SELINA Solution Class 9 Chapter 21 Solids [Surface area and Volume of 3-D Solids ] Exercise 21 B

Question 1

The following figure shows a solid of uniform cross-section. Find the volume of the solid. All measurements are in centimeters.
Assume that all angles in the figures are right angles.

Sol:

The given figure can be divided into two cuboids of dimensions 6 cm, 4 cm, 3 cm, and 9 cm respectively. Hence, volume of solid

= 9 x 4 x 3 + 6 x 4 x 3
= 108 + 72 
= 180 cm³.

Question 2

A swimming pool is 40 m long and 15 m wide. Its shallow and deep ends are 1.5 m and 3 m deep respectively. If the bottom of the pool slopes uniformly, find the amount of water in liters required to fill the pool.

Sol:

Area of cross-section of the solid = ${\displaystyle \frac{1}{2}(1.5+3)\times \left(40\right)}$cm²

= ${\displaystyle \frac{1}{2}}$( 4.5 ) x ( 40 ) cm²
= 90 cm²

Volume of solid
= Area of cross section x Length
= 90 x 15 cm³
= 1350 cm³
= 1350000 liters     ....( Since 1cm³ = 1000L )

Question 3

The cross-section of a tunnel perpendicular to its length is a trapezium ABCD as shown in the following figure; also given that:

AM = BN; AB = 7 m; CD = 5 m. The height of the tunnel is 2.4 m. The tunnel is 40 m long. Calculate:

(i) The cost of painting the internal surface of the tunnel (excluding the floor) at the rate of Rs. 5 per m² (sq. meter).

(ii) The cost of paving the floor at the rate of Rs. 18 per m².

Sol:

The cross-section of a tunnel is of the trapezium-shaped ABCD in which  AB = 7 m, CD = 5 m and AM = BN. The height is 2.4 m and its length is 40 m.

(i) AM = BN =${\displaystyle \frac{7-5}{2}=\frac{2}{2}=1\text{m}}$

∴ In ΔADM,

AD² = AM² + DM²      ...[ Using pythsgor as theorem ]
       = 1² + (2 . 4)² 
       = 1  + 5.76
       = 6.67 
       = (2.6)² 
AD  = 2.6 m 

Perimeter of the cross- section of the tunnel = ( 7+ 2.6 + 2.6 + 5 ) m = 17.2 m

Length = 40 m

∴ The internal surface area of the tunnel ( except the floor ) 

= ( 17.2 x 40 - 40 x 7) m²
= ( 688 - 280 ) m²
= 408 m² 

Rate of painting = Rs. 5 per m²
Hence, total cost of painting = Rs. 5 x  5408 = Rs. 2040

(ii) Area of floor of tunnel = l x b = 40 x 7 = 280 m²

Rate of cost of paving = Rs. 18 per m² 

Total cost = 280 x 18 = Rs. 5040

Question 4

Water is discharged from a pipe of a cross-section area 3.2 cm² at the speed of 5m/s. Calculate the volume of water discharged: 
(i) In cm³ per sec.
(ii) In liters per minute.

Sol:

(i) The rate of speed  = ${\displaystyle 5\frac{\text{m}}{s}=500\frac{\text{cm}}{s}}$

The volume of water flowing per sec = 3.2 x 500cm³ = 1600 cm³ 

(ii) Vol. of water flowing per min = 1600 x 60 cm³ = 96000 cm³ 

Since 1000 cm³ = 1 litre

Therefore, Vol. of water flowing per min= ${\displaystyle \frac{96000}{1000}}$ = 96 litres

Question 5

A hose-pipe of cross-section area 2 cm² delivers 1500 liters of water in 5 minutes. What is the speed of water in m/s through the pipe? 

Sol:

Vol. of water flowing in 1 sec= ${\displaystyle \frac{1500\times 1000}{5\times 60}=5000{\text{cm}}^3}$

Vol. of water flowing =area of cross-section speed of water

${\displaystyle 5000\frac{{\text{cm}}^3}{s}=2{\text{cm}}^2\times \text{speed of water}}$

⇒ speed of water = ${\displaystyle \frac{5000}{2}\frac{\text{cm}}{s}}$

⇒ speed of water = ${\displaystyle 2500\frac{\text{cm}}{s}}$

⇒ speed of water = ${\displaystyle 25\frac{\text{m}}{s}}$

Question 6

The cross-section of a piece of metal 4 m in length is shown below. Calculate :


(i) The area of the cross-section;
(ii) The volume of the piece of metal in cubic centimeters.

If 1 cubic centimeter of the metal weighs 6.6 g, calculate the weight of the piece of metal to the nearest kg.

Sol:

(i) Area of total cross section = Area of rectangle abce + area of  Δdef

= ( 12 x 10 ) + ${\displaystyle \frac{1}{2}}$ ( 16 - 10 )( 12 - 7.5 )

= 120 + ${\displaystyle \frac{1}{2}}$ (6)( 4.5 ) cm²

= 120 + 13.5 cm²

= 133.5 cm² 

(ii) The volume of the piece of metal in cubic centimeters = Area of total cross section x length

= 133.5 cm² x 400 cm² = 53400 cm³

1 cubic centimetre of the metal weighs 6.6 g

53400 cm³ of the metal weighs 6.6 x 53400 g = ${\displaystyle \frac{6.6\times 53400}{1000}}$ kg

= 352.440 kg

The weight of the piece of metal to the nearest Kg is 352 Kg.

Question 7

A rectangular water-tank measuring 80 cm x 60 cm is filled form a pipe of cross-sectional area 1.5 cm², the water emerging at 3.2 m/s. How long does it take to fill the tank?

Sol:

Vol. of the rectangular tank = 80 x 60 x 60 cm³ = 288000 cm³

One liter = 1000 cm³

Vol. of water flowing in per sec =
${\displaystyle 1.5{\text{cm}}^2\times 3.2\frac{\text{m}}{s}=1.5{\text{cm}}^2\times \frac{(3.2\times 100)\text{cm}}{\text{s}}}$

                                     = ${\displaystyle 480\frac{{\text{cm}}^3}{\text{s}}}$

Vol. of water flowing in 1 min= 480 x 60 = 28800cm³ 

Hence,

28800 cm³  can be filled = 1 min

288000cm³ can be filled =${\displaystyle (\frac{1}{28800}\times 288000)min=10min}$ 

Question 8

A rectangular cardboard sheet has length 32 cm and breadth 26 cm. Squares each of side 3 cm, are cut from the corners of the sheet and the sides are folded to make a rectangular container. Find the capacity of the container formed.

Sol:

Length of sheet = 32 cm

Breadth of sheet = 26 cm 

Side of each square = 3cm

∴  Inner length = 32 - 2 x 3 = 32 - 6 = 26 cm

Inner breadth = 26 - 2 x 3 = 26 - 6 = 20 cm

By folding the sheet, the length of the container = 26 cm

Breadth of the container = 20 cm and height of the container = 3 cm

∴ Vol. of the container = l x b x h

= 26 cm x 20 cm x 3 cm = 1560 cm³  

Question 9

A swimming pool is 18 m long and 8 m wide. Its deep and shallow ends are 2 m and 1.2 m respectively. Find the capacity of the pool, assuming that the bottom of the pool slopes uniformly. 

Sol:

Length of pool = 18 m

Breadth of pool = 8 m

Height of one side = 2m

Height on second side = 1.2 m

∴ Volume of pool = 18 x 8 x ${\displaystyle \frac{(2+1.2)}{2}}$ m³

= ${\displaystyle \frac{18\times 8\times 3.2}{2}}$

= 230.4 m³ 

Question 10

The following figure shows a closed victory-stand whose dimensions are given in cm.

Find the volume and the surface area of the victory stand. 

Sol:

Consider the box 1

Thus, the dimensions of box 1 are 60 cm, 40 cm, and 30 cm. 

Therefore, the volume of box 1 = 60 x 40 x 30 = 72000 cm³ 
Surface area of box 1 = 2 ( lb + bh + lh )
Since the box is open at the bottom and from the given figure, we have,
Surface area of box 1 = 40 x 40 + 40 x 30 + 40 x 30 + 2 ( 60 x 30 )
                                  = 1600 + 1200 + 1200 + 3600
                                 = 7600 cm²   

Consider the box 2 

Thus, the dimensions of box 2 are 40 cm, 30 cm, and 30 cm.

Therefore, the volume of box 2 = 40 x 30 x 30 = 36000 cm³ 
Surface area of box 2 = 2 ( lb + bh + lh )
Since the box is open at the bottom and from the given figure, we have,
Surface area of box 2 = 40 x 30 + 40 x 30  + 2 ( 30 x 30 )
                                  = 1200 + 1200 + 1800 
                                 = 4200 cm²   

Consider the box 3

Thus, the dimensions of the box 3 are 40 cm, 30 cm, and 20 cm.

Therefore, the volume of box 3 = 40 x 30 x 20 = 24000 cm³ 
Surface area of box 3 = 2 ( lb + bh + lh )
Since the box is open at the bottom and from the given figure, we have,
Surface area of box 3 = 40 x 30 + 40 x 20  + 2 ( 30 x 20 )
                                  = 1200 + 800 + 1200 
                                 = 3200 cm²   

Total volume of the box
= volume of box 1 + volume of box 2 + volume of box 3
= 72000 + 36000 + 24000
= 132000 cm³ 

Similarly, total surface area of the box
= surface area of box 1 + surface area  of box 2 + surface area of box 3
= 7600 + 4200 + 3200
= 15000 cm²