}^3$

$$\begin{gathered} \mathrm{Now}, \mathrm{volume} \mathrm{of} \mathrm{each} \mathrm{brick}=22\times 12.5\times 7.5 {\mathrm{cm}}^3 \\ =2062.5 {\mathrm{cm}}^3 \end{gathered}$$

Also, volume of the mortar = $\frac{1}{12}\times \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{wall}$
                                        $=\frac{18000000}{12}=1500000 {\mathrm{cm}}^3$
Total volume of the bricks in the wall = volume of the wall − volume of the mortar
                                                           = (18000000 − 1500000) cm³= 16500000 cm³

∴ Number of bricks = $\frac{\mathrm{volume} \mathrm{of} \mathrm{bricks}}{\mathrm{volume} \mathrm{of} \mathrm{one} \mathrm{brick}}=\frac{16500000}{2062.5}=8000 \mathrm{bricks}$

Question 11:

How many cubic centimetres of iron are there in an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm, the iron being 1.5 cm thick throughout? If 1 cm³ of iron weighs 15 g, find the weight of the empty box in kilograms.

Answer 11:

The external dimensions of the box are 36 cm, 25 cm and 16.5 cm.

Thickness of the iron = 1.5 cm

∴ Inner length of the box = 36 − 1.5 −1.5 = 33 cm

Inner breadth of the box = 25 − 1.5 −1.5 = 22 cm

Inner height of the box = 16.5 − 1.5 = 15 cm

Now,

Volume of iron in the open box

= Volume of the outer box − Volume of the inner box

= 36 × 25 × 16.5 − 33 × 22 × 15

= 14850 − 10890

= 3960 cm³

It is given that 1 cm³ of iron weighs 15 g.

∴ Weight of the empty box = 3960 × 15 = 59400 g = $\frac{59400}{1000}$ = 59.4 kg         (1 kg = 1000 g)

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Question 12:

A box made of sheet metal costs ₹ 6480 at ₹ 120 per square metre. If the box is 5 m long and 3 m wide, find its height.

Answer 12:

Length of the box =5 m
Breadth of the box =3 m

Area of the sheet required = $\frac{\mathrm{total} \mathrm{cost}}{\mathrm{cost} \mathrm{per} \mathrm{metre} \mathrm{square} }$

Let h m be the height of the box.
Then area of the sheet = total surface area of the box    
                                 $$\begin{gathered} =2(lb+lh +bh) {\mathrm{m}}^2 \\ =2(5\times 3+5\times h+3\times h) {\mathrm{m}}^2 \\ =2(15+8h) =( 30+16h ){\mathrm{m}}^2 \end{gathered}$$

$$\begin{gathered} \mathrm{Now}, 30+16h=\frac{6480}{120} \\ \Rightarrow 30+16h=54 \\ \Rightarrow 16h = 24 \\ \Rightarrow h= 1.5 \mathrm{m} \end{gathered}$$

∴ The height of the box is 1.5 m.

Question 13:

The volume of a cuboid is 1536 m³. Its length is 16 m, and its breadth and height are in the ratio 3 : 2. Find the breadth and height of the cuboid.

Answer 13:

Length of the cuboid = 16 m
Suppose that the breadth and height of the cuboid are 3x m and 2x m, respectively.
$$\begin{gathered} \mathrm{Then} 1536=16\times 3x\times 2x \\ \Rightarrow 1536=16\times 6x^2 \\ \Rightarrow x^2=\frac{1536}{96}=16 \\ \Rightarrow x=\sqrt{16}=4 \end{gathered}$$

∴ The breadth and height of the cuboid are 12 m and 8 m, respectively.

Question 14:

How many persons can be accommodated in a dining hall of dimensions (20 m × 16 m × 4.5 m), assuming that each person requires 5 cubic metres of air?

Answer 14:

Volume of the dining hall = $$\begin{gathered} (20\times 16\times 4.5) {\mathrm{m}}^3 \\ =1440 {\mathrm{m}}^3 \end{gathered}$$

Volume of air required by each person = 5 m³

∴ Capacity of the dining hall = $\frac{\mathrm{volume} \mathrm{of} \mathrm{dining} \mathrm{hall}}{\mathrm{volume} \mathrm{of} \mathrm{air} \mathrm{required} \mathrm{by} \mathrm{each} \mathrm{person}}=\frac{1440}{5}=288 \mathrm{persons}$

Question 15:

A classroom is 10 m long, 6.4 m wide and 5 m high. If each student be given 1.6 m² of the floor area, how many students can be accommodated in the room? How many cubic metres of air would each student get?

Answer 15:

Length of the classroom = 10 m
Breadth of the classroom = 6.4 m
Height of the classroom = 5 m
Area of the floor = length $\times$ breadth = 10 $\times$ 6.4 m²
$$\begin{gathered} \mathrm{No}. \mathrm{of} \mathrm{students}=\frac{\mathrm{area} \mathrm{of} \mathrm{the} \mathrm{floor}}{\mathrm{area} \mathrm{given} \mathrm{to} \mathrm{one} \mathrm{student} \mathrm{on} \mathrm{the} \mathrm{floor}} \\ =\frac{10\times 6.4}{1.6} \\ =\frac{640}{16}=40 \mathrm{students} \end{gathered}$$

Now, volume of the classroom=$10\times 6.4\times 5 {\mathrm{m}}^3$

$$\begin{gathered} \therefore \mathrm{Air} \mathrm{required} \mathrm{by} \mathrm{each} \mathrm{student}=\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{room}}{\mathrm{number} \mathrm{of} \mathrm{students}} \\ =\frac{10\times 6.4\times 5}{40} {\mathrm{m}}^3 \\ =8 {\mathrm{m}}^3 \end{gathered}$$

Question 16:

The surface area of a cuboid is 758 cm². Its length and breadth are 14 cm and 11 cm respectively. Find its height.

Answer 16:

Length of the cuboid = 14 cm
Breadth of the cuboid = 11 cm
Let the height of the cuboid be x cm.
Surface area of the cuboid = 758 cm²
$$\begin{gathered} \mathrm{Then} 758 = 2(14\times 11+14\times x +11\times x) \\ \Rightarrow 758 = 2(154 +14x +11x) \\ \Rightarrow 758 =2(154+25x) \\ \Rightarrow 758 = 308 +50x \\ \Rightarrow 50x = 758-308 = 450 \\ \Rightarrow x=\frac{450}{50}=9 \end{gathered}$$

∴ The height of the cuboid is 9 cm.

Question 17:

In a shower, 5 cm of rain falls. Find the volume of water that falls on 2 hectares of ground.

Answer 17:

Volume of the water that falls on the ground = $\mathrm{area} \mathrm{of} \mathrm{ground}\times \mathrm{depth}$
                                                                 $$\begin{gathered} =20000\times 0.05 {\mathrm{m}}^3 \\ =1000 {\mathrm{m}}^3 \end{gathered}$$

Question 18:

Find the volume, the lateral surface area, the total surface area and the diagonal of a cube, each of whose edges measures 9 m. (Take $\sqrt{3}=1.73$.)

Answer 18:

Here, a = 9 m
Volume of the cube = $a^3=9^3 {\mathrm{m}}^3=729 {\mathrm{m}}^3$
Lateral surface area of the cube = $4a^2=4\times 9^2{\mathrm{m}}^2=4\times 81 {\mathrm{m}}^2=324 {\mathrm{m}}^2$
Total surface area of the cube = $6a^2= 6\times 9^2{\mathrm{m}}^2=6\times 81 {\mathrm{m}}^2=486 {\mathrm{m}}^2$
∴ Diagonal of the cube = $\sqrt{3}a=\sqrt{3}\times 9=15.57 \mathrm{m}$
 

Question 19:

The total surface area of a cube is 1176 cm². Find its volume.

Answer 19:

Suppose that the side of cube is x cm.
Total surface area of cube = 1176 sq cm
$$\begin{gathered} \mathrm{Then} 1176 =6x^2 \\ \Rightarrow x^2=\frac{1176}{6}=196 \\ \Rightarrow x =\sqrt{196}=14 \end{gathered}$$
i.e., the side of the cube is 14 cm.

∴ Volume of the cube = $x^3 ={14}^3 {\mathrm{cm}}^3=2744 {\mathrm{cm}}^3$

Question 20:

The lateral surface area of a cube is 900 cm². Find its volume.

Answer 20:

Suppose that the side of cube is x cm.
Lateral surface area of the cube = 900 cm²
$$\begin{gathered} \mathrm{Then} 900= 4x^2 \\ \Rightarrow x^2=\frac{900}{4}=225 \\ \Rightarrow x=\sqrt{225}=15 \end{gathered}$$
i.e., the side of the cube is 15 cm.
∴ Volume of the given cube = $x^3 {\mathrm{cm}}^3={15}^{3 }{\mathrm{cm}}^3=3375 {\mathrm{cm}}^3$

Question 21:

The volume of a cube is 512 cm³. Find its surface area.

Answer 21:

Suppose that the side of the given cube is x cm.
Volume of the cube = 512 cm³
$$\begin{gathered} \mathrm{Then} 512 =x^3 \\ \Rightarrow x =\sqrt[3]{512} =8 \\ \mathrm{i}.\mathrm{e}., \mathrm{the} \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{cube} \mathrm{is} 8 \mathrm{cm}. \end{gathered}$$

∴ Surface area of the cube = $6x^2 {\mathrm{cm}}^2=6\times 8^2 {\mathrm{cm}}^2=384 {\mathrm{cm}}^2$

Question 22:

Three cubes of metal with edges 3 cm, 4 cm and 5 cm respectively are melted to form a single cube. Find the lateral surface area of the new cube formed.

Answer 22:

Three cubes of metal with edges 3cm, 4cm and 5 cm are melted to form a single cube.

∴ Volume of the new cube = sum of the volumes the old cubes
                                        $$\begin{gathered} =\left(3^3+4^3+5^3\right){\mathrm{cm}}^3 \\ =(27+64+125) {\mathrm{cm}}^3=216 {\mathrm{cm}}^3 \end{gathered}$$

Suppose the edge of the new cube = x cm
Then we have:
$$\begin{gathered} \mathrm{Then} 216=x^3 \\ \Rightarrow x=\sqrt[3]{216} =6 \\ \mathrm{i}.\mathrm{e}., \mathrm{the} \mathrm{edge} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{cube} \mathrm{is} 6 \mathrm{cm}. \end{gathered}$$

∴ Lateral surface area of the new cube = $4x^2 {\mathrm{cm}}^2=4\times 6^2 {\mathrm{cm}}^2=144 {\mathrm{cm}}^2$

Question 23:

Find the length of the longest pole that can be put in a room of dimensions (10 m × 10 m × 5 m).

Answer 23:

Length of the longest pole = length of the diagonal of the room
                                      $$\begin{gathered} =\sqrt{l^2+b^2+h^2} \mathrm{m} \\ =\sqrt{{10}^2+{10}^2+5^2} \mathrm{m} \\ =\sqrt{100+100+25} \\ =\sqrt{225} =15 \mathrm{m} \end{gathered}$$

Question 24:

The sum of length, breadth and depth of a cuboid is 19 cm and the length of its diagonal is 11 cm. Find the surface area of the cuboid.

Answer 24:

Let the length, breadth and height (or depth) of the cuboid be l cm, b cm and h cm, respectively.

∴ l + b + h = 19           .....(1)

Also,

Length of the diagonal = 11 cm

$$\begin{gathered} \Rightarrow \sqrt{l^2+b^2+h^2}=11 \\ \Rightarrow l^2+b^2+h^2=121 .....\left(2\right) \end{gathered}$$

Squaring (1), we get

(l + b + h)² = 19²

⇒ l² + b² + h² + 2(lb + bh + hl) = 361

⇒ 121 + 2(lb + bh + hl) = 361                      [Using (2)]

⇒ 2(lb + bh + hl) = 361 − 121 = 240 cm²

Thus, the surface area of the cuboid is 240 cm².

Question 25:

Each edge of a cube is increased by 50%. Find the percentage increase in the surface area of the cube.

Answer 25:

Let the initial edge of the cube be a units.

∴ Initial surface area of the cube = 6a² square units

New edge of the cube = a + 50% of a = $a+\frac{50}{100}a$ = 1.5a units

∴ New surface of the cube = 6(1.5a)² = 13.5a² square units

Increase in surface area of the cube = 13.5a² − 6a² = 7.5a² square units

∴ Percentage increase in the surface area of the cube

$$\begin{gathered} =\frac{\mathrm{Increase} \mathrm{in} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cube}}{\mathrm{Initial} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cube}}\times 100\% \\ =\frac{7.5a^2}{6a^2}\times 100\% \\ =125\% \end{gathered}$$

Question 26:

If V is the volume of a cuboid of dimensions a, b, c and S is its surface area then prove that $\frac{1}{V}=\frac{2}{S}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$.

Answer 26:

Let the length, breadth and height of the cuboid be a, b and c, respectively.

∴ Surface area of the cuboid, S = 2(ab + bc + ca) 

Volume of the cuboid, V = abc

Now,

$$\begin{gathered} \frac{S}{V}=\frac{2\left(ab+bc+ca\right)}{abc} \\ \Rightarrow \frac{{\displaystyle S}}{{\displaystyle V}}=2\left(\frac{ab}{abc}+\frac{{\displaystyle bc}}{{\displaystyle abc}}+\frac{{\displaystyle ca}}{{\displaystyle abc}}\right) \\ \Rightarrow \frac{{\displaystyle S}}{{\displaystyle V}}=2\left(\frac{1}{c}+\frac{{\displaystyle 1}}{{\displaystyle a}}+\frac{{\displaystyle 1}}{{\displaystyle b}}\right) \\ \Rightarrow \frac{1}{V}=\frac{2}{S}\left(\frac{1}{a}+\frac{{\displaystyle 1}}{{\displaystyle b}}+\frac{{\displaystyle 1}}{{\displaystyle c}}\right) \end{gathered}$$

Question 27:

Water in a canal, 30 dm wide and 12 dm deep, is flowing with a velocity of 20 km per hour. How much area will it irrigate, if 9 cm of standing water is densired?

Answer 27:

Width of the canal = 30 dm = 3 m                (1 m = 10 dm)

Depth of the canal = 12 dm = 1.2 m

Speed of the water flow = 20 km/h = 20000 m/h

∴ Volume of water flowing out of the canal in 1 h = 3 × 1.2 × 20000 = 72000 m³

Height of standing water on field = 9 cm = 0.09 m            (1 m = 100 cm)

Assume that water flows out of the canal for 1 h. Then,

Area of the field irrigated

$$\begin{gathered} =\frac{\mathrm{Volume} \mathrm{of} \mathrm{water} \mathrm{flowing} \mathrm{out} \mathrm{of} \mathrm{the} \mathrm{canal}}{\mathrm{Height} \mathrm{of} \mathrm{standing} \mathrm{water} \mathrm{on} \mathrm{the} \mathrm{field}} \\ =\frac{72000}{0.09} \\ =800000 {\mathrm{m}}^2 \end{gathered}$$
$$\begin{gathered} =\frac{800000}{10000} \left(1 \mathrm{hectare} = 10000 {\mathrm{m}}^2\right) \\ =80 \mathrm{hectare} \end{gathered}$$

Thus, the area of the field irrigated is 80 hectares.

Disclaimer: In this question time is not given, so the question is solved assuming that the water flows out of the canal for 1 hour.

Question 28:

A solid metallic cuboid of dimensions (9 m × 8 m × 2 m) is melted and recast into solid cubes of edge 2 m. Find the number of cubes so formed.

Answer 28:

Volume of the solid metallic cuboid = 9 m × 8 m × 2 m = 144 m³

Volume of each solid cube = (Edge)³ = (2)³ = 8 m³

∴ Number of cubes formed = $\frac{\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{solid} \mathrm{metallic} \mathrm{cuboid}}{\mathrm{Volume} \mathrm{of} \mathrm{each} \mathrm{solid} \mathrm{cube}}=\frac{144}{8}$ = 18

Thus, the number of cubes so formed is 18.