Chapter 19 Volume and Surface Area of Solids RS Aggarwal Solutions Exercise 19A Class 10 Maths
Chapter Name  RS Aggarwal Chapter 19 Volume and Surface Area of Solids 
Book Name  RS Aggarwal Mathematics for Class 10 
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Related Study  NCERT Solutions for Class 10 Maths 
Volume and Surface Area of Solids Exercise 19A Solutions
1. Two cubes each of volume 27 cm^{3} are joined end to end to form a solid. Find the surface area of the resulting cuboid.
Solution
Let the length of the side of cube be ‘a’ cm
Volume of each cube = 27 cm^{3}
Volume of cube = a^{3}
∴ a^{3} = 27 cm^{3}
⇒ a = (27 cm^{3})^{1/3c}
⇒ a = 3 cm
Length of a side of cube = 3 cm
Since, two cubes are joined and a cuboid is formed so,
Length of cuboid = l = 2a = 2 × 3 cm = 6 cm
Breadth of cuboid = b = a = 3 cm
Height of cuboid = h = a = 3 cm
Surface area of cuboid = 2 × (l × b + b × h + l × h)
∴ Surface area of resulting cuboid = 2 × (6 × 3 + 3 × 3 + 6 × 3) cm^{2}
= 2 × (18 + 9 + 18) cm^{2}
= 2 × 45 cm^{2}
= 90 cm^{2}
So, surface area of resulting cuboid is 90 cm^{2}
2. The volume of a hemisphere is 2425.1/2 cm^{3}. Find its curved surface area.
Solution
Let the radius of hemisphere be r cm
Volume of hemisphere is given by 2/3Ï€r^{3}
Given, volume of hemisphere = 24251/2 cm^{3}
∴ 2/3Ï€r^{3} = 24251/2
⇒ r^{3} = 4851 × 1/2 × 3/2 × 7/22
⇒ r^{3} = 1157.625 cm^{3}
⇒ r = (1157.625)^{1/3} cm
⇒ r = 10.5 cm
Curved surface Area of hemisphere = 2Ï€r^{2}
Curved surface Area of hemisphere = 2 × 22/7 × (10.5)^{2} cm^{2}
= 693 cm^{2}
∴ Curved surface area of hemisphere = 693 cm^{2}
3. If the total surface area of a solid hemisphere is 462 cm^{2}, find its volume.
Solution
Let the radius of solid sphere be r cm
Total surface area of solid hemisphere = 3Ï€r^{2}
Given, total surface area of solid hemisphere = 462 cm^{2}
∴ 3Ï€r^{2} = 462 cm^{2}
⇒ 3 × 22/7 × r^{2} = 462 cm^{2}
⇒ r^{2} = 462 × 1/3 × 7/22 cm^{2} = 49 cm^{2}
⇒ r = 7 cm
Volume of solid hemisphere = 4/3.Ï€r^{3}
= 2/3 × 22/7 × 7^{3} cm^{3}
= 718.67 cm^{3}
∴ Volume of solid hemisphere is 718.67 cm^{3}
4. A 5mwide cloth is used to make a conical tent of base diameter 14 m and height 24 m. Find the cost of cloth used at the rate of Rs 25 per metre.
Solution
Width of cloth used = 5 m
Diameter of conical tent to be made = 14 m
Let the radius of the conical tent be r m
Radius of conical tent = r = diameter ÷ 2 = 142/2 m = 7 m
Height of conical tent = h = 24 m
Let the slant height of conical tent be l
Area of cloth required to make a conical tent = Curved surface area of conical tent
= Ï€rl
= 22/7 × 7 × 25 m^{2}
= 550 m^{2}
Length of cloth used = Area of cloth required ÷ Width of cloth
= 550/5 m
= 110 m
∴ Length of cloth used = 110 m
Cost of cloth used = Rs 25 per meter
Total Cost of cloth required to make a conical tent = 110 × Rs 25
= Rs 2750
∴ Total cost of cloth required to make a conical tent = 2750
5. If the volume of two cones are in the ratio of 1 : 4 and their diameters are in the ratio 4 : 5, find the ratio of their heights.
Solution
Let V_{1} be the volume of first cone and V_{2} be the volume of second cone.
Then, V_{1} : V_{2} = 1 : 4
Let d_{1} be the diameter of first cone and d_{2} be the diameter of second cone.
Then d_{1} : d_{2} = 4 : 5
Let h_{1} be the height of first cone and h_{2} be the height of second cone.
We know that volume of cone is given by V = 1/3 × Ï€(d^{2}/4)h
V_{1}/V_{2} = 1/4
⇒ 16/25 × h_{1}/h_{2} = 1/4
⇒ h_{1}/h_{2} = 1/4 × 25/26
⇒ h_{1}/h_{2} = 25/64
∴ h_{1}/h_{2} = 25 : 64
∴ Ratio of height of two cones is 25 : 64.
6. The slant height of a conical mountain is 2.5 km and the area of its base is 1.54 km^{2}. Find the height of mountain.
Solution
Let the radius of base be ‘r’ km and slant height be ‘l’ km
Slant height of conical mountain = 2.5 km
Area of its base = 1. 54 km^{2}
Area of base is given by Ï€r^{2}
∴ Ï€r^{2} = 1.54 km^{2 }
⇒ 22/7 × r^{2 }= 1.54 km^{2}
⇒ r^{2} = 1.54 × 7/22 km^{2} = .49 km^{2}
⇒ r = 0.7 km
Let ‘h’ be the height of the mountain
We know,
l^{2} = r^{2} + h^{2}
Substituting the values of l and r in the above equation
2.5^{2} = 0.7^{2} + h^{2}
h^{2} = 2.5^{2} – 0.7^{2}
= 6.25 – 0.49 km^{2}
⇒ h^{2} = 5.76 km^{2}
⇒ h = 2.4 km
∴ Height of the mountain = 2.4 km
7. The sum of the radius of the base and the height of a solid cylinder is 37 metres. If the total surface area of the cylinder be 1628 sq metres, find its volume.
Solution
Let the radius of the solid cylinder be ‘r’ m and its height be ‘h’ m.
Given,
Sum of radius and height of solid cylinder = 37 m
r + h = 37 m
⇒ r = 37 – h
Total surface area of solid cylinder = 1628 m^{2}
Total surface area of solid cylinder is given by 2Ï€r (h + r)
∴ 2Ï€r (h + r) = 1628 m^{2}
Substituting the value of r + h in the above equation
⇒ 2Ï€r × 37 = 1628 m^{2}
⇒ r = 1628 × 7/22 × 1/2 × 1/37 m
⇒ r = 7 m
Since, r + h = 37 m
h = 37 – r m
⇒ h = 37 – 7 m = 30 m
Volume of solid cylinder = Ï€r^{2}h
= 22/7 × 7^{2} × 30 m^{2}
= 4620 m^{2}
8. The surface area of a sphere is 2464 cm^{2}. If the radius be doubled, what will be the surface area of the new sphere?
Solution
Let the radius of sphere ne ‘r’ cm
Surface area of sphere = 2464 cm^{2}
Surface area of sphere is given by 4Ï€r^{2}
∴ 4Ï€r^{2} = 2664 cm^{2}
⇒ 4 × 22/7 × r^{2} = 2464 cm^{2}
⇒ r^{2} = 2464 × 1/4 × 7/22 cm^{2} = 196 cm^{2}
⇒ r = 14 cm
Radius of new sphere is double the radius of given sphere.
Let the radius of new sphere be ‘r’ cm
∴ r’ = 2r
r’ = 2 × 14 cm = 28 cm
Surface area of new sphere = 4Ï€r’^{2}
= 4 × 22/7 × 28^{2} cm^{2}
= 9856 cm^{2}
∴ Surface area of new sphere is 9856 cm^{2}.
9. A military tent of height 8.25 m is in the form of a right circular cylinder of base diameter 30 m and height 5.5 m surmounted by a right circular cone of a same base radius. Find the length of canvas used in making the tent, if the breadth of the canvas is 1.5 m.
Solution
The military tent is made as a combination of right circular cylinder and right circular cone oh top.
Total Height of tent = h = 8.25 m
Base diameter of tent = 30 m
Base radius of tent = r = 30/2 = 15 m
Height of right circular cylinder = 5.5 m
Curved surface area of right circular cylindrical part of tent = 2Ï€rh
Height of conical part = total height of tent – height of cylindrical part
h_{cone} = 8.25 – 5.5 m = 2.75 m
Base radius of cone = 15 m
Let l be the slant height of cone
Then, l^{2} = h_{cone}^{2} + r^{2} = 2.75^{2} + 15^{2} m^{2}
⇒ l^{2} = 7.5625 + 225 m^{2} = 232.5625
⇒ l = 15.25
Curved surface area of conical part of the tent = Ï€rl
Total surface area of the tent = Curved surface area of cylindrical part + curved surface area of conical part
Total surface area of tent = 2Ï€rh + Ï€rl
= Ï€r(2h + l)
= 22/7 × 15 × (2 × 5.5 + 15.25) m^{2}
= 22/7 × 15 × (11 + 15.25) m^{2}
= 22/7 × 15 × 26.25 m^{2}
= 1237.5 m^{2 }
Breadth of canvas used = 1.5 m
Length of canvas used = Total surface are of tent ÷ breadth of canvas used
Length of canvas used = (1237.5)/(1.5)m = 825 m
∴ Length of canvas used is 825 m
10. A tent is in the shape of a right circular cylinder up to a height of 3 m and conical above it. The total height of the tent is 13.5 m and the radius of its base is 14 m. Find the cost of cloth required to make the tent at the rate of Rs 80 per square metre. [Take Ï€ = 22/7.]
Solution
The tent is made as a combination of right circular cylinder and right circular cone on top.
Height of cylindrical part of the tent = h = 3 m
Radius of its base = r = 14 m
Curved surface area of cylindrical part of tent = 2Ï€rh
= 2 × 22/7 × 14 × 3 m^{2}
= 264 m^{2}
Total height of the tent = 13.5 m
Height of conical part of the tent = 13.5 – 3 m = 10.5 m
Let the slant height of the conical part be l
l^{2} = h_{cone}^{2} + r^{2}
⇒ l^{2} = 10.5^{2} + 14^{2} = 110.25 + 196 m^{2} = 306.25 m^{2}
⇒ l = 17.5 m
Curved surface area of conical part of tent = Ï€rl
= 22/7 × 14 × 17.5 m^{2}
= 770 m^{2}
Total surface area of tent = Curved surface area of cylindrical part of tent + Curved surface area of conical part of tent
Total surface area of tent = 264 m^{2 }+ 770 m^{2} = 1034 m^{2}
Cloth required = Total Surface area of tent = 1034 m^{2}
Cost of cloth = Rs 80/m^{2}
Total cost of cloth required = Total surface area of tent × Cost of cloth
= 1034 × Rs 80
= Rs 82720
Cost of cloth required to make the tent is Rs 82720.
11. A circus tent is cylindrical to a height of 3 m and conical above it. If its base radius is 52.5 m and the slant height of the conical potion is 53 m, find the area of canvas needed to make the tent. [Take Ï€ = 22/7]
Solution
The Circus tent is made as a combination of cylinder and cone on top.
Height of cylindrical part of tent = h = 3 m
Base radius of tent = r = 52.5 m
Area of canvas required for cylindrical part of tent = 2Ï€rh
= 2 × 22/7 × 52.5 × 3 m^{2}
= 990 m^{2}
Slant height of cone = l = 53 m
Area of canvas required for conical part of the tent = Ï€rl
= 22/7 × 52.5 × 53 m^{2}
= 8745 m^{2 }
Area of canvas required to make the tent = Area of canvas required for cylindrical part of tent + Area of canvas required for conical part of tent
Are of canvas required to make the tent = 990 + 8745 m^{2}
= 9735 m^{2}
12. A rocket is in the form of a circular cylinder closed at the lower and a cone of the same radius is attached to the top. The radius of the cylinder is 2.5 m, its height is 21 m and the slant height of the cone is 8 m. Calculate the total surface area of the rocket.
Solution
The rocket is in the form of cylinder closed at the bottom and cone on top.
Height of cylindrical part rocket = h = 21 m
Base radius of rocket = r = 2.5 m
Surface area of cylindrical part of rocket = 2Ï€rh + Ï€r^{2}
= 2 × 22/7 × 2.5 × 21 + 22/7 × 2.5 × 2.5 m^{2}
= 330 + 19.64 m^{2} = 349.64 m^{2}
Slant height of cone = l = 8 m
Surface area of conical part of the rocket = Ï€rl
= 22/7 × 2.5 × 8 m^{2}
= 62.86 m^{2}
Total surface area of the rocket = Surface Area of cylindrical part of rocket + surface area of conical part of rocket
Total surface area of the rocket = 349.64 + 62.86 m^{2}
= 412.5 m^{2}
13. A solid is in the shape of a cone surmounted on a hemisphere, the radius of each of them being 3.5 cm and the total height of the solid is 9.5 cm. Find the volume of the solid.
Solution
The solid is in the form of a cone surmounted on a hemisphere.
Total height of solid = h = 9.5 m
Radius of solid = r = 3.5 m
Volume of hemispherical part solid = 2/3 × Ï€r^{3}
= 2/3 × 22/7 × 3.5^{3} m^{3}
= 89.83 m^{3}
Height of conical part of solid = h_{cone} = Total height of solid – Radius of solid
Height of conical part of solid = h_{cone} = 9.5 – 3.5 = 6 m
Volume of conical part of solid = 1/3 × Ï€r^{2}h cone
= 1/3 × 22/7 × 3.52 × 6 m^{3}
= 77 m^{3}
Volume of solid = Volume of hemispherical part solid + Volume of conical part solid
Volume of solid = 89.83 + 77 m^{3 }= 166.83 m^{3}
14. A toy is in the form of a cone mounted on a hemisphere of same radius 7 cm. If the total height of the toy is 31 cm, find its total surface area.
Solution
The toy is in the form of a cone mounted on a hemisphere.
Total height of toy = h = 31 cm
Radius of toy = r = 7 cm
Surface area of hemispherical part toy = 2Ï€r^{2}
= 2 × 22/7 × 7^{2} cm^{2 }
= 308 cm^{2}
Height of conical part of toy = h_{cone} = Total height of toy – Radius of toy
Height of conical part of toy = h_{cone} = 31 – 7 = 24 cm
Let the slant height of the conical part be l
l^{2} = h_{cone}^{2} + r^{2}
⇒ l^{2} = 24^{2} + 7^{2} = 576 + 49 cm^{2} = 625 cm^{2}
⇒ l = 25 cm
Surface area of conical part of toy = Ï€rl
= 22/7 × 7 × 25 cm^{2}
= 550 cm^{2}
Total surface area of toy = Surface area of hemispherical part of toy + Surface area of conical part of toy
Total surface area of toy = 308 cm2 + 550 cm^{2 }
= 858 cm^{2}
15. A toy is in the shape of a cone mounted on a hemisphere of same base radius. If the volume of the toy is 231 cm^{3} and its diameter is 7 cm, find the height of the toy.
Solution
A toy is in the shape of a cone mounted on a hemisphere of same base radius.
Volume of Toy = 231 cm^{3}
Base diameter of toy = 7 cm
Base radius of toy = 7/2 cm = 3.5 cm
Volume of hemisphere = 2/3Ï€r^{3}
= 2/3 × 22/7 × 3.5^{3} cm^{3}
= (2 × 22 × 35 × 35 × 35)/(3 × 7 × 10 × 10 × 10)
= 539/6
Volume of cone = Volume of hemisphere – Volume of toy
= 231 – 539/6
= (1386 – 539)/6
= 847/6
Volume of cone is given by 1/3Ï€r^{2}h
Where h is the height of cone
⇒ 1/3Ï€r^{2}h = 847/6
⇒ 22/7(7/2)^{2}h = 847/2
⇒ 77/2.h = 747/2
⇒ h = 11
Height of cone = 11 cm
Height of toy = Height of cone + Height of hemisphere
= 11 cm + 3.5 cm
= 14.5 cm
[Height of hemisphere = Radius of hemisphere]
∴ Height of toy is 14.5 cm.
16. A cylindrical container of radius 6 cm and height 15 cm is filled with icecream. The whole icecream has to be distributed to 10 children in equal cones with hemispherical tops. If the height of the conical portion is 4 times the radius of its base, find the radius of the icecream cone.
Solution
Radius of cylindrical container = r = 6 cm
Height of cylindrical container = h = 15 cm
Volume of cylindrical container = Ï€r^{2}h
= 22/7 × 6 × 6 × 15 cm^{3}
= 1697.14 cm^{3}
Whole icecream has to be distributed to 10 children in equal cones with hemispherical tops.
Let the radius of hemisphere and base of cone be r’
Height of cone = h = 4 times the radius of its base
h’ = 4r’
Volume of Hemisphere = 2/3Ï€(r’)^{3}
Volume of cone = 1/3Ï€(r’)^{2}h’ = 1/3Ï€(r’)^{2} × 4r’
= 2/3 Ï€(r’)^{3}
Volume of icecream = Volume of Hemisphere + Volume of cone
= 2/3Ï€(r’)^{3 }+ 4/3Ï€(r’)^{3} = 6/3Ï€(r’)^{3}
Number of icecreams = 10
∴ Total volume of icecream = 10 × Volume of icecream
= 10 × 6/3 Ï€(r’)^{3}
= 60/3 Ï€(r’)^{3}
Also, total volume of ice cream = Volume of cylindrical container
⇒ 60/3 Ï€(r’)^{3} = 1697.14 cm^{3}
⇒ 60/3 × 22/7 × (r’)^{3} = 1697.14 cm^{3 }
⇒ (r’)^{3} = 1697.14 × 3/60 × 7/22 = 27 cm^{3 }
⇒ r = 3 cm
∴ Radius of icecream cone = 3 cm
17. A vessel is in the form of a hemispherical bowl surmounted by a hallow cylinder. The diameter of the hemispherical is 21 cm and the total height of the vessel is 14.5 cm. Find its capacity.
Solution
Vessel is in the form of hemispherical bowl surmounted by a hallow cylinder.
Diameter of hemisphere = 21 cm
Radius of hemisphere = 10.5 cm
Volume of hemisphere = 2/3Ï€r^{3} = 2/3 × 22/7 × 10.5^{3} cm^{3}
= 2425.5 cm^{3}
Total height of vessel = 14.5 cm
Height of cylinder = h = Total height of vessel – Radius of hemisphere
= 14.5 – 10.5
= 4 cm
Volume of cylinder = Ï€r^{2}h = 22/7 × 10.5 × 10.5 × 4 cm^{3}
= 1386 cm^{3}
Volume of vessel = Volume of hemisphere + Volume of cylinder
= 2425.5 cm^{3} + 1386 cm^{3}
= 3811.5 cm^{3}
∴ Capacity of vessel = 3811.5 cm^{3}
18. A toy is in the form of a cylinder with hemisphere ends. If the whole length of the toy is 90 cm and its diameter is 42 cm, find the cost of painting the toy at the rate of 70 paise per sq cm.
Solution
Toy is in the form of a cylinder with hemisphere ends
Total length of toy = 90 cm
Diameter of toy = 42 cm
Radius of toy = r = 21 cm
Length of cylinder = l = Total length of toy – 2 × Radius of toy
= 90 – 2 × 21 cm
= 48 cm
For cost of painting we need to find out the curved surface area of toy
Curved surface area of cylinder = 2Ï€rl
= 2 × 22/7 × 21 × 48 cm^{2}
= 6336 cm^{2 }
Curved surface area of hemispherical ends = 2 × 2Ï€r^{2}
= 2 × 2 × 22/7 × 21 × 21 cm^{2 }
= 5544 cm^{2}
Surface area of toy = Curved surface area of cylinder + Curved surface area of hemispherical ends
Surface area of toy = 6336 cm^{2} + 5544 cm^{2} = 11880 cm^{2}
Cost of painting = Rs (0.70) cm^{2}
Total cost of painting = Surface area of toy × Cost of painting
= 11880 cm^{2} × Rs 0.70 cm^{2} = Rs 8316.00
Total cost of painting the toy = Rs 8316.00
19. A medicine capsule is in the shape of a cylinder with two hemisphere stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Solution
A medicine capsule is in the shape of a cylinder with two hemisphere stuck to each of its ends.
Total length of entire capsule = 14 mm
Diameter of capsule = 5 mm
Radius of capsule = r = Diameter ÷ 2 = 5/2 mm = 2.5 mm
Length of cylindrical part of capsule = l = Total length of entire capsule – 2 × Radius of capsule
= 14 – 2 × 2.5 mm
= 14 – 5 mm
= 9 mm
Curved surface area of cylindrical part of capsule = 2Ï€rl
= 2 × 3.14 × 9 × 2.5 mm^{2}
= 141.3 mm^{2}
Curved surface area of hemispherical ends = 2 × 2Ï€r^{2 }
= 2 × 2 × 3.14 × 2.5 × 2.5 cm^{2}
= 78.5 cm^{2}
Surface area of capsule = Curved surface area of cylindrical part of capsule + Curved surface area of hemispherical ends
Surface area of capsule = 141.3 mm^{2} + 78.5 mm^{2} = 219.8 mm^{2}
20. A wooden article was made by scooting out a hemisphere from each end of cylinder, as shown in the figure. If the height of the cylinder is 20 cm and its base is of diameter 7 cm, find the total surface area of the article when it is ready.
SolutionThe wooden article was made by scooting out a hemisphere from each end of a cylinder
Let the radius of cylinder be r cm and height be h cm.
Height of cylinder = h = 20 cm
Base diameter of cylinder = 7 cm
Base radius of cylinder = r = diameter ÷ 2 = 7/2 cm = 3.5 cm
Lateral Surface area of cylinder = 2Ï€rh
= 2 × 22/7 × 3.5 × 20 cm^{2}
= 2 × 22/7 × 3.5 × 20 cm^{2}
= 440 cm^{2}
Since, the wooden article was made by scooting out a hemisphere from each end of a cylinder
∴ Two hemisphere are taken out in total
Radius of cylinder = radius of hemisphere
∴ Radius of hemisphere = 3.5 cm
Lateral Surface area of hemisphere = 2Ï€r^{2}
= 2 × 22/7 × 3.5 × 3.5 cm^{2}
= 77 cm^{2}
Total Surface area of two hemispheres = 2 × 77 cm^{2} = 154 cm^{2}
Total surface area of the article when it is ready = Lateral Surface area of cylinder + Lateral Surface area of hemisphere
Total surface area of the article when it is ready = 440 cm^{2} + 154 cm^{2}
= 594 cm^{2}
21. solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is 2.1 cm and the height of the cone is 4 cm. The solid is placed in a cylindrical tub full of water in such a way that the whole solid is submerged in water. If the radius of the cylinder is 5 cm and its height is 9.8 cm, find the volume of the water left in the tub.
Solution
A solid is in the form of a right circular cone mounted on a hemisphere
Let r be the radius of hemisphere and cone
Let h be the height of the cone
Radius of hemisphere = 2/3Ï€r^{3}
= 2/3 × 22/7 × 2.1 × 2.1 × 2.1 cm^{3}
= 19.404 cm^{3}
Height of cone = h = 4 cm
Radius of cone = r = 2.1 cm
Volume of cone = 1/3Ï€r^{2}h
= 1/3 × 22/7 × 2.1 × 2.1 × 4 cm^{3}
= 18.48 cm^{3}
Volume of solid = Volume of hemisphere + Volume of cone
= 19.404 cm^{3} + 18.48 cm^{3}
= 37.884 cm^{3}
The solid is placed in a cylindrical tub full of water in such a way that the whole solid is submerged in water, so, to find the volume of water left in the tub we need to subtract volume of solid from cylindrical tub.
Radius of cylinder = r’ = 5 cm
Height of cylinder = h’ = 9.8 cm
Volume of cylindrical tub = Ï€r’^{2}h = 22/7 × 5 × 5 × 9.8 cm^{3}
= 770 cm^{3}
Volume of water left in the tub = Volume of cylindrical tub – Volume of solid
Volume of water left in the tub = 770 cm^{3} – 37.884 cm^{3}
= 732.116 cm^{3}
∴ Volume of water left in the tub is 732.116 cm^{3}
22. From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of height 8 cm and of base radius 6 cm is hollowed out. Find the volume of the remaining solid. Also, find the total surface area of the remaining solid. [Take Ï€ = 3.14]
Solution
Height of solid cylinder = h = 8 cm
Radius of solid cylinder = r = 6 cm
Volume of solid cylinder = Ï€r^{2}h
= 3.14 × 6 × 6 × 8 cm^{3}
= 904.32 cm^{3}
Curved Surface area of solid cylinder = 2Ï€rh
Height of conical cavity = h = 8 cm
Radius conical cavity = r = 6 cm
Volume of conical cavity = 1/3Ï€r^{2}h
= 1/3 × 3.14 × 6 × 6 × 8 cm^{3}
= 301.44 cm^{3}
Let l be the slant height of conical cavity
l^{2} = r^{2} + h^{2}
⇒ l^{2} = (6^{2} + 8^{2}) cm^{2}
⇒ l^{2} = (36 + 64) cm^{2}
⇒ l^{2} = 100 cm^{2}
⇒ l = 10 cm
Curved surface area of conical cavity = Ï€rl
Since, conical cavity is hollowed out from solid cylinder, So, volume and total surface area of remaining solid will be found out by subtracting volume and total surface area of conical cavity from volume and total surface area of solid cylinder.
Volume of remaining solid = Volume of solid cylinder – Volume of conical cavity
Volume of remaining solid = 904.32 cm^{3} – 301.44 cm^{3}
= 602.88 cm^{3}
Total surface area of remaining solid = Curved surface area of solid cylinder + Curved surface area of conical cavity + Area of circular base
Total surface area of remaining solid = 2Ï€rh + Ï€rl + Ï€r^{2}
= Ï€r × (2h + l + r)
= 3.14 × 6 × (2 × 8 + 10 + 6) cm^{2}
= 3.14 × 6 × 32 cm^{2}
= 602.88 cm^{2}
23. From a solid cylinder of height 2.8 cm and diameter 4.2 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid.
Solution
Height of solid cylinder = h = 2.8 cm
Diameter of solid cylinder = 4.2 cm
Radius of solid cylinder = r = Diameter ÷ 2 .2 cm
Curved surface area of solid cylinder = 2Ï€rh
= 2 × 22/7 × 2.1 × 2.8 cm^{2}
= 2 × 22/7 × 2.1 × 2.8 cm^{2}
= 36.96 cm^{2}
Height of conical cavity = h = 2.8 cm
Radius conical cavity = r = 2.1 cm
Let l be the slant height of conical cavity
l^{2} = r^{2} + h^{2}
⇒ l^{2} = (2.8^{2} + 2.1^{2}) cm^{2}
⇒ l^{2} = (7.84 + 4.41) cm^{2}
⇒ l^{2} = 12.25 cm^{2}
⇒ l = 3.5 cm
Curved Surface area of conical cavity = Ï€rl
= 22/7 × 2.1 × 3.5
= 23.1 cm^{2}
Total surface area of remaining = Curved surface area of solid cylinder + Curved surface area of conical cavity + Area of circular base
Total surface area of remaining solid = (36.96 + 23.1 + 22/7 × 2.1^{2}) cm^{2}
= (36.96 + 23.1 + 13.86) cm^{2}
= 73.92 cm^{2}
24. From a solid cylinder of height 14 cm and base diameter 7 cm, two equal conical holes each of radius 2.1 cm and height 4 cm are cut off. Find the volume of the remaining solid.
Solution
Height of solid cylinder = h = 14 cm
Diameter of solid cylinder = 7 cm
Radius of solid cylinder = r = Diameter ÷ 2 = 7/2 cm = 3.5 cm
Volume of solid cylinder = Ï€r^{2}h
= 22/7 × 3.5 × 3.5 × 14 cm^{3}
= 539 cm^{3}
Height of conical cavity = h’ = 4 cm
Radius conical cavity = r’ = 2.1 cm
Volume of conical cavity = 1/3Ï€r’^{2}h’
= 1/3 × 22/7 × 2.1 × 2.1 × 4 cm^{3}
= 18.48 cm^{3}
Since, there are two conical cavities
∴ Volume of two conical cavities = 2 × 18.48 cm^{3} = 36.96 cm^{3}
Volume of remaining solid = Volume of solid cylinder – Volume of two conical cavity
Volume of remaining solid = 539 cm^{3} – 36.96 cm^{3}
= 502.04 cm^{3}
25. A metallic cylinder has radius 3 cm and height 5 cm. To reduce its weight, a conical hole is drilled in the cylinder. The conical hole has a radius of 3/2 cm and its depth is 8/9 cm. Calculate the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape.
Solution
Height of metallic cylinder = h = 5 cm
Radius of metallic cylinder = r = 3 cm
Volume of solid cylinder = Ï€r^{2}h
= Ï€ × 3 × 3 × 5 cm^{3}
= 45Ï€ cm^{3}
Height of conical hole = h’ = 8/9 cm.
Radius conical hole = r’ = 3/2 cm
Volume of conical hole = 1/3Ï€r’^{2}h’
= 1/3 × Ï€ × 3/2 × 3/2 × 8/9 cm^{3}
= 2/3Ï€ cm^{3}
Volume of metal left in cylinder = Volume of metallic cylinder – Volume of conical hole
Volume of metal left in cylinder = 45Ï€ – 2/3Ï€ = 133Ï€/3
Ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape = Volume of metal; left in cylinder/Volume of conical hole
Volume of metal left in cylinder: Volume of conical hole = 133Ï€/3 : 2/3Ï€
Volume of metal left in cylinder: Volume of conical hole = 133 : 2
Ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape is 339: 4
26. A spherical glass vessel has a cylindrical neck 7 cm long and 4 cm in diameter. The diameter of the spherical part is 21 cm. Find the quantity of water it can hold. [Use Ï€= 22/7]
Solution
Length of cylindrical neck = l = 7 cm
Diameter of cylindrical neck = 4 cm
Radius of cylindrical neck = r = Diameter ÷ 2 = 4/2 cm = 2 cm
Volume of cylindrical neck = Ï€r^{2}l
= 22/7 × 2 × 2 × 7 cm^{3}
= 88 cm^{3}
Diameter of spherical part = 21 cm
Radius of spherical part = r’ = Diameter ÷ 2 = 21/2 cm = 10.5 cm
Volume of spherical part = 4/3Ï€r’^{3}
= 4/3 × 22/7 × 10.5 × 10.5 × 10.5 cm^{3}
= 4851 cm^{3}
Quantity of water spherical glass vessel with cylindrical neck can hold = Volume of spherical part + Volume of cylindrical neck
Quantity of water spherical glass vessel with cylindrical neck can hold = 4851 cm^{3} + 88 cm^{3} = 4939 cm^{3}
Quantity of water spherical glass vessel with cylindrical neck can hold is 4939 cm^{3}.
27. The given figure represent a solid consisting of cylinder surmounted by a cone at one end a hemisphere at the other. Find the volume of the solid.
Solution
The solid consisting of a cylinder surmounted by a cone at one end a hemisphere at the other.
Length of cylinder = l = 6.5 cm
Diameter of cylinder = 7 cm
Radius of cylinder = r = Diameter ÷ 2 = 7/2 cm = 3.5 cm
Volume of cylinder = Ï€r^{2}l
= 22/7 × 3.5 × 3.5 × 6.5 cm^{3}
= 250.25 cm^{3}
Length of cone = l’ = 12.8 cm – 6.5 cm
= 6.3 cm
Diameter of cone = 7 cm
Radius of cone = r = Diameter ÷ 2 = 7/2 cm = 3.5 cm
Volume of cone = 1/3Ï€r^{2}’l
= 1/3 × 22/7 × 3.5 × 3.5 × 6.3 cm^{3}
= 80.85 cm^{3}
Diameter of hemisphere = 7 cm
Radius of hemisphere = r = Diameter ÷ 2 = 7/2 cm = 3.5 cm
Volume of hemisphere = 2/3Ï€r^{3}
= 2/3 × 22/7 × 3.5 × 3.5 × 3.5 cm^{3}
= 89.83 cm^{3}
Volume of the solid = Volume of cylinder + Volume of cone + Volume of hemisphere
Volume of solid = 250.25 cm^{3} + 80.85 cm^{3} + 89.83 cm^{3}
= 420.93 cm^{3}
28. From a cubical piece of wood of side 21 cm, a hemisphere is carved out in such a way that the diameter of the hemisphere is equal to the side of the cubical piece. Find the surface area and volume of the remaining piece.
Solution
Length of cubical piece of woof = a = 21 cm
Volume of cubical piece of wood = a^{3}
= 21 × 21 × 21 cm^{3 }
= 9261 cm^{3}
Surface area of cubical piece of wood = 6a^{2}
= 6 × 21 × 21 cm^{2}
= 2646 cm^{2 }
Since, a hemisphere is carved out in such a way that the diameter of the hemisphere is equal to the side of the cubical piece.
So, diameter of hemisphere = length of side of the cubical piece
Diameter of hemisphere = 21 cm
Radius of hemisphere = r = Diameter ÷ 2 = 21/2 cm = 10.5 cm
Volume of hemisphere = 2/3Ï€r^{3}
= 2/3 × 22/7 × 10.5 × 10.5 × 10.5 cm^{3 }
= 2425.5 cm^{3}
Surface area of hemisphere = 2Ï€r^{2}
= 2 × 22/7 × 10.5 × 10.5 cm^{2}
= 693 cm^{2}
A hemisphere is carved out from cubical piece of wood
Volume of remaining solid = Volume of cubical piece of wood – Volume of hemisphere
Volume of remaining solid = 9261 cm^{3} – 2425.5 cm^{3} = 6835.5 cm^{3}
Surface area remaining piece of solid = Surface area of cubical piece of wood – Area of circular base of hemisphere + Curved Surface area of hemisphere
Surface area remaining piece of solid = 6a^{2} – Ï€r^{2} + 2Ï€r^{2}
= (2646 – 22/7 × 10.5 2 + 693) cm^{2}
= 2992.5 cm^{2}
29. A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of Rs 5 per sq cm. [use Ï€ = 3.14]
Solution
Length of side of cubical block = a = 10 cm
Since a cubical block is surmounted by a hemisphere, so, the largest diameter of hemisphere = 10 cm
Since, hemisphere will be touching the sides of cubical block.
Radius of hemisphere = r = Diameter ÷ 2 = 10/2 cm = 5 cm
Surface area of solid = Surface area of cube – Area of circular part of hemisphere + Curved surface area of hemisphere
Total surface area of solid = 6a^{2} – Ï€r^{2} + 2Ï€r^{2} = 6a^{2} + Ï€r^{2 }
= 6 × 10 × 10 cm^{2} + 3.14 × 5.5 cm^{2}
= 678.5 cm^{2}
Rate of painting = Rs 5/100 cm^{2}
Cost of painting the total surface area of the solid so formed = Total Surface area of solid × Rate of painting
Cost of painting the total surface area of the solid = Rs 5/100 × 678.5
= Rs 33.925
30. A toy is in the shape of a right circular cylinder with a hemisphere on one end a cone on the other. The radius and height of the cylindrical part area 5 cm and 13 cm respectively. The radii of the hemispherical and the conical parts are the same as that of the cylindrical part. Find the surface area of the toy, if the total height of the toy is 30 cm.
Solution
The toy is in the shape of a right circular cylinder surmounted by a cone at end a hemisphere at the other.
Total height of toy = 30 cm
Height of cylinder = h = 13 cm
Radius of cylinder = r = 5 cm
Curved surface area of cylinder = 2Ï€rh
= 2 × 22/7 × 5 × 13 cm^{2}
Height of cone = h’ = total height of toy – Height of cylinder – Radius of hemisphere
Height of cone = h’ = 30 cm – 13 cm – 5 cm = 12 cm
Radius of cone = r = radius of cylinder
Radius of cone = r = 5 cm
Let the slant height of cone be l
l^{2} = h’^{2} + r^{2}
⇒ l^{2} = 12^{2} + 52 cm^{2} = 144 + 25 cm^{2} = 169 cm^{2}
⇒ l = 13 cm
Curved surface area of cone = Ï€rl
= 22/7 × 5 × 13 cm^{2}
Radius of hemisphere = r = Radius of cylinder
Radius of hemisphere = r = 5 cm
Curved surface area of hemisphere = 2Ï€r^{2}
= 2 × 22/7 × 5 × 5 cm^{2 }
Surface area of the toy = Surface area of cylinder + Surface area of cone + Surface area of hemisphere
Surface area of toy = 2Ï€rrh + Ï€rrl + 2Ï€r^{2}
= Ï€r(2h + l + 2r)
= 22/7 × 5 × (2 × 13 + 13 + 2 × 5) cm^{2}
= 22/7 × 5 × 49 cm^{2}
= 770 cm^{2}
Surface area of toy is 770 cm^{2}
31. The inner diameter of a glass is 7 cm and it has a raised portion in the bottom in the shape of a hemisphere, as shown in the figure. If the height of the glass is 16 cm, find the apparent capacity and the actual capacity of the glass.
SolutionInner diameter of a glass = 7 cm
Inner radius of glass = r = 7/2 cm = 3.5 cm
Height of glass = h = 16 cm
Apparent capacity of glass = Ï€r^{2}h
= 22/7 × 3.5 × 3.5 × 16 cm^{3}
= 616 cm^{3}
Volume of the hemisphere in the bottom = 2/3Ï€r^{3}
= 2/3 × 22/7 × 3.5^{3} cm^{3}
= 89.83 cm^{3}
Actual capacity of the class=Apparent capacity of glass – Volume of the hemisphere
Actual capacity of the class=616 cm^{3} – 89.83 cm^{3} = 526.17 cm^{3}
32. A wooden toy is in the shape of a cone mounted on a cylinder, as shown in the figure. The total height of the toy is 26 cm, while the height of the conical part is 6 cm. The diameter of the base of the conical part is 5 cm and that of the cylindrical part in 4 cm. the conical part and the cylindrical part are respectively painted red and white. Find the area to be painted by each of them colours. [Take Ï€ = 22/7]
SolutionThe wooden toy is in the shape of a cone mounted on a cylinder
Total height of the toy = 26 cm
Height of conical part = H = 6 cm
Height of cylindrical part = Total height of the toy – Height of conical part
h = 26 cm – 6 cm = 20 cm
Diameter of conical part = 5 cm
Radius of conical part = R = Diameter/2 = 5/2 cm = 2.5 cm
Let L be the slant height of the cone
L^{2} = H^{2} + R^{2}
⇒ L^{2} = 6^{2} + 2.5^{2} cm^{2} = 36 + 6.25 cm^{2} = 42.25 cm^{2}
⇒ L = 6.5 cm
Diameter of cylindrical part = 4 cm
Radius of cylindrical part = r = Diameter/2 = 4/2 cm = 2 cm
Area to be painted red = Curved surface area of cone + Base are of cone – base area of cylinder
Area to be painted red = Ï€RL + Ï€R^{2} – Ï€r^{2} = Ï€(RL + R^{2} – r^{2})
= 22/7 × (2.5 × 6.5 + 2.5 × 2.5 – 2 × 2) cm^{2}
= 22/7 × (16.25 + 6.25 – 4) cm^{2}
= 22/7 × 18.5 cm^{2}
= 58.143 cm^{2}
Area to be painted white = Curved Surface area of cylinder + Base area of cylinder
Area to be painted White = 2Ï€rh + Ï€r^{2} = Ï€r(2h + r)
= 22/7 × 2 × (2 × 20 + 2) cm^{2}
= 22/7 × 2 × (40 + 2) cm^{2}
= 22/7 × 2 × 42 cm^{2 }
= 264 cm^{2}
∴ Area to be painted red is 58.143 cm^{2 }and area to be painted white is 264 cm^{2}.