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

 MULTIPLE CHOICE QUESTIONS

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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 1:

(c) 810 cm3

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

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 2:

(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

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Question 3:

The length, breadth and height of a cuboid are 15 m, 6 m and 5 dm respectively. The lateral surface area of the cuboid is
(a) 45 m2
(b) 21 m2
(c) 201 m2
(d) 90 m2

Answer 3:


Length of the cuboid, l = 15 m

Breadth of the cuboid, b = 6 m

Height of the cuboid, h = 5 dm = 0.5 m           (1 m = 10 dm)

∴ Lateral surface area of the cuboid = 2h(lb) = 2 × 0.5 × (15 + 6) = 2 × 0.5 × 21 = 21 cm2

Hence, the correct answer is option (b).

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 4:

(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

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 5:

(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

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 6:

(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

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 7:

(b)  192

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

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 8:

(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

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 9:

(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 

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 10:

(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.

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 11:

(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

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 12:

(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 

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 13:

(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 

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 14:

(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

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 15:

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

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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 16:

(b)  384 cm2
We have:
3a=83a =8 cm

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

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 17:

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%

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 18:

(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

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 19:

(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

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 20:

(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

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 21:

(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.

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 22:

(b)  396 cm3

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

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 23:

(b)  1760 cm2

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

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 24:

(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

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 25:

(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

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 26:

(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

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 27:

(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

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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 28:

(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

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 29:

(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

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 30:

(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

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 31:

(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

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 32:

(c) halved

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

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 33:

(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  

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 34:

(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.

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 35:

(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

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 36:

(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

Question 37:

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

Answer 37:

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

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 38:

(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

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 39:

(d) (144π) cm3

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

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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 40:

(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

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 41:

(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

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 42:

(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

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 43:

(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

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 44:

(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

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 45:

(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%.

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 46:

(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

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 47:

(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

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 48:

(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

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 49:

(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

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 50:

(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

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 51:

(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-609

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 52:

(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.

Question 53:

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

Answer 53:

(a) 32πr33

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

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 54:

(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

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 55:


(d)  5544 cm2

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

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 56:

(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

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 57:

(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

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 58:

(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

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 59:

(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

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 60:

(d) 64

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

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 61:

(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

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 62:

(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

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 63:

(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

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 64:

(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-610

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 65:

(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

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 66:

(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

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 67:

(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

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 68:

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

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 69:

(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

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 70:

(c) 54

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

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 71:

(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

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 72:

(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

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 73:

(c) 3 units

We have:

43πr3= 4πr213r =1r = 3 units

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