Class 10 Maths Chapter 13 Surface Areas and Volumes all MCQs

 

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Chapter 13 Surface Areas and Volumes 

1. The shape of an ice-cream cone is a combination of:

(a)Sphere+cylinder

(b)Sphere+cone

(c)Hemisphere+cylinder

(d)Hemisphere+cone

Answer: d

2. If a cone is cut parallel to the base of it by a plane in two parts, then the shape of the top of the cone will be a:

(a)Sphere

(b)Cube

(c)Cone itself

(d)Cylinder

Answer: c

Explanation: If we cut a cone into two parts parallel to the base, then the shape of the upper part remains the same.

3. If we cut a cone in two parts by a plane parallel to the base, then the bottom part left over is the:

(a)Cone

(b)Frustum of cone

(c)Sphere

(d)Cylinder

Answer: b

Explanation: See the figure below

4. If r is the radius of the sphere, then the surface area of the sphere is given by;

(a)4 π r2

(b)2 π r2

(c)π r2

(d)4/3 π r2

Answer: a

5. If we change the shape of an object from a sphere to a cylinder, then the volume of cylinder will

(a)Increase

(b)Decrease

(c)Remains unchanged

(d)Doubles

Answer: c

Explanation: If we change the shape of a three-dimensional object, the volume of the new shape will be same.

6. Fifteen solid spheres are made by melting a solid metallic cone of base diameter 2cm and height 15cm. The radius of each sphere is:

(a)½

(b)¼

(c)1/3√2

(d)1/3√4

Answer: d

Explanation: Volume of 15 spheres = Volume of a cone

15 x (4/3) π r3= ⅓ πr2h

5×4 π r3=⅓ π 12(15)

20r3=5

r3=5/20=¼

r=1/3√4

7. The radius of the top and bottom of a bucket of slant height 35 cm are 25 cm and 8cm. The curved surface of the bucket is:

(a)4000 sq.cm

(b)3500 sq.cm

(c)3630 sq..cm

(d)3750 sq.cm

Answer: c

Explanation: Curved surface of bucket = π(R1+R2) x slant height (l)

Curved Surface = (22/7) x (25+8) x 35

C.S = 22 x 33 x 5 = 3630 sq. cm.

8. If a cylinder is covered by two hemispheres shaped lid of equal shape, then the total curved surface area of the new object will be

(a)4πrh+2πr2

(b)4πrh-2πr2

(c)2πrh+4πr2

(d)2πrh+4πr

Answer: c

Explanation: Curved surface area of cylinder = 2πrh

The curved surface area of hemisphere = 2πr2

Here, we have two hemispheres.

So, total curved surface area = 2πrh+2(2πr2) = 2πrh+4πr2

9. A tank is made of the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and radius is 30cm. The total surface area of the tank is:

(a)30m

(b)3.3m

(c)30.3m

(d)3300m

Answer: b

Explanation: Total surface area of tank = CSA of cylinder + CSA of hemisphere

= 2πrh + 2πr2= 2π r(h + r)

= 2 x 22/7 x 30(145+30) cm2

=33000 cm2

= 3.3 m2

10. If we join two hemispheres of same radius along their bases, then we get a;

(a)Cone

(b)Cylinder

(c)Sphere

(d)Cuboid

Answer: c

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Basics

1.The lateral surface area of a right circular cone of height 28 cm and base radius 21 cm(in sq. cm) is:

1. 2310
2. 2110
3. 1055
4. 1155

Answer: (A) 2310

Solution: h = 28 cm; r = 21 cm

Therefore, slant height (i) =

= 35 cm

Lateral surface area = πrl

= 22/7 × 21 × 35

= 2310 cm2 

2.If the ratio of the radius of a cone and a cylinder of equal volume is 3:5, then find the ratio of their heights.

1. 25/3
2. 28/3
3. 23/3
4. 7

Answer: (A) 25/3

Solution: Let r1 and h1 be the radius and height of the cone and r2 and h2 be the radius and height of the cylinder.

It is given that the volume of cone is equal to the volume of cylinder.

Thus, the ratio of their height is 25: 3.

3. An iron rod of diameter 1cm and length 8cm is drawn into a wire of length 18m of uniform thickness. Find the thickness of the wire?

1. 0.09cm
2. 0.08cm
3. 0.06cm
4. 0.05cm

Answer: (C) 0.06cm

Solution: Volume of the rod = πr2h= (π) × (1/2)2 × 8= 2πcm3

Volume of the wire = πr2h= (π) × (r)2 × 1800 = 1800πr2cm3

Volume of the rod (old solid shape) = Volume of the wire (New solid shape)

2 π=1800πr2

r2=1/900

r = 1/30

, Diameter = 1/15=0.06cm

4. What do you understand by the quantity called ‘area’?

1. It is the height of an object
2. It is the quantity that expresses the extent of a planar 2-D surface
3. It is the length of an object
4. It is the quantity of an object

Answer: (B) It is the quantity that expresses the extent of a planar 2-D surface

Solutions: The first thing that needs to be understood is that area is a 2 dimensional quantity. Area is the quantity that expresses the extent of a two-dimensional figure or shape, or planar lamina, in the plane. It is only possible to measure area for 2-D surfaces. There is a different quantity to deal with areas for 3-D surfaces which we will look at slightly later in the upcoming questions.

Examples of 2-D surfaces are rectangles, circles, ellipse etc. It is possible to find area for all these 2-D surfaces. But area of objects such as cubes, cylinders, spheres etc. are not defined. But intuitively we know that there is some area associated with such 3-D objects. How come? The answer is coming soon!

5. A solid metallic sphere of diameter 21 cm is melted and recast into a number of smaller cones, of diameter 3.5 cm and height 3cm. The number of cones so formed is:

1. 254
2. 504
3. 540
4. 405

Answer: (B) 504

Solution: Radius of the sphere= 21/2 cm

Volume of the sphere= (4/3) π (21/2)3cm3

Radius of the cone=7/4 cm and height=3 cm

Volume of cone=1/3πr2h=1/3π (7/4)2×3 cm3

Let the number of cones formed be n. Then,

6. How many dimensions are required to make a cuboid?

1. 3
2. 1
3. 15
4. 100

Answer: (A) 3

Combination of solids

7.There are 2 identical cubes each having a total surface area equal to ‘A’. Let ‘S’ be the surface area of the solid obtained by joining these 2 cubes end to end. Which of the following statements is true?

1.  Cannot be determined
2. S < 2A
3. S > 2A
4. S = 2A

Answer: (B) S < 2A

Solution: When 2 cubes are joined end to end it can be easily figured out that the ends of the 2 cubes which are joined together will not be visible after joining. But all the other faces will remain visible.

The total surface area of combined solid can be obtained by adding surface areas of individual cubes and then subtracting the surface areas of one face of each cube which have become hidden due to joining. The combined solid will be a cuboid whose height and breadth will be same as cube’s side. The length of cuboid will be summation of lengths sides of each cube.

 

8. The figure consists of 2 cylinders, the inner cylinder is a solid cylinder whose radius is r and the outer cylinder is a hollow cylinder whose radius is R and height is h, the volume of fluid it can hold is:

1. πr2h
2. πR2h
3. π(R2−r2)h
4. π(R2+r2)h

Answer: (C) π (R2−r2) h

Solution: The inner cylinder is solid whereas the outer is hollow. If the inner cylinder was not there, then the volume of fluid the outer cylinder can hold would be πR2h but since the inner cylinder is solid and is occupying some space, it is limiting the volume of the outer cylinder.

So, the volume of fluid the given shape can hold is the difference in volume of the outer and inner cylinders.

Volume = π (R2−r2) h

9. Find the volume of the figure.

1. 3181.2
2. 5162.5
3. 7142.8
4. 8527.2

Answer: (D) 8527.2

Solution: Volume of the figure = volume of cone + volume of cylinder + volume of frustum

= 594 + 5346 + 2587.2

= 8527.2

10. A piece of cloth is required to completely cover a solid object. The solid object is composed of a hemisphere and a cone surmounted on it. If the common radius is 7 m and height of the cone is 1 m, what is the area of cloth required?

1. 262.39m2
2. 463.39m2
3. 662.39m2
4. 563m2

Answer: (B) 463.39m2

Solution: Surface area of hemisphere = 2πr2

=2×22/7× (7)2 = 308 m2.

For calculating the surface area of a cone we need to calculate its slant height,

So, area of cloth required = (308 + 155.39) m2 = 463.39 m2

11.An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, the diameter of the cylindrical portion is 8 cm and diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.

1. 525.25 cm2
2. 724.25!cm2
3. 781.86 cm2
4. 700 cm2

Answer: (c) 781.86 cm2

Solution: Curved surface area of cylinder

2πrh=π×8×10=80×π

The slant height of frustum can be calculated as follows:

Curved surface area of frustum

Total curved surface area

=169π+80π

=249×3.14

=781.86 cm2

12.Ram has a semicircular disc. He rotates it about its diameter by 360 degrees. When he rotates the disc, a volume of air in his room gets swept. What is the name of the object/shape that exactly occupies this volume?

1. Cylinder
2. Hemisphere
3. Sphere
4. Cuboid

Answer: (C) Sphere

Solution:

It is clear that by seeing 2nd diagram we can know how semicircle sweeps and what is the shape obtained. The shape obtained is a sphere. If rotation had been done for only 180 degrees instead of 360 degrees, we get a hemisphere.  The line segment AB which acted as axis of rotation will also be diameter of sphere formed.

Shape conversion of solids

13.A bucket is in the form of a frustum of a cone, its depth is 15 cm and the diameters of the top and the bottom are 56 cm and 42 cm respectively. How many liters of water can the bucket hold?

1. 28.49
2. 7.5
3. 2.5
4. 10

Answer: (A) 28.49

Solution:

R = 28 cm

r = 21 cm

h = 15 cm

Capacity of the bucket = 1/3πh (R2+r2+Rr)

= 28.49 liters

14. A 20 m deep well of diameter 7 m is dug and the earth taken out is evenly spread out to form a platform of 22 m by 14 m. Find the height of the platform (in m).

1. 7.5
2. 2.5
3. 10
4. 5

Answer: (B) 2.5

Solution: Diameter of the well = d = 7 m

⇒ Radius of the well = r = 7/2=3.5m

Height of the well = h =20 m

Length of the platform = 1 = 22 m

Breadth of the platform = b = 14 m

Let height of the platform =h1 

According to given condition we have:

Volume of Well = Volume of Earth Dug out

⇒ π. r2.h= l × b × h1

⇒22/7 × 3.5 × 3.5 × 20=22×14×h1

⇒h1=22/7 × 3.5 × 3.5 × 20 × 122 × 114

⇒h1=2.5m

15. A cylindrical tank is filled by pumping water from a cuboidal tank of dimensions 200cm × 150cm × 95 cm. The radius of the cylindrical tank is 60cm and height is 95cm. Find the height (in m) of the water left in the cuboidal tank after the cylindrical tank is completely filled. (Take π = 3.14)

1. 0.76 m
2. 0.69 m
3. 0.59 m
4. 0.45 m

Answer: (C) 0.59 m

Solution: Volume of the cylindrical tank = πr2h= (3.14) × (0.6)2×0.95m=1.07m3

Volume of the Cuboidal tank when full = l × b × h = (2m × 1.5m × 0.95m) = 2.85 m3

Volume of water left in the cuboidal tank after completely filling the cylindrical tank = 2.85 – 1.07) = 1.78 m3

Height of water left in cuboidal tank = Volume of water left in cuboidal tank / (l × b)

= 1.78/ (2 × 1.5) = 0.59m

16. A cylinder is moulded into the shape of a sphere. Which of the following factors will be same for both the shapes?

1. None of these
2. Curved surface area
3. Surface area
4. Volume

Answer: (D) Volume

Solution: Volume is a factor which does not differ with change of shape. A cylinder can be moulded into a sphere or a cube or a cuboid of varying dimensions keeping the volume constant.

17. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate (in m2) in 30 minutes, if 8 cm of standing water is needed?

1. 256500
2. 526500
3. 625500
4. 562500

Answer: (D) 562500

Solution: Speed of water flowing through canal =10 km/h =10, 000 m/h

Volume of water flowing through canal in 1 hour = 6 × 1.5 × 10,000 = 90,000 m3

Volume of water flowing through canal in 30 minutes = 90000/2=45,000m3

Standing water = 8 cm =0.08 m

According to given condition we have:

Area which can be irrigated × 0.08 = 45000

⇒ Area which can be irrigated = 45000/ 0.08=562500m2

18. 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:

1. 66m
2. 36m
3. 18m
4. 12m

Answer: (B) 36m

Solution: Diameter of metallic sphere=6 cm

∴ Radius of metallic sphere=3 cm

Also, we have diameter of cross-section of cylindrical wire=0.2 cm

∴ Radius of cross-section of cylindrical wire=0.1 cm

Let the length of the wire be h cm

Since metallic sphere is converted into a cylinder shaped wire of length h cm

∴ Volume of the metal used in wire = Volume of the sphere

19. How many gold coins of 1.75cm in diameter and 2mm in thickness can be melted to form a cuboid of dimensions 5.5cm x 10cm x 3.5cm?

1. 400
2. 500
3. 350
4. 550

Answer: (A) 400

Solution: Radius of the coin = 0.875cm

Height of coin = 0.2 cm

Volume of the cylinder = 

Volume of cuboid = l × b × h = 5.5 × 10 × 3.5

= 192.5 cm3

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1. A cylindrical pencil sharpened at one edge is the combination of

(A) a cone and a cylinder

(B) frustum of a cone and a cylinder

(C) a hemisphere and a cylinder

(D) two cylinders.

Answer: (A)

Explanation: The shape of a sharpened pencil is:

2. A cone is cut through a plane parallel to its base and then the cone that is for medon one side of that plane is removed. The new part that is left over on the other side of the plane is called

(A) a frustum of a cone

(B) cone

(C) cylinder

(D) sphere

Answer:  (A)

Explanation: Observe figure

3. During conversion of a solid from one shape to another, the volume of the new shape will

(A) increase

(B) decrease

(C) remain unaltered

(D) be doubled

Answer:  (C)

Explanation: During conversion of one solid shape to another, the volume of the new shape will remain unaltered.

4. A right circular cylinder of radius r cm and height h cm (h>2r) just encloses a sphere of diameter

(A) r cm

(B) 2r cm

(C) h cm

(D) 2h cm

Answer: (B)

Explanation: Because the sphere is enclosed inside the cylinder, therefore the diameter of sphere is equal to the diameter of cylinder which is 2r cm.

5. A hollow cube of internal edge 22cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that 1/8th space of the cube remains unfilled. Then the number of marbles that the cube can accommodate is

(A) 142244

(B) 142396

(C) 142496

(D) 142596

Answer:  (A)

Explanation:

6. A metallic spherical shell of internal and external diameters 4 cm and 8 cm respectively, is melted and recast into the form of a cone with base diameter 8cm. The height of the cone is

(A) 12cm

(B) 14cm

(C) 15cm

(D) 18cm

Answer:  (B)

Explanation:Since volume will remain same, therefore,

7. A solid piece of iron in the form of a cuboid of dimensions 49cm × 33cm × 24cm, is moulded to form a solid sphere. The radius of the sphere is

(A) 21cm

(B) 23cm

(C) 25cm

(D) 19cm

Answer:  (A)

Explanation: Since volume will remain the same, therefore,

8. If two solid hemispheres of same base radii r, are joined together along their bases, then curved surface area of this new solid is

(A) 4πr²

(B) 6πr²

(C) 3πr²

(D) 8πr²

Answer:  (A)

Explanation: Because curved surface area of a hemisphere is and here we join two solid hemispheres along their bases of radii r, from which we get a solid sphere.

Hence the curved surface area of new solid = 2πr² + 2πr² = 4πr²

9. A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. The total surface area of the shape so formed is

(A) 4πrh + 4πr²

(B) 4πrh − 4πr²

(C) 4πrh + 2πr²           

(D) 4πrh − 2πr²

Answer:  (C)

Explanation: Since the total surface area of cylinder of radius r and height h = 2πrh + 2πr².

When one cylinder is placed over the other cylinder of same height and radius,

Then height of new cylinder = 2h

And radius of the new cylinder = r

Therefore total surface area of new cylinder

= 2πr (2h) + 2πr²

= 4πrh + 2πr²

10. The radii of the top and bottom of a bucket of slant height 45cm are 28cm and 7 cm respectively. The curved surface area of the bucket is:

(A) 4950 cm²

(B) 4951 cm²

(C) 4952 cm²

(D) 4953 cm²

Answer: (A)

Explanation:

Curved Surface area of the bucket = π (R + r) l

⇒ Curved surface area of the bucket = π (28 + 7) X 45

⇒ Curved surface area of the bucket = 4950 cm²

11. A medicine-capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of entire capsule is 2 cm. The capacity of the capsule is

(A) 0.36 cm³

(B) 0.35 cm³

(C) 0.34 cm³

(D) 0.33 cm³

Answer:  (A)

Explanation:

Since diameter of the cylinder = diameter of the hemisphere = 0.5cm

Radius of cylinder r = radius of hemisphere r = 0.5/2 = 0.25 cm

Observe the figure,

Total length of capsule = 2cm

Capacity of capsule is:

12. Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is

(A) 4 cm

(B) 3 cm

(C) 2 cm

(D) 6 cm

Answer:  (C)

Explanation:

Therefore diameter of each solid sphere = 2cm

13. The diameters of the two circular ends of the bucket are 44 cm and 24 cm. The height of the bucket is 35 cm. The capacity of the bucket is

(A) 32.7 litres

(B) 33.7 litres

(C) 34.7 litres

(D) 31.7 litres

Answer:  (A)

Explanation: Since shape of bucket is like Frustum,

Therefore, volume of bucket

14. Volumes of two spheres are in the ratio 64:27. The ratio of their surface areas is:

(A) 3 : 4

(B) 4 : 3

(C) 9 : 16

(D) 16 : 9

Answer: (D)

Explanation: According to question,

 

Therefore ratio of surface area is:

15. A mason constructs a wall of dimensions 270cm× 300cm × 350cm with the bricks each of size 22.5cm × 11.25cm × 8.75cm and it is assumed that 1/8 space is covered by the mortar. Then the number of bricks used to construct the wall is:

(A) 11100

(B) 11200

(C) 11000

(D) 11300

Answer:  (B)

Explanation: According to question,