MCQ Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes

Surface Areas and Volumes Class 10 MCQ Questions with Answers

Question 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) a cone and a cylinder

Explanation:
The sharpened part of the pencil is cone and unsharpened part is cylinder.

Question 2.

A surahi is the combination of:

(A) a sphere and a cylinder
(B) a hemisphere and a cylinder
(C) two hemispheres
(D) a cylinder and a cone
Answer:
(A) a sphere and a cylinder

Explanation:
A surahi is the combination of a sphere and a cylinder.

Question 3.

A plumbline (Sahul) is the combination of:

(A) a cone and a cylinder
(B) a hemisphere and a cone
(C) frustum of a cone and a cylinder
(D) sphere and cylinder
Answer:
(B) a hemisphere and a cone

Explanation:
Plumbline is an instrument used to check the vertically of an object. It is a combination of a hemisphere and a cone.

Question 4.

The shape of a gilli, in the gilli-danda game (see in Figure) is a combination of:

(A) two cylinders
(B) a cone and a cylinder
(C) two cones and a cylinder
(D) two cylinders and a cone
Answer:
(C) two cones and a cylinder

Explanation:
The shape of gilli, in the gilli- danda game is a combination of two cones and a cylinder.

Question 5.

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

(A) 142296
(B) 142396
(C) 142496
(D) 142596
Answer:
(A) 142296

Explanation:
Let the spherical marble has radius r. Diameter of the marble = 0.5 cm
⇒ r = \(\frac{0.5}{2}\) cm = 0.25 cm
Length of side of l = 22 cm
Let n marbles can fill the cube.
⇒Volume of n marbles = (1 – \(\frac{1}{8}\))
part of volume of cube
⇒ n. \(\frac{4}{3}\) πr³ = \(\frac{7}{8}\) × l³
n = \(\frac{7 l^{3}}{8}\)× \(\frac{3}{4 \pi r^{3}}\)
⇒ = \(\frac{7 \times 3 \times 22 \times 22 \times 22 \times 7}{8 \times 4 \times 22 \times 0.25 \times 0.25 \times 0.25}\)
⇒ π n = 7 × 3 × 22 × 22 × 2 × 7
= 42 × 484 × 7
n = 142296
So, cube can accommodate up 142296 marbles.

Question 6.

A medicine-capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each to 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) 0.36 cm³

Explanation:
Capsule consists of 2 hemispheres and a cylinder.
⇒ r = \(\frac{0.5}{2}\) cm = 0.25 cm
⇒ r = 0. 25 cm
Total length of capsule = r + h + r
⇒ 2 cm = 2r + h
⇒ 2 = 2 × 0.25 + h
⇒ h = 2 – 0.5 = 1.5 cm
Volume of capsule = Volume of two
hemispheres + Volume of cylinder
= 2  × \(\left(\frac{4}{3} \pi r^{3} \times \frac{1}{2}\right)\) + πr²h
= \(\frac{4}{3}\) πr² + πr²h
= πr² \(\left(\frac{4}{3} r+h\right)\)
= \(=\frac{22}{7}\)  × 0. 25  × 0. 25 \(\left(\frac{4}{3} \times 0.25+\frac{15}{10}\right)\)
= \(\frac{22}{7}\)  × 0. 25  × 0. 25 \(\left(\frac{1}{3}+\frac{3}{2}\right)\)
= \(\frac{22}{7}\)  × \(\frac{25}{100}\)  × \(\frac{25}{100}\) × \(\frac{11}{6}\) = \(\frac{121}{336}\)
⇒Volume of capsule = 0. 3601 cm³ = 0. 36 cm³

Question 7.

If two solid hemispheres of same base radius V are joined together along their bases, then curved surface area of this new solid is :

(A) 4 πr²
(B) b πr²
(C) 3 π²
(D) 8 πr²
Answer:
(A) 4 πr²

Explanation:
When two hemispheres of equal radii are joined base to base, new solid becomes sphere and curved surface area of sphere is im².

Question 8.

A right circular cylinder of radius r cm and height h cm (where h > 2r) just encloses of sphere of diameter :

(A) r cm
(B) 2r cm
(C) h cm
(D) 2h cm
Answer:
(B) 2r cm

Explanation:
As the cylinder just enclosed the sphere so the radius or diameter of cylinder and sphere are equal, i.e., 2r and height h > 2r.

Question 9.

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 of base diameter 8 cm. The height of the cone is:

(A) 12 cm
(B) 14 cm
(C) 15 cm
(D) 18 cm
Answer:
(B) 14 cm

Explanation:
During recasting a shape into another shape it’s volume does not change. For Spherical shell
r₁ = \(\frac{4}{2}\) = 2 cm
r₂ = \(\frac{8}{2}\) = 4 cm
For Cone r = \(\frac{8}{2}\) = 4 cm
h = π
During recasting voleme remains same so, Volume of cone = Volume of hollow spherical shell
⇒ \(\frac{1}{3}\)πr² h = \(\frac{4}{3} \pi r_{2}^{3}-\frac{4}{3} \pi r_{1}^{3}\)
⇒ \(\frac{1}{3}\) = \(\frac{4}{3} \pi\left(r_{2}^{3}-r_{1}^{3}\right)\)
⇒ r² h = 4\(\left(r_{2}^{3}-r_{1}^{3}\right)\)
⇒ 4 × 4h = 4[(4)3 – (2)3]
⇒ 4h = 64 – 8
⇒ h = \(\frac{56}{4}\)
⇒ h = 14 cm

Question 10.

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

(A) 21 cm
(B) 23 cm
(C) 25 cm
(D) 19 cm
Answer:
(A) 21 cm

Explanation:
Solid cuboid of iron is moulded into solid sphere. For hence, volume of cuboid and sphere are equal.
For sphere
r = π
cuboid
l = 49 cm
b = 33 cm
h = 24 cm
⇒Volume of sphere (solid) = Volu me of cuboid
⇒ \(\frac{4}{3 \pi r^{3}}\) = l x b x h
πr³ = \(\frac{(l \times b \times h \times 3)}{(4 \times \pi)}\)
= \(\frac{(49 \times 33 \times 24 \times 3 \times 7)}{(4 \times 22)}\)
πr³ = 7 × 7 × 7× 3 × 3 × 3
r = 21 cm

Question 11.

A mason constructs a wall of dimensions 270 cm × 300 cm × 350 cm with the bricks each of size 22.5 cm × 11.25 cm × 8.75 cm 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) 11200

Explanation:
The volume of the wall covered mortar = \(\frac{1}{8}\) part So, the volume covered by bricks of wall = \(\left(1-\frac{1}{8}\right)\) volume of wall
= \(\frac{7}{8}\)
volume of wall
Bricks (Cuboid) Wall (Cuboid)
l₁= 22.5 cm l – 270 cm
b₁ = 11.25 cm b = 300 cm
h1 = 8.75 cm 350 cm
Bricks (Cuboid) Wall (Cuboid)
l₁= 22.5 cm l – 270 cm
b₁ = 11.25 cm b = 300 cm
h1 = 8.75 cm 350 cm

Let n be the number of bricks.
According to the question, we have
Volume of n bricks = \(\frac{7}{8}\) Volume of wall (Cuboid)
⇒ n × l₁ × b₁ × h₁ = \(\frac{7}{8}\) × l × b × h
⇒ n = \(\frac{(7 \times l \times b \times h)}{\left(8 \times l_{1} \times b_{1} \times h_{1}\right)}\) = \(\frac{(7 \times 270 \times 300 \times 350)}{(8 \times 22.5 \times 11.25 \times 8.75)}\)
⇒ n = \(\frac{(7 \times 270 \times 300 \times 350 \times 100 \times 10 \times 100)}{(8 \times 225 \times 1125 \times 875)}\)
⇒ n = 2 × 4 × 350 × 4 = 32 × 350 = 11,200 bricks.

Question 12.

Twelve solid sphere 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) 2 cm

Explanation:
Solid cylinder is recasted into 12 spheres. So, the volume of 12 spheres will be equal to the volume of the cylinder.
12 For spheres
R = π
For cylinder
r = \(\frac{2}{2}\) = 1 cm
h – 16 cm
⇒ Volume , diameter = 2R = 2 × 1 = 2 cm
⇒ \(\frac{4}{3 \pi r^{3}}\) = πr²h
⇒ R³ = \(\frac{\left(3 r^{2} h\right)}{(4 \times 12)}\)
= \(\frac{(3 \times 1 \times 1 \times 16)}{(4 \times 12)}\) = 1
⇒ R = 1 cm
Hence, diameter = 2R = 2 × 1 = 2 cm.

Question 13.

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

(A) increase
(B) decrease
(C) remains unaltered
(D) be doubled
Answer:
(C) remains unaltered

Explanation:
During reshaping a solid, the volume of new solid will be equal to old one or remains unaltered.

Question 14.

A rectangular sheet of paper 40 cm × 22 cm, is rolled to form a hollow cylinder of height 40 cm. The radius of the cylinder (in cm) is:

(A) 3.5
(B) 7
(C) \(\frac{80}{7}\)
(D) 5
Answer:
(A) 3.5

Explanation:
Circumference = 22 cm
2πr = 22
2 × \(\frac{22}{7}\) × r = 22
r = 3.5 cm

Question 15.

The number of solid spheres, each of diameter 6 cm that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm, is:

(A) 3
(B) 5
(C) 4
(D) 6
Answer:
(B) 5

Explanation:
No. of solid spheres = \(\frac{\text { Volume of cylinder }}{\text { Volume of sphere }}\)
= \(\frac{\pi R^{2} h}{\frac{4}{3} \pi r^{3}}\)
= \(\frac{\pi(2)^{2} \times 45 \times 3}{4 \times \pi \times(3)^{3}}\)
= 5.

Assertion and Reason Based MCQs

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is True

Question 1.

Assertion (A): In a right circular cone, the cross-section made by a plane parallel to the base is a circle.
Reason (R): If the volume and the surface area of a solid hemisphere are numerically equal, then the diameter of hemisphere is 9 units.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
In case of assertion:
In a right circular cone, if any cut is made parallel to its base, we get a circle.
∴ Assertion is correct.
In case of reason:
Let radius of sphere be r.
Given, volume of hemisphere = Surface area of hemisphere
or, \(\vec{a}\) πr³ = 3πr²
or, r = \(\vec{a}\) units
⇒ Diameter = \(\vec{a}\) × 2 = 9 units
∴ Reason is correct :
Hence, both assertion and reason are correct but reason is the correct explanation for assertion.

Question 2.

Assertion (A): If the volumes of two spheres are in the ratio 64 : 27, then the ratio of their surface areas is 4 : 3.
Reason (R): If the surface areas of two spheres are in the ratio 16 : 9, then the ratio of their volumes is 64 : 27.

Answer:
(D) A is false and R is True

Explanation:
In case of assertion:
\(\frac{V_{1}}{V_{2}}\) = \(\frac{64}{27}\)
\(\frac{\frac{4}{3} \pi r_{1}{ }^{3}}{\frac{4}{3} \pi r_{2}{ }^{3}}\) = \(\frac{64}{27}\)
\(\left(\frac{r_{1}}{r_{2}}\right)^{3}\) = \(\frac{64}{27}\)
\(\frac{r_{1}}{r_{2}}\) = \(\frac{4}{3}\)
Now the ratio of their surface areas,
\(\frac{4 \pi r_{1}^{2}}{4 \pi r_{2}^{2}}\) = \(\left(\frac{r_{1}}{r_{2}}\right)^{2}\) = \(\left(\frac{4}{3}\right)^{2}\) = \(\frac{16}{9}\)
∴ Assertion is correct.
In case of reason:
Given,
\(\frac{A_{1}}{A_{2}}\) = \(\frac{4 \pi r_{1}^{2}}{4 \pi r_{2}^{2}}\) = \(\frac{16}{9}\)
\(\left(\frac{r_{1}}{r_{2}}\right)^{2}\) = \(\frac{16}{9}\)
\(\frac{r_{1}}{r_{2}}\) = \(\sqrt{\frac{16}{9}}\) = \(\frac{4}{3}\)
Now, volumes of two spheres,
\(\frac{V_{1}}{V_{2}}\) = \(\frac{\frac{4}{3} \pi r_{1}^{2}}{\frac{4}{3} \pi r_{2}^{2}}\)
= \(\left(\frac{r_{1}}{r_{2}}\right)^{3}\) = \(\left(\frac{4}{3}\right)^{3}\) = \(\frac{64}{27}\)
= 64 : 27
∴ Reason is correct:
Hence, assertion is incorrect but reason is correct.

Question 3.

Assertion (A): The volume of a right circular cylinder of base radius 7 cm and height 10 cm is 1540 cm3
Reason (R): If the curved surface area of a cylinder is 264m2 and its volume is 924m3, then the ratio of its height to its diameter is 4 : 7.

Answer:
(C) A is true but R is false

Explanation:
In case of assertion:
Here r = 7 cm,
h = 10 cm,
Volume of cylinder = πrh
= \(\frac{22}{7}\) × (7)² × 10
= 1540 cm³
∴ Assertion is correct.
In case of reason:
Curved Surface area of cylinder = 2πrh
Volume of cylinder = 2π2rh
\(\frac{\pi r^{2} h}{2 \pi r h}\) =\(\frac{924}{264}\)
\(\frac{y^{*}}{2}\) = \(\frac{7}{2}\)
r = 7 m
2πrh = 264
or, 2 × \(\frac{22}{7}\) × 7 × h = 264
h = 6 m
∴ \(\frac{h}{2 r}\) = \(\frac{6}{14}\) = \(\frac{3}{7}\)
Hwnce, h : r = 3 : 7
∴ Reason is correct:
Hence, assertion is incorrect but reason is correct.

Question 4.

Assertion (A): The number of solid spheres of diameter 6 cm can be made by melting a solid metallic cylinder of height 45 cm and diameter 4 cm, is 5.
Reason (R): If three solid metallic spherical balls of radius 3 cm, 4 cm and 5 cm are melted into a single spherical ball, then its radius is 6 cm.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
In case of assertion:
Let the number of spheres be n.
Radius of sphere r₁ = 3 cm, radius of cylinder r₂ = 2 cm
Volume of spheres = Volume of cylinder
n × \(\frac{4}{3} \pi r^{3}\) πr³ = \(\pi r_{1}^{2} h\)
or, n × \(\frac{4}{3}\) x \(\frac{22}{7}\) × (3)³ = \(\frac{22}{7}\) × (2)² × 45
or, 36 n = 180
or, n = \(\frac{180}{36}\) = 5
Thus, the number of solid spheres = 5 r. Assertion is correct.
In case of reason:
Let the radius of spherical ball be R.
Volume of spherical ball = Volume of three
\(\frac{4}{3} \pi R^{3}\) = \(\frac{4}{3} \pi\)[(3)³ + (4)³ + (5)³]
or, R³ = 27 + 64 + 125
or, R³ = 216
or, R = 6 cm
∴ Reason is correct:
Hence, both assertion and reason are correct but reason is not the correct explantion for assertion.

Question 5.

Assertion (A): If 12 solid spheres of the same size are made by melting a solid metallic cone of base radius 1 cm and height of 48 cm, then radius of each sphere is 2 cm.
Reason (R): If three cubes of iron whose edges are 3 cm, 4 cm and 5 cm respectively are melted and formed into a single cube, then the edge of the single cube is 6 cm.

Answer:
(D) A is false and R is True

Explanation:
In case of assertion:
No. of spheres = 12
Radius of cone, r = 1 cm
Height of the cone = 48 cm
∴ Volume of 12 spheres = Volume of cone
Let the raddius of sphere be R cm
12 × \(\frac{4}{3} \pi R^{3}\) = \(\frac{1}{3} \pi r^{2} h\)
or, 12 × \(\frac{4}{3} \pi R^{3}\) = \(\frac{1}{3} \pi\) × (1)² × 48
16R³ = 16
R³ = 1
or, R = 1 cm
∴ Assertion is incorrect.
In case of reason:
Let the edge of single cube be x cm Volume of single cube = Volume of three cubes
x³ = (3)³ + (4)³ + (5)³
= 27 + 64 + 125
= 216
= 6 cm
∴ Reason is correct:
Hence, both assertion and reason are correct but reason is not the correct explantion for assertion.

Question 6.

Assertion (A): If a solid sphere of radius r is melted and recast into the shape of a solid cone of height r, then the radius of the base of a cone is 2r.
Reason (R): If a metallic sphere of total volume n is melted and recast into the shape of a right circular cylinder of radius 0.5 cm, then the height of cylinder is 8 cm.

Answer:
(C) A is true but R is false

Explanation:
In case of assertion:
Volume of sphere = Volume of cone Let the radius of cone be R cm.
∴  \(\frac{4}{3} \pi r^{3}\)
= \(\frac{1}{3} \pi R^{2} \) × r
or, 4r3 = R2r
or, R2 = 4r2
or, R = 2r
∴ Assertion is correct.
In case of reason:
Volume of cylinder = Volume of sphere,
πr²h = π
where r and h are radius of base and height of cylinder
(0.5)² h = 1
\(\left(\frac{1}{2}\right)\) h = 1
h = 4 cm.
∴ Reason is incorrect:
Hence, assertion is correct but reason is incorrect.

Case – Based MCQs

Attempt any four sub-parts from each question. Each sub-part carries 1 mark.
I. Read the following text and answer the following question on the basis of the same:
Adventure camps are the perfect place for the children to practice decision making for themselves without parents and teachers guiding their every move. Some students of a school reached for adventure at Sakleshpur. At the camp, the waiters served some students with a welcome drink in a cylindrical glass and some students in a hemispherical cup whose dimensions are shown below. After that they went for a jungle trek. The jungle trek was enjoyable but tiring. As dusk fell, it was time to take shelter. Each group of four students was given a canvas of area 551 m2. Each group had to make a conical tent to accommodate all the four students. Assuming that all the stitching and wasting incurred while cutting, would amount to 1 m2, the students put the tents. The radius of the tent is 7 m.

Question 1.

The volume of cylindrical cup is

(A) 295.75 cm3
(B) 7415.5 cm3
(C) 384.88 cm3
(D) 404.25 cm3
Answer:
(D) 404.25 cm3

Explanation:
diameter = 7 cm
radius = 3.5 cm
height = 10.5 cm
Volume of cylindrical cup
= πr²h
\(\frac{22}{7}\) × 3.5 × 3.5 × 10. 5
= 404 . 25 cm²

Question 2.

The volume of hemispherical cup is

(A) 179.67 cm³
(B) 89.83 cm³
(C) 172.25 cm³
(D) 210.60 cm³
Answer:
(B) 89.83 cm³

Question 3.

Which container had more juice and by how much?

(A) Hemispherical cup, 195 cm3
(B) Cylindrical glass, 207 end
(C) Hemispherical cup, 280.85 cm3
(D) Cylindrical glass, 314.42 cm3
Answer:
(D) Cylindrical glass, 314.42 cm3

Question 4.

The height of the conical tent prepared to accommodate four students is

(A) 18 m
(B) 10 m
(C) 24 m
(D) 14 m
Answer:
(C) 24 m

Explanation:
Radius = 7m
Area of conical tent = 551 m² – 1 m²
= 550 m²
πrl = 551
\(\frac{22}{7}\) x 7\(\sqrt{r^{2}+h^{2}}\) = 550
\(\frac{22}{7}\) x 7\(\sqrt{7^{2}+h^{2}}\) =550
\(\sqrt{7^{2}+h^{2}}\) = \(\frac{550}{22}\)
\(\sqrt{7^{2}+h^{2}}\) = \(\frac{50}{2}\)
\(\sqrt{7^{2}+h^{2}}\) = 25
72 + h2 = (25)2
h2 = 625 – 49
h2 = 576
h = √576
= 24 m

Question 5.

How much space on the ground is occupied by each student in the conical tent

(A) 54 m²
(B) 38.5 m²
(C) 86 m²
(D) 24 m²
Answer:
(B) 38.5 m²

Explanation:
Area of Base of conical tent = πr²
\(\frac{22}{7}\)  ×7 × 7
= 154 m²
Area of occupied by each
student = \(\frac{1}{4}\) × 154 m²
= 38.5 m²

II. Read the following text and answer the following question on the basis of the same:
A The Great Stupa at Sanchi is one of the oldest stone structures in India, and an important monument of Indian Architecture. It was originally commissioned by the emperor Ashoka in the 3rd century BCE. Its nucleus was a simple hemispherical brick structure built over the relics of the Buddha. It is a perfect example of combination of solid figures. A big hemispherical dome with a cuboidal structure
\(\left(\text { Take } \pi=\frac{22}{7}\right)\)

Question 1.

Calculate the volume of the hemispherical dome if the height of the dome is 21 m:

(A) 19404 sq. m
(B) 2000 sq. m
(C) 15000 sq. m
(D) 19000 sq.. m
Answer:
(A) 19404 sq.. m

Explanation:
height of hemispherical dome = Radius of hemispherical dome = 21 m.
Volume of dome = \(\frac{2}{3}\)πr³
= \(\frac{2}{3}\) x \(\frac{22}{7}\) ×21 × 21 × 21
= 19,404 m³

Question 2.

The formula to find the Volume of Sphere is:

(A)\(\frac{2}{3}\){7}πr³
(B) \(\frac{4}{3}\)πr³
(C) 4πr²
(D) 2πr²
Answer:
(B) \(\frac{4}{3}\)πr³

Question 3.

The cloth require to cover the hemispherical dome if the radius of its base is 14m is:

(A) 1222 sq. m
(B) 1232 sq. m
(C) 1200 sq. m
(D) 1400 sq. m
Answer:
(B) 1232 sq. m

Question 4.

The total surface area of the combined figure i.e. hemispherical dome with radius 14 m and cuboidal shaped top with dimensions 8m × 6m × 4mis

(A) 1200 sq. m
(B) 1232 sq. m
(C) 1392 sq. m
(D) 1932 sq. m
Answer:
(C) 1392 sq. m

Explanation:
Total surface Area of Combined figure
= 2πr² + 2(lb + bh + hl) – lb
= 2 x \(\frac{22}{7}\) × 14 × 4 + 2( 8 × 6 + 6 × 4 + 4 x 8) – 8 × 6lm²
= [1232 + 208 – 48] m²
= 1392 m²

Question 5.

The volume of the cuboidal shaped top is with dimensions mentioned in question 4.

(A) 182.45 m³
(B) 282.45 m³
(C) 292 m³
(D) 192 m³
Answer:
(D) 192 m³

Explanation:
Volume of the cuboidal shaped 1 top
= l × b × h
= 8m × 6m × 4m
= 192 m³.

III. Read the following text and answer the following question on the basis of the same:
On a Sunday, your Parents took you to a fair. You could see lot of toys displayed, and you wanted them to buy a RUBIK’s cube and strawberry ice-cream for you.
Observe the figures and answer the questions:

Question 1.

The length of the diagonal if each edge measures 6 cm is

(A) 3√3
(B) 3√6
(C) √l2
(D) 6√3
Answer:
(D) 6√3

Question 2.

Volume of the solid figure if the length of the edge is 7 cm is:

(A) 256 cm³
(B) 196 cm³
(C) 343 cm³
(D) 434 cm³
Answer:
(C) 343 cm³

Question 3.

What is the curved surface area of hemisphere (ice cream) if the base radius is 7 cm ?

(A) 309 cm²
(B) 308 cm²
(C) 803 cm²
(D) 903 cm²
Answer:
(B) 308 cm²

Question 4.

Slant height of a cone if the radius is 7 cm and the height is 24 cm……..

(A) 26 cm
(B) 25 cm
(C) 52 cm
(D) 62 cm
Answer:
(B) 25 cm

Question 5.

The total surface area of cone with hemispherical ice cream is

(A) 858 cm²
(B) 885 cm²
(C) 588 cm²
(D) 855 cm²
Answer:
(A) 858 cm²

IV. Read the following text and answer the following question on the basis of the same:
A carpenter made a wooden pen stand. It is in the shape of cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. (See Figure).

Question 1.

What is the volume of cuboid?

(A) 525 cm³
(B) 225 cm³
(C) 552 cm³
(C) 255 cm³
Answer:
(A) 525 cm³

Explanation:
For cuboid
l = 15 cm, b = 10 cm and h = 3.5 cm
Volume of the cuboid = l × b × h
= 15 × 10 × 3.5
= 525 cm³

Question 2.

What is the volume of a conical depression ?

(A) \(\frac{11}{3}\) cm³
(B) \(\frac{11}{30}\) cm³
(C) \(\frac{3}{11}\) cm³
(D) \(\frac{30}{11}\) cm³
Answer:
(B) \(\frac{11}{30}\) cm³

Explanation:
For conical depression:
r = 0.5 cm,
h = 1.4 cm
Volume of conical depression
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 0. 5 × 0.5 × 1.4
= \(\frac{11}{30}\) cm³

Question 3.

What is the total volume of conical depressions?

(A) 1.74 cm³
(B) 1.44 cm³
(C) 1.47 cm³
(D) 1.77 cm³
Answer:
(C) 1.47 cm³

Explanation:
Volume of four conical depressions
= 4 × \(\vec{a}\) = 1.47 cm³

Question 4.

What is the volume of wood in the entire stand?

(A) 522.35 cm³
(B) 532.53 cm³
(C) 523.35 cm³
(D) 523.53 cm³
Answer:
(D) 523.53 cm³

Explanation:
Volume of the wood in the entire stand = Volume of cuboid – Volume of 4 conical depressions
= 525 – 1.47
= 523.53 cm3

Question 5.

The given problem is based on which mathematical concept?

(A) Triangle
(B) Surface Area and Volumes
(C) Height and Distances
(D) None of these
Answer:
(B) Surface Area and Volumes

MCQ Questions for Class 10 Maths with Answers