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Page No 482:

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:

(i)
Here, l = 12 cm, b = 8 cm, h = 4.5 cm
Volume of the cuboid = l×b×h
                                  =(12×8×4.5) cm3= 432 cm3

Total Surface area = 2(lb + lh+ bh)
                              =2(12×8 + 12×4.5 +8×4.5) cm2=2(96 +54 + 36) cm2=2× 186  cm2=372  cm 2

Lateral surface area = 2l+b×h
                              =212+8×4.5 cm2=2(20)×4.5  cm2=40×4.5  cm2=180 cm2

(ii)
Here, l = 26 m; b = 14 m; h =6.5 m
Volume of the cuboid =  l×b×h
                                =26×14×6.5m3=2366 m3

Total surface area = 2(lb + lh+ bh)
                             =2(26×14+26×6.5+6.5×14) m2=2(364+169+91) m2=2×624 m2=1248 m2

Lateral surface area = 2l+b×h 
                              =226+14×6.5 m2=2×40×6.5 m2=520 m2

(iii)
Here, l = 15 m; b = 6 m; h = 5 dm = 0.5 m
Volume of the cuboid =  l×b×h
                                =(15×6×0.5) m3=45 m3


Total surface area = 2(lb + lh+ bh
                          =2(15×6+15×0.5+6×0.5) m2=2(90+7.5 +3) m2=2×100.5 m2= 201m2

Lateral surface area = 2l+b×h 
                               =215+6×0.5 m2=2×21×0.5 m2=21 m2

(iv)
Here, l = 24 m; b = 25 cm = 0.25 m; h =6 m
Volume of the cuboid =  l×b×h 
                                =24×0.25×6 m3=36 m3

Total Surface area = 2(lb + lh+ bh)
                            =2(24×0.25 +24×6 +0.25×6) m2=2(6+144 + 1.5) m2=2×151.5 m2=303 m2

Lateral surface area = 2l+b×h 
                             =224+0.25×6 m2=2×24.25×6 m2=291 m2

Page No 482:

Question 2:

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:

Here, l = 8 m; b = 6 m; h = 2.5 m
Capacity of the closed rectangular cistern = volume of the cistern = l×b×h
∴ Capacity of the cistern =8×6×2.5m3=120 m3

Also, area of the iron sheet required to make the cistern = total surface area of the cistern
                                                                                     = 2(lb+lh +bh)                       
                                                                                    =2(8×6+8×2.5 +6×2.5) m2=2(48+20 +15) m2=2×83 m2=166 m2

Page No 482:

Question 3:

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 6.40 per square metre.

Answer:

We need to white wash the four walls of the room
Here, l = 9 m; b = 8 m; h = 6.5 m
Area of the four walls = 2l+b×h 
                                 =29+8×6.5 m2=2×17×6.5 m2=221 m2
Now, the door and the windows are not to be white washed.
Area of the door = l×b=2×1.5=3 m2
Area of a window = l×b=1.5 ×1=1.5 m2

Area to be white washed = area of four walls − (area of door + area of two windows)
                                      = 221 − (3 + 2 × 1.5)
                                     = 221 − 6 = 215 m2                    
Cost of whitewashing = Rs 6.40/m2
∴ Cost of whitewashing 215 m2 = Rs 6.40 × 215 = Rs 1376

Page No 482:

Question 4:

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:

Number of planks = volume of the pit in cm3volume of 1 plank in cm3

Volume of one plank = (l×b×h) cm3
                                  =500×25×10  cm3= 125000 cm3
Volume of the pit=(l×b×h) cm3Here, l=20 m=2000 cm; b=6 m=600 cm; h=80 cmi.e., volume of the pit =2000×600×80 cm3                                  =96000000 cm3

∴ Number of planks = 96000000125000 = 96000125=768

Page No 482:

Question 5:

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:

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=800×22.5×600 cm3=10800000 cm3

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 =25×11.25×6 =1687.5 cm3

∴ Number of bricks = volume of the wallvolume of one brick = 108000001687.5=6400

Page No 482:

Question 6:

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 112 of the total volume of the wall consists of mortar, how many bricks are there in the wall?

Answer:

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 =  =1500×30×400 cm3= 18000000 cm3

Now, volume of each brick=22×12.5×7.5 cm3                                         =2062.5 cm3

Also, volume of the mortar = 112×volume of the wall
                                        =1800000012=1500000 cm3
Total volume of the bricks in the wall = volume of the wall − volume of the mortar
                                                           = (18000000 − 1500000) cm3= 16500000 cm3

∴ Number of bricks = volume of bricksvolume of one brick=165000002062.5=8000 bricks



Page No 483:

Question 7:

An open rectangular cistern when measured from outside is 1.35 m long, 1.08 m broad and 90 cm deep. It is made up of iron, which is 2.5 cm thick. Find the capacity of the cistern and the volume of the iron used.

Answer:

External length of the cistern = 1.35 m = 135 cm
External breadth of the cistern = 1.08 m = 108 m
External height of the cistern = 90 cm
∴ External volume of  the cistern=135×108×90 cm3 =1312200 cm3

In order to find the internal volume of the cistern, we need to determine the internal dimensions of the cistern.
Internal length of the cistern = 135 − (2.5 + 2.5) cm  = 130 cm
Internal breadth of the cistern = 108 − (2.5 + 2.5) = 103 cm                    [ Subtracting the thickness of iron from both sides ]
Internal height of the cistern = 90 − 2.5 cm = 87.5 cm
∴ Internal volume of the cistern = =130×103×87.5 =1171625 cm3

i.e., capacity of the cistern = 1171625 cm3
∴ Volume of iron used  = external volume of the cistern − internal volume of the cistern
                                    = (1312200 − 1171625) cm3 = 140575 cm3

Page No 483:

Question 8:

A river 2 m deep and 45 m wide is flowing at the rate of 3 km per hour. Find the volume of water that runs into the sea per minute.

Answer:

Breadth of the river = 45 m
Depth of the river = 2 m
Rate of flow of the river = 3 km/h =300060=50 m per min
∴ Volume of the water that runs into the sea per minute = 50×45×2=4500 m3

Page No 483:

Question 9:

A box made of sheet metal costs Rs 1620 at Rs 30 per square metre. If the box is 5 m long and 3 m wide, find its height.

Answer:

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

Area of the sheet required = total costcost per metre square 

Let h m be the height of the box.
Then area of the sheet = total surface area of the box    
                                 =2(lb+lh +bh) m2=2(5×3+5×h+3×h) m2=2(15+8h) =( 30+16h )m2

Now, 30+16h=16203030+16h=5416h = 24h= 1.5 m

∴ The height of the box is 1.5 m.

Page No 483:

Question 10:

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

Answer:

Length of the longest pole = length of the diagonal of the room
                                      =l2+b2+h2 m=102+102+52 m=100+100+25=225 =15 m

Page No 483:

Question 11:

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:

Volume of the dining hall = (20×16×4.5) m3=1440 m3

Volume of air required by each person = 5 m3

∴ Capacity of the dining hall = volume of dining hallvolume of air required by each person=14405=288 persons

Page No 483:

Question 12:

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:

Length of the classroom = 10 m
Breadth of the classroom = 6.4 m
Height of the classroom = 5 m
Area of the floor = length × breadth = 10 × 6.4 m2
No. of students=area of the floorarea given to one student on the floor =10×6.41.6=64016=40 students

Now, volume of the classroom=10×6.4×5 m3

 Air required by each student=volume of the roomnumber of students=10×6.4×540 m3=8 m3

Page No 483:

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:

Length of the cuboid = 16 m
Suppose that the breadth and height of the cuboid are 3x m and 2x m, respectively.
Then 1536=16×3x×2x1536=16×6x2x2=153696=16x=16=4

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

Page No 483:

Question 14:

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

Answer:

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
Then 758 = 2(14×11+14×x +11×x)758 = 2(154 +14x +11x)758  =2(154+25x) 758  = 308 +50x50x = 758-308 = 450x=45050=9

∴ The height of the cuboid is 9 cm.

Page No 483:

Question 15:

Find the volume, the lateral surface area, the total surface area and the diagonal of a cube, each of whose edges measures (a) 9 m, (b) 6.5 cm.

Answer:

(a)  Here, a = 9 m
Volume of the cube = a3=93 m3=729 m3
Lateral surface area of the cube = 4a2=4×92m2=4×81 m2=324 m2
Total surface area of the cube = 6a2= 6×92m2=6×81 m2=486 m2
∴ Diagonal of the cube = 3a=3×9=15.57 m

(b)  Here, a = 6.5 cm

Volume of the cube = a3=6.53 cm3=274.625 cm3
Lateral surface area of the cube = 4a2=4×6.52 cm2=169 cm2
Total surface area of the cube = 6a2= 6×6.52 cm2=253.5 cm2
∴ Diagonal of the cube = 3a cm=3×6.5 cm=11.245 cm

Page No 483:

Question 16:

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

Answer:

Suppose that the side of cube is x cm.
Total surface area of cube = 1176 sq cm
Then 1176 =6x2x2=11766=196x =196=14 
i.e., the side of the cube is 14 cm.

∴ Volume of the cube = x3 =143 cm3=2744 cm3

Page No 483:

Question 17:

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

Answer:

Suppose that the side of cube is x cm.
Lateral surface area of the cube = 900 cm2
Then 900= 4x2x2=9004=225x=225=15
i.e., the side of the cube is 15 cm.
∴ Volume of the given cube = x3 cm3=153 cm3=3375 cm3

Page No 483:

Question 18:

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

Answer:

Suppose that the side of the given cube is x cm.
Volume of the cube = 512 cm3
Then 512 =x3x =5123 =8i.e., the side of the cube is 8 cm.

∴ Surface area of the cube = 6x2 cm2=6×82 cm2=384 cm2

Page No 483:

Question 19:

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:

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
                                        =33+43+53cm3=(27+64+125) cm3=216 cm3

Suppose the edge of the new cube = x cm
Then we have:
Then 216=x3x=2163 =6 i.e., the edge of the new cube is 6 cm.

∴ Lateral surface area of the new cube = 4x2 cm2=4×62 cm2=144 cm2

Page No 483:

Question 20:

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

Answer:

Volume of the water that falls on the ground = area of ground×depth
                                                                 =20000×0.05   m3=1000 m3



Page No 492:

Question 1:

Find the volume and curved surface area of a right circular cylinder of height 21 cm and base radius 5 cm.

Answer:

Here, r = 5 cm; h = 21 cm

Volume of the cylinder=πr2h                                      =227×52×21 cm3                                     =22×25×3                                      = 1650 cm3

Curved surface area of the cylinder=2πrh                                                             =2×227×5×21  cm2                                                            =2×22×5×3  cm2                                                            =660 cm2

Page No 492:

Question 2:

The diameter of a cylinder is 28 cm and its height is 40 cm. Find the curved surface area, total surface area and the volume of the cylinder.

Answer:

Here, r = 28/2 = 14 cm; h = 40 cm

Curved surface area of the cylinder=2πrh                                                             =2×227×14×40  cm2                                                            =2×22×2×40 cm2                                                            =3520 cm2


Total surface area of the cylinder=2πrh + 2πr2                                                        = 3520 +2×227×142 cm2                                                       =(3520+1232 )=4752 cm2

 Volume of the cylinder=πr2h                                          =227×142×40 cm3                                         =22×14×2×40 cm3                                         =24640 cm3

Page No 492:

Question 3:

Find the weight of a solid cylinder of radius 10.5 cm and height 60 cm if the material of the cylinder weighs 5 g per cm2.

Answer:

Here, r = 10.5 cm; h = 60 cm

Now, volume of the cylinder=πr2h                                       =227×(10.5)2×60 cm3                                      =22×10.5×1.5×60 cm3=20790 cm3

∴ Weight of cylinder = volume of cylinder ×weight of cylinder per gram
                              =20790×5  g=103950 g=103.95 kg 

Page No 492:

Question 4:

The curved surface area of a cylinder is 1210 cm2 and its diameter is 20 cm. Find its height and volume.

Answer:

Curved surface area = 1210 cm2
Suppose that the height of cylinder is h cm.
We have r = 10 cm
Now, 1210 =2πrh1210 =2×227×10×h h =1210×72×22×10=11×72×2=19.25 cm

 Volume of the cylinder=πr2h                                            =227×102×19.25 cm3                                           = 2200×2.75 cm3 =6050 cm3

Page No 492:

Question 5:

The curved surface area of a cylinder is 4400 cm2 and the circumference of its base is 10 cm. Find the height and the volume of the cylinder.

Answer:

Suppose that the radius is r cm and the height is h cm.
Here, curved surface area = 4400 cm2; circumference = 10 cm

Then 2πrh=4400; 2πr = 110i.e., 110×h=4400h=4400÷110=40 cmAlso, 2×227×r=110r =110×72×22=5×72=352cm

 Volume of the cylinder=πr2h                                            =227×352×352×40                                           =22×35×5×10=38500 cm3



Page No 493:

Question 6:

The radius of the base and the height of a cylinder are in the ratio 2 : 3. If its volume is 1617 cm3, find the total surface area of the cylinder.

Answer:

Suppose that the radius of the base and the height of the cylinder are 2x cm and 3x cm, respectively.
Then 1617=πr2h =227×(2x)2×3x                   = 227×12x3x3=1617×722×12=42.875x=42.8753 = 3.5 cm

Hence, radius = 7 cm; height of the cylinder = 10.5 cm

 Total surface area of the cylinder=2πrh+2πr2                                                         =2×227(7×10.5+7×7) cm2                                                       =447×(73.5+49) cm2 =770  cm2

Page No 493:

Question 7:

The total surface area of a cylinder is 462 cm2. Its curved surface area is one-third of its total surface area. Find the volume of the cylinder.

Answer:

Total surface area = 462 cm2

Given: Curved surface area =13×total surface area = 13×462=154 cm2

Now, total surface area − curved surface area = 2πrh +2πr2 -2πrh
462 -154 =2πr2308 =2×227×r2r2=308×744=49r = 7 cm

Now, curved surface area = 154 cm2

2πrh = 1542×227×7×h=154h =15444=3.5 cm

  Volume of the cylinder=πr2h                                              =227×72×3.5                                           =539 cm3

Page No 493:

Question 8:

The total surface area of a solid cylinder is 231 cm2 and its curved surface area is 23 of the total surface area. Find the volume of the cylinder.

Answer:

Curved surface area = 23×total surface area = 23×231 = 2×77 =154 cm2

Now, total surface area − curved surface area = 2πrh +2πr2-2πrh

Then 231-154=2πr22×227×r2=77r2=77×744=12.25r=3.5 cmAlso, curved surface area=154 cm22πrh = 1542×227×3.5×h =154h=154×744×3.5=7 cm

Volume of the cylinder=πr2h                                           =227×(3.5)2×7=269.5 cm3

Page No 493:

Question 9:

The sum of the height and radius of the base of a solid cylinder is 37 m. If the total surface area of the cylinder be 1628 m2, find its volume.

Answer:

We have  r + h = 37, where r m is the radius and h m is the height of the given cylinder.
i.e., h = 37 − r
Total surface area = 1628 m2
2πrh+2πr2=16282×227r37-r+r2=162837r-r2+r2=1628×74437r=37×7r=7 m

Then h = 37 − 7 = 30 m

Volume of the cylinder=πr2h                                           =227×72×30=4620 m3

Page No 493:

Question 10:

The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. Find the volume of the cylinder if its total surface area is 616 cm2.

Answer:

Suppose that the curved surface area and the total surface area of the right circular cylinder are x cm2 and 2x cm2.
Then we have:
2x = 616
x = 308 sq cm
Hence, the curved surface area of the cylinder is 308 sq cm.
Let  r cm and h cm be the radius and height of the cylinder, respectively.
Then 2πrh +2πr2 =616 cm2  and 2πrh = 308 cm2 2πrh +2πr2 - 2πrh =616-3082πr2=3082×227×r2=308r2=308×744=49r= 7 cmNow, 2πrh = 308 cm22×227×7×h =308h =30844=7 cm

 Volume of the cylinder=πr2h  cubic cm                                          =227×72×7                                           =1078 cm3

Page No 493:

Question 11:

1 cm3 of gold is drawn into a wire 0.1 mm in diameter. Find the length of the wire.

Answer:

Let r cm be the radius of the wire and h cm be the height of the wire.
Volume of the gold = 1 cm3
1 cm3 of gold is drawn into a wire of diameter 0.1 mm.
Here, r 0.1 mm=0.120cm=1200cm
 πr2h=1227×1200×1200×h =1h=40000×722=12727.27 cm

Hence, length of the wire = 127.27 m

Page No 493:

Question 12:

The radii of two cylinder are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of their volumes and the ratio of their curved surfaces.

Answer:

Suppose that the radii of the cylinders are 2r and 3r and the respective heights are 5h and 3h.
Now, ratio of their volumes = volume of first cylindervolume of second cylinder=π(2r)2×5hπ3r2×3h= 4×59×3=2027=20 : 27

Also, ratio of their curved surface areas = 2π×2r×5h2π×3r×3h=109=10:9

Page No 493:

Question 13:

A powder tin has a square base with side 12 cm and height 17.5 cm. Another is cylindrical with diameter of its base 12 cm and height 17.5 cm. Which has more capacity and by how much?

Answer:

Length of the powder tin =12 cm
Breadth of the powder tin =12 cm
Height of the powder tin =17.5 cm

 Volume of the powder tin =l×b×h  =12×12×17.5 cm3=2520 cm3

Now, radius of the cylinder = 6 cm
Height of the cylinder = 17.5 cm
Volume of the cylinder=πr2h                                      =227×62×17.5                                    =1980  cm3

∴ The capacity of the powder tin is more than that of the cylinder by (2520 -1980) cm3 = 540 cm3

Page No 493:

Question 14:

A cylindrical bucket, 28 cm in diameter and 72 cm and high, is full of water. The water is emptied into a rectangular tank, 66 cm long and 28 cm wide. Find the height of the water level in the tank.

Answer:

 Given: Diameter of the cylindrical bucket = 28 cm
i.e., radius = 14 cm
Height of the cylindrical bucket, h1 = 72 cm
Length of the rectangular tank, l = 66 cm
Breadth of the rectangular tank, b = 28 cm
Let the height of the rectangular tank be h cm.
The water from the cylindrical bucket is emptied into the rectangular tank.
i.e., volume of the bucket = volume of the tank
πr2h1 =l×b×h227×142×72=66×28×hh =22×14×2×7266×28=24 cm

∴ Height of the rectangular tank = 24 cm

Page No 493:

Question 15:

If 1 cm3 of case iron weighs 21 g, find the weight of a cast iron pipe of length 1 m with a bore of 3 cm in which the thickness of the metal is 1 cm.

Answer:

Internal radius of the pipe = 1.5 cm
External radius of the pipe = (1.5 + 1) cm = 2.5 cm
Height of the pipe = 1 m = 100 cm
Volume of the cast iron = total volume of the pipe − internal volume of the pipe
                                   =π×(2.5)2×100-π×1.52×100=227×100×6.25-2.25=22007×4=88007cm3
1 cm3 of cast iron weighs 21 g.

∴ Weight of 88007cm3cast iron = 88007×211000 kg=88×0.3=26.4 kg

Page No 493:

Question 16:

A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal is 8 mm everywhere. Calculate the volume of the metal.

Answer:

Inner radius of the cylindrical tube, r = 5.2 cm
Height of the cylindrical tube, h = 25 cm
Outer radius of the cylindrical tube, R = (5.2 + 0.8) cm = 6 cm

 Volume of the metal=external volume - internal volume                                         = πR2h -πr2h       (where R and r are the outer and inner radii, respectively)                                        =227×25×62-5.22                                         =227×25×36-27.04                                           =227×25×8.96=704  cm3

Page No 493:

Question 17:

The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up on writing 330 words on an average. How many words would use up a bottle of ink containing one-fifth of a litre?

Answer:

Height of the barrel = h = 7 cm
Radius of the barrel = r = 2.5 mm = 0.25 cm
Volume of the barrel = πr2h 
                               =227×0.25×0.25×7 =22×0.25×0.25=1.375  cm3

i.e., 1.375 cm3 of ink is used for writing 330 words
Now, number of words that could be written with one-fifth of a litre, i.e., 15×1000 cm3 = 330×11.375×15×1000=48000   
∴ 48000 words would use up a bottle of ink containing one-fifith of a litre.  

Page No 493:

Question 18:

A lead pencil consists of a cylinder of wood with a solid cylinder of graphite fitted into it. The diameter of the pencil is 7 mm, the diameter of the graphite is 1 mm and the length of the pencil is 10 cm. Calculate the weight of the whose pencil, if the specific gravity of the wood is 0.7 g/cm2 and that of the graphite is 2.1 g/cm2.

Answer:

Here, r = 72 mm=0.35 cm; h = 10 cm, where r and h are the radius and height of the pencil, respectively.

 Volume of graphite=πr2h                                =227×0.052×10 =0.557cm3

 Weight of graphite=volume×specfic gravity of graphite                                      =0.557×2.1=0.165 g

Similarly, weight of the wood = volume of the wood ×specific gravity of wood
 Now, volume of the wood = volume of the pencil - volume of graphite                            =227×0.352-0.052×10×0.7                              =227×0.1225-0.0025×7                            =22×0.12 = 2.64  g

∴ Total weight of the pencil = weight of the wood + weight of graphite
                                          = (0.165 + 2.64)  g =  2.805 g                               



Page No 500:

Question 1:

Find the volume, curved surface area and the total surface area of a cone having base radius 35 cm and height 84 cm.

Answer:

Here, h = 84 cm and r = 35 cm
Volume of the cone = 13πr2h 
                              =13×227×352×84  cm3= 22×5×35×28=107800 cm3

Also,  l = h2+r2 =352+842=1225+7056 =8281= 91 cm

Curved surface area = πrl
                                =227×35×91 cm2=10010 cm2

∴ Total surface area = (πrl +πr2 )sq units
                               =10010 +227×352=10010 +3850=13860 cm2

Page No 500:

Question 2:

Find the volume, curved surface area and the total surface area of a cone whose height and slant height are 6 cm and 10 cm respectively.

Answer:

Height of the cone, h = 6 cm
Slant height of the cone, l = 10 cm
Radius, r = l2-h2=100-36 =64 =8 cm

Volume of the cone = πr2h 
                              =13×3.14×82×6=401.92 cm3

Curved surface area of the cone = πrl 
                                                =3.14×8×10=251.2 cm2

∴ Total surface area = πrl+πr2 
                               =251.2 + 3.14×82=452.16 cm2

Page No 500:

Question 3:

The volume of a right circular cone is (100π) cm3 and its height is 12 cm. Find its slant height and its curved surface area.

Answer:

Volume of the cone = (100π) cm3  and  h = 12 cm

 13πr2h = 100π13r2×12 = 100r2 =30012=1004r =25=5 cm

Now, l =r2+h2    =52+122   =25+144   =169l = 13 cm


 Curved surface area of the cone =πrl                                                   =π×5×13                                                    =65π cm2

Page No 500:

Question 4:

The circumference of the base of a cone is 44 cm and its slant height is 25 cm. Find the volume and curved surface area of the cone.

Answer:

Slant height, l = 25 cm
Circumference of the base of the cone = 44 cm

i.e., 2πr = 44r =442π=44×72×22=7 cm

Now, h =l2-r2=252-72     =625-49      =576      =24 cm


Also, volume of the cone = 13πr2h                                           =13×227×72×24                                             =1232 cm3

 Curved surface area of the cone =πrl                                                 =227×7×25                                                   =550 cm2

Page No 500:

Question 5:

A cone of slant height 25 cm has a curved surface are 550 cm2. Find the height and volume of the cone.

Answer:

Slant height of the cone = 25 cm
Curved surface area of the = 550 cm2

i.e., πrl =550227×r×25 = 550r =550×722×25r =7cm

Also, h=l2-r2  =252-72  =625-49  =576  =24 cm


 Volume of the cone =13πr2h                                 =13×227×72×24                                  =1232 cm3

Page No 500:

Question 6:

Find the volume of a cone having radius of the base 35 cm and slant height 37 cm.

Answer:

Radius of the base of the cone, r = 35 cm
Slant height of the cone, l = 37 cm

Now, h=l2-r2   =372-352    =1369-1225    =144    =12 cm

∴ Volume of the cone = 13πr2h 
                                 =13×227×352×12=22×5×35×4=15400 cm3

Page No 500:

Question 7:

The curved surface area of a cone is 4070 cm2 and its diameter is 70 cm. Find its slant height.

Answer:

Diameter of the cone = 70 cm
i.e., radius, r = 35 cm
Curved surface area = 4070 cm2

 πrl=4070227×35×l=4070l=4070×722×35 l=37 cm

Hence, slant height of the cone = 37 cm

Page No 500:

Question 8:

How many metres of cloth, 2.5 m wide, will be required to make a conical tent whose base radius is 7 m and height 24 metres?

Answer:

Radius of the conical tent, r = 7 m
Height of the conical tent, h = 24 m
 Now, l=r2+h2 =49 +576  =625   =25 m

 Curved surface area of the cone =πrl                                                       =227×7×25                                                 = 550 m2Here, area of the cloth =curved surface area of the cone = 550 m2

Width of the cloth = 2.5 m
∴ Length of the cloth = area of the clothwidth of the cloth=5502.5=220 m

Page No 500:

Question 9:

A right circular cone is 3.6 cm high and the radius of its base is 1.6 cm. It is melted and recast into a right circular cone having base radius 1.2 cm. Find its height.

Answer:

Let the cone which is being melted be denoted by cone 1 and let the cone into which cone 1 is being melted be denoted by cone 2.

Height of cone 1 = 3.6 cm
Radius of the base of cone 1 = 1.6 cm
Radius of the base of cone 2 = 1.2 cm
Let h cm be the height of cone 2.
The volumes of both the cones should be equal.

 i.e., 13π×1.62×3.6 =13π×1.22×hh  = 1.6×1.6×3.61.2×1.2=6.4 cm

∴ Height of cone 2 = 6.4 cm

Page No 500:

Question 10:

Two cones have their heights in the ratio 1 : 3 and the radii of their bases in the ratio 3 : 1. Show that their volumes are in the ratio 3 : 1.

Answer:

Let the heights of the first and second cones be h  and 3h, respectively .
Also, let the radius of the first and second cones be 3r and r,  respectively.

∴ Ratio of volumes of the cones = volume of the first conevolume of the second cone
                            
                                               =13π×3r2×h13π×r2×3h=9r2h3r2h=3 : 1 

Page No 500:

Question 11:

A circus tent is cylindrical to a height of 3 metres and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m wide to make the required tent.

Answer:

Radius of the tent, r = 52.5 m
Height of the cylindrical portion of the tent, H = 3 m
Slant height of the conical portion of the tent, l = 53 m
The tent is a combination of a cylindrical and a conical portion.
i.e., area of the canvas = curved surface area of the cone + curved surface area of the cylinder
                                   = πrl +2πrH  
                                  =227×105253 +2×3 =227×1052×53+6=9735 m2

But area of the canvas = length × breadth

∴ Length of the canvas = areabreadth=97355=1947 m
                       

Page No 500:

Question 12:

A conical tent is to accommodate 11 persons. Each person must have 4 m2 of the space on the ground and 20 m3 of air to breath. Find the height of the cone.

Answer:

Suppose that the radius of base is r m and the height is h m.
Area of base of the tent = number of people × area required by 1 person
                                    = 11×4 = 44 m2

Now, volume of the cone = number of people ×volume of air to breadth for 1 person
                                      = 11×20 = 220 m3

 πr2 = 44r2=44×722=14

Also, 13πr2h =22013×227×14×h=220h = 220×2122×14=15 m

∴ Height of the tent = 15 m



Page No 501:

Question 13:

A cylindrical bucket, 32 cm high and 18 cm of radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and the slant height of the heap.

Answer:

Let the radius of the heap and slant height be R cm and l cm, respectively.
Radius of the cylindrical bucket = 18 cm
Height of the cylindrical bucket=32 cm
Height of the conical heap =24 cm
The volume of sand remains the same in the bucket and in the heap.
π×182×32=13π×R2×24R2=3×18×18×3224=1296R =1296=36 cm

 Slant height of the conical heap=R2+H2                                                            =362+242                                                            =1296+576                                                            =1872                                                             =43.27 cm

Page No 501:

Question 14:

A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5., show that the radius and height of each has the ratio 3 : 4.

Answer:

Suppose that the respective radii and height of the cone and the cylinder are r and h.
Then ratio of curved surface areas = 8 : 5
Let the curved surfaces areas be 8x and 5x.

   i.e., 2πrh = 8x and πrl = 5x πrh2+r2=5xHence 4π2r2h2=64x2 and π2r2(h2+r2)=25x2 Ratio of curved surface areas=4π2r2h2π2r2(h2+r2) =6425  4π2r2h2π2r2(h2+r2) =64254h2(h2+r2)=6425     25h2 =16h2 +16r2   9h2=16r2   r2h2 =916    rh=34                                             

∴ The ratio of the radius and height of the cone is 3 : 4.

Page No 501:

Question 15:

An iron pillar consists of a cylindrical portion 2.8 m high and 20 cm in diameter and a cone 42 cm high is surmounting it. Find the weight of the pillar, given that 1 cm3 of iron weights 7.5 g.

Answer:

Height of the cylindrical portion of the iron pillar, h = 2.8 m = 280 cm
Radius of the cylindrical portion of the iron pillar, r = 20 cm
Height of the cone which is surmounted on the cylindrical portion, H = 42 cm
Now, volume of the pillar = volume of the cylindrical portion + volume of the conical portion
                                       = πr2h +13πr2H   
                                      =227×102280+13×42=227×280+14=227×100×294=92400 cm3

∴ Weight of the pillar = volume of the pillar × weight per cubic cm
                                 =92400×7.51000=693 kg

Page No 501:

Question 16:

The height of a cone is 30 cm. A small cone  is cut off at the top by a plane parallel to the base. If its volume be 127 of the volume of the given cone, at what height above the base, the section has been made?

Answer:




Suppose that the smaller cone has radius r cm, height h cm and the original cone has radius R cm.
Height of the original cone = 30 cm
Considering ABC & ADE, we have:
 BAC=DAE         (Common)ABC=ADE          (90° each)i.e., ABC~ADE   (AA similarity condition)
Therefore, we can say that
rR=h30r=hR30
Also, volume of a small cone = 127 volume of the original cone
13πr2h =127×13πR2×30r2h =127×30R2hR302h=109R2h3=10×9009=1000h=10 cm
i.e., height of the small cone = 10 cm.

∴ The section has been made at a height of (30 − 10) cm = 20 cm above the base of original cone.

Page No 501:

Question 17:

From a solid right circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and base is removed. Find the volume of the remaining solid.

Answer:

Height of the cylinder = 10 cm
Radius of the cylinder = 6 cm
The respective heights and radii of the cone and the cylinder are the same.
∴ Volume of the remaining solid = volume of the cylinder − volume of the cone
                                                 =πr2h -13πr2h  =23πr2h=23×3.14×62×10=753.6 cm3

Page No 501:

Question 18:

Water flows at the rate of 10 metres per minute through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter art the surface 40 cm and depth 24 cm?

Answer:

Radius of the cylindrical pipe = 2.5 mm = 0.25 cm
Water flowing per minute = 10 m = 1000 cm
Volume of water flowing per minute through the cylindrical pipe = π(0.25)21000 cm3 = 196.4 cm3
Radius of the the conical vessel = 40 cm
Depth of the vessel = 24 cm
Volume of the vessel = 13π(20)224=10057.1 cm3
Let the time taken to fill the conical vessel be x min.
Volume of water flowing per minute through the cylindrical pipe × x = volume of the conical vessel
x=10057.1196.4=51 min 12 sec
∴ The cylindrical pipe would take 51 min 12 sec to fill the conical vessel.



Page No 508:

Question 1:

Find the volume and surface area of a sphere whose radius is:
(i) 3.5 cm
(ii) 4.2 cm
(iii) 5 m

Answer:

(i) Radius of the sphere = 3.5 cm
Now, volume = 43πr3 
                     =43×227×3.5×3.5×3.5=179.67 cm3

∴ Surface area = 4πr2
                       =4×227×3.5×3.5=154 cm2

(ii) Radius of the sphere=4.2 cm
Now, volume = 43πr3 
                      =43×227×4.2×4.2×4.2=310.46 cm3

∴ Surface area = 4πr2 
                         =4×227×4.2×4.2=221.76 cm2

(iii) Radius of sphere=5 m
Now, volume = 43πr3 
                      =43×227×53=523.81 cm3

∴ Surface area = 4πr2 
                       =4×227×52= 314.29 cm2

Page No 508:

Question 2:

The volume of a sphere is 38808 cm3. Find its radius and hence its surface area.

Answer:

Volume of the sphere = 38808 cm3
Suppose that r cm is the radius of the given sphere.

 43πr 3=38808r3=38808×3×74×22=9261r =92613 =21 cm

∴ Surface area of the sphere =4πr2 
                                            =4×227×21×21=5544 cm2

Page No 508:

Question 3:

Find the surface area of a sphere whose volume is 606.375 m3.

Answer:

Volume of the sphere = 606.375 m3

Then 43πr3=606.375r3=606.375×3×74×22=144.703r =5.25 m

∴ Surface area = 4πr2 
                      =4×227×5.25×5.25=346.5 m2

Page No 508:

Question 4:

The surface area of a sphere is 394.24 cm2. Find its radius and volume.

Answer:

Surface area = 394.24 m2
Suppose that r cm is the radius of the sphere.

Then 4πr2=394.24r2=394.24×74×22=31.36r = 5.6 m

  Volume of the sphere=43πr3                                           =43×227×5.6×5.6×5.6                                           =735.91 m3

Page No 508:

Question 5:

The surface area of a sphere is (576π) cm2. Find its volume.

Answer:

Surface area of the sphere = (576π) cm2
Suppose that r cm is the radius of the sphere.
Then 4πr2=576πr2=5764=144r=12 cm

  Volume of the sphere=43×π×12×12×12 cm3                                           =2304π cm3

Page No 508:

Question 6:

The outer diameter of a spherical shell is 12 cm and its inner diameter is 8 cm. Find the volume of metal contained in the shell. Also, find its outer surface area.

Answer:

Outer radius of the spherical shell = 6 cm
Inner radius of the spherical shell = 4 cm

Volume of metal contained in the shell =43×22763-43                                                            =8821×216-64                                                             =8821×152                                                             =636.95 cm3


 Outer surface area =4×227×6×6                                 =452.57 cm2

Page No 508:

Question 7:

How many lead shots, each 3 mm in diameter, can be made from a cuboid with dimensions (12 cm × 11 cm × 9 cm)?

Answer:

Here, l = 12 cm, b = 11 cm and h = 9 cm

Volume of the cuboid =l×b×h                                         =12×11×9                                        = 1188 cm3

Radius of one lead shot = 3 mm= 0.32cm

 Volume of one lead shot =43×227×0.323                                                         =11×97000                                          =0.014 cm3

 Number of lead shots =volume of the cuboidvolume of one lead shot                                           =11880.014                                            =84857.1484857 

Page No 508:

Question 8:

How many lead balls, each of radius 1 cm, can be made from a sphere of radius 8 cm?

Answer:

Radius of the sphere = 8 cm

Volume of the sphere = 43πr3 =43×227×83=2145.52 cm3
Radius of one lead ball = 1 cm
Volume of one lead ball = 43×227×13=4.19 cm3

∴ Number of lead balls = volume of the spherevolume of one lead ball=2145.524.19=512.05512

Page No 508:

Question 9:

A solid sphere of radius 3 cm is melted and then cast into smaller spherical balls, each of diameter 0.6 cm. Find the number of small balls thus obtained.

Answer:

Radius of the solid sphere = 3 cm

Volume of the solid sphere = 43πr3 
                                         = 43×227×33 cm3

Diameter of the spherical ball = 0.6 cm
Radius of the spherical ball = 0.3 cm

Volume of the spherical ball = 43×227×0.33 cm3

Now, number of small spherical balls = volume of the spherevolume of the spherical ball
                                                        =43π×2743π×0.33 =1000

∴ The number of small balls thus obtained is 1000

Page No 508:

Question 10:

A metallic sphere of radius 10.5 cm is melted and then recast into smaller cones, each of radius 3.5 cm and height 3 cm. How many cones are obtained?

Answer:

Radius of the metallic sphere, r = 10.5 cm
Volume of the sphere = 43πr3=43π(10.5)3 cm3
Radius of each smaller cone, r2 = 3.05 cm
Height of each smaller cone = 3 cm
Volume of each smaller cone =13πr22h=13π(3.05)2×3 cm3
Number of cones obtained = volume of the spherevolume of each smaller cone
                                        =43πr313πr22h=4×10.5×10.5×10.53.5×3.5×3=126.006126

∴ 126 cones are obtained.

Page No 508:

Question 11:

How many spheres 12 cm in diameter can be made from a metallic cylinder of diameter 8 cm and height 90 cm?

Answer:

Diameter of each sphere, d=12 cm
Radius of each sphere, r=6 cm
Volume of each sphere=43πr3=43π(6)3 cm3
Diameter of base of  the cylinder, D=8 cm
Radius of base of cylinder, R=4 cm
Height of the cylinder, h = 90 cm
Volume of the cylinder=πR2h=π(4)2×90 cm3

Number of spheres= volume of the cylindervolume of the sphere
                              =πR2h43πr3=42×90×34×63=12×90216 =5

∴ Five spheres can be made.



Page No 509:

Question 12:

The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 mm. Find the length of the wire.

Answer:

Let the length of the wire be h cm.
Radius of the wire, r = 1 mm = 0.1 cm
Radius of the sphere, R = 3 cm
Now, volume of the sphere = volume of the cylindrical wire
43πR3=πr2h4×32=0.12×hh=4×90.1×0.1=3600 cm= 36 m

∴ Length of the wire = 36 m

Page No 509:

Question 13:

The diameter of a copper sphere is 18 cm. It is melted and drawn into a long wire of uniform cross section. If the length of the wire is 108 m, find its diameter.

Answer:

Radius of the copper sphere, R = 9 cm
Length of the wire, h = 108 m = 10800 cm
Volume of the sphere = volume of the wire
Suppose that r cm is the radius of the wire.

Then 43πR3=πr2h43×93=r2×10800r2=4×7293×10800=4×813×1200=9100r=310=0.3 cm

∴ Diameter of the wire = 0.6 cm

Page No 509:

Question 14:

A sphere of diameter 15.6 cm is melted and cast into a right circular cone of height 31.2 cm. Find the diameter of the base of the cone.

Answer:

Radius of the sphere, r = 7.8 cm
Height of the cone, h = 31.2 cm
Suppose that R cm is the radius of the cone.
Now, volume of the sphere = volume of the cone
43πr3=13πR2h4×7.83=R2×31.2R2=4×7.8×7.8×7.831.2=60.84R = 7.8 cm

∴ The diameter of the cone is 15.6 cm.

Page No 509:

Question 15:

A spherical cannonball 28 cm in diameter is melted and cast into a right circular cone would, whose base is 35 cm in diameter. Find the height of the cone.

Answer:

Radius of the spherical cannonball, R = 14 cm
Radius of the base of the cone, r = 17.5 cm
Let h cm be the height of the cone.
Now, volume of the sphere = volume of the cone
43πR3=13πr2h4×143=17.52×hh=4×14×14×1417.5×17.5=35.84 cm

∴ The height of the cone is 35.84 cm.

Page No 509:

Question 16:

A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5 cm and 2 cm. Find the radius of the third ball.

Answer:

Radius of the original spherical ball = 3 cm
Suppose that the radius of third ball is r cm.
Then volume of the original spherical ball = volume of the three spherical balls
43π×33=43π×1.53+43π×23+43π×r327 =3.375+8+r3r3=27-11.375 = 15.625r = 2.5 cm

∴ The radius of the third ball is 2.5 cm.

Page No 509:

Question 17:

The radii of two spheres are in the ratio 1 : 2. Find the ratio of their surface areas.

Answer:

Suppose that the radii are r and 2r.
Now, ratio of the surface areas  = 4πr24π2r2=r24r2=14
                                                     = 1:4

∴ The ratio of their surface areas is 1 : 4.

Page No 509:

Question 18:

The surface areas of two spheres are in the ratio 1 : 4. Find the ratio of their volumes.

Answer:

Suppose that the radii of the spheres are r and R.
We have:
4πr24πR2=14rR=14=12

Now, ratio of the volumes = 43πr343πR3=rR3=123 =18

∴ The ratio of the volumes of the spheres is 1 : 8.

Page No 509:

Question 19:

A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball?

Answer:

Suppose that the radius of the ball is r cm.
Radius of the cylindrical tub =12 cm
Depth of the tub= 20 cm
Now, volume of the ball = volume of water raised in the cylinder
43πr3=π×122×6.75r3=144×6.75×34=36×6.5×3=729r = 9 cm

∴ The radius of the ball is 9 cm.

Page No 509:

Question 20:

A cylindrical bucket with base radius 15 cm is filled with water up to a height of 20 cm. A heavy iron spherical ball of radius 9 cm is dropped into the bucket to submerge completely in the water. Find the increase in the level of water.

Answer:

Let h cm be the increase in the level of water.
Radius of the cylindrical bucket = 15 cm
Height up to which water is being filled = 20 cm
Radius of the spherical ball = 9 cm
Now, volume of the sphere = increased in volume of the cylinder
43π×93=π×152×hh=4×7293×15×15=4×2725=4.32 cm

∴ The increase in the level of water is 4.32 cm.

Page No 509:

Question 21:

A hemisphere of lead of radius 9 cm is cast into a right circular cone of height 72 cm. Find the radius of the base of the cone.

Answer:

Radius of the hemisphere = 9 cm
Height of the right circular cone = 72 cm
Suppose that the radius of the base of the cone is r cm.
Volume of the hemisphere = volume of the cone

23π×93=13π×r2×72r2=2×9×9×972=814r =92=4.5 cm

∴ The radius of the base of the cone is 4.5 cm.

Page No 509:

Question 22:

A hemispherical bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical shaped small bottles of diameter 3 cm and height 4 cm. How many bottles are required to empty the bowl?

Answer:

Internal radius of the hemispherical bowl = 9 cm
Radius of a cylindrical shaped bottle = 1.5 cm
Height of a bottle = 4 cm

Number of bottles required to empty the bowl  = volume of the hemispherical bowlvolume of a cylindrical shaped bottle
                                                                        =23π×93π×1.52×4=2×9×9×93×1.5×1.5×4=54

∴ 54 bottles are required to empty the bowl.

Page No 509:

Question 23:

A hollow spherical shell is made of a metal of density 4.5 g per cm3. If its internal and external radii are 8 cm and 9 cm respectively, find the weight of the shell.

Answer:

Internal radius of the hollow spherical shell, r= 8 cm
External radius of the hollow spherical shell, R= 9 cm

Volume of the shell =  43πR3-r3 
                            =43π93-83=43×227×729-512=4×22×21721=88×313=27283cm3

Weight of the shell = volume of the shell × density per cubic cm
                             = 27283×4.54092 g = 4.092 kg

∴ Weight of the shell = 4.092 kg

Page No 509:

Question 24:

A hemispherical bowl is made of steel 0.5 cm thick. The inside radius of the bowl is 4 cm. Find the volume of steel used in making the bowl.

Answer:

Internal radius of the hemispherical bowl = 4 cm
Thickness of a the bowl = 0.5 cm
Now, external radius of the bowl = (4 + 0.5 ) cm = 4.5 cm

Now, volume of steel used in making the bowl =  volume of the shell
                                                                       =23π4.53-43 =23×227×91.125-64=23×227×27.125= 56.83 cm3

∴ 56.83 cm3 of steel is used in making the bowl .



Page No 512:

Question 1:

The length, breadth and height of a cuboid are 15 cm, 12 cm and 4.5 cm respectively. Its volume is
(a) 243 cm3
(b) 405 cm3
(c) 810 cm3
(d) 603 cm3

Answer:

(c) 810 cm3

 Volume of the cuboid=l×b×h                                          = 15×12×4.5  cm3                                     =810 cm3

Page No 512:

Question 2:

A cuboid is 12 cm long, 9 cm broad and 8 cm high. Its total surface area is
(a) 864 cm2
(b) 552 cm2
(c) 432 cm2
(d) 276 cm2

Answer:

(b) 552 cm2

Total surface area of the cuboid =2lb+bh+lh cm2                                                 =2(12×9+8×9+12×8) cm2                                                 =2108+72+96 cm2                                                 =552 cm2



Page No 513:

Question 4:

A beam 9 m long, 40 cm wide and 20 cm high is made up of iron which weighs 50 kg per cubic metre. The weight of the beam is
(a) 27 kg
(b) 48 kg
(c) 36 kg
(d) 56 kg

Answer:

(c) 36 kg
Length = 9 m
Breadth = 40 cm = 0.4 m
Height = 20 cm = 0.2 m
∴ Weight of the beam = volume of the beam ×weight of iron per cubic metre
                              =9×0.4×0.2×50=36 kg

Page No 513:

Question 5:

The length of the longest rod that can be placed in a room of dimensions (10 m × 10 m × 5 m) is
(a) 15 m
(b) 16 m
(c) 105 m
(d) 12 m

Answer:

(a) 15 m

Length of longest rod = diagonal of the room
                                 = diagonal of a cuboid
                                 =l2+b2+h2 =100+100+25 m=225 m= 15 m

Page No 513:

Question 6:

What is the maximum length of a pencil that can be placed in a rectangular box of dimensions (8 cm × 6 cm × 5 cm)?
(a) 8 cm
(b) 9.5 cm
(c) 19 cm
(d) 11.2 cm

Answer:

(d) 11.2 cm
Maximum length of the pencil =  diagonal of the box
                                             =l2+b2+h2 =82+62+52 cm=64+36+25 cm=125 cm=11.2 cm

Page No 513:

Question 7:

The number of planks of dimensions (4 m × 5 m × 2 m) that can be stored in a pit which is 40 m long, 12 m wide and 16 m deep, is
(a) 190
(b) 192
(c) 184
(d) 180

Answer:

(b)  192

Number of planks =  volume of the pitvolume of 1 plank
                             =40×12×164×5×2=768040=192  
        

Page No 513:

Question 8:

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 50 cm deep?
(a) 480
(b) 450
(c) 320
(d) 360

Answer:

(a)  480

Length of the pit = 20 m
Breadth of the pit = 6 m
Height of the pit = 50 cm = 0.5 m

Length of the plank = 5m
Breadth of the plank = 25 cm = 0.25 m
Height of the plank = 10 cm = 0.1 m

∴ Number of planks = volume of the pitvolume of 1 plank

                                  =20×6×0.55×0.25×0.1=60×1000125=480

Page No 513:

Question 9:

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)?
(a) 4800
(b) 5600
(c) 6400
(d) 5200

Answer:

(c)  6400

Length of the wall = 8 m = 800 cm
Breadth of the wall = 6 m = 600 cm
Height of the wall = 22.5 cm

Length of the brick = 25 cm
Breadth of the brick = 11.25 cm
Height of the brick = 6 cm

∴ Number of bricks required = volume of the wallvolume of 1 brick

                                             =800×600×22.525×11.25×6=108000001687.5=6400 

Page No 513:

Question 10:

How many persons can be accommodated in a dining hall of dimensions (20 m × 15 m × 4.5 m), assuming that each person requires 5 m3 of air?
(a) 250
(b) 270
(c) 320
(d) 300

Answer:

(b) 270

Number of persons  = volume of the hallvolume of air required by 1 person

                               =20×15×4.55=20×3×4.5=270 

∴ 270 persons can be accommodated.

Page No 513:

Question 11:

A river 1.5 m deep and 30 m wide is flowing at the rate of 3 km per hour. The volume of water that runs into the sea per minute is
(a) 2000 m3
(b) 2250 m3
(c) 2500 m3
(d) 2750 m3

Answer:

(b) 2250 m3

Length of the river = 1.5 m
Breadth of the river = 30 m
Depth of the river = 3 km = 3000 m

  Now, volume of water that runs into the sea =1.5×30×3000 m3                                                                           =135000 m3

∴ Volume of water that runs into the sea per minute = 13500060=2250 m3

Page No 513:

Question 12:

The lateral surface area of a cube is 256 m2. The volume of the cube is
(a) 64 m3
(b) 216 m3
(c) 256 m3
(d) 512 m3

Answer:

(d) 512  m3

Suppose that a m be the edge of the cube.
We have:
4a2=256a2=2564= 64a =8 m

∴ Volume of the cube = a3  m3  =83 m3 = 512  m3 

Page No 513:

Question 13:

The total surface area of a cube is 96 cm2. The volume of the cube is
(a) 8 cm3
(b) 27 cm3
(c) 64 cm3
(d) 512 cm3

Answer:

(c)  64 cm3

Let a cm be the edge of the cube.
We have:
6a2=96a2=16a=4 cm

∴ Volume of the cube = a3 cm3 =43 cm3 = 64 cm3 

Page No 513:

Question 14:

The volume of a cube is 512 cm3. Its total surface area is
(a) 256 cm2
(b) 384 cm2
(c) 512 cm2
(d) 64 cm2

Answer:

(b) 384 cm2

Suppose that a cm is the edge of the cube.
We have:
a3 = 512a =5123 = 8 cm

Total surface area of cube =6a2 cm2=6×8×8 cm2= 384 cm2



Page No 514:

Question 15:

The length of the longest rod that can fit in a cubical vessel of side 10 cm, is
(a) 10 cm
(b) 20 cm
(c) 102 cm
(d) 103 cm

Answer:

(d)  103 cm103 cm
Length of the longest rod = body diagonal of the vessel
                                 =3a =3×10 =103 cm

Page No 514:

Question 16:

If the length of diagonal of a cube is 83 cm, then its surface area is
(a) 192 cm2
(b) 384 cm2
(c) 512 cm2
(d) 768 cm2

Answer:

(b)  384 cm2
We have:
3a=83a =8 cm

∴ Surface area of the cube = 6a2=6×8×8= 384 cm2

Page No 514:

Question 17:

If each edge of a cube is increased by 50%, then the percentage increase in its surface area is
(a) 50%
(b) 75%
(c) 100%
(d) 12%

Answer:

Let a be the edge of the cube.
Then the surface area is 6a2=S (say)

Now, increased edge = a+50100a = 150100a = 32a

Then, new surface area=632a2 =6×94a2 =94SIncrease in surface area =94S - S  = 54S Percentage increase in surface area =54SS = 54×100% = 125%

Page No 514:

Question 18:

Three cubes of metal with edges 3 cm, 4 cm and 5 cm respectively are melted to form a single cube. The lateral surface area of the new cube formed is
(a) 72 cm2
(b) 144 cm2
(c) 128 cm2
(d) 256 cm2

Answer:

(b)  144  cm2
Volume of the new cube formed = total volume of the three cubes
Suppose that a cm is the edge of the new cube, then

a3=33+43+53     =27+64+125      =216     = 6 cm

∴ Lateral surface area of the new cube = 4a2 = 4×6×6 = 144 cm2

Page No 514:

Question 19:

In a shower, 5 cm of rain falls. What is the volume of water that falls on 2 hectares of ground?
(a) 500 m3
(b) 750 m3
(c) 800 m3
(d) 1000 m3

Answer:

(d) 1000 m3

 Area of the land=2 sq hec=2000 sq mAmount of rainfall=5 cm=0.05 m Volume of the water=area of the land ×amount of rainfall                              =2000×0.05                             =1000 m3

Page No 514:

Question 20:

Two cubes have their volumes in the ratio 1 : 27. The ratio of their surface areas is
(a) 1 : 3
(b) 1 : 8
(c) 1 : 9
(d) 1 : 18

Answer:

(c) 1 : 9
Suppose that the edges of the cubes are a and b.
We have:
a3b3=127ab3=127ab=13

∴ Ratio of the surface areas = 6a26b2=ab2=132 =19

Page No 514:

Question 21:

If each side of a cube is doubled, then its volume
(a) is doubled
(b) becomes 4 times
(c) becomes 6 times
(d) becomes 8 times

Answer:

(d) becomes 8 times

Suppose that the side of the cube is a.
When it is doubled, it becomes 2a.
New volume of the cube = 2a3=8a3  

Hence, the volume becomes 8 times the original volume.

Page No 514:

Question 22:

The diameter of the base of a cylinder is 6 cm and its height is 14 cm. The volume of the cylinder is
(a) 198 cm3
(b) 396 cm3
(c) 495 cm3
(d) 297 cm3

Answer:

(b)  396 cm3

 Volume of the cylinder=πr2h                                      =227×32×14                                      =22×9×2                                       = 396 cm3

Page No 514:

Question 23:

If the diameter of a cylinder is 28 cm and its height is 20 cm, then its curved surface area is
(a) 880 cm2
(b) 1760 cm2
(c) 3520 cm2
(d) 2640 cm2

Answer:

(b)  1760 cm2

Curved surface area of the cylinder=2πrh                                                             =2×227×14×20                                                            =44×40                                                             =1760 cm2

Page No 514:

Question 24:

If the curved surface area of a cylinder is 1760 cm2 and its base radius is 14 cm, then its height is
(a) 10 cm
(b) 15 cm
(c) 20 cm
(d) 40 cm

Answer:

(c) 20 cm

Curved surface area = 1760 cm2
Suppose that h cm is the height of the cylinder.
Then we have:
2πrh = 17602×227×14×h = 1760h=1760×744×14=20 cm

Page No 514:

Question 25:

The height of a cylinder is 14 cm and its curved surface area is 264 cm2. The volume of the cylinder is
(a) 308 cm3
(b) 396 cm3
(c) 1232 cm3
(d) 1848 cm3

Answer:

(b)  396 cm3

Curved surface area = 264 cm2.
Let r cm be the radius of the cylinder.
Then we have:

2πrh =2642×227×r×14 =264r=264×744×14=3 cm

  Volume of the cylinder=227×32×14                                =22×9×2                               = 396 cm3

Page No 514:

Question 26:

The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. The height of the pillar is
(a) 4 m
(b) 5 m
(c) 6 m
(d) 7 m

Answer:

(c)  6 m

Curved surface area = 264 m2 
Volume = 924 m3
Let r m be the radius and h m be the height of the cylinder.
Then we have:
2πrh = 264  and  πr2h =924rh =2642πh =2642r×πNow, πr2h =π×r2×2642r×π=924r=924×2264r=7 m h =264×72×7×22=6 m



Page No 515:

Question 27:

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their curved surface area is
(a) 2 : 5
(b) 8 : 7
(c) 10 : 9
(d) 16 : 9

Answer:

(c)  10 : 9

Suppose that the radii of the cylinders are 2r and 3r and their respective heights are 5h and 3h.
Then, ratio of the curved surface areas = 2π(2r)(5h)2π(3r)(3h)=10 : 9

Page No 515:

Question 28:

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their volumes is
(a) 27 : 20
(b) 20 : 27
(c) 4 : 9
(d) 9 : 4

Answer:

(b)  20 : 27

Suppose that the radii of the cylinders are 2r and 3r and their respective heights are 5h and 3h..
 Ratio of their volumes=π2r2×5hπ3r2×3h=4×59×3=2027

Page No 515:

Question 29:

The ratio between the radius of the base and the height of a cylinder is 2 : 3. If its volume is 1617 cm3, then its total surface area is
(a) 308 cm2
(b) 462 cm2
(c) 540 cm2
(d) 770 cm2

Answer:

(d) 770 cm2
We have:
  r: h = 2 : 3
 rh=23h= 32r
Now, volume = 1617 cm3
πr2h = 1617227×r2×32r =1617r3=1617×1466=343r= 7 cm

h = 10.5 cm

Hence, total surface area=2πrh+2πr2                                           =227(2×7×10.5 +2×72)                                          =227147+2×49                                          =227×245                                           =770 cm2

Page No 515:

Question 30:

Two circular cylinders of equal volume have their heights in the ratio 1 : 2. The ratio of their radii is
(a) 1 : 2
(b) 2 : 1
(c) 1 : 2
(d) 1 : 4

Answer:

(b) 2 : 1

Suppose that the heights of two cylinders are h and 2h whose radii are r and R, respectively.
Since the volumes of the cylinders are equal, we have:

πr2h=πR2×2hr2R2=21rR=21r:R = 2 :1

Page No 515:

Question 31:

The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. If the total surface area is 616 cm2, then the volume of the cylinder is
(a) 1078 cm3
(b) 1232 cm3
(c) 1848 cm3
(d) 924 cm3

Answer:

(a) 1078 cm3

We have:

2πrh2πrh+2πr2=124πrh = 2πrh +2πr22πrh =2πr2rh=11

Also, 2πrh +2πr2=6162πr2+2πr2=6164πr2=616πr2=154r2=154πcm2

 Volume of the cylinder=πr2h  =π×154π×r=154×154×722=154×7=1078 cm3

Page No 515:

Question 32:

In a cylinder, if the radius is halved and the height is doubled, then the volume will be
(a) the same
(b) doubled
(c) halved
(d) four times

Answer:

(c) halved

Suppose that the new radius is 12r and the height is 2h.
 Volume =π×12r2×2h             =π×r24×2h             =12πr2h 

Page No 515:

Question 33:

The number of coins 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm is
(a) 540
(b) 450
(c) 380
(d) 472

Answer:

(b) 450

Number of coins = volume of cylinder to be formedvolume of a coin
 
                          =π×2.25×2.25×10π×0.75×0.75×0.2=225×225×575×75=450  

Page No 515:

Question 34:

The radius of a wire is decreased to one-third. If volume remains the same, the length will become
(a) 2 times
(b) 3 times
(c) 6 times
(d) 9 times

Answer:

(d) 9 times

Let the new radius be 13r.
 Suppose that the new height is H.
The volume remains the same.
i.e., πr2h =π×13r2×Hh = 19H H =9h  The new height becomes nine times the original height.

Page No 515:

Question 35:

The diameter of a roller, 1 m long, is 84 cm. If it takes 500 complete revolutions to level a playground, the area of the playground is
(a) 1440 m2
(b) 1320 m2
(c) 1260 m2
(d) 1550 m2

Answer:

(b)  1320  m2

Area covered by the roller in 1 revolution = 2πrh
 =2×227×42×100=44×600=26400 cm2  Area covered in 50 revolutions =26400 ×500 cm2                                                           =1320 m2

Page No 515:

Question 36:

2.2 dm3 of lead is to be drawn into a cylindrical wire 0.50 cm in diameter. The length of the wire is
(a) 110 m
(b) 112 m
(c) 98 m
(d) 124 m

Answer:

(b) 112 m

Volume of the lead = volume of the cylindrical wire
Suppose that h m is the length of the wire.
As 2.2 dm3 of lead is to be drawn into a cylindrical wire of diameter 0.50 cm, we have:
π0.252×h =2.2×10×10×10h =2200×722×0.0625    =7000.0625    =11200 cm     = 112 m

Page No 515:

Question 37:

The lateral surface area of a cylinder is
(a) πr2h
(b) πrh
(c) 2πrh
(d) 2πr2

Answer:

(c) 2πrh
The lateral surface area of a cylinder is equal to its curved surface area.
∴ Lateral surface area of a cylinder = 2πrh  

Page No 515:

Question 38:

The height of a cone is 24 cm and the diameter of its base is 14 cm. The curved surface area of the cone is
(a) 528 cm2
(b) 550 cm2
(c) 616 cm2
(d) 704 cm2

Answer:

(b) 550 cm2

   l =r2+h2  =72+242  =49+576   =625   =25 cm Curved surface area of the cone =πrl                                                  = 227×7×25                                                   =550 cm2



Page No 516:

Question 39:

The volume of a right circular cone of height 12 cm and base radius 6 cm, is
(a) (12π) cm3
(b) (36π) cm3
(c) (72π) cm3
(d) (144π) cm3

Answer:

(d) (144π) cm3

Volume of the cone = 13πr2h 
                       =13π×62×12 =π×36×4=144π  cm3

Page No 516:

Question 40:

How much cloth 2.5 m wide will be required to make a conical tent having base radius 7 m and height 24 m?
(a) 120 m
(b) 180 m
(c) 220 m
(d) 550 m

Answer:

(c)  220 m
Let the length of the required cloth be L m and its breadth be B m.
Here, B = 2.5 m
Radius of the conical tent =7 m
Height of the tent = 24 m
Area of cloth required = curved surface area of the conical tent
L×B=πrlL×2.5=227×7×72+2422.5L=22×49 + 576L=22×252.5=220 m

Page No 516:

Question 41:

The volume of a cone is 1570 cm3 and its height is 15 cm. What is the radius of the cone?
(a) 10 cm
(b) 9 cm
(c) 12 cm
(d) 8.5 cm

Answer:

(a) 10 cm
Let r cm be the radius of the cone.
Volume = 1570  cm3

Then 13×3.14×r2×15 = 1570r2 =15703.14×5=100r = 10 cm

Page No 516:

Question 42:

The height of a cone is 21 cm and its slant height is 28 cm. The volume of the cone is
(a)  7356 cm3
(b) 7546 cm3
(c) 7506 cm3
(d) 7564 cm3

Answer:

(b)  7546 cm3

Radius of the cone, r=l2-h2                                 =282-212                                 =784-441                                 =343 cm

 Volume of the cone =13πr2h                            =13×227×343×21                             =22×343                             = 7546 cm3

Page No 516:

Question 43:

The volume of a right circular cone of height 24 cm is 1232 cm3. Its curved surface area is
(a)  1254 cm2
(b) 704 cm2
(c) 550 cm2
(d) 462 cm2

Answer:

(c) 550 cm2
Let r cm be the radius of the cone.
Volume of the right circular cone  = 1232 cm3
Then we have:
13×227×r2×24=1232r2=1232×2122×24r2=49r = 7 cm

 Curved surface area of the cone =πrl                                                        =227×7×72+242                                                         =22×49+576                                                         =22×25                                                         =550 cm2

Page No 516:

Question 44:

If the volumes of two cones be in the ratio 1 : 4 and the radii of their bases be in the ratio 4 : 5, then the ratio of their heights is
(a) 1 : 5
(b) 5 : 4
(c) 25 : 16
(d) 25 : 64

Answer:

(d) 25 : 64

Suppose that the radii of the cones are 4r and 5r and their heights are h and H, respectively .
It is given that the ratio of the volumes of the two cones is 1:4
Then we have:
13π4r2h13π5r2H=1416r2h25r2H=14hH=2564 h : H = 25 : 64

Page No 516:

Question 45:

If the height of a cone is doubled, then its volume is increased by
(a) 100%
(b) 200%
(c) 300%
(d) 400%

Answer:

(a) 100 %

Suppose that height of the cone becomes 2h and let its radius be r.
Then new volume of the cone = 13πr2(2h) = 23πr2h = 2×volume of the cone
Increase in volume = 23πr2h-13πr2h=13πr2h
 Percentage increase=increase in volume initial volume ×100 %=13πr2h13πr2h×100%=100%

Hence, the volume increases by 100%.

Page No 516:

Question 46:

The curved surface area of one cone is twice that of the other while the slant height of the latter is twice that of the former. The ratio of their radii is
(a) 2 : 1
(b) 4 : 1
(c) 8 : 1
(d) 1 : 1

Answer:

(b) 4 : 1
If the slant height of the first cone is l, then the slant height of the second cone will be 2l.
Let the radii of the first and second cones be r and R, respectively.
Then we have:

πrl=2×(πR×2l)r=4RrR=41

Page No 516:

Question 47:

The ratio of the volumes of a right circular cylinder and a right circular cone of the same base and the same height will be
(a) 1 : 3
(b) 3 : 1
(c) 4 : 3
(d) 3 : 4

Answer:

(b)  3 : 1

It is given that the right circular cylinder and the right circular cone have the same base and height.
Suppose that the respective radii of bases and heights are equal to r and h.
Then ratio of their volumes = πr2h13πr2h=31

Page No 516:

Question 48:

A right circular cylinder and a right circular cone have the same radius and the same volume. The ratio of the height of the cylinder to that of the cone is
(a) 3 : 5
(b) 2 : 5
(c) 3 : 1
(d) 1 : 3

Answer:

(d)  1 : 3
It is given that the right circular cylinder and the right circular cone have the same radius and volume.
Suppose that the radii of their bases are equal to r and the respective heights of the cylinder and the cone are h and H.
As the volumes of the cylinder and the cone are the same, we have:
πr2h=13πr2HhH=13

Page No 516:

Question 49:

The radii of the bases of a cylinder and a cone are in the ratio 3 : 4 and their heights are in the ratio 2 : 3. Then, their volumes are in the ratio
(a) 9 : 8
(b) 8 : 9
(c) 3 : 4
(d) 4 : 3

Answer:

(a)  9 : 8

Suppose that the respective radii of the cylinder and the cone are 3r and 4r and their respective heights are 2h and 3h.

 Ratio of the volumes=π3r2×2h13π4r2×3h=3×9×216×3=98

Page No 516:

Question 50:

If the height and the radius of a cone are doubled, the volume of the cone becomes
(a) 3 times
(b) 4 times
(c) 6 times
(d) 8 times

Answer:

(d) 8 times

Let the new height and radius be 2h and 2r, respectively.

New volume of the cone=13π2r2×2h=83πr2h =8×initial volume of the cone



Page No 517:

Question 51:

A solid metallic cylinder of base radius 3 cm and height 5 cm is melted to make n solid cones of height 1 cm and base radius 1 mm. The value of n is
(a) 450
(b) 1350
(c) 4500
(d) 13500

Answer:

(d) 13500
Radius of the cylinder = 3 cm
Height of the cylinder = 5 cm
Radius of the cone = 1 mm = 0.1 cm
Height of the cone = 1 cm

∴ Number of cones, n = volume of the cylindervolume of 1 cone
                             =π×32×513π×0.12×1=3×9×50.01=13500 

Page No 517:

Question 52:

A conical tent is to accommodate 11 persons such that each person occupies 4 m2 of space on the ground. They have 220 m3 of air to breathe. The height of the cone is
(a) 14 m
(b) 15 m
(c) 16 m
(d) 20 m

Answer:

(b) 15 m

Suppose that the height of the cone is h m.
Area of the ground = 11×4 = 44 m2

 πr2=44r2=44×722=14Also, 13πr2h =22013×227×14h = 220h = 220×2122×14=15 m
Hence, the height of the cone is 15 m.

Page No 517:

Question 53:

The volume of a sphere of radius 2r is
(a) 32πr33
(b) 16πr33
(c) 8πr33
(d) 64πr33

Answer:

(a) 32πr33

Volume of the sphere of radius 2r=43π2r3 =323πr3

Page No 517:

Question 54:

The volume of a sphere of radius 10.5 cm is
(a) 9702 cm3
(b) 4851 cm3
(c) 19404 cm3
(d) 14553 cm3

Answer:

(b)  4851 cm3
Volume of the sphere = 43πr3  
                                 =43×227×10.5×10.5×10.5=88×0.5×10.5×10.5 = 4851 cm3

Page No 517:

Question 55:

The surface area of a sphere of radius 21 cm is
(a) 2772 cm2
(b) 1386 cm2
(c) 4158 cm2
(d) 5544 cm2

Answer:


(d)  5544 cm2

Surface area of sphere = 4πr2 
                                     = 4×227×21×21=5544 cm2

Page No 517:

Question 56:

The surface area of a sphere is 1386 cm2. Its volume is
(a) 1617 cm3
(b) 3234 cm3
(c) 4851 cm3
(d) 9702 cm3

Answer:

(c)  4851 cm3

Surface area = 1386 cm2
Let r cm be the radius of the sphere.
Then we have:
4πr2=1386r2=1386×74×22=110.25r = 10.5 cm Volume of the sphere=43πr3                                          =43×227×10.5×10.5×10.5                                             =4851 cm3

Page No 517:

Question 57:

If the surface area of a sphere is (144π) m2, then its volume is
(a) (288π) m3
(b) (188π) m3
(c) (300π) m3
(d) (316π) m3

Answer:

(a) (288π) m3

Surface area = (144π) m2
Ler r m be the radius of the sphere.
Then we have:

4πr2=144πr2=1444=36r =6 m Volume of the sphere = 43πr3                                                  =43π×6×6×6                                                   =288π  m3

Page No 517:

Question 58:

The volume of a sphere is 38808 cm3. Its curved surface area is
(a) 5544 cm2
(b) 8316 cm2
(c) 4158 cm2
(d) 1386 cm2

Answer:

(a) 5544 cm2
Let r cm be the radius of the sphere.
Then we have:

43πr3=38808r3=38808×3×74×22=9261r=21 cm Curved surface area=4πr2 = 4×227×21×21                                                         = 5544 cm3

Page No 517:

Question 59:

If the ratio of the volumes of two spheres is 1 : 8, then the ratio of their surface area is
(a) 1 : 2
(b) 1 : 4
(c) 1 : 8
(d) 1 : 16

Answer:

(b) 1 : 4
Suppose that r and R are the radii of the spheres.
Then we have:

43πr343πR3=18rR3=18rR=12
∴ Ratio of surface area of spheres =4πr24πR2=rR2=122=14

Page No 517:

Question 60:

A solid metal ball of radius 8 cm is melted and cast into smaller balls, each of radius 2 cm. The number of such balls is
(a) 8
(b) 16
(c) 32
(d) 64

Answer:

(d) 64

Number of balls = volume of solid metal ballvolumeof 1 small ball
                        = 43π×8343π×23=5128=64

Page No 517:

Question 61:

A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. The radius of the sphere is
(a) 4.2 cm
(b) 2.1 cm
(c) 2.4 cm
(d) 1.6 cm

Answer:

(b) 2.1 cm

Let r cm be the radius of the sphere.
Volume of the cone = volume of the sphere
13π×2.1×2.1×8.4 =43πr3r3 =2.1×2.1×2.1r = 2.1 cm

Page No 517:

Question 62:

A solid lead ball of radius 6 cm is melted and then drawn into a wire of diameter 0.2 cm. The length of wire is
(a) 272 m
(b) 288 m
(c) 292 m
(d) 296 m

Answer:

(b)  288 m

Let h m be the length of wire.
Volume of the lead ball = volume of the wire
43π×63=π×0.12hh = 4×2163×0.01=28800 cm =288 m

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Question 63:

A metallic sphere of radius 10.5 cm is melted and then recast into small cones, each of radius 3.5 cm and height 3 cm. Then number of such cones will be
(a) 21
(b) 63
(c) 126
(d) 130

Answer:

(c)  126

Number of cones =volume of the spherevolume of 1 cone
                          =43π×10.5×10.5×10.513π×3.5×3.5×3=4×3×3×3.5 =126



Page No 518:

Question 64:

How many lead shots, each 0.3 cm in diameter, can be made from a cuboid of dimensions 9 cm × 11 cm× 12 cm?
(a) 7200
(b) 8400
(c) 72000
(d) 84000

Answer:

(d) 84000

Number of lead shots =volume of cuboidvolume of 1 lead shot                                     =9×11×1243×227×0.15×0.15×0.15                                     =9×11×3×3×722×0.15×0.15×0.15                                     =84000

Page No 518:

Question 65:

The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 mm. The length of the wire is
(a) 12 m
(b) 18 m
(c) 36 m
(d) 66 m

Answer:

(c)  36 m
Let h m be the length of the wire.
Volume of the sphere = volume of the wire
43π×33 =π×0.12×hh= 4×90.001   =3600 cm    =36 m

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Question 66:

A sphere of diameter 12.6 cm is melted and cast into a right circular cone of height 25.2 cm. The radius of the base of the cone is
(a) 6.3 cm
(b) 2.1 cm
(c) 6 cm
(d) 4 cm

Answer:

(a)  6.3 cm
Let r cm be the radius of base of the cone.
Volume of the sphere = volume of the cone
43π×6.33=13π×r2×25.2r2=4×6.3×6.3×6.325.2=39.69r = 39.69= 6.3 cm

Page No 518:

Question 67:

A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5 cm and 2 cm. The radius of the third ball is
(a) 1 cm
(b) 1.5 cm
(c) 2.5 cm
(d) 0.5 cm

Answer:

(c) 2.5 cm

Let r cm be the radius of the third ball.
Volume of the original ball = volume of the three balls
43π×33 =43π×1.53+43π×23+43πr327 =3.375+8 +r3r3 = 27- 11.375 = 15.625r = 2.5 cm

Page No 518:

Question 68:

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratio of the surface areas of the balloons in two cases is
(a) 1 : 4
(b) 1 : 3
(c) 2 : 3
(d) 1 : 2

Answer:

(a)  1 : 4
Ratio of the surface areas of balloon = 2π×622π×122=36144=14

Page No 518:

Question 69:

The volumes of the two spheres are in the ratio 64 : 27 and the sum of their radii is 7 cm. The difference of their total surface areas is
(a) 38 cm2
(b) 58 cm2
(c) 78 cm2
(d) 88 cm2

Answer:

(d) 88 cm2

Suppose that the radii of the spheres are r cm and (7 − r) cm. Then we have:
43π7-r343πr3=6427(7-r)r=4321-3r=4r21=7rr=3 cm

Now, the radii of the two spheres are 3 cm and 4 cm.

Required difference=4×227×42  -4×227×32=4×227×16-9 = 4×227×7=88 cm2

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Question 70:

A hemispherical bowl of radius 9 cm contains a liquid. This liquid is to be filled into cylindrical small bottles of diameter 3 cm and height 4 cm. How many bottles will be needed to empty thee bowl?
(a) 27
(b) 35
(c) 54
(d) 63

Answer:

(c) 54

Number of bottles  = volume of bowlvolume of 1 bottle
                         
                           =23×π×93π×1.52×4=2×7292.25×12=54 

Page No 518:

Question 71:

A cone and a hemisphere have equal bases and equal volumes. The ratio of their heights is
(a) 1 : 2
(b) 2 : 1
(c) 4 : 1
(d) 2 : 1

Answer:

(b) 2 : 1

Let the radii of the cone and the hemisphere be r and their respective heights be h and H.
Then 13π×r2×h = 23π×r2×HhH=21

Page No 518:

Question 72:

A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is
(a) 1 : 2 : 3
(b) 2 : 1 : 3
(c) 2 : 3 : 1
(d) 3 : 2 : 1

Answer:

(a) 1 : 2 : 3
The cone, hemisphere and the cylinder stand on equal bases and have the same height.
We know that radius and height of a hemisphere are the same.
Hence, the height of the cone and the cylinder will be the common radius.
i.e., r = h
Ratio of the volumes of the cone, hemisphere and the cylinder:

13πr2h23πr3πr2h =13πr323πr3πr3  =123  =   1 : 2 : 3

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Question 73:

If the volume and the surface area of a sphere are numerically the same, the nits radius is
(a) 1 unit
(b) 2 units
(c) 3 units
(d) 4 units

Answer:

(c) 3 units

We have:

43πr3= 4πr213r =1r = 3 units

Page No 518:

Question 74:

Which is false in case of a hollow cylinder?
(a) Curved surface area of a hollow cylinder = 2πh(R + r).
(b) Total surface area of a hollow cylinder = 2π(R + r)(h + Rr).
(c) Inner curved surface area of a hollow cylinder = 2 πh(Rr).
(d) Area of each end of a hollow cylinder = π(R2r2).

Answer:

(c) Inner curved surface area of a hollow cylinder = 2 πh(Rr)

Inner curved surface area of a hollow cylinder is 2πrh.



Page No 519:

Question 75:

Which of false?
(a) Volume of a hollow sphere=43π(R3 - r3)
(b) Volume of a hemisphere=23πR3
(c) Total surface area of a hemisphere = 3 πR2
(d) Curved surface area of a hemisphere = πR2

Answer:

(d) Curved surface area of a hemisphere = πR2

Curved surface area of hemisphere is =2πR2, where R is the radius of hemisphere

Page No 519:

Question 76:

For a right circular cylinder of base radius = 7 cm and height = 14 cm, which is false?
(a) Curved surface area = 616 cm2.
(b) Total surface area = 924 cm2.
(c) Volume = 2156 cm3.
(d) Total area of the end faces = 154 cm2.

Answer:

(d) Total area of the end faces = 154 cm2

(a) CSA of the cylinder=2πrh=2×227×7×14=616 cm2(b) Total surface area of the cylinder=2πr(h+r)=2×227×7×(14+7)=1232 cm2(c) Volume of the cylinder=πr2 h=227×72×14=2156 cm2

 (d) Total area of end faces =πr2 +πr2                                            =2×227×7×7                                               = 44×7                                              = 308  cm2
Hence, (d) is false.

Page No 519:

Question 77:

Which is false?
A metal pipe is 63 cm long. Its inner diameter is 4 cm and the outer diameter is 4.4 cm. Then,
(a) its inner curved surface area = 792 cm2
(b) its outer curved surface area = 871.2 cm2
(c) surface area of each end = 2.64 cm2
(d) its total surface area = 1665.84 cm2

Answer:

(d) its total surface area = 1665.84 cm2

Height of the metal pipe = 63 cm
Its inner radius = 2 cm
Its outer radius = 2.2 cm

(a) Inner curved surface area=2π×2×63=792 cm2
Hence, (a) is true.

(b)  Outer curved surface area =2π×2.2×63=871.2 cm3
Hence, (b) is true.

(c) Surface area of each end=π[(2.2)2-22]=2.64 cm2
Hence, (c) is true.
(d)
 Total surface area of the metal pipe =2π4.4-4×63 +2π(4.42 -42)                                                        =2×227×0.4×63  +19.36-16                                                        =447×25.2+3.36                                                         = 447×28.56                                                          =179.52 cm2

Hence, (d) is false.

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Question 78:

Assertion: The base radius of a cone is 7 cm and its slant height is 25 cm. The volume of the cone is 1232 cm3.
Reason: The volume of a right circular cone of base radius r and height h is 13πr2h.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true and Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Reason (R) is clearly true.
Let r cm be the radius of base of the cone .
Then h=l2-r2 =252-72   = 625-49 = 576=24 cmVolume of the cone = 13πr2h                                =13×227×7×7×24                                  = 1232 cm3

Reason (R) and assertion (A) are both true and R is the correct explanation of A.
Hence, the correct option is (a).



Page No 520:

Question 79:

Assertion: The surface are of a sphere is 2464 cm2. Its volume is 1149823cm3π=227.
Reason: The volume of a sphere of radius r is 43πr3.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true and Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(b) Both Assertion and Reason are true and Reason is not a correct explanation of Assertion.

   Surface area of sphere=4πr2 =2464 cm2  r = 2464×74×22=14 cm
Now, its volume =43π(14)3=11498.66 cm3 =1498823cm3
Reason (R) is clearly true.
Assertion (A) is true but  reason (R)  is not the correct explanation of assertion (A).

Page No 520:

Question 80:

Assertion: The outer and inner radii of a hollow cylinder 2 m 10 cm long are 5 cm and 3 cm respectively. If it is made up of copper, then volume of copper in it is 10560 cm3.
Reason: The volume of a hollow cylinder of length h, external radius R and internal radius r is given by V=πh(R2 - r2).
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true and Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
Outer radius of the hollow cylinder, R = 5 cm
Inner radius of the hollow cylinder, r = 3 cm
Height of the hollow cylinder, h = 2 m 10 cm = 210 cm

∴ Volume of the hollow cylinder = πR2-r2h  
                                                 = 227×52-32×210=22×25-9×30=22×16×30 = 10560 cm3

Reason (R) is clearly true.
Assertion (A) is true and reason (R)  is a correct explanation of assertion (A).

Page No 520:

Question 81:

Assertion: If the radius of a sphere is doubled then the ratio of the volume of the first sphere to that of the second is 1 : 8.
Reason: A cone and a hemisphere have equal bases and equal volumes. The ratio of their heights is 1 : 2.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true and Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(c)  Assertion is true and Reason is false.
Assertion (A):
Let r be the radius of the sphere. On doubling it, it becomes 2r.
Ratio of the volumes=43πr3 : 43π2r3= 1 : 8

Reason (R):
It is given that the cone and the hemisphere have equal bases and equal volumes.


Let their radii be r and let h be the height of the cone.  Volume of coneVolume of hemisphere=13πr2h23πr3Since volume of the cone = volume of the hemisphere, we have:13πr2h = 23πr3 h = 2r h:r  = 2:1

Hence, reason (R) is false. 

Assertion (A) is true but reason (R) is clearly false and (R) is not the explanation for (A).

Page No 520:

Question 82:

Assertion: The curved surface area of a cone is 550 cm2 and its diameter is 14 cm. Then, its slant height is 25 cm.
Reason: The curved surface area of a cone having base radius r and slant height l is πrl.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true and Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Reason (R) is clearly true.

 Curved surface area of the cone  = πrl                                                           =227×7×25                                                          =550 cm2
Reason (R) is clearly true.
Assertion (A) is true and reason(R)  is a correct explanation of assertion (A) .

Page No 520:

Question 83:

A right circular cylinder just encloses a sphere of radius r (as shown in the figure). Then the surface area of the sphere is equal to the curved surface area of the cylinder.

Answer:

True.

Radius of the cylinder = radius of the sphere = r
Height of the cylinder = diameter of the sphere = 2r

∴ Surface area of the cylinder = 2πrh=2πr×2r=4πr2=surface area of the sphere

Page No 520:

Question 84:

The largest possible right circular cone is cut out of a cube of edge r cm. The volume of the cone is 112πr3.

Answer:

True.

Radius of the cone = r2
Height of the cone = r
∴ Volume of the cone  = 13π×r22×r=112πr3



Page No 521:

Question 85:

If a sphere is inscribed in a cube, then the ratio of the volume of the cube to the volume of the sphere will be 6 : π.

Answer:

True.
Let the side of the cube be a, then the radius of the sphere will be a2.
Ratio of the volume of the cube and the sphere:

a343πa23 = 3a3×84πa3=6π

= 6 : π

Page No 521:

Question 86:

If the length of diagonal of a cube is 63 cm, then the length of each edge of the cube is 3 cm.

Answer:

False.
Let a cm be the side of the cube.
Then we have:

3a = 63a =6 cm



Page No 532:

Question 1:

The radii of two cylinders are in the ratio of 2 : 3 and their heights are in the ratio of 5 : 3. Then the ratio of their volumes is
(a) 10 : 17
(b) 20 : 27
(c) 17 : 27
(d) 20 : 37

Answer:

(b)  20 : 27

Volume of a cylinder = πR2H, where R = radius and H = height
Suppose that the radii of the cylinders are 2r and 3r and their respective heights are 5h and 3h.
∴ Ratio of volumes = π2r2×5hπ3r2×3h=20r2h27r2h=2027

Page No 532:

Question 2:

The total surface area of a cone whose radius is r2 and slant height 2l is
(a) 2πr(l + r)
(b) πrl+r4
(c)πr(l + r)
(d) 2πrl

Answer:

 (b) πrl+r4

 Total surface area of the given cone =πr2×2l +πr22                                                               =πrl+r24                                                                =πrl+r4

Page No 532:

Question 3:

A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. The radius of the sphere is
(a) 1.6 cm
(b) 2.1 cm
(c) 2.4 cm
(d) 4.2 cm

Answer:

(b) 2.1 cm

Let r cm be the radius of the sphere.
Radius of the base of the cone = 2.1 cm
Height of the cone = 8.4 cm
The cone is melted and recast into a sphere.
Then volume of sphere = volume of cone

43πr3=13π×2.12×8.4r3=2.1×2.1×8.44=2.1×2.1×2.1r = 2.1 cm

Page No 532:

Question 4:

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratio of the surface areas of the balloon in the two cases is
(a) 1 : 4
(b) 1 : 3
(c) 2 : 3
(d) 2 : 1

Answer:

(a)  1: 4

Ratio of the surface area of the balloon with radii 6 cm and 12 cm =    4r124r22

                                                                                                   =4624122=36144=14

Page No 532:

Question 5:

A copper sphere of diameter 6 cm is melted and drawn into 36 cm long wire of uniform circular cross-section. Then, its radius is
(a) 2 cm
(b) 1.5 cm
(c) 1.2 cm
(d) 1 cm

Answer:

(d)  1 cm

Ler r cm be the radius of the wire.
Diameter of the copper sphere = 6 cm
i.e.,  radius of the copper sphere = 3 cm
Length of the wire= 36 cm
The sphere is melted and drawn into a wire.
Then volume of wire = volume of copper sphere
πr2×36=43π×33r2=4×936=1r = 1 cm

Hence, the radius of the wire is 1 cm.

Page No 532:

Question 6:

Find the lateral surface area and the total surface area of a cube of side 8 cm.

Answer:

Side of the cube = 8 cm
Lateral surface area = 4a2 
                              =4×8×8=256 cm2

∴ Total surface area of the cube = 6a2 
                                                =6×8×8=384 cm2

Page No 532:

Question 7:

Find the lateral surface area and the total surface area of a cuboid of dimensions 40 cm × 30 cm × 20 cm.

Answer:

Here, l = 40 cm, b = 30 cm, h = 20 cm

Lateral surface area of the cuboid = 2l+b×h 
                                                   =240+30×20=2×70×20=2800 cm2

∴ Total surface area of the cuboid = 2lb+bh+lh 
                                                   =240×30+30×20+40×20=21200+600+800=2×2600=5200 cm2

Page No 532:

Question 8:

The total surface area of a cylinder is 462 cm2 and its curved surface area is one-third of its total surface area. Find the volume of the cylinder.

Answer:

Let r cm be the radii and h cm be the height of the cylinder.

Total surface area of the cylinder= 462 cm2Curved surface area=13 ×total surface area of the cylinder2πrh +2πr2 = 462Also, 2πrh=13(462) = 154Then 154 +2πr2 = 4622πr2 = 308r2= 308×72×22=49r = 7cmNow, 2π×7×h= 154 h =1542×22=3.5 cm

 Hence, volume of the cylinder =πr2h                                                        =227×7×7×3.5                                                        =539 cm3

Page No 532:

Question 9:

The length and breadth of a room are in a ratio 3 : 2. The cost of carpeting the room at Rs 25 per m2 is Rs 1350 and the cost of papering the four walls at Rs 15 per m2 is Rs 2580. If one door and two windows occupy 8 m2, find the dimensions of the room.

Answer:

Let the length, breadth and height be 3x m, 2x m and h m

Now, area of the floor =  cost of carpetingcost per sq. meter=135025=54 m2

i.e., 2x ×3x  = 54x2 =546=9x  = 3 Length, l = 6 m and breadth, b = 9 mCost of papering the four walls at Rs. 15 per m2=Rs 2580Area occupied by one door and two windows=8 m2Now, area of the walls=258015+8  m2=172+8 m2=180 m2 2l+b×h=180h=180215=18030=6 m 

Page No 532:

Question 10:

If the radius of a sphere is increased by 10%, prove that its volume will be increased by 33.1%.

Answer:

Suppose that r is the radius of the sphere.
Volume of the original sphere = 43πr3=V (say)
Let us increase the radius of the sphere by 10%.
Then new radius = r+10100r = 110100×r=11r10

Now, new volume of the sphere=43π11r103                                                       =43π×13311000r3                                                         =13311000VHence, increase in volume =1331V1000-V                                                 = 331V1000 Increase %= 331V1000V×100=33.1%

Hence, proved

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Question 11:

The surface area of a sphere of radius 5 cm is five times the area of the curved surface of a cone of radius 4 cm. Find the height and volume of the cone.

Answer:

Let h cm be the height of cone and l cm be its slant height.
Radius of the sphere=5 cm
Radius of the cone = 4 cm

Now, surface area of the sphere = 5(Curved surface area of the cone)
4π×52=5×(π×4×l)l=5 cmHence, height  h=l2-r2=25-16=9=3 cm Volume of the cone =13π×42×3                                        =22×167                                        =50.3  cm3



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Question 12:

A rectangular tank measuring 5 m × 4.5 m × 2.1 m is dug in the centre of the field measuring 13.5 m × 2.5 m. The earth dug out is spread over the remaining portion of the field. How much is the level of the field raised?

Answer:

Length of the tank = 5 m
Breadth of the tank = 4.5 m
Height of the tank = 2.1 m
Volume of the earth dug out = 5×4.5×2.1 =47.25 m3

Now, area over which the earth is spread = area of the field − area of the earth dug out
                                                              =13.5×2.5 - 5×4.5=33.75 -22.5=11.25 m2

∴ Height of the raised field = volume of earth dug outarea over which the earth is spread=47.2511.25=4.2 m

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Question 13:

A joker's cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Answer:

Area of sheet required for one cap = πrl
                                                    =227×7×72+242=22×49+576=22×25 = 550 cm2

∴ Area of sheet for 10 such caps = 550×10 = 5500 cm2

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Question 14:

The volume of a right circular cone is 9856 cm3. If the diameter of its base is 28 cm, find the height of the cone.

Answer:

Let h cm be the height of the cone.
Diameter of the base of the cone = 28 cm
∴ Radius of the base of the cone = 14 cm

We have:
13π×142×h = 9856h = 9856×322×14×2=48 cm

Hence, the height of the given cone is 48 cm.

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Question 15:

Into a circular drum of radius 4.2 m and height 3.5 m, how many full bags of wheat can be emptied if the space required for wheat in each bag is 2.1 m3?

Answer:

Radius of the circular drum = 4.2 m
Height of the drum = 3.5 m
Number of bags = volume of drumspace required for each bag

                         =227×4.2×4.2×3.52.1=22×2×4.2×0.5 = 92.4

The number of bags cannot be in decimals.
Hence, the number of  bags that can be emptied is 92.

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Question 16:

A well with 10 m inside diameter is dug 14 m deep. Earth taken out of it is spread all around to a width of 5 m to form an embankment. Find the height of the embankment.

Answer:

Let h m be the height of the embankment.
Internal radius of the well, r =5 m
 Volume of earth dug out =πr2h                                         =227×5×5×14                                         =22×5                                         = 1100 m3

Here, external radii = 10 m; internal radii = 5 m

Now, area of the embankment = π102-52=227×75=16507m2

∴ Height of the embankment = volume of earth dug outarea of the embankment=110016507=4.67 m

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Question 17:

How many metres of cloth 5 m wide will be required to make a conical tent, the radius of whose base is 7 m and whose height is 24 m?

Answer:

Let l  m be the length of the cloth.
Width of the cloth =5 m
Radius of the base of the tent =7 m
Height of the tent =24 m
Now, area of the required cloth = curved surface area of the tent
l×5 =πrL5l = 227×7×72+242l = 22×49+5765  =22×255  =110 m

∴ The length of the cloth is 110 m.

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Question 18:

The volume of a solid cylinder is 1584 cm3 and its height is 14 cm. Find its total surface area.

Answer:

Let r cm be the radius of the base of the cylinder.
Volume of  the solid cylinder = 1584 cm3
Height of the solid cylinder, h = 14 cm
We have:

227×r2×14 = 1584r2=158444=36r = 6 cm


 Total surface area of the cylinder=2πr(r + h )                                                                    =2×227×6×(6+14)                                                                    =44×6×207                                                                    =754.29 cm2

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Question 19:

The volume of two spheres are in the ratio 64 : 27. Find the difference of their surface areas if the sum of their radii is 7 cm.

Answer:

Let the radii of the spheres be R cm and r cm.
Then we have:
43πR343πr3=6427Rr3=6427Rr=43Now, r +R = 7Then 7-rr=4321-3r=4r7r=21i.e., r=3 cm  and R= 4 cm

Difference of their surface areas=4πR2-4πr2=4×227×42-32=887×7 =88 cm2

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Question 20:

The radius and height of a right circular cone are in the ratio 4 : 3 and its volume is 2156 cm3. Fin the curved surface area of the cone.

Answer:

Let the radius and height of the cone be 4x cm and 3x cm, respectively.
Volume of the right circular cone =2156 cm3
We have:
13π4x2×3x=215616x3π = 2156x3=2156×716×22=42.875x = 3.5

Now, radius of the cone = 14 cm; height = 10.5 cm


 Curved surface area of the cone =πrl                                                  =227×14×142+10.52                                                  =22×2×196 +110.25                                                  =44×17.5                                                    = 770 cm2

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Question 21:

The radius of the base of a cone is 14 cm and its height is 24 cm. Find the volume, curved surface area and the total surface area of the cone.

Answer:

Here, r = 14 cm  and h = 24 cm

Volume of the cone = 13πr2h 
                             =13×227×14×14×24=4928 cm3

Curved surface area of the cone = πrl
                                                =227×14×72+242=22×2×25 = 1100 cm2

∴ Total surface area of the cone = πrl +πr2 
                                                =1100+227×14×14=1100 + 616=1716  cm2   

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Question 22:

Two cylindrical vessels are filled with oil. Their radii are 15 cm and 10 cm respectively and their heights are 25 cm and 18 cm respectively. Find the radius of the cylindrical vessel. 33 cm in height which will just contain the oil of the two given vessels.

Answer:

The radii of two cylindrical vessels that are filled with oil are 15 cm and 10 cm .
Let r cm be the radius of the new cylindrical vessel.
Height of the new vessel = 33 cm
According to the question, we have:
Volume of two cylindrical vessels = volume of the vessel that contains liquid of both the vessels
i.e., π×152×25 +π×102×18=πr2×335625+1800=33r2r2=742533=225r=15 cm

Hence, the radius of the new cylindrical vessel is 15 cm.

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Question 23:

The ratio of the curved surface area and the total surface area of a circular cylinder is 1 : 2 and the total surface area is 616 cm2. Find its volume.

Answer:

Total surface area of cylinder = 616 cm2
The ratio of the curved surface area and the total surface area of cylinder is 1 : 2.
∴ 2(Curved surface area) = total surface area

22πrh =2πrh +2πr22πrh = 2πr2rh = r2r =hAlso, we have:2πrh +2πr2= 6164πr2 =616r2=616×788=49r=7 cmi.e., h =7cm Volume of the given cylinder = πr2h                                                        =227×7×7×7                                                         = 1078 cm3



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