volume and surface areas of solids - · pdf file · 2009-08-24volume and surface...

Math Class IX 1 Question Bank

VOLUME AND SURFACE

AREAS OF SOLIDS Q.1. Find the total surface area and volume of a rectangular solid (cuboid)

measuring 1 m by 50 cm by 0.5 m.

Ans. Length of cuboid 1 m,l = Breadth of cuboid, 50 1

50 cm m100 2

b = = =

Height of cuboid, 5 1

0.5 m m= m10 2

h = =

∴ Volume of cuboid l b h= × ×2 31 1 1

1 m 0.25 m2 2 4

= × × = = .

Total surface area of cuboid 2 ( )lb bh hl= + +1 1 1 1

2 1 12 2 2 2

= × + × + ×

1 1 1 2 1 2

2 22 4 2 4

+ + = + + =

25 5

2 2.50 m4 2

= × = =

Hence, volume of cuboid is 0.25 3m and total surface area of cuboid is 2

2.50 m

Q.2. The length, breadth and height of a rectangular solid are in the ratio 5 : 4 :

2. If the total surface area is 1216 2cm , find the length, breadth and height

of the solid.

Ans. Let length, 5 ,l x= breadth, 4b x= and height, 2 .h x=

∴ Total surface area 21216 cm=

⇒ 2 ( ) 1216lb bh hl+ + =

⇒ 2 (5 4 4 2 2 5 ) 1216x x x x x x× + × + × =

⇒ 2 2 22 (20 8 10 ) 1216x x x+ + = ⇒ 2

2 (38 ) 1216x =

⇒ 276 1216x = ⇒ 2 121616

76x = = ⇒ 2 16x = ⇒ 4x = ∴ 5 5 4 20 cml x= = × =

4 4 4 16 cm,b x= = × = and 2 2 4 8 cmh x= = × =

Hence, length, 20 cm,l = breadth, 16 cm,b = and height 8 cm.h =


Q.3. The volume of a rectangular wall is 33 3m . If its length is 16.5 m and height

8 m, find the width of the wall.

Ans. Volume of rectangular wall 233 m=

Length of wall ( ) 16.5 m,l = Height of wall ( ) 8 mh =

Let width of wall m,b= then volume of rectangular wall l b h= × ×

⇒ 16.5 8 33b× × =

⇒ 33 1

m=0.25 m16.5 8 4

b = =×

Hence, width of wall is 0.25 m.

Q.4. A class room is 12.5 m long, 6.4 m broad and 5 m high. How many students

can accommodate if each student needs 1.6 2

m of floor area ? How many

cubic metres of air would each student get ?

Ans. Length of room ( ) 12.5 m,l = width of room ( ) 6.4 mb = and height of room

( ) 5 mh =

∴ Volume of air inside the room 312.5 6.4 5 ml b h= × × = × ×

3400 m=

Area of floor of the room l b= ×2

12.5 6.4 m= ×2

80 m=

For each student area required 21.6 m=

∴ Number of students 80 80 10

501.6 16

×= = =

The required air received by each student Volume of air

Number of students=

34008 m

50= =

Q.5. Find the length of the longest rod that can be placed in a room measuring 12

m × 9 m × 8 m.

Ans. Length of room ( ) 12 m,l = breadth of room ( ) 9 m,b = and height of room

( ) 8 mh =

Hence, the longest rod required to place in the room

2 2 2l b h= + +

2 2 2(12) (9) (8)= + + 144 81 64 m= + + 289 17 m= =

Q.6. The volume of a cuboid is 14400 3cm and its height is 15 cm. The cross-

section of the cuboid is a rectangle having its sides in the ratio 5 : 3. Find the

perimeter of the cross-section.

Ans. Volume of cuboid 314400 cm ,= and Height of cuboid ( ) 15 cmh =

∴ Length × Breadth 2volume 14400 cm

15h= =

2960 cm=

Ratio of remaining sides 5 : 3=

Thus, length 5x= and breadth 3x=


∴ 5 3 960x x× = ⇒ 215 960x = ⇒ 2 2

64 (8)x = = ∴ 8x =

Length 5 8 5 40 cmx= = × = and breadth 3 8 3 24 cmx= = × =

Hence, perimeter of rectangular cross-section 2 ( ) 2 (40 24) cml b= + = +

2 64 128 cm.= × =

Q.7. The area of path is 6500 2m . Find the cost of covering it with gravel 14 cm

deep at the rate of Rs 5.60 per cubic metre.

Ans. Area of path 26500 m= and Depth of gravel

1414 cm m

100= =

∴ Volume of gravel Area Depth= ×314

6500 910 m100

= × =

Rate of covering the gravel 3Rs 5.60 per m=

∴ Total cost Rs 910 5.60= ×910 560

Rs Rs 5096.100

×= =

Q.8. The cost of papering the four walls of a room 12 m long at Rs 6.50 per

square metre is Rs 1638 and the cost of matting the floor at Rs 3.50 per

square metre is Rs 378. Find the height of the room.

Ans. Rate of papering the walls 2Rs 6.50 per m= and Total cost Rs 1638=

∴ Area of four walls 1638

6.50=

2 21638 100m 252 m

650

×= =

Rate of matting the floor 2Rs 3.50 per m= and Total cost Rs 378=

∴ Area of floor 378

3.50=

2 2378 100m 108 m

350

×= =

Length of room 12 m= ∴ Breadth of room Area of floor

Length=

1089 m

12= =

Area of four walls 2 ( ) 252l b h= + =

⇒ 2 (12 9) 252h+ = ⇒ 2 21 252h× =

⇒ 252

2 21h =

×⇒ 6h =

Hence, height of the room is 6 m.


Q.9. The dimensions of a field are 15 m × 12 m. A pit 7.5 m × 6 m × 1.5 m is dug

in one corner of the field and the earth removed from it, is evenly spread

over the remaining area of the field, calculate, by how much the level of the

field is raised ?

Ans. Length of field ( ) 15 ml =

and breadth of field ( ) 12 mb =

∴ Area of total field l b= ×2

15 12 180 m= × =

Length of pit 7.5 m= breadth of pit 6 m=

and depth of pit 1.5 m=

∴ Area of pit l b= ×2

7.5 6 45 m= × =

and volume of earth removed l b h= × ×37.5 6 1.5 m= × ×

367.5 m=

∴ Area of field leaving pit 2(180 45) m= −2

135 m=

and volume of earth removed l b h= × ×3

7.5 6 1.5 m= × ×3

67.5 m=

Area of field leaving pit 2180 45) m= −2

135 m=

∴ Level (height) of the earth in the field Volume of earth

Area of remaining part=

67.5 675

m135 10 135

= =×

0.5 m= 50 cm=

Q.10. 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.

Ans. Let l, b and h be the length, breadth and depth of the cuboid, then

19 cml b h+ + =

and 2 2 211 cml b h+ + = ⇒ 2 2 2 2

(11) 121l b h+ + = =

The surface area of the cuboid 2 ( )lb bh hl= + +

We know that 2 2 2 2( ) 2 ( )l b h l b h lb bh hl+ + = + + + + +

⇒ 2(19) 121 2 ( )lb bh hl= + + + ⇒ 2

2 ( ) (19) 121lb bh hl+ + = −

2361 121 240 cm= − =

Hence, surface area of the cuboid 2240 cm=

Q.11. Find the number of bricks, each measuring 25 cm × 12.5 cm × 7.5 cm,

required to construct a wall 6 m long, 5 m high and 50 cm thick, while the

cement and the sand mixture occupies 1

th20

of the volume of the wall.

Ans. Volume of one brick 25 cm 12.5 cm 7.5 cm= × ×

325 12.5 7.5m

100 100 100= × ×

31 1 3m

4 8 40= × ×

33m

1280=


Length of wall ( ) 6 ml = and Height of wall ( ) 5 mh =

and thickness of wall 50 1

( ) m m100 2

b = =

∴ Volume of wall l b h= × ×31

6 5 15 m2

= × × =

Volume of cement and sand 3 31 3of 15 m m

20 4= =

∴ Volume of bricks Volume of wall Volume of cement and sand= −

3 360 3 57m m

4 4

− = =

∴ Number of bricks Volume of total bricks

Volume of one brick=

5757 12804

3 3 4

1280

= = × 19 320= × 6080=

Q.12. The total surface area of a cube is 726 2cm . Find its volume.

Ans. Total surface area of cube 2726 cm=

Let edge of cube cma=

∴ Total surface area of cube 26 726a= =

⇒ 2 726121

6a = = ⇒ 121 11 cma = =

∴ Edge of cube 11 cm=

Volume of cube 3 3 3(side) (11 cm) 11 11 11 cm= = = × ×3

1331 cm=

Q.13. The volume of a cube is 729 3cm . Find its total surface area.

Ans. Volume of cube 3729 cm=

Let edge of cube a=

∴ 3729 9 9 9a = = × × ∴ 9 cma =

∴ Total surface area of cube 26a=

26 9 cm 9 cm 486 cm .= × × =

Q.14. The edges of three cubes of metal are 3 cm, 4 cm and 5 cm. They are

melted and formed into a single cube. Find the edge of the new cube.

Ans. The edges of three cubes are 3 cm, 4 cm and 5 cm.

Volume of three cubes are 3 3(3) , (4) and 3

(5) i.e. 3 327 cm , 64 cm and 3

125 cm .

∴ Volume of new cube 327 64 125 216 cm= + + =

Let edge of new cube be a


∴ 3 216 6 6 6a = = × × , 6 cma =

Hence, edge of new cube 6 cm=

Q.15. Three cubes, whose edges are x cm, 8 cm and 10 cm respectively, are

melted and recasted into a single cube of edge 12 cm. Find ‘x’.

Ans. Edges of three cubes are x cm, 8 cm and 10 cm respectively

∴ Volume of these cubes are 3 3 3, (8) and (10)x i.e. 3 3 3 cm , 512 cmx and

31000 cm

Edge of new cube formed 12 cm=

∴ Volume of new cube 3 3(side) (12) 12 12 12= = = × ×3

1728 cm=

According to the given problem

3512 1000 1728x + + = ⇒ 3

1512 1728x + =

⇒ 31728 1512 216 6 6 6x = − = = × ×

⇒ 6 cmx =

Hence, edge of cube 6 cm.x= =

Q.16. Three equal cubes are placed adjacently in a row. Find the ratio of the

total surface area of the resulting cuboid to that of the sum of the total

surface areas of the three cubes.

Ans. Let edge of each of three cubes be ‘x’ when three cubes are placed end to end, a

cuboid is formed whose dimensions are :

3 , ,l x x x x b x h x= + + = = =

∴ Total surface area of a resulting cuboid 2 ( )lb bh hl= + +

2 (3 3 )x x x x x x= × + × + ×

2 2 22 (3 3 )x x x= + +2 2

2 7 14x x= × =

Sum of total surface areas of three cubes 2 23 6 18x x= × =

∴ Ratio of suface areas of cubes 2

2

14 14 7

18 918

x

x

= = =

Hence, the required ratio is 7 : 9.

Q.17. A metal cube of edge 12 cm is melted and formed into three smaller cubes.

If the edges of two smaller cubes are 6 cm and 8 cm, find the edge of third

smaller cube.

Ans. Edge of metal cube 12 cm=

∴ Volume of metal cube 3 3(12) 1728 cm= =

Edge of first smaller cube 6 cm=

∴ Volume of smaller cube 3 3(6) 216 cm= =

Edge of second smaller cube 8 cm=


∴ Volume of second smaller cube 3 3(8) 512 cm= =

∴ Volume of third smaller cube 1728 (216 512)= − +3

1728 728 1000 cm= − =

Hence, edge of third cube

1

3(1000) cm=

1

3 3[(10) ] cm= 10 cm=

Q.18. The dimensions of a metallic cuboid are 100 cm × 80 cm × 64 cm. It is

melted and recast into a cube. Find (i) the edge of the cube (ii) the surface

area of the cube.

Ans. Length of the metallic cuboid ( ) 100 cml =

breadth of the metallic cuboid ( ) 80 cmb =

and height of metallic cuboid ( ) 64 cmh =

∴ Volume of metallic cuboid 100 80 64l b h= × × = × ×3

512000 cm=

∴ Volume of cube so formed 3512000 cm=

(i) Edge of the cube 1/3( ) (512000)a =3 1/3[(80) ] 80 cm= =

(ii) Surface area of cube 2 26 6 (80)a= =

26 80 80 cm= × ×

238400 cm=

Q.19. The square on the diagonal of cube has an area of 1875 2cm , Calculate (i)

the side of the cube (ii) the total surface area of the cube.

Ans. 2(Diagonal) of cube 21875 cm=

(i) Let side of cube a=

∴ 2( 3 ) 1875a = ⇒ 23 1875a = ⇒ 2 1875

3a =

⇒ 2 2625 (25)a = = ∴ 25a =

Hence, side of the cube is 25 cm.

(ii) Total surface area of the cube 2 2 2 2 26 side 6 6 (25) 6 625 cm 3750 cma= = = = × =

Q.20. Four identical cubes are joined end to end to form a cuboid. If the total

surface area of the resulting cuboid is 648 2cm . Find the length of edge of

each cube. Also find the ratio between the surface area of resulting cuboid

and the surface area of a cube.

Ans. Surface area of cuboid 2648 cm= , let edge of each cube cmx=

∴ Length of cuboid 4 cm,x= width of cuboid cmx= and height of

cuboid cmx=

∴ Surface area of cuboid 2 ( )lb bh hl= + +

⇒ 648 2 (4 4 )x x x x x x= × + × + ×2 2 2

2 (4 4 )x x x= + +

⇒ 2648 18x= ⇒ 2 2648

36 (6)18

x = = = ∴ 6x =


Thus, length of edge of each cube 6 cm=

Surface area of cube 2 2 2

6 (side) 6 6 (6)x= = = ×2

6 36 216 cm= × =

Hence, ratio between the surface area of cuboid and that of cube 648 : 216 3 :1= =

Q.21. A rectangular container has base of length 12 cm and width 9 cm. A cube

of edge 6 cm is placed in the container and then sufficient water is filled into

it, so that the cube is just submerged. Find the fall in level of water in the

container, when the cube is removed.

Ans. Length of the base of container 12 cm= and width of container 9 cm=

Let fall in water level be x cm. ∴ Side of cube 6 cm=

∴ Volume of cube 36 6 6 216 cm= × × = ∴ Volume of water 3216 cm=

According to the given problem, we get

12 9 216x× × = ⇒ 216

212 9

x = =×

Hence, fall in water level is 2 cm.

Q.22. A rectangular container whose base is a square of side 12 cm, contains

sufficient water to submerge a rectangular solid 8 cm × 6 cm × 3 cm. Find

the rise in level of water in the container when the solid is in it.

Ans. Let the rise in water level cmx=

Length of solid cuboid 8 cm,=

width of solid cuboid 6 cm=

Height of solid cuboid 3 cm=

∴ Volume of cuboid l b h= × ×

38 6 3 144 cm= × × =

∴ Volume of water 3144 cm=

Side of the square base 12 cm=

∴ Area of base 212 12 144 cm= × =

According to given problem, we get

144 144x× =

⇒ 1x =

Hence, rise in water level is 1 cm.

Q.23. A field is 120 m long and 50 m broad. A tank 24 m long, 10 m broad and 6

m deep is dug any where in the field and the earth taken out of the tank is

evenly spread over the remaining part of the field. Find the rise in level of

the field.

Ans. Length of rectangular field 120 m,= Breadth of rectangular field 50 m=

Area of of the total field 2 2120 50 m 6000 m= × =

Length of tank 24 m,= breadth of tank 10 m= and depth of tank 6 m=

∴ Volume of earth dug out 324 10 6 m= × ×

31440 m=


Area of surface of tank l b= ×224 10 240 m= × =

∴ Area of remaining part of the field

Area of total field Surface area of tank= −

26000 240 5760 m= − =

Let height of rise of field mx= ∴ 5760 1440x× =

1440 1

0.255760 4

x = = = . Hence, rise in level of the field 0.25 m or 25 cm.=

Q.24. The following figure shows a solid of uniform

cross-section. Find the volume of the solid. All

measurements are in centimetres. Assume that

all angles in the figure are right-angles.

Ans. 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

6 4 3 4 3 9= × × + × ×372 108 180 cm= + =

Q.25. 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 litres required to fill the pool.

Ans. The cross-section of the swimming pool is a

trapezium with parallel sides 1.5 m and 3 m

(i.e. depths of both ends) and 40 m (i.e. length of pool) as the r⊥ distance

between || sides.

The width of pool i.e. 15 m can be taken as the length of cross-section.

Length of swimming pool ( ) 40 ml =

Breadth of swimming pool ( ) 15 mb =

Depth of its ends 1.5 m= and 3.5 m

∴ Volume of water 31[(1.5 3.0) 40] 15 m

2= + × ×

314.5 40 15 m

2= × × ×

31350 m=

∴ Capacity of water in litres 1350 1000 litres= × 3(1 m 1000 litres)=

1350000 litres=


Q.26. The cross-section of a piece of metal 2 m in length is shown in the adjoining

figure. Calculate :

(i) The area of its cross-section.

(ii) The volume of piece of metal.

(iii) The weight of piece of metal to the nearest kg,

if 31 cm of the metal weighs 6.5 g.

Ans. From C, draw CF || AB

then CF AB 13 cm= =

AF CB 8 cm= =

∴ EF EA FA 12 8 4 cm= − = − =

(i) Now area of figure ABCDE (cross-section)

ar. (Rectangle ABCF) ar. (Trapezium FCDE)= +

2113 cm 8 cm (13 16) 4 cm

2= × + + ×

104 29 2 104 58= + × = +

2162 cm=

(ii) Length of the piece 2 m 200 cm= =

∴ Volume of the metal piece Area Length= ×

3162 200 cm= ×

332400 cm=

3324003.24 m

100 100= =

×

(iii) Weight of 31 cm metal 6.5 g=

∴ Total weight of the metal piece 32400 6.5 g= ×

210600 g=

211 kg (approx)=

Q.27. A square brass plate of side x cm is 1 mm thick and weighs 5.44 kg. If 3

1 cm of brass weighs 8.5 g, find the value of x.

Ans. Side of square plate cmx=

Thickness of square plate 1 mm=

Total weight of square plate 5.44 kg=

Weight of 31 cm 8.5 g=

∴ Total volume of the plate 35.44 1000cm

8.5

×=

3640 cm=


∴ According to the given problem 1

64010

x x= × × =

⇒ 26400x =

⇒ 6400 80 cmx = =

Q.28. The area of cross-section of a rectangular pipe is 25.4 cm and water is

pumped out of it at the rate of 27 kmph. Find, in litres, the volume of water

which flows out of the pipe in 1 minute.

Ans. Area of cross-section of rectangular pipe 25.4 cm=

Speed of water pumped out 27 kmph=

Time 1 minute=

∴ Length of water flow 27

1000 m60

= × 450 m=

∴ Volume of water Area Length= ×

35.4

450 m100 100

= ××

3450 54m

100 100 10

×=

× ×

324300m

100000=

3243m

1000=

∴ Volume of water in litres 243

10001000

= × 3(1 m 1000 litres)=

243 litres=

Q.29. The cross-section of a tunnel, perpendicular

to its length is a trapezium ABCD in which

AB = 8 m, DC = 6 m and AL = BM. The

height of the tunnel is 2.4 m and its length is

40 m.

Find :

(i) The cost of paving the floor of the tunnel at Rs 16 per 2m .

(ii) The cost of painting the internal surface of the tunnel, excluding the floor

at the rate of Rs 5 per 2m .

Ans. The cross-section of a tunnel is of the trapezium shaped ABCD in which

AB 8 m, DC 6 m= = and AL BM.= The height is 2.4 m and its length is 40 m.

(i) Area of floor of tunnel 240 8 320 ml b= × = × =


Rate of cost of paving 2Rs 16 per m=

Total cost 320 16 Rs 5120= × =

(ii) 8 6 2

AL BM 1 m2 2

−= = = =

∴ In ADL,∆

2 2 2

AD AL DL= + (Using Pythagoras Theorem)

2 2(1) (2.4) 1 5.76 6.76= + = + =

2(2.6)=

∴ AD 2.6 m=

Perimeter of the cross-section of the tunnel (8 2.4 2.4 6) m 18.8 m= + + + =

Length 40 m=

∴ Internal surface area of the tunnel (except floor)

2(18.8 40 40 8) m= × − ×

2(752 320) m= −

2432 m=

Rate of painting 2Rs 5 per m=

Hence, total cost of painting Rs 5 432 Rs 2160= × =

Q.30. A stream, which flows at a uniform rate of 4 km/hr, is 10 metres wide and

1.2 m deep at a certain point. If its cross-section is rectangular is shape find,

in litres, the volume of water that flows in a minute.

Ans. Speed of stream 4 km/hr=5

4 m/s18

= ×10

m/s9

=

Area of cross-section of stream 210 1.2 12 m= × =

∴ Volume of water discharged in 1 second 310 4012 m

9 3= × =

But 31 m 1000 litres=

∴ 340 40m 1000

3 3= ×

40000litres

3=

∴ Volume of water discharged in 1 minutes i.e. 60 sec.

40000

60 800000 litres.3

= × =


Q.31. If 100.8 cubic metres of sand be thrown into a rectangular tank 14 m long

and 5 m wide; find the rise in level of the water. Ans. Length of rectangular tank 14 m=

Width of rectangular tank 5 m=

Let rise in level mx=

∴ Volume 314 5 70 mx x= × × =

But volume 3100.8 m= (Given)

∴ 70 100.8x =

100.8 1008 1

70 70 10x x= ⇒ = ×

⇒ 144

100x = ⇒ 1.44 mx =

Q.32. A rectangular card-board sheet has length

32 cm and breadth 26 cm. Squares of each

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.

Ans. Length of sheet 32 cm=

Breadth of sheet 26 cm=

Side of each square 3 cm=

∴ Inner length 32 2 3 32 6 26 cm= − × = − =

Inner breadth 26 2 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 container 3 cm=

∴ Volume of the container 26 cm 20 cm 3 cml b h= × × = × ×3

1560 cm=

Q.33. 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.

Ans. Length of pool 18 m=

Breadth of pool 8 m=

Height of one side 2 m=

Height on second side 1.2 m=

∴ Volume of pool 3(2 1.2)18 8 m

2

+= × ×

318 8 3.2230.4 m

2

× ×= =


Q.34. A plot is 83 m long and 35 m broad. A

tank 15 m long, 7 m broad and 4 m deep

is dug anywhere in the plot and the earth

so removed is evenly spread over the

remaining part of the plot. Find the rise in

level of the plot.

Ans. Length of plot 83 m= and width of plot 35 m=

∴ Area of plot 2 283 35 m 2905 m= × =

Area of tank 215 7 ml b= × = ×

2105 m=

∴ Area of remaining plot 2 2(2905 105) m 2800 m= − =

Volume of earth dug out 3 315 7 4 m 420 ml b d= × × = × × =

Height of level of earth spread over the remaining part Volume of earth

Area of plot=

420 3 3

m m 100 15 cm2800 20 20

= = = × =

Q.35. Find the volume of the cylinder in which :

(i) Height = 21 cm and Base radius = 5 cm

(ii) Diameter = 28 cm and Height = 40 cm.

Ans. (i) Height of cylinder ( ) 21 cmh = and radius of the base ( ) 5 cmr =

∴ Volume of cylinder 2r hπ=

3225 5 21 cm

7= × × ×

31650 cm=

(ii) Radius of the base of cylinder, 28

14 cm2

r = =

Height of cylinder ( ) 40 cmh =


32214 14 40 cm

7= × × ×

322 28 40 24640 cm= × × =

Q.36. Find the weight of the solid cylinder of radius 10.5 cm and height 60 cm, if

the material of the cylinder weighs, 5 grams per cu. cm.

Ans. Radius of solid cylinder ( ) 10.5 cmr =

Height of cylinder ( ) 60 cmh =


32210.5 10.5 60 cm

7= × × ×

322 1.5 10.5 60 20790 cm= × × × =

Weight of 1 cubic cm (i.e. 31 cm ) 5 g=


∴ Total weight of the cylinder 20790 5 g= × 103950 g=103950

kg1000

= 103.95 kg=

Q.37. A cylindrical tank has a capacity of 6160 m3. Find its depth if its radius is

14 m. Also find the cost of painting its curved surface at Rs 3 per m2.

Ans. Volume of cylinder 36160 m= and radius of cylinder ( ) 14 mr =

Let depth of cylinder be h m.

∴ Volume of cylinder 26160r hπ= =

⇒ 22

14 14 61607

h× × × = ⇒6160 7

10 m22 14 14

h×

= =× ×

Curved surface area of cylinder 2 rhπ=222

2 14 10 m7

= × × ×2

880 m=

Rate of painting its curved surface 2Rs 3 per m=

∴ Total cost 880 3 Rs 2640.= × =

Q.38. The curved surface area of a cylinder is 4400 cm2 and the circumference of

its base is 110 cm. Find the height and the volume of the cylinder.

Ans. Curved surface area 24400 cm= and circumference of its base 110 cm=

Let radius be r and height be h

Circumference of base 2 110rπ= =

⇒ 22

2 1107

r× = ⇒110 7 35

cm22 2 2

r×

= =×

Curved surface area of cylinder 2 4400rhπ= =

⇒ 22 35

2 44007 2

h× × × = ⇒4400 7 2

40 cm2 22 35

h× ×

= =× ×

Volume of the cylinder 2r hπ=

322 35 3540 cm

7 2 2= × × ×

338500 cm=

Q.39. The total surface area of a solid cylinder is 462 2cm and its curved surface

area is one-third of its total surface area. Find the volume of the cylinder.

Ans. Total surface area of cylinder 2462 cm=

and curved surface area 1

of total surface area3

=21

462 154 cm3

= × =

Let r be the radius and h be its height, then

2 154rhπ = ⇒22

2 1547

rh× = ⇒7 49

15444 2

rh = × = ...(i)

and Area of two bases 22 462 154 308rπ = − =

⇒ 2222 308

7r× × = ⇒ 2 2308 7

49 (7)2 22

r×

= = =×

∴ 7 cmr =


Now putting the value of r in (i), we get

49

72

h× = ⇒49 7

cm2 7 2

h = =×

Hence, volume of cylinder 2r hπ=

322 77 7 cm

7 2= × × ×

3539 cm=

Q.40. The sum of the radius of the base and the height of a solid cylinder is 37 m.

If the total surface area of the cylinder be 1628 m2, find its volume.

Ans. Let r be the radius and h be the height of the cylinder. Then

total surface area of cylinder 21628 m=

⇒ 2 ( ) 1628r r hπ + = ⇒22

2 37 16287

r× × =

⇒ 1628 7

72 22 37

r×

= =× ×

. But 37 7 37r h h+ = ⇒ + =

∴ 37 7 30 mh = − =

Volume of the cylinder 2r hπ=

3227 7 30 m

7= × × ×

34620 m=

Q.41. A road roller is cylindrical in shape. Its circular end has a diameter 0.7 m

and its width is 2 m. Find the least number of complete revolutions that the

roller must make in order to level a play-ground measuring 80 m by 44 m.

Ans. Diameter of cylinder 0.7 m= ∴ Radius of cylinder, 0.7

0.35 m2

r = =

Width of roller, 2 mh =

Curved surface area of roller 2 rhπ=222

2 0.35 2 4.40 m7

= × × × =

∴ In 1 revolution, area covered by roller 24.40 m=

Area of play-ground 280 44 3520 m= × =

∴ Number of revolutions 3520

4.40=

3520100 8 100 800.

440= × = × =

Q.42. How many cubic metres of earth must be dug out to make a well 28 m deep

and 2.8 m in diameter ? Also, find the cost of plastering its inner surface at

Rs 4.50 per sq. metre.

Ans. Depth of well, 28 mh = and diameter of well 2.8 m=

∴ Radius of well, 2.8

m 1.4 m2

r = =

∴ Volume of earth dug out 2r hπ=

3221.4 1.4 28 172.48 m

7= × × × =


Curved surface of well 2 rhπ=222

2 1.4 28 246.4 m7

= × × × =

Cost of platering at the rate of Rs 4.50 per sq. metre

246.4 4.50 Rs 1108.800= × =

Q.43. Find the length of 11 kg copper wire of diameter 0.4 cm. Given one cubic

cm of copper weighs 8.4 gram.

Ans. Let length of copper wire be h cm.

Diameter of copper wire 0.4 cm= then radius of copper wire ( ) 0.2 cmr =

∴ Volume of copper wire (cylinder) 2r hπ=

220.2 0.2

7h= × × ×

322 2 2 88

cm7 10 10 700

h h= × × × =

Weigth of 31 cm of wire 8.4 grams=

∴ Total weight of wire 88

8.4700

h= ×88 84

grams700 10

h= ×

But, weight of wire 11 kg= 11 1000 11000 gm= × =

∴ 88 84

11000700 10

h × =

11000 700 10

88 84h

× ×=

×⇒

125 100 10 125 25 10

12 3h

× × × ×= =

⇒ 31250

cm3

h =31250

m3 100

=×

625m

6= ∴ 104.17 mh =

Q.44. A cylinder has a diameter of 20 cm. The area of curved surface is 100 2cm

(sq. cm). Find :

(i) the height of the cylinder correct to one decimal place.

(ii) the volume of the cylinder correct to one decimal place.

Ans. (i) Diameter of cylinder 20 cm= then radius 20

10 cm2

r = =

Let height of cylinder be h cm.

Curved surface area of cylinder 2100 cm=

2 100rhπ =

22

2 10 1007

h× × × =

100 7

2 22 10h

×=

× ×

35cm

22= 1.6=


(ii) If h is taken as 1.6 cm

Then volume of cylinder 2r hπ= 322

10 10 1.6 502.9 cm7

= × × × =

Hence, (i) height of cylinder 1.6 cm= (ii) volume of cylinder 3502.9 cm .

Q.45. A metal pipe has a bore (inner diameter) of 5 cm. The pipe is 2 mm thick

all round. Find the weight in kilogram, of 2 metres of the pipe, if 1 cm3 of the

metal weighs 7.7 g.

Ans. Inner diameter 5 cm= ∴ Inner radius 1

52.5 cm

2r = =

Thickness 2 mm 0.2 cm= =

∴ Outer radius, 2 2.5 cm 0.2 cm 2.7 cmr = + =

Length of pipe 2 metres 200 cm= =

∴ Volume of metal in the pipe 2 22 1r h r hπ π= −

2 2 2 22 1

22[ ] 200 [(2.7) (2.5) ]

7h r rπ= − = × −

22200 [7.29 6.25]

7= × −

22

200 1.047

= × ×22 104

2007 100

= × ×344 104

cm7

×=

31 cm of metal weighs 7.7 gm=

∴ 44 104

7

× of metal weighs

44 1047.7

7

×= × 44 104 1.1 5033.6 gm= × × =

5033.6

kg1000

= 5.0336 kg= 5.034 kg=

Q.46. Find the total surface area of a hollow cylinder open at both ends, if its

length is 12 cm, external diameter is 8 cm and the thickness is 2 cm.

Ans. Length of hollow cylinder ( ) 12 cmh =

External diameter of cylinder 8 cm=

∴ External radius of cylinder 8

( ) cm 4 cm2

r = =

Thickness of the cylinder 2 cm=

Inner radius of cylinder 4 2 2 cm= − =

Thus, total surface area of cylinder

2 2 2[2 2 2 ( )] cmRh rh R rπ π π π= + + −

22 22

2 4 12 2 2 127 7

= × × × + × × ×

222 222 4 4 2 2 cm

7 7

+ × × − × ×


22112 1056 352 882 cm

7 7 7 7

= + + −

23168 2642 cm

7 7

= + ×

23168 528 3696528 cm

7 7 7= + = =

Q.47. Water is flowing at the rate of 8 m per second through a circular pipe

whose internal diameter is 2 cm, into a cylindrical tank, the radius of whose

base is 40 cm. Determine the increase in the water level in 30 minutes.

Ans. Diameter of pipe 2 cm= , ∴ Radius of pipe 1 cm=

Rate of water flow 8 m/sec=

Time taken 30 minutes=

∴ Length of water flow into the pipe ( ) 30 60 8 14400 mh = × × =

∴ Volume of water 2r hπ=

322 1 114400 m

7 100 100= × × ×

3144 22 m

700

×=

Radius of the base of the cylindrical tank 40 2

40 cm m m100 5

= = =

Let h be the water increased in the tank

∴ 2 144 22

700r hπ

×= ⇒

22 2 2 144 22

7 5 5 700h

×× × =

∴ 144 22 7 5 5

9 m700 22 2 2

h× × × ×

= =× × ×

Hence, height of water is 9 m.

Q.48. The sum of the inner and the outer curved surfaces of a hollow metallic

cylinder is 1056 cm2, and the volume of material in it is 1056 cm

3. Find the

internal and external radii. Given that the height of the cylinder is 21 cm.

Ans. Let r and R be the inner and outer radii and h be their height

∴ 2 2 1056Rh rhπ π+ =

⇒ 2 ( ) 1056h R rπ + =

⇒ 22

2 21 ( ) 10567

R r× × + =

⇒ 132 ( ) 1056R r+ = ⇒1056

8132

R r+ = = ...(i)

Again 2 21056R h r hπ π− = ⇒ 2 2( ) 1056h R rπ − =

⇒ 2 22221 ( ) 1056

7R r× − =

⇒ 2 266 ( ) 1056R r− =


⇒ 2 2 105616

66R r− = =

⇒ ( ) ( ) 16R r R r+ − =

⇒ 8 ( ) 16R r− =

⇒ 16

28

R r− = =

8R r+ = ...(ii)

2R r− = ...(iii)

Adding (ii) and (iii), we get

2 10 5R R= ⇒ =

Subtracting eqn. (iii) from (ii), we get

2 6 3r r= ⇒ =

Hence, outer radius of cylinder 5 cm= and inner radius of cylinder 3 cm.=

volume and surface areas of solids - · pdf file · 2009-08-24volume and surface...

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