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This mock test of Test: Surface Area & Volumes- 3 for Class 10 helps you for every Class 10 entrance exam.
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QUESTION: 1

If the volume of a cube is 1728 cm^{3}, the length of its edge is equal to

Solution:

Volume of the cube=a^{3}=1728cm^{3}

QUESTION: 2

Two cubes each of 10 cm edge are joined end to end. The surface area of the resulting cuboid is

Solution:

The length of the resulting cuboid=2 x 10 cm = 20 cm

Its width= 10 cm and its height = 10 cm

i.e. l=20cm, b=10cm, h=10 cm

∴The total surface area of the resulting cuboid

QUESTION: 3

A rectangular sheet of paper 44 cm × 18 cm is rolled along its length and a cylinder is formed. The volume of the cylinder so formed is equal to (Take π = 22/7)

Solution:

QUESTION: 4

If the radius and height of a cylinder are in the ratio 5 : 7 and its volume is 550 cm^{3}, then its radius is equal to (Take π = 22/7)

Solution:

Let the radius = r = 5x and height = h = 7x

So, the volume of the cylinder = π r^{2} h

= 22/7 x (5x)^{2} x 7x

= 22/7 x 25x^{2} x 7x

= 22 x 25x (cube) (7 gets cancelled)

But it is given that volume = 550 cm cube

So, 22 x 25x (cube) = 550

x (cube) = 550 / 22 x 25

x(cube) = 1 ( as 22 x 25=550 and 550 is both in nm and dm)

x = 3**√**1

x = 1

so radius = 5x = 5 x 1 cm = 5cm

QUESTION: 5

If the curved surface area of a solid right circular cylinder of height h and radius r is one-third of its total surface area, then

Solution:

QUESTION: 6

A hollow cylindrical pipe is 21 cm long. If its outer and inner diameters are 10 cm and 6 cm respectively, then the volume of the metal used in making the pipe is (Take π = 22/7)

Solution:

The height of the cylindrical pipe=21cm

inner radius=6cm

outer radius=10cm

volume of the metal used to make the Pipe=Π(R^{2}-r^{2})h

=22/7×(10^{2}-6^{2})

=1056cm^{3}

QUESTION: 7

If the radius and slant height of a cone are in the ratio 4 : 7 and its curved surface area is 792 cm^{2}, then its radius is (Take π = 22/7)

Solution:

QUESTION: 8

If the radius of the base and the height of a right circular cone are respectively 21 cm and 28 cm, then the curved surface area of the cone is (Take π = 22/7)

Solution:

QUESTION: 9

A conical tent with base-radius 7 m and height 24 m is made from 5 m wide canvas. The length of the canvas used is (Take π = 22/7)

Solution:

QUESTION: 10

The total surface area of a solid hemisphere of radius 3·5 m is covered with canvas at the rate of Rs. 20 per m^{2}. The total cost to cover the hemisphere is (Take π = 227)

Solution:

Total Surface Area of a hemisphere of radius 'r' =3πr^{2}

Hence, total surface area of this hemisphere =3×22/7×3.5×3.5=115.5sqm

∴ cost of covering the hemisphere with canvas at the rate of Rs. 20 per m^{2}=115.5×20=Rs2310

QUESTION: 11

If the volume of a vessel in the form of a right circular cylinder is 448 cm^{3} and its height is 7 cm, then the curved surface area of the cylinder is

Solution:

Volume of a Cylinder of Radius R and height h =πR^{2}h

∴ volume of the given cylinder =π×R^{2}×7=448πcm^{3}

R^{2}=64

R=8cm

Curved surface area of a cylinder of radius "R" and height "h" =2πRh

∴ curved surface area of the given cylinder =2×π×8×7=112π cm^{2}

QUESTION: 12

If the curved surface area of a right circular cone is 12320 cm^{2} and its base-radius is 56 cm, then its height is (Takeπ = 22/7)

Solution:

QUESTION: 13

If a solid metallic sphere of radius 8 cm is melted and recasted into n spherical solid balls of radius 1 cm, then n =

Solution:

In order to find the number of balls carved from larger sphere we need to divide the larger ball to small ball

QUESTION: 14

If the diameter of a metallic sphere is 6 cm, it is melted and a wire of diameter 0·2 cm is drawn, then the length of the wire made shall be

Solution:

We have

Diameter of metallic sphere =6cm

Radius of metallic sphere =3cm

Also we have

Diameter of cross-section of cylindrical wire=0.2cm

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

Let the length of the wire be h cm

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

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

QUESTION: 15

If n coins each of diameter 1·5 cm and thickness 0·2 cm are melted and a right circular cylinder of height 10cm and diameter 4·5 cm is made, then n =

Solution:

QUESTION: 16

A tent is in the form of a cylinder of diameter 8 m and height 2 m, surmounted by a cone of equal base and height 3 m. The canvas used for making the tent is equal to

Solution:

QUESTION: 17

A toy is in the form of a cone mounted on a hemisphere with same radius. The diameter of the base of the conical portion is 6 cm and its height is 4 cm. The surface area of the toy is

Solution:

QUESTION: 18

A frustum of a right circular cone is of height 16 cm with radii of its ends as 8 cm and 20 cm. The volume of the frustum is

Solution:

QUESTION: 19

A frustum of a right circular cone is of height 16 cm with radii of its ends as 8 cm and 20 cm has lateral surface area equal to

Solution:

QUESTION: 20

A solid metal cone with base-radius 12 cm and height 24 cm, is melted to form solid spherical balls, each of diameters 6 cm. The number of such balls made is

Solution:

QUESTION: 21

One cubic metre piece of copper is melted and recast into a square cross-section bar, 36 m long. An exact cube is cut off from this bar. If cubic metre of copper cost Rs. 108, then the cost of this cube is :

Solution:

QUESTION: 22

If the surface areas of two spheres are in the ratio 4 : 9, then the ratio of their volumes is :

Solution:

QUESTION: 23

In a shower 10 cm of rain fall. The volume of water that falls on 1.5 hectares of ground is :

Solution:

QUESTION: 24

The radius of base and the volume of a right circular cone are doubled. The ratio of the length of the larger cone to that of the smaller cone is :

Solution:

QUESTION: 25

A cone and a hemisphere have equal base diameter and equal volumes. The ratio of their heights is :

Solution:

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