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RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2

January 30, 2023

RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2

Other Exercises

RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere MCQS

Question 1.
Find the volume of a sphere whose radius is
(i) 2 cm
(ii) 3.5 cm
(iii) 10.5 cm
Solution:
(i) Radius of sphere (r) = 2 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 1.1

Question 2.
Find the volume of a sphere whose diameter is,
(i) 14 cm
(ii) 3.5 dm
(iii) 2.1 m
Solution:
(i) Diameter of a sphere = 14 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 2.1

Question 3.
A hemspherical tank has inner radius of 2.8 m. Find its capacity in litres.
Solution:
Radius of hemispherical tank (r) = 2.8 m
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 3.1

Question 4.
A hemispherical bowl is made of steel 0.25 cm thick. The inside radius of the bowl is 5 cm. Find the volume of steel used in making the bowl.
Solution:
Thickness of steel = 0.25 cm = \(\frac { 1 }{ 4 }\)cm
Inside radius of the hemispherical bowl (r) = 5 cm
∴ Outer radius (R) = 5 + 0.25 = 5.25 cm
∴ Volume of the steel used = \(\frac { 1 }{ 4 }\)π(R³ – r³)
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 4.1

Question 5.
How many bullets can be made out of a cube of lead, whose edge measures 22 cm, each bullet being 2 cm in diameter?
Solution:
Edge of cube (r) = 22 cm
∴ Volume = a³ = (22)³ cm³
= 22 x 22 x 22 = 10648 cm³
Diameter of a bullet = 2 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 5.1

Question 6.
A shopkeeper has one laddoo of radius 5 cm. With the same material how many laddoos of radius 2.5 cm can be made?
Solution:
Radius of bigger laddoo (R) = 5 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 6.1

Question 7.
A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. It the diameters of two balls be \(\frac { 3 }{ 2 }\) cm and 2 cm, find the diameter of the third ball.
Solution:
Diameter of a spherical ball of lead = 3 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 7.1

Question 8.
A sphere of radius 5 cm is immersed in water filled in a cylinder, the level of water rises \(\frac { 5 }{ 3 }\) cm. Find the radius of the cylinder.
Solution:
Radius of sphere (r1) = 5 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 8.1
Level of water rises in the cylinder after immersing the sphere in it
∴ Height of water level = \(\frac { 5 }{ 3 }\) cm
Let r be radius of the cylinder, then Volume of water = Volume of the sphere

Question 9.
If the radius of a sphere is doubled, what is the ratio of the volumes of the first sphere to that of the second sphere?
Solution:
Let r2 be the radius of the given sphere
then volume = \(\frac { 4 }{ 3 }\) πr³
By doubling the radius the radius of the new sphere = 2r
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 9.1

Question 10.
A vessel in the form of a hemispherical bowl is full of water. Its contents are emptied in a right circular cylinder. The internal radii of the bowl and the cylinder are 3.5 cm and 7 cm respectively. Find the height to which the water will rise in the cylinder.
Solution:
Radius of hemispherical bowl (r) = 3.5 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 10.1

Question 11.
A cylinder whose height is two thirds of its diameter, has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder.
Solution:
Radius of a sphere (r) = 4 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 11.1

Question 12.
A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of water in the cylinder.
Solution:
Radius of hemispherical bowl (r) = 6 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 12.1

Question 13.
The diameter of a copper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is 108 m, find its diameter.
Solution:
Diameter of a copper sphere = 18 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 13.1

Question 14.
The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm. Find the length of the wire.
Solution:
Diameter of a sphere = 6 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 14.1

Question 15.
The radius of the internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm respectively. If it is melted and recast into a solid cylinder of height 2 \(\frac { 2 }{ 3 }\) cm. Find the diameter of the cylinder.
Solution:
Internal radius of the hollow spherical shell (r) = 3 cm
and external radius (R) = 5 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 15.1

Question 16.
A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base.
Solution:
Radius of hemisphere (r) = 7 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 16.1

Question 17.
A hollow sphere of internal and external radius 2 cm and 4 cm respectively is melted into a cone of base radius 4 cm. Find the height and slant height of the cone.
Solution:
Internal radius of a hollow sphere (r) = 2 cm
and external radius (R) = 4 cm
∴ Volume of the metal used
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 17.1

Question 18.
A metallic sphere of radius 10.5 cm is melted and thus recast into small cones each of radius 3.5 cm and height 3 cm. Find how many cones are obtained.
Solution:
Radius of a metallic sphere (R) = 10.5 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 18.1

Question 19.
A cone and a hemisphere have equal bases and equal volumes. Find the ratio Of their heights.
Solution:
Let r be the radius and h be the height of the cone, hemisphere
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 19.1

Question 20.
The largest sphere is carved out of a cube of side 10.5 cm. Find the volume of the sphere.
Solution:
By carving a largest sphere out of the cube, the diameter of the sphere = 10.5
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 20.1

Question 21.
A cube, of side 4 cm, contains a sphere touching its sides. Find the volume of the gap in between.
Solution:
Side of cube = 4 cm
∴ Volume = (side)³ = 4x4x4 = 64 cm³
Diameter of the largest sphere touching its sides = 4 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 21.1

Question 22.
A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1m, then find the volume of the iron used to make the tank. (NCERT)
Solution:
Thickness of hemispherical tank = 1 cm
Inner radius (r) = 1 m = 100 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 22.1

Question 23.
A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to fill this capsule? (NCERT)
Solution:
Diameter of a medicine spherical capsule = 3.5 mm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 23.1

Question 24.
The diameter of the moon is approximately one fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon? (NCERT)
Solution:
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 24.1

Question 25.
A cone and a hemisphere have equal bases and equal volumes. Find the ratio in their heights.
Solution:
Let r be the radius of cone and hemisphere and let h be the height of the cone then
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 25.1

Question 26.
A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the tub and thus level of water is raised by 9 cm. What is the radius of the ball?
Solution:
Radius of cylinderical tub (r) = 16 cm
Height of water in it (h) = 30 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 26.1

Question 27.
A cylinder of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball. (Use π = 22/7).
Solution:
Radius of cylinder (r) = 12 cm
Depth of water in it (h) = 20 cm
By dropping a ball, the water level rose by 6.75 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 27.1

Question 28.
A cylindrical jar of radius 6 cm contains oil. Iron spheres each of radius 1.5 cm are immersed in the oil. How many spheres are necessary to raise the level of the oil by two centimetres?
Solution:
Radius of cylinderical jar (r) = 6 cm
Level of oil in it (h) = 2 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 28.1

Question 29.
A measuring jar of internal diameter 10 cm is partially filled with water. Four equal spherical balls of diameter 2 cm eacfy are dropped in it and they sink down in water completely. What will be the change in the level of water in the jar?
Solution:
Diameter of measuring jar = 10 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 29.1
Now after swing the ball in the water of jar Let volume of water raised, by h cm

Question 30.
A cone, a hemisphere and a cylinder stand on equal bases and have the same height. Show that their volumes are in the ratio 1 : 2:3.
Solution:
∵ Bases and heights of a cones hemisphere and a cylinder are equal
Let r be the radius and h be their heights
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 30.1

Question 31.
A cylinderical tub of radius 12 cm contains water to a depth of 20 cm. A spherical form 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?
Solution:
Radius of the cylinderical tub (r) = 12 cm
Depth of water in it (h) = 20 cm
By dropping a spherical ball in it, the water raised by 6.75 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 31.1

Question 32.
A sphere, a cylinder and a cone have the same diameter. The height of the cylinder and also the cone are equal to the diameter of the sphere. Find the ratio of their volumes.
Solution:
Diameter of a sphere, cylinder and a cone are equal
Let each as diameter = 2r
Then radius of each = r
Height of cylinder = diameter = 2r
and height of cone = 2r
Now volume of sphere = \(\frac { 4 }{ 3 }\)πr³
Volume of cylinder = πr²h
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 32.1

Hope given RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C

January 30, 2023

RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20C.

Other Exercises

Tick the correct answer in each of the following:

Question 1.
Solution:
Answer = (b)
Length (l) = 12 cm
Breadth (b) = 9cm
height (h) = 8 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 1.1

Question 2.
Solution:
Total surface area of cube = 150 cm²
Side = \( \sqrt { \frac { 150 }{ 6 } } \)
= √25
= 5 cm
Volume = (side)³
= (5)³
= 125 cm³ (b)

Question 3.
Solution:
Volume of cube = 343 cm²
Side = \( \sqrt [ 3 ]{ 343 } =\sqrt [ 3 ]{ 7\times 7\times 7 } \)
= 7 cm
Total surface area = 6 (side)²
= 6 x (7)²
= 6 x 49 cm²
= 294 cm² (c)

Question 4.
Solution:
Rate of painting = 10 paise per cm²
Total cost = Rs. 264.60
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 4.1

Question 5.
Solution:
Answer = (c)
Length of wall (l) = 8m = 800 cm
Breadth (b) = 22.5 cm
Height (h) = 6 m
= 600 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 5.1

Question 6.
Solution:
Answer = (c)
Edge of cube = 10 cm
Volume = a³ = (10)³ = 1000 cm³
Edge of box = 1 m = 100 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 6.1

Question 7.
Solution:
Answer = (a)
Ratio in sides of a cuboid = 1 : 2 : 3
Surface area = 88 cm²
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 7.1

Question 8.
Solution:
Ratio in the two volumes = 1 : 27
Let volume of first volume = x³
and volume of second volume = 27x³
Side of first cube = x
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 8.1

Question 9.
Solution:
Surface area of a brick of measure 10 cm x 4 cm x 3 cm
= 2 (l x b + b x h + h x l)
= 2 [10 x 4 + 4 x 3 + 3 x 10] cm²
= 2 [40 + 12 + 30]
= 82 x 2
= 164 cm² (c)

Question 10.
Solution:
Length of beam (l) = 9 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 10.1

Question 11.
Solution:
Water in rectangular reservoir = 42000
Volume = \(\\ \frac { 42000 }{ 1000 } \) = 42 m³
Length (l) = 6 m
Breadth (b) = 3.5 m
Depth = \(\\ \frac { volume }{ l\times b } \)
= \(\\ \frac { 42 }{ 6\times 3.5 } \)
= 2 m (c)

Question 12.
Solution:
Dimensions of a room are 10 m, 8 m, 3.3 m
Volume of air in it = lbh
= 10 x 8 x 3.3 = 264 m³
Air required for one man = 3 m³
No. of men = \(\\ \frac { 264 }{ 3 } \)
= 88 (b)

Question 13.
Solution:
Length of water tank (l) = 3 m
Width (b) = 2 m
and height (h) = 5 m
Volume = lbh = 3 x 2 x 5 = 30 m³
Water in it = 30 x 1000
= 30000 (a)

Question 14.
Solution:
Size of box = 25 cm, 15 cm, 8 cm
Surface area = (lb + bh + hl)
= 2 ( 25 x 15 + 15 x 8 + 8 x 25) cm²
= 2 (375 + 120 + 200) cm²
= 2(695)
= 1390 cm²(b)

Question 15.
Solution:
Diagonal of cube = 4√3
Side = \( \frac { 4\sqrt { 3 } }{ \sqrt { 3 } } \)
= 4 cm
Volume = a³ = (4)³
= 64 cm³ (d)

Question 16.
Solution:
Diagonal of cube = 9√3 cm
Side = \( \frac { 9\sqrt { 3 } }{ \sqrt { 3 } } \)
= 9 cm
Surface area = 6a²
= 6 (9)² = 6 x 81 cm²
= 486 cm² (b)

Question 17.
Solution:
Let side of cube in first case = a
Then volume = a³
If side of cube is doubled, then side = 2a
Volume (2a)³ = 8a³
Becomes 8 times (d)

Question 18.
Solution:
Let side of cube in first case = a
Then surface area = 6a²
and side of second cube = 2a
Surface area = 6 (2a)² = 6 x 4a² = 24a²
Ratio = \(\frac { { 24a }^{ 2 } }{ { 6a }^{ 2 } } \) = 4
Becomes 4 times (b)

Question 19.
Solution:
Sides (edges) of 3 cubes are 6 cm, 8 cm, and 10 cm respectively
Volume of first cube = (6)³ = 216 cm³
Volume of second cube = (8)³ = 512 cm³
and volume of third cube
= (10)³ = 1000 cm³
Sum of volumes of 3 cubes = 216 + 512 + 1000
= 1728 cm³
Volume of new single cube = 1728 cm³
Edge = \(\sqrt [ 3 ]{ 1728 } \)
\(\sqrt [ 3 ]{ { \left( 12 \right) }^{ 3 } } \)
= 12 cm (a)

Question 20.
Solution:
Each edge of 5 cubes = 5 cm
Placing than adjacent to each other
Length of new cuboid (l)
= 5 x 5 = 25 cm
Breadth (b) = 5 cm
and height (h) = 5 cm
Volume of new cuboid = lbh
= 25 x 5 x 5 cm³
= 625 cm³ (d)

Question 21.
Solution:
Diameter of circular well = 2n
Radius = \(\\ \frac { 2 }{ 2 } \) = 1 m
Depth(h) = 14 m
Volume of earth dug out = πr²h
= \(\\ \frac { 22 }{ 7 } \) x 1 x 1 x 14
= 44 m (d)

Question 22.
Solution:
Capacity of cylindrical tank = 1848 m³
Diameter = 14 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 22.1

Question 23.
Solution:
Radius of a cylinder (r) = 20 cm
and height (h) = 60 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 23.1

Question 24.
Solution:
Radius of each coin (r) = 0.75 cm
and thickness (h) = 0.2 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 24.1

Question 25.
Solution:
Volume of silver = 66 cm³
Diameter of wire = 1 mm = \(\\ \frac { 1 }{ 10 } \)
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 25.1

Question 26.
Solution:
Diameter of cylinder = 10 cm
Radius (r) = \(\\ \frac { 10 }{ 2 } \) = 5 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 26.1

Question 27.
Solution:
Diameter of cylinder = 7 cm
Radius (r) = \(\\ \frac { 7 }{ 2 } \) cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 27.1

Question 28.
Solution:
Curved surface area of a cylinder = 264 cm³
Height (h) = 14 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 28.1

Question 29.
Solution:
Diameter of cylinder = 14 cm
Radius (r) = 7 cm
Curved surface area = 220 cm²
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 29.1

Question 30.
Solution:
Ratio in radii of two cylinder = 2 : 3
and ratio in their height = 5 : 3
Let radii of two cylinder = 2x and 3x
and corresponding heights = 5y, 3y
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 30.1

Hope given RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1

January 30, 2023

RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1

Other Exercises

Question 1.
Find the surface area of a sphere of radius.
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm
Solution:
In a sphere,
(i) Radius (r) = 10.5 cm
Surface area = 4πr²
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 1.1

Question 2.
Find the surface area of a sphere of diameter
(i) 14 cm
(ii) 21 cm
(iii) 3.5 cm
Solution:
(i) Diameter of a sphere = 14 cm
Radius (r) = \(\frac { 14 }{ 2 }\) = 7 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 2.1

Question 3.
Find the total surface area of a hemisphere and a solid hemisphere each of radius 10 cm. [Use π = 3.14]
Solution:
(i) Radius of hemisphere = 10 cm
∴ Total surface area of hemisphere = 2πr²
= 2 x 3.14 x 10 x 10 cm²
= 628 cm²
(ii) Total surface area of solid hemisphere
= 3πr² = 3 x 3.14 x 10 x 10 cm²
= 942 cm²

Question 4.
The surface area of a sphere in 5544 cm², find the diameter.
Solution:
Let r be the radius of a sphere, then Surface area = 4πr²
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 4.1

Question 5.
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹4 per 100 cm². [NCERT]
Solution:
Inner diameter of a hemispherical bowl = 10.5 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 5.1

Question 6.
The dome of a building is in the form of a hemisphere. Its radius is 63 dm. Find the cost of painting it at the rate of ₹2 per sq. m.
Solution:
Radius of dome (hemispherical) = 63 dm
Area of curved surface
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 6.1

Question 7.
Assuming the earth to be a sphere of radius 6370 km, how many square kilometres is area of the land, if three-fourth of the earth’s surface is covered by water?
Solution:
Radius of earth (sphere) = 6370 km
Water on the earth = \(\frac { 3 }{ 4 }\) % total area
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 7.1

Question 8.
A cylinder of same height and radius is placed on the top of a hemisphere. Find the curved surface area of the shape if the length of the shape be 7 cm.
Solution:
Total height of the so formed shape = 7 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 8.1

Question 9.
The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Solution:
Diameter of moon = \(\frac { 1 }{ 4 }\) of diameter of earth
Let radius of earth = r km
Then radius of moon = \(\frac { 1 }{ 4 }\) r km
Now surface area of earth = 4πr²
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 9.1

Question 10.
A hemi-spherical dome of a building needs to be painted. If the circumference of the base of the dome is 17.6 m, find the cost of painting it, given the cost of painting is ₹5 per 100 cm². [NCERT]
Solution:
Circumference of the base of dome (r) = 17.6 m
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 10.1

Question 11.
A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 16 cm and its height is 15 cm. Find the cost of painting the toy at ₹7 per 100 cm².
Solution:
Diameter of toy = 16 cm
Radius (r) = \(\frac { 16 }{ 2 }\) = 8 cm
Height of conical part (h) = 15 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 11.1

Question 12.
A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m, find the cost of painting it on the outside at the rate of ₹10 per m².
Solution:
Diameter of the tank = 1.4 m
∴ Radius (r) = \(\frac { 1.4 }{ 2 }\) m = 0.7 m
and height of cylindrical portion = 8m
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 12.1

Question 13.
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in the figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm² and black paint costs 5 paise per cm². [NCERT]
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 13.1
Solution:
Diameter of each spheres = 21 cm
∴ Radius (R) = \(\frac { 21 }{ 2 }\) cm
Radius of each cylinder (r) = 1.5 cm
and height (h) = 7 cm

Now surface area of one sphere = 4πR²

Hope given RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B

January 30, 2023

RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20B.

Other Exercises

Question 1.
Solution:
(i) Radius of the base of the cylinder (r) = 7 cm.
Height (h) = 50 cm.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 1.1

Question 2.
Solution:
Radius of cylindrical tank (r) = 1.5 m
and height (h) = 10.5 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 2.1
= 74.25 x 1000l
= 74250 l

Question 3.
Solution:
Radius of the base of pole (r)
= 10 dm
= \(\\ \frac { 10 }{ 100 } \) m
= \(\\ \frac { 1 }{ 10 } \) m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 3.1

Question 4.
Solution:
Volume of cylinder = 1.54 m³
= 1540000 cm³
Diameter of its base = 140 cm
Radius (r) = 70 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 4.1

Question 5.
Solution:
Volume of cylindrical rod = 3850 cm³
Length of rod (h) = 1 m = 100 cm
Let radius of the base of the rod = r
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 5.1

Question 6.
Solution:
Diameter of closed cylinder = 14 m
Radius = \(\\ \frac { 14 }{ 2 } \)
= 7 m
Height = 5
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 6.1

Question 7.
Solution:
Circumference of the base of cylinder = 88 cm.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 7.1

Question 8.
Solution:
Lateral surface of cylinder = 220 m²
Height (h) = 14 m
Let radius of cylinder = r
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 8.1

Question 9.
Solution:
Volume of cylinder = 1232 cm³
height (h) = 8cm
Let r be the radius, then
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 9.1

Question 10.
Solution:
Ratio in radius and height of a cylinder = 7 : 2
Let radius = 7x
then height = 2x
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 10.1

Question 11.
Solution:
Curved surface area = 4400 cm²
circumference of base = 110 cm
Let r be the radius
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 11.1

Question 12.
Solution:
In first case,
Side of square base (a) = 5 cm.
and height (h) = 14 cm.
Volume = 5 x 5 x 14 = 350 cm³
In second case,
Radius of the circular base (r) = 3.5 cm.
Height (h) = 12 cm.
Volume = πr²h
= \(\\ \frac { 22 }{ 7 } \) x 3.5 x 3.5 x 12 cm³
= 462 cm²
Hence second type of circular plastic can has greater capacity.
Difference = 462 – 350
= 112 cm³

Question 13.
Solution:
Diameter of a cylindrical pillar = 48 cm.
Radius (r) = \(\\ \frac { 48 }{ 2 } \) = 24 cm.
\(\\ \frac { 24 }{ 100 } \) m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 13.1

Question 14.
Solution:
Length of rectangular vessel (l) = 22 cm.
Breadth (A) = 16 cm.
and height (A) = 14 cm.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 14.1

Question 15.
Solution:
Diameter of cylindrical metal = 1 cm.
Radius (r) = \(\\ \frac { 1 }{ 2 } \) cm.
Length. (A) = 11 cm.
Volume = πr²h
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 15.1

Question 16.
Solution:
Side of a solid cube = 2.2 cm
Volume = (side)³
= (2.2)³
= 10.648 cm³
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 16.1

Question 17.
Solution:
Diameter of a well = 7 m
Radius (r) = \(\\ \frac { 7 }{ 2 } \) m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 17.1

Question 18.
Solution:
Inner diameter of well = 14 m
Inner radius = \(\\ \frac { 14 }{ 2 } \) = 7 m
Depth (h) = 12 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 18.1

Question 19.
Solution:
No. of revolutions = 750
Diameter of road roller = 84 cm
Length (h) = 1 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 19.1

Question 20.
Solution:
Thickness of the metal = 1.5 cm.
External diameter = 12 cm.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 20.1

Question 21.
Solution:
Inner diameter of tube = 12 cm.
Inner radius (r) = \(\\ \frac { 12 }{ 2 } \) = 6 cm.
Thickness of metal = 1 m.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 21.1

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Value Based Questions in Science for Class 9 Chapter 16 Floatation

January 30, 2023

Value Based Questions in Science for Class 9 Chapter 16 Floatation

These Solutions are part of Value Based Questions in Science for Class 9. Here we have given Value Based Questions in Science for Class 9 Chapter 16 Floatation

VALUE BASED QUESTIONS

Question 1.
Anita and Suneeta were good friends. They had gone to a coastal side for walking. Anita was wearing flat surfaces chapal and Suneeta was wearing a high heel chapal. On the sandy surface, Suneeta was feeling uncomfortable while walking. Anita helped her to walk for some time on the sandy surface. Thereafter, both of them returned back. Anita asked Suneeta not to wear high heel chapal while walking on a sandy surface.
Answer the following questions based on the above paragraph.

Why was Suneeta feeling uncomfortable while walking on the sandy surface ?
Comment on the attitude of Anita.

Answer:

The pressure exerted by Suneeta on the sandy surface was large as pressure = weight of Suneeta/area of heel of the chapal. Due to large pressure, her feet were sinking in sand. Therefore, she was feeling uncomfortable while walking.
Anita is concerned about Suneeta. She was a good friend. Anita had high degree of general awareness.

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Question 2.
Amit is a student of class IX. His neighbour used to complain that he was not getting pure milk. But the man, who was supplying the milk said that the milk was pure. Amit knew how to test the purity of milk. He brought the lactometer and tested the milk. He told the man supplying milk not to cheat his neighbour.
Answer the following questions based on the above paragraph.

On what principle, lactometer works ?
What values are shown by Amit ?

Answer:

Archimedes principle.
Amit has high degree of general awareness. He is concerned about his neighbour. On the basis of his knowledge, he proved that the man supplying milk was wrong.

Question 3.
Some students were swimming in a swimming pool. Aayan, a student of class V was sitting sad near the swimming pool. Suneel came to Aayan and asked the reason of his sadness. Aayan told Suneel that he did not know how to swim. However, he also wanted to swim. Suneel asked Aayan to wear the life saving jacket. Then Suneel helped Aayan to swim.
Answer the following questions based on the above paragraph.

What is the basic principle, on which life saving jacket works ?
Comment on the attitude of Suneel.

Answer:

When a person wears a life saving jacket and enters into water, the weight of water displaced by jacket is more than the weight of the person. Hence, the person can float in water easily.
Suneel feels concerned for others. He could not see the sadness of Aayan. He is helpful. He used his knowledge to fulfill the desire of Aayan.

Question 4.
A milkman used to sell milk in the city and always carried lactometer with him. The customers trusted him and his business flourished.

What is lactometer ?
What is the principle of lactometer ?
Carrying a lactometer by a milkman shows high values. List any two. (CBSE 2015)

Answer:

Lactometer is a device used to test the purity of milk.
Lactometer is based on Archimedes principle.
1. Milkman is honest,
2. He is a good business man.
3. He is concerned with the health of his customers.

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NCERT Solutions for Class 9 Science Chapter 16 Floatation

January 30, 2023

NCERT Solutions for Class 9 Science Chapter 16 Floatation

These Solutions are part of NCERT Solutions for Class 9 Science. Here we have given NCERT Solutions for Class 9 Science Chapter 16 Floatation

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NCERT TEXT BOOK QUESTIONS

IN TEXT QUESTIONS

Question 1.
Why is it difficult to hold a school bag having a strap made of a thin and strong string ?
Answer:
It is painful to hold a heavy bag having strap made of a strong and thin string.
NCERT Solutions for Class 9 Science Chapter 16 Floatation image - 1
When we hold a heavy bag having strap made of a strong and thin string, then the area under the strap is small. Hence, large pressure is exerted by the strap on our fingers or shoulder. Due to this large pressure, the strap tends to cut the skin and hence pain is caused.

Question 2.
What do you mean by buoyancy ? (CBSE 2011, 2012, 2016, 2017)
Answer:
The tendency of an object to float in a liquid or the power of liquid to make an object to float in it is called buoyancy.

Question 3.
Why does an object float or sink when placed on the surface of water ? (CBSE 2012)
Answer:
When an object is placed on the surface of water, two forces act on the object.

The gravitational force in the downward direction and
Upthrust of the water on the object.

When the gravitational force acting on the object is greater than the upthrust of the water, then the object sinks in water. On the other hand, if the upthrust of water on the object is greater than the gravitational force, then the object floats.

If density of an object is less than the density of water, it floats on the water. On the other hand, if density of the object is greater than the density of water, it sinks in water.

Question 4.
You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier and why ?
(CBSE 2011)
Answer:
Iron bar is heavier than a bag of cotton. This is because iron bar experiences a less upward force i.e. upthrust than the bag of cotton.

NCERT CHAPTER END EXERCISE

Question 1.
In what direction does the buoyant force on an object immersed in a liquid act ?
Answer:
Buoyant force acts in the upward direction on an object immersed in a liquid.

Question 2.
Why does a block of plastic released under water come upto the surface of water ?(CBSE 2011, 2016)
Answer:
This is because upthrust or buoyant force acting on the block of plastic is greater than the gravitational force acting on the block in the downward direction.

Question 3.
The volume of 50 g of a substance is 20 cm³. If the density of water is 1 g cm^-3, will the substance float or sink ? Justify your answer. (CBSE 2011, 2012)
Answer:
NCERT Solutions for Class 9 Science Chapter 16 Floatation image - 2
Since density of substance is more than the density of water, so the substance will sink.

Question 4.
The volume of a 500 g sealed packet is 350 cm³. Will the packet float or sink in water if density of water is 1 g cm^-3 ? What wall be the mass of the water displaced by the packet ?
Answer:
NCERT Solutions for Class 9 Science Chapter 16 Floatation image - 3
Since the density of the packet is greater than the density of water, so the packet will sink.
Mass of water displaced= Volume of Packet x density of water = 350 x 1 = 350 g.

PRACTICAL SKILL BASED QUESTIONS

Question 1.
Write two main precautions to be taken to read the water level in the graduated cylinder. (CBSE 2015)
Answer:

The cylinder must be placed on a flat surface in exactly vertical positoin.
The level of eyes must be exactly at the level of lower end of the meniscus of water.

Question 2.
Write precuations to be taken to measure the weight of a body using a spring balance. (CBSE 2015)
Answer:

The least count of the spring balance must be calculated.
Zero error of the spring balance must be calculated and added or substracted from the observed value of the . weight of the body.
The level of eyes must be exactly at the level of the pointer of the spring balance.

Question 3.
A 500 g mass body is immersed in two liquids X and Y in succession. The extent to which the body sinks in liquid Y is less than that in liquid X. From such observation, compare the densities of liquids X and Y. Justify your answer. (CBSE 2015)
Answer:
Weight of body immersed in liquid = Weight of body in air -upthrust of liquid on the body.
Since the extent to which the body sinks in liquid Y is less than that in liquid X, therefore, upthrust of liquid Y is greater than the upthrust of liquid X.
NCERT Solutions for Class 9 Science Chapter 16 Floatation image - 4
Thus, density of liquid Y is greater than the density of liquid X.

Question 4.
In a spring balance, the space between 0 to 25 g wt. is divided into 5 equal parts. Find its least count. What is the weight of the object in the diagram given below ? (CBSE 2015)
NCERT Solutions for Class 9 Science Chapter 16 Floatation image - 5
Answer:
5 divisions = 25 gwt
∴1 divisions = 5 gwt
Hence, least count of the spring balance = 5 gwt.
Weight of object in the diagram = 125 x 5 gwt = 625 g wt

Question 5.
While determining the density of a solid of weight 58 g wt, a student is provided with four different combinations of mesuring cylinder and spring balance as shown below :
The student should prefer to choose which combination and why ? (CBSE 2015)
Answer:

Density can be determined accurately if mass and volume of solid are measured accurately. Therefore, mass must be determined by a spring balance of minimum least count and volume must be determined by a measuring cylinder of minimum least count. Hence, student should prefer the combination 1.

Question 6.
Observe the two spring balances shown in the figures given aside :

What are the least count and range of the spring balances ?
Find the weights of objects in the two spring balances. (CBSE 2015)

Answer:

In both spring balances, 5 divisions = 10 kg
1 division = 2 kg
Hence, least count of both spring balances = 2kg.
Range of both spring balances = 0-50 kg
Weight of object measured by spring balance A = 10 kg
Weight of object measured by spring balance B = 40 kg
A spring balance used for measuring mass of the cuboid has a least count of 2 gwt.

Question 7.
Two students using the same spring balance noted two different readings, 46 gwt and 47 gwt. Which reading is correct and why ? (CBSE 2015)
Answer:
The reading which is the multiple of least count (i.e. 2 gwt) is correct. Therefore, 46 gwt is the correct reading.

Question 8.
A 10 mL measuring cylinder has 100 divisions of equal spacing. Find the least count of the measuring cylinder.
Answer:
Least count of measuring cylinder = Value of 1 division marked on the cylinder 100 divisions =10 mL .

Therefore, least count of the measuring cylinder = 0.1 mL

Question 9.
In which of the two, glycerine or kerosene, the loss in weight of a solid when fully immersed in them will be more and why ? (CBSE 2017)
Answer:
The loss in weight of a solid in a liquid will be more if buoyant force acting on the solid in a liquid is more. The buoyant force is directly proportional to the density of the liquid. Since density of glycerine is more than the density of kerosene, therefore, the loss in weight of the solid in glycerine is more than in kerosene.

Question 10.
When a body is immersed in a liquid, name the two forces acting on it and state their direction of action.
Answer:

Weight of the body in vertically downward direction.
Buoyant force acting on the body due to the liquid in vertically upward direction.

Hope given NCERT Solutions for Class 9 Science Chapter 16 are helpful to complete your science homework.

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RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS

January 30, 2023

RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS

Other Exercises

Mark the correct alternative in each of the following:
Question 1.
The number of surfaces of a cone has, is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
Number of surfaces of a cone are 2 (b)

Question 2.
The area of the curved surface of a cone of radius 2r and slant height \(\frac { 1 }{ 2 }\), is
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q2.1
Solution:
Radius of a cone = 2r

Question 3.
The total surface area of a cone of radius \(\frac { r }{ 2 }\) and length 2l, is
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q3.1
Solution:

Question 4.
A solid cylinder is melted and cast into a cone of same radius. The heights of the cone and cylinder are in the ratio
(a) 9 : 1
(b) 1 : 9
(c) 3 : 1
(d) 1 : 3
Solution:
Let r be the radius and h be the height of cylinder, then volume = πr²h
Now volume of cone = πr²h
r is the radius
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q4.1

Question 5.
If the radius of the base of a right circular cone is 3r and its height is equal to the radius of the base, then its volume is
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q5.1
Solution:
Radius of the base of a cone (R) = 3r
and height (H) = 3r

Question 6.
If the volumes of two cones are in the ratio 1 : 4 and their diameters are in the ratio 4 : 5, then the ratio of their heights, is
(a) 1 : 5
(b) 5 : 4
(c) 5 : 16
(d) 25 : 64
Solution:
Ratio in the volumes of two cones =1:4
and ratio in their diameter = 4:5
Let h₁, h₂ be their heights
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q6.1

Question 7.
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
Solution:
Let r be the radius and l be the slant height
∴ Curved surface area of first cone = πr₁l₁
and let curved surface area of second cone = πr₂l₂
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q7.1

Question 8.
If the height and radius of a cone of volume V are doubled, then the volume of the cone, is
(a) 3V
(b) 4V
(c) 6V
(d) 8V
Solution:
Let r and h be the radius and height of a cone, then
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q8.1

Question 9.
The ratio of the volume of a right circular cylinder and a right circular cone of the same base and height, is
(a) 1 : 3
(b) 3 : 1
(c) 4 : 3
(d) 3 : 4
Solution:
Let r be the radius and h be the height of a right circular cylinder and a right circular cone, and V₁ and V₂ are their volumes, the V₁ = πr²h and
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q9.1

Question 10.
A right cylinder and a right circular cone have the same radius and same volumes. 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
Solution:
Let r be the radius of cylinder and cone and volumes are equal
and h₁, and h₂ be their have h₂ is respectively
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q10.1

Question 11.
The diameters of two cones are equal. If their slant heights are in the ratio 5 : 4, the ratio of their curved surface areas, is
(a) 4 : 5
(b) 25 : 16
(c) 16 : 25
(d) 5 : 4
Solution:
∵ Diameters of two cones are equal
∴ Their radii are also be equal
Let r be their radius of each cone,
and ratio in their slant heights = 5:4
Let slant height of first cone (h₁) = 5x
Then height of second cone (h₂) = 4x
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q11.1

Question 12.
If the heights of two cones are in the ratio of 1 : 4 and the radii of their bases are in the ratio 4 : 1, then the ratio of their volumes is
(a) 1 : 2
(b) 2 : 3
(c) 3 : 4
(d) 4 : 1
Solution:
Ratio in the heights of two cones =1 : 4
and ratio in their radii of their bases = 4 : 1
Let height of the first cone = x
and height of the second cone = 4x
Radius of the first cone = 4y
and radius of the second cone = y
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q12.1

Question 13.
The slant height of a cone is increased by 10%. If the radius remains the same, the curved surface area is increased by
(a) 10%
(b) 12.1%
(c) 20%
(d) 21%
Solution:
Let r be radius and l be the slant height of a cone, then curved surface area = πrl
If slant height is increased by 10%, then
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q13.1

Question 14.
The height of a solid cone is 12 cm and the area of the circular base is 6471 cm². A plane parallel to the base of the cone cuts through the cone 9 cm above the vertex of the cone, the area of the base of the new cone so formed is
(a) 9π cm²
(b) 16π cm²
(c) 25π cm²
(d) 36π cm²
Solution:
Height of a solid cone (h) = 12 cm
Area of circular base = 64π cm²
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q14.1

Question 15.
If the base radius and the height of a right circular cone are increased by 20%, then the percentage increase in volume is approximately
(a) 60
(b) 68
(c) 73
(d) 78
Solution:
In first case,
Let r be radius and h be height, in volume
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q15.1

Question 16.
If h, S and V denote respectively the height, curved surface area and volume of a right circular cone, then 3πVh³ – S2h² + 9V² is equal to
(a) 8
(b) 0
(c) 4π
(d) 32π²
Solution:
h = height, S = curved surface area
V = volume of a cone
Let r be the radius of the cone, then
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q16.1

Question 17.
If a cone is cut into two parts by a horizontal plane passing through the mid¬point of its axis, the ratio of the volumes of upper and lower part is
(a) 1 : 2
(b) 2 : 1
(c) 1 : 7
(d) 1 : 8
Solution:
∴ ∆PDC ~ ∆PBA (AA axiom)
and O’ is mid point of PO
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q17.1

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NCERT Exemplar Solutions for Class 9 Science Chapter 12 Sound

January 30, 2023

NCERT Exemplar Solutions for Class 9 Science Chapter 12 Sound

These Solutions are part of NCERT Exemplar Solutions for Class 9 Science . Here we have given NCERT Exemplar Solutions for Class 9 Science Chapter 12 Sound

Question 1.
Note is a sound
(a) of mixture of several frequencies
(b) of mixture of two frequencies only
(c) of a single frequency
(d) always unpleasant to listen
Answer:
(a).

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Question 2.
A key of a mechanical piano struck gently and then struck again but much harder this time. In the second case
(a) sound will be louder but pitch will not be different
(b) sound will be louder and pitch
(c) sound will be louder but pitch will be lower
(d) both loudness and pitch will remain unaffected
Answer:
(a). Loudness depends on the amplitude of vibration.

Question 3.
In SONAR, we use
(a) ultrasonic waves
(b) infrasonic waves
(c) radio waves
(d) audible sound waves
Answer:
(a).

Question 4.
Sound travels in air if
(a) particles of medium travel from one place to another
(b) there is no moisture in the atmosphere
(c) disturbance moves
(d) both particles as well as disturbance travel from one place to another.
Answer:
(c).

Question 5.
When we change feeble sound to loud sound we increase its
(a) frequency
(b) amplitude
(c) velocity
(d) wavelength
Answer:
(b). Loudness depends on the amplitude of vibration.

Question 6.
In the curve (Fig. 1) half the wavelength is
NCERT Exemplar Solutions for Class 9 Science Chapter 12 Sound image - 1
(a) A B
(b) B D
(c) D E
(d) A E
Answer:
(b).

Question 7.
Earthquake produces which kind of sound before the main shock wave begins
(a) ultrasound
(b) infrasound
(c) audible sound
(d) none of the above
Answer:
(b).

Question 8.
Infrasound can be heard by
(a) dog
(b) bat
(c) rhinoceros
(d) human beings.
Answer:
(c).

Question 9.
Before playing the orchestra in a musical concert, a sitarist tries to adjust the tension and pluck the string suitably. By doing so, he is adjusting
(a) intensity of sound only
(b) amplitude of sound only
(c) frequency of the sitar string with the frequency of other musical instruments
(d) loudness of sound.
Answer:
(c).

SHORT ANSWER QUESTIONS

Question 10.
The given graph (Fig. 2) shows the displacement versus time relation for a disturbance travelling with velocity of 1500 the wavelength of the disturbance.
NCERT Exemplar Solutions for Class 9 Science Chapter 12 Sound image - 2
Answer:

Question 11.
Which of the above two graphs (a) and (b) (Fig. 3) representing the human voice is likely to be the male voice ? Give reason for your answer.
NCERT Exemplar Solutions for Class 9 Science Chapter 12 Sound image - 4
Answer:
Frequency of male voice is lesser than that of female voice. So, graph (a) represents the human voice.

Question 12.
A girl is sitting in the middle of a park of dimension 12 m x 12 m. On the left side of it, there is a building adjoining the park and on the right side of the park, there is a road adjoining by park. A sound is produced on the road by a cracker. Is it possible for the girl to hear the echo of this sound ? Explain your answer.
Answer:
To hear echo, the distance between the source of sound and the obstacle (i.e., building) should be more than 17 m. In this case, distance between source of sound (i.e., cracker) and the obstacle (i.e., building) is 12 m i.e., less than 17 m. Hence, girl cannot hear the echo of the sound of the cracker.

Question 13.
Why do we hear the sound produced by the humming bees while the sound of vibrations of pendulum is not heard ?
Answer:
This is because the frequency of sound produced by humming bees lies in audible range (20 Hz to 20,000 Hz) and frequency of the sound of vibrations of pendulum lies in infrasonic region. ( 1 Hz to 19 Hz)

Question 14.
If any explosion takes place at the bottom of a lake, what type of shock waves in water will take place ?
Answer:
Longitudinal wave as transverse wave cannot travel in water.

Question 15.
Sound produced by a thunderstorm is heard 10 s after the lightning is seen. Calculate the approximate distance of the thunder cloud. (Given speed of sound = 340 m s^-1).
Answer:
Distance of thunder cloud = speed of sound x time = 340 x 10 = 3400 m = 3.4 km.

Question 16.
For hearing the loudest ticking sound heard by the ear, find the angle x in the Fig. 4.
NCERT Exemplar Solutions for Class 9 Science Chapter 12 Sound image - 5
Answer:

Question 17.
Why is the ceiling and wall behind the stage of good conference halls or concert halls made curved ?
Answer:
So that sound after reflection reach to every audience.

Question 18.
Represent graphically by two separate diagrams in each case :
(i) Two sound waves having the same amplitude but different frequencies ?
(ii) Two sound waves having the same frequency but different amplitudes.
(iii) Two sound waves having the different amplitude but different wavelengths.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 12 Sound image - 6

Question 19.
Establish the relationship between speed of sound, its wavelength and frequency. If velocity of sound in air is 340 m s^-1, calculate
(i) wavelength when frequency is 256 Hz.
(ii) frequency when wavelength is 0.85 m. (CBSE 2012)
Answer:
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 12 Sound image - 7

Question 20.
Draw a curve showing density or pressure variations with respect to distance for a disturbance produced by sound. Mark the position of compression and rarefaction on this curve. Also define wavelengths and time period using this curve. (CBSE 2012)
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 12 Sound image - 8
Wavelength : Distance between two successive compressions or two successive rarefactions.
Time Period : Time taken by the disturabnce to travel between two successive compressions or between two successive rarefaction.

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RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A

January 30, 2023

RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20A.

Other Exercises

Question 1.
Solution:
(i)
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 1.1
Length of cuboid (l) = 22 cm.
Breadth (b) = 12 cm
and height (h) = 7.5 cm.

Question 2.
Solution:
Length of water tank (l) = 2 m
75cm = 2.75 m
breadth (b) = 1 m 80cm = 1.80 m
and height (h) = 1 m 40 cm = 1.40 m
Volume of water filled in it = l.b.h = 2.75 x 1.80 x 1.40 m³
= 6.93 m³
Water in litres = 6.93 x 1000
= 6930 litres (1 m³ = 1000 litres) Ans.

Question 3.
Solution:
Length of iron (l) = 1.05 m
= 105 cm
breadth (b) = 70 cm and height (h) = 1.5 cm
volume of iron = l x b x h = 105 x 70 x 1.5 cm³
= 11025 cm³
weight of 1cm³ iron = 8 gram
Total weight = 11025 x g = 88200 g
= \(\\ \frac { 88200 }{ 1000 } \) kg
= 88.2 kg Ans.

Question 4.
Solution:
Area of courtyard = 3750 m²
Height of gravel = 1 cm.
Volume of gravel = 3750 x \(\\ \frac { 1 }{ 100 } \) m³
= 37.50 m³
Cost of 1 m³ gravel = Rs. 6.40
Total cost = Rs. 6.40 x 37.50
= Rs. 240 Ans.

Question 5.
Solution:
Length of hall (l) = 16 m
Breadth (b) = 12.5 m
height (h) = 4.5 m
Volume of air in it = l x b x h
= 16 x 12.5 x 4.5 m3
= 900 m³
Air for one person is required = 3.6 m³
Number of person which can be accommodated in the hall = 900 ÷ 3.6
= \(\\ \frac { 900\times 10 }{ 36 } \)
= 250

Question 6.
Solution:
Length of cardboard box (l)
= 1.2 m = 120 cm.
breadth (b) = 72 cm.
Height (h) = 54 cm.
Volume of box = l x b x h
= 120 x 72 x 54 cm³
= 466560 cm³
Volume of one soap bar = 6 x 4.5 x 4 cm³
= 108 cm³
No. of bars to be kept in it = \(\\ \frac { 466560 }{ 108 } \)
= 4320 Ans.

Question 7.
Solution:
Volume of one match box = 4 x 2.5 x 1.5 cm³ = 15 cm³
Volume of 144 matchboxes = 15 x 144 cm³
or volume of one packet = 2160 cm³
Length of carton (l) = 1.5 m = 150 cm
Breadth (b) = 84 cm
and height (h) = 60 cm.
Volume of one carton = l x b x h
= 150 x 84 x 60 cm³
= 756000 cm³
No. of packets = 756000 ÷ 2160
= 350 Ans.

Question 8.
Solution:
Length of one plank = 2m
= 200 cm
Breadth (b) = 25 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 8.1

Question 9.
Solution:
Length of wall (l) = 8m = 800 cm
Height (h) = 5.4 m = 540 cm
Width (b) = 33 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 9.1

Question 10.
Solution:
Length of wall (l) = 15 m
Width (b) = 30 cm = 0.3 m
Height (h) = 4 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 10.1

Question 11.
Solution:
Length of rectangular cistern (l) = 11.2 m
Breadth (b) = 6 m
Height (h) = 5.8 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 11.1

Question 12.
Solution:
Volume of block of gold = 0.5 m³
= 0.5 x 1000000 cm³
= 500000 cm³
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 12.1

Question 13.
Solution:
Area of field = 2 hectare
= 20000 m²
Rainfall = 5 cm. = 0.05 m
Volume of water of rainfall
= Area of field x height of rainfall water
= 20000 x 0.05 m³
= 1000 m³ Ans.

Question 14.
Solution:
Speed of water = 3 km/h
Length of water flow in 1 minute
= \(\\ \frac { 3km }{ 60m } \)
= \(\\ \frac { 3000 }{ 60 } \)
= 50 m
Width of river = 45 m
Depth of river = 2 m
Volume of water in 1 minute
= 45 x 2 x 50 m³
= 4500 m³ Ans.

Question 15.
Solution:
Length of pit (l) = 5m
Width (b) = 3.5 m
Let depth of pit = h
then volume of earth dug out
= l.b.h = 5 x 3.5 x h = 17.5 h m³
But volume of earth = 14 m³
17.5 h = 14
h = \(\\ \frac { 14 }{ 17.5 } \) = \(\\ \frac { 140 }{ 175 } \)
=> h = \(\\ \frac { 4 }{ 5 } \) m
= \(\\ \frac { 4 }{ 5 } \) x 100
= 80 cm Ans.

Question 16.
Solution:
Width of tank = 90 cm = \(\\ \frac { 90 }{ 100 } \) m
Depth = 40 cm = \(\\ \frac { 40 }{ 100 } \) m
Water = 576 litre
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 16.1

Question 17.
Solution:
Volume of wood = 1.35 m³
Length of beam = 5m
Thickness = 36 cm = \(\\ \frac { 36 }{ 100 } \) m.
Width = \(\\ \frac { Volume }{ length\times thickness } \)
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 7.1

Question 18.
Solution:
Volume of a room = 378 m³
Area of its floor = 84 m²
Height = \(\\ \frac { Volume }{ Area } \)
= \(\\ \frac { 378 }{ 84 } \) m
= 4.5 m Ans.

Question 19.
Solution:
Length of pool = 260 m
and width = 140 m.
Volume of water = 54600 m³
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 19.1

Question 20.
Solution:
Outer length of wooden box (L) = 60 cm
Width (B) = 45 cm
and height (H) = 32 cm.
Thickness of wood used = 2.5 cm.
Inner length (l) = 60 – 2 x 2.5
= 60 – 5
= 55 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 20.1

Question 21.
Solution:
Outer length of open box = 36 cm
breadth = 25 cm
and height = 16.5 cm
thickness of iron = 1.5 cm.
∴ Inner length = 36 – 2 x 1.5
= 36 – 3
= 33 cm
breadth = 25 – 2 x 1.5
= 25 – 3
= 22 cm
and height = 16.5 – 1.5
= 15 cm .
∴ Volume of iron used in it = Outer volume – Inner volume
= 36 x 25 x 16.5 cm3 – 33 x 22 x 15 cm³
= 14850 – 10890
= 3960 cm³
weight of 1 cm³ = 8.5 gram
∴ Total weight = 3960 x 8.5 g
= 33660 g
= 33.660 kg
= 33.66 kg Ans.

Question 22.
Solution:
Outer length of the box = 56 cm
Width = 39 cm
and height = 30 cm
Volume = 56 x 39 x 30
= 65520 cm³
Thickness of wood used = 3cm.
∴ Inner length = 56 – 2 x 3
= 56 – 6
= 50 cm
Width = 39 – 2 x 3
= 39 – 6
= 33 cm
and height = 30 – 2 x 3
= 30 – 6
= 24 cm
∴ Inner volume of the box = 50 x 33 x 24 cm³
= 39600 cm³
and volume of wood used = Outer volume – Inner volume
= (65520 – 39600)cm³
= 25920 cm³ Ans.

Question 23.
Solution:
Outer length of box = 62 cm.
Outer width = 30 cm.
Outer height = 18 cm.
Thickness of wood = 2 cm.
∴ Internal length = 62 – 2 x 2
= 58 cm.
Internal width = 30 – 2 x 2
= 26 cm.
Internal height =18 – 2 x 2
= 14 cm.
Capacity of the box = lbh
= 58 x 26 x 14 cm³
= 21112 cm³ Ans.

Question 24.
Solution:
Outer length = 80 cm.
Outer width = 65 cm.
Outer height = 45 cm.
Total volume = 80 x 65 x 45 cm³
= 234000 cm³
Thickness of wood = 2.5 cm.
∴ Inner length = 80 – 2 x 2.5 = 75 cm.
Inner width = 65 – 2 x 2.5 = 60 cm.
Inner height = 45 – 2 x 2.5 = 40 cm.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 24.1

Question 25.
Solution:
(i) Edge of cube (a) = 7 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 25.1
(a) Volume = a³ = (7)³
= 7 x 7 x 7 m³
= 343 m³ Ans.

Question 26.
Solution:
Surface area of a cube = 1176 cm²
Let edge of the cube = a
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 26.1

Question 27.
Solution:
Volume of a cube = 729 cm³
Let edge of cube = a
then a³ = 729 = (9)³
a = 9 cm.
Hence surface area = 6a² = 6 (9)² cm²
= 6 x 81
= 486 cm² Ans.

Question 28.
Solution:
Length of metal block (l) = 2.25 m = 225 cm
Width (b) = 1.5 m = 150 cm
and height (h) = 27 cm
Volume of block = l x b x h
= 225 x 150 x 27 cm³
= 911250 cm³
Side of each cube = 45 cm.
Volume of each cube = a³
= 45 x 45 x 45
= 94125 cm³
Number of cubes = \(\\ \frac { 911250 }{ 91125 } \)
= 10 Ans.

Question 29.
Solution:
Let edge of given cube = a
Volume = a³
and surface area = 6a²
By doubling the edge of cube3 the side of new cube = a x 2 = 2a
Volume (2a)³ = 8a³
and surface area = 6 (2a)² = 6 x 4a²
= 24a² = 4 x 6a²
It is clear from the above that
Volume is increased 8 times and surface area is 4 times. Ans.

Question 30.
Solution:
Total cost of wood = Rs. 256
Rate = Rs.,500 per m³
Volume of wood = \(\\ \frac { 256 }{ 500 } \) = 0.512 m³
= 0.512 x 100 x 100 x 100 cm³
= 512000 cm³
Let length of each side = a
then a³ = 512000 = (80)³
a = 80
Hence length of each side = 80 cm. Ans.

Hope given RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS

January 30, 2023

RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS

Other Exercises

Question 1.
The height of a cone is 15 cm. If its volume is 500π cm³, then find the radius of its base.
Solution:
Volume of cone = 500π cm³
and height (h) = 15 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q1.1

Question 2.
If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.
Solution:
Volume of a cone = 48π cm³
Height (h) = 9 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q2.1

Question 3.
If the height and slant height of a cone are 21 cm and 28 cm respectively. Find its volume.
Solution:
Height of a cone (h) = 21 cm
and slant height (l) = 28 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q3.1

Question 4.
The height of a conical vessel is 3.5 cm. If its capacity is 3.3 litres of milk. Find the diameter of its base.
Solution:
Capacity of conical vessel = 3.3 litres
Volume = 3.3 m³
= 3.3 x 1000 = 3300 cm²
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q4.1

Question 5.
If the radius and slant height of a cone are in the ratio 7 : 13 and its curved surface area is 286 cm2, find its radius.
Solution:
Two ratio in radius and slant height of a cone = 7 : 13
Let radius (r) = 7x
and slant height (1) = 13x
Curved surface area = πrl
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q5.1

Question 6.
Find the area of canvas required for a conical tent of height 24 m and base radius 7 m.
Solution:
Base radius of the closed cone (r) = 7 cm
and vertical height (h) = 24 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q6.1

Question 7.
Find the area of metal sheet required in making a closed hollow cone of base radius 7 cm and height 24 cm. making a closed hollow cone of base radius 7 cm and height 24 cm.
Solution:
Base radius of the closed cone (r) = 7 cm
and vertical height (h) = 24 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q7.1

Question 8.
Find the length of cloth used in making a conical pandal of height 100 m and base radius 240 m, if the cloth is 100π m wide.
Solution:
Height of conical pandal (A) = 100 m
Base radius (r) = 240 m
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q8.1

Hope given RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 19 Three-Dimensional Figures Ex 19B

January 30, 2023

RS Aggarwal Class 8 Solutions Chapter 19 Three-Dimensional Figures Ex 19B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 19 Three-Dimensional Figures Ex 19B.

Other Exercises

Question 1.
Solution:
Euler’s Relation
= F – E + V = 2
Where F is no. of faces
E is no. of edges
V is no. of vertices

Question 2.
Solution:
Edges of
(i) Cuboid are 12
(ii) Tetrahedron are 6
(iii) Triangular prism are 9
(iv) Square pyramid are – 8

Question 3.
Solution:
Faces of
(i) Cube are 6
(ii) Pentagonal prism are 7 (5 + 2)
(iii) Tetrahedron are 4
(iv) Pentagonal pyramid are 6

Question 4.
Solution:
Vertices of
(i) Cuboid are 8
(ii) Tetrahedron are 4
(iii) Pentagonal prism are 10
(iv) Square pyramid are 5

Question 5.
Solution:
(i) A cube
F – E + V = 2
=> 6 – 12 + 8 = 2
=> 14 – 12 = 2
=> 2 = 2
(ii) A tetrahedron
F – E + V = 2
=> 4 – 6 + 4 = 2
=> 8 – 6 = 2
=> 2 = 2
(iii) A triangular prism
F – E + V = 2
=> 5 – 9 + 6 = 2
=> 11 – 9
=> 2 = 2
(iv) A square pyramid
F – E + V = 2
=> 5 – 8 + 5 = 2
=> 10 – 8 = 2
=> 2 = 2

Hope given RS Aggarwal Solutions Class 8 Chapter 19 Three-Dimensional Figures Ex 19B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.