Surface Areas And Volumes Class 10th Mathematics Gujarat Board Solution

Class 10th Mathematics Gujarat Board Solution
Exercise 14.1
  1. A toy is made by mounting a cone onto a hemisphere. The radius of the cone and…
  2. A show - piece shown in figure 14.10 is made of two solids - a cube and a…
  3. A vessel is in the form of a hemisphere mounted on a hollow cylinder. The…
  4. Chirag made a bird - bath for his garden in the shape of a cylinder with a…
  5. A solid is composed of a cylinder with hemispherical ends on both the sides.…
  6. The radius of a conical tent is 4 m and slant height is 5 m. How many meters of…
  7. If the radius of a cone is 60 cm and its curved surface area is 23.55 m^2 ,…
  8. The cost of painting the surface of sphere is Rs. 1526 at the rate of Rs. 6 per…
Exercise 14.2
  1. The curved surface area of a cone is 550 cm^2 . If its diameter is 14 cm, find…
  2. A solid is in the form of cone with hemispherical base. The radius of the cone…
  3. How many litres of milk can be stored in a cylindrical tank with radius 1.4 m…
  4. The spherical balloon with radius 21 cm is filled with air. Find the volume of…
  5. A solid has hemi - spherical base with diameter 8.5 cm and it is surmounted by…
  6. A playing top is made up of steel. The top is shaped like a cone surmounted by…
  7. How many litres of petrol will be contained in a closed cylindrical tank with…
  8. The capacity of a cylindrical tank at a petrol pump is 57750 litres. If its…
  9. A hemispherical pond is filled with 523.908 m^3 of water. Find the maximum…
  10. A gulab - jamun contain 40% sugar syrup in it. Find how much syrup would be…
  11. The height and the slant height of a cone are 12 cm and 20 cm respectively.…
  12. Find the total volume of a cone having a hemispherical base. If the radius of…
  13. If the slant height of a cone is 18.7 cm and the curved surface area is 602.8…
  14. If the surface area of a spherical ball is 1256 cm^2 , then find the volume of…
Exercise 14.3
  1. A hemispherical bowl of internal radius 12 cm contains some liquid. This liquid…
  2. A cylindrical container having diameter 16 cm and height 40 cm is full of ice -…
  3. A cylindrical tank of diameter 3 m and height 7 m is completely filled with…
  4. A cylinder of radius 2 cm and height 10 cm is melted into small spherical balls…
  5. A metallic sphere of radius 15 cm is melted and a wire of diameter 1 cm is…
  6. There are 45 conical heaps of wheat, each of them having diameter 80 cm and…
  7. A cylindrical bucket, 44 cm high and having radius of base 21 cm, is filled…
Exercise 14.4
  1. A metal bucket is in the shape of a frustum of a cone, mounted on a hollow…
  2. A container, open from the top and made up of a metal sheet is the form of…
Exercise 14
  1. A tent is in the shape of cylinder surmounted by a conical top. If the height…
  2. A metallic sphere of radius 5.6 cm is melted and recast into the shape of a…
  3. How many spherical balls of radius 2 cm can be made out of a solid cube of lead…
  4. A hemispherical bowl of internal radius 18 cm contains an edible oil to be…
  5. A hemispherical tank of radius 2.4 m is full of water. It is connected with a…
  6. A shuttle cock used for playing badminton has the shape of a frustum of a cone…
  7. A fez, the headgear cap used by the trucks is shaped like the frustum of a…
  8. A bucket is in the form of a frustum of a cone with capacity of 12308.8 cm^3 of…
  9. The volume of sphere with diameter 1 cm is …….. cm^3 .A. 2/3 π B. 1/6 π C. π…
  10. The volume of hemisphere with radius 1.2 cm is ……….. cm^3 .A. 1.152π B. 0.96π…
  11. The volume of sphere is in 4/3 π cm^3 . Then its diameter is ………. cm.A. 0.5 B.…
  12. The volume of cone with radius 2 cm and height 6 cm is ……….. cm^3 .A. 8π B.…
  13. The diameter of the base of cone is 10 cm and its slant height is 17 cm. Then…
  14. The diameter and the height of the cylinder are 14 cm and 10 cm respectively.…
  15. The ratio of the radii of two cones having equal height is 2 : 3. Then, the…
  16. If the radii of a frustum of a cone are 7 cm and 3 cm and the height is 3 cm,…
  17. The radii of a frustum of a cone are 5 cm and 9 cm and height is 6 cm, then…

Exercise 14.1
Question 1.

A toy is made by mounting a cone onto a hemisphere. The radius of the cone and a hemisphere is 5 cm. The total height of the toy is 17 cm. Find the total surface area of the toy.


Answer:

Given.


A toy is made by mounting a cone onto a hemisphere


The radius of the cone and a hemisphere is 5 cm.


The total height of the toy is 17 cm


Formula used/Theory.


Curve surface area of Cone = πrl 


Curve surface area of Hemisphere = 2πr2


⇒ As we put Cone on hemisphere


The circle part of both cone and hemisphere will attach


∴ TSA of toy is sum of CSA of both cone and hemisphere


TSA of toy = CSA of cone + CSA of Hemisphere




As height of hemisphere is equal to radius of hemisphere


Then;


Height of cone = Height of Toy – Radius


= 17cm – 5cm


= 12 cm


TSA of toy 








Question 2.

A show - piece shown in figure 14.10 is made of two solids - a cube and a hemisphere. The base of the block is a cube with edge 7 cm and the hemisphere fixed on the top has diameter 5.2 cm. Find the total surface area of the piece.



Answer:

Given.


A show - piece is made by mounting a hemisphere on cube


The Diameter of hemisphere is 5.2 cm.


Length of side of cube is 7 cm


Formula used/Theory.


Total surface area of Cube = 6 × side2


Curve surface area of Hemisphere = 2πr2


⇒ As we put hemisphere on Cube


The circle part of hemisphere will attach to cube 1 plane


∴ TSA of show - piece is sum of CSA of hemisphere and TSA of cube subtracted by Area of circle of hemisphere


TSA of show - piece =


CSA of Hemisphere + TSA of cube–Area of circle


= 2πr2 + 6 × side2–πr2


= πr2 + 6 × side2


Diameter = 5.2 cm


Radius =  = 2.6


TSA of show - piece = 3.14 × 2.6 × 2.6 + 6 × 7 × 7


= 21.22 + 294


= 315.22 cm2



Question 3.

A vessel is in the form of a hemisphere mounted on a hollow cylinder. The diameter of the hemisphere is 21 cm and the height of vessel is 25 cm. If the vessel is to be painted at the rate of Rs. 3.5 per cm2, then find the total cost to paint the vessel from outside.


Answer:

Given.


A vessel is made by mounting hemisphere on cylinder


The radius of the hemisphere is  cm = 10.5 cm.


The total height of the cylinder is 25 cm


Rates of painting = 3.5 per cm2


Formula used/Theory.


Curve surface area of Cylinder = 2πrh


Curve surface area of Hemisphere = 2πr2


⇒ As we put hemisphere on hollow Cylinder


∴ Area of Vessel is sum of CSA of both hemisphere and cylinder


Area of vessel = CSA of cylinder + CSA of Hemisphere


= 2πrh + 2πr2


= 2πr[h + r]


As height of hemisphere is equal to radius of hemisphere


Then;


Area of vessel = 2 ×  × 10.5 × [25 + 10.5]


= 2 ×  × 10.5 × [35.5]


= 2343 cm2


Rates for 1cm2 = Rs.3.5


Rates of 1648.5 cm2 = Rs.3.5 × 2343


= Rs. 8200.5



Question 4.

Chirag made a bird - bath for his garden in the shape of a cylinder with a hemispherical depression at one end, (see the figure 14.11). The height of the cylinder is 1.5 m and its radius is 50 cm. Find the total area of the bird - bath. (π = 3.14)



Answer:

Given.


A vessel is made by depressing hemisphere on cylinder


The radius of the hemisphere is 50 cm


Height of the cylinder is 150 cm


Formula used/Theory.


Curve surface area of Cylinder = 2πrh


Curve surface area of Hemisphere = 2πr2


⇒ As we put hemisphere in Cylinder


∴ Area of bird - bath is sum of CSA of both hemisphere and cylinder


Area of Bird - bath = CSA of cylinder + CSA of Hemisphere


= 2πrh + 2πr2


= 2πr[h + r]


TSA of toy = 2 × 3.14 × 50 × [150 + 50]


= 2 × 3.14 × 50 × [200]


= 62800 cm2



Question 5.

A solid is composed of a cylinder with hemispherical ends on both the sides. The radius and the height of the cylinder are 20 cm and 35 cm respectively. Find the total surface area of the solid.


Answer:

Given.


A Solid is made by mounting 2 hemisphere on cylinder


The radius of the hemisphere is 20 cm.


The total height of the cylinder is 35 cm


Formula used/Theory.


Curve surface area of Cylinder = 2πrh


Curve surface area of Hemisphere = 2πr2


⇒ As we put 2 hemisphere on both sides of Cylinder


∴ Area of Solid is sum of CSA of both hemisphere and of cylinder


Area of Solid = CSA of cylinder + 2 × CSA of Hemisphere


= 2πrh + 2 × (2πr2)


= 2πr[h + 2r]


Area of solid 





Question 6.

The radius of a conical tent is 4 m and slant height is 5 m. How many meters of canvas of width 125 cm will be used to prepare 12 tents? If the cost of canvas is Rs. 20 per meter, then what is total cost of 12 tents (π = 3.14)


Answer:

Given.


Radius of conical tent is 4 m


Slant height of conical tent is 5m


Formula used/Theory.


CSA of cone = πrl


⇒ As the conical tent is not on ground


∴ only CSA is required to construct the tent


CSA of conical tents = 3.14 × 5 × 4


= 62.8 m2


Let length of cloth be x cm


∴ Area of Cloth = 1.25 × x m2


If 12 tents are to be constructed


Then area of cloth required to make 12 conical tents is


12 × 62.8 m2


= 753.6 m2


1.25 m × x m = 753.6 m2


x =  = 602.88 m


If 1m length of canvas cost Rs. 20


Then 602.88 m length of canvas cost Rs. 20 × 602.88


= Rs. 12056.7



Question 7.

If the radius of a cone is 60 cm and its curved surface area is 23.55 m2, then find its slant height. (π = 3.14)


Answer:

Given.


The radius of a cone is 60 cm


Curved surface area is 23.55 m2


Formula used/Theory.


CSA of cone = πrl


Radius is in cm but area is given in m2


Then convert radius into m


Radius = m = 0.6m


23.55 m2 = 3.14 × 0.6 × l


L × 1.884 m = 23.55 m2


L =  = 12.5 m


∴ The slant height of cone is 12.5 m



Question 8.

The cost of painting the surface of sphere is Rs. 1526 at the rate of Rs. 6 per m2. Find the radius of sphere.


Answer:

Given.


Cost of painting the surface of sphere is Rs. 1526


Rate of Rs. 6 per m2


Formula used/Theory.


TSA of sphere = 4πr2


Let radius of sphere is r


Then area of sphere is 4πr2 m2


Rate of painting = Rs. 6 per m2


Cost of painting of sphere =


4πr2 m2 × Rs. 6 per m2


Rs. 24πr2


Rs. 24πr2 = Rs. 1526


24 ×  × r2 = 1526


r2 =  = 20.23


r = √20.23 = 4.49 m


= 4.5 m




Exercise 14.2
Question 1.

The curved surface area of a cone is 550 cm2. If its diameter is 14 cm, find its volume.


Answer:

Given.


CSA of cone = 550 cm2


Diameter is 14 cm


Formula used/Theory.


CSA of cone = πr]


Volume of cone = πr2h


If diameter is 14 cm


Then radius is half of diameter


Radius =  = 7 cm


CSA of cone = πr[]


 × 7 × []


= 22 × 


 = 


Squaring both sides


h2 + 49 = 252


h2 = 625 – 49


h2 = 576


h = √576 = 24 cm


Volume of cone = πr2h


 × 7 × 7 × 24


= 8 × 7 × 22


= 1232 cm2



Question 2.

A solid is in the form of cone with hemispherical base. The radius of the cone is 15 cm and the total height of the solid is 55 cm. Find the volume of the solid. (π = 3.14)


Answer:

Given.


A solid is made by mounting a cone onto a hemisphere


The radius of the cone and a hemisphere is 15 cm.


The total height of the solid is 55 cm


Formula used/Theory.


Volume of Cone = πr2h


Volume of hemisphere = πr3


⇒ as we put Cone on hemisphere


The circle part of both cone and hemisphere will attach


∴ Volume of Solid is sum of volume of both cone and hemisphere


Volume of solid = Volume of cone + Volume of Hemisphere


πr2h + πr3


πr2[h + 2r]


As height of hemisphere is equal to radius of hemisphere


Then;


Height of cone = Height of Solid – Radius


= 55cm – 15cm


= 40 cm


Volume of solid =  × 3.14 × 15 × 15 × [40 + 2 × 15]


 × 15 × 15 × [40 + 30]


 × 5 × 15 × 70


= 22 × 5 × 15 × 10


= 16500 cm3



Question 3.

How many litres of milk can be stored in a cylindrical tank with radius 1.4 m and height 3 m?


Answer:

Given.


Radius of cylindrical tank = 1.4 m


Height of cylindrical tank = 3 m


Formula used/Theory.


Volume of Cylinder = πr2h


1m3 = 1000 Litres


Volume of Cylinder = πr2h


 × 1.4 × 1.4 × 3


= 22 × 0.2 × 1.4 × 3


= 18.48 m3


1m3 = 1000 Litres


18.48m3 = 18.48 × 1000 Litres


= 18480 Litres



Question 4.

The spherical balloon with radius 21 cm is filled with air. Find the volume of air contained in it.


Answer:

Given.


Radius of spherical balloon = 21 cm


Formula used/Theory.


Volume of Sphere = πr3


1cm3 =  Litres


Volume of Balloon = πr3


 × 21 × 21 × 21


= 22 × 4 × 21 × 21


= 38808 cm3


1cm3 =  Litres


38808 cm3 = 38808 ×  Litres


= 38.808 Litres



Question 5.

A solid has hemi - spherical base with diameter 8.5 cm and it is surmounted by a cylinder with height 8 cm and diameter of cylinder is 2 cm. Find the volume of this solid. (π = 3.14)


Answer:

Given


Diameter of hemisphere = 8.5


Diameter of cylinder = 2cm


Height of cylinder = 8cm


Formula used


Volume of Cylinder = πr2h


Volume of hemisphere = πr3


Radius of cylinder = 


 = 1 cm


⇒ Volume of Cylinder = πr2h


 × (1)2 × 8



= 25.14 cm3


Radius of hemisphere = 


 = 4.25 cm


⇒ Volume of hemisphere = πr3


 × (4.25)3


= 160.84 cm3


Volume of solid = Volume of cylinder + volume of hemisphere


= 25.14 cm3 + 160.84 cm3


= 185.98 cm3



Question 6.

A playing top is made up of steel. The top is shaped like a cone surmounted by a hemisphere. The total height of top is 5 cm and the diameter of the top is 3.5 cm. Find the volume of the top.


Answer:

Given.


Diameter of cone and hemisphere = 3.5cm


Total height of top = 5cm


Formula used/Theory.


Volume of Cone = πr2h


Volume of hemisphere = πr3


⇒ As we put Cone on hemisphere


The circle part of both cone and hemisphere will attach


∴ Volume of top is sum of volume of both cone and hemisphere


Volume of top = Volume of cone + Volume of Hemisphere


πr2h + πr3


πr2[h + 2r]


Radius of hemisphere = 


 = 1.75


As height of hemisphere is equal to radius of hemisphere


Then;


Height of cone = Height of top – Radius


= 5cm – 1.75cm


= 3.25 cm


Volume of top =  × 1.75 × 1.75 × [3.25 + 2 × 1.75]


 × 1.75 × 1.75 × [3.25 + 3.5]


 × 1.75 × 1.75 × [6.75]


= 22 × 0.25 × 1.75 × 2.25


= 21.65 cm3



Question 7.

How many litres of petrol will be contained in a closed cylindrical tank with hemisphere at one end having radius 4.2 cm and total height 27.5 cm?


Answer:

Formula used/Theory.


Volume of Cylinder = πr2h


Volume of hemisphere = πr3


1 cm3 =  litres


⇒ As we put Cylinder on hemisphere


The circle part of both cylinder and hemisphere will attach


∴ Volume of container is sum of volume of both cylinder and hemisphere


Volume of container = Volume of cylinder + Volume of Hemisphere


= πr2h + πr3


= πr2[h + r]


As height of hemisphere is equal to radius of hemisphere


Then;


Height of cylinder = Height of Container – Radius


= 27.5cm – 4.2cm


= 23.3 cm


Volume of container =  × 4.2 × 4.2 × [23.3 +  × 4.2]


= 22 × 0.6 × 4.2 × [23.3 + 2.8]


= 22 × 0.6 × 4.2 × [26.1]


= 1446.98 cm3


For Volume in litres


1 cm3 =  litres


1446.98 cm3 = 1446.98 ×  litres


1.44698 litres


1.45 litres (approx.)



Question 8.

The capacity of a cylindrical tank at a petrol pump is 57750 litres. If its diameter is 3.5 m, find the height of cylinder.


Answer:

Given.


Capacity of petrol tank = 57750 litres


Diameter of tank = 3.5 m


Formula used/Theory.


Volume of cylinder = πr2h


1m3 = 1000 litres


Radius of tank = 


 = 1.75


Let’s take height of tank to be h


Volume of cylinder = πr2h


 × 1.75 × 1.75 × h


9.625 × h m3


Volume in litres = 57750 litres


1 m3 = 1000 litres


⇒ Volume = 9.625 × h × (1 m3)


= 9.625 × h × 1000 litres


57750 = 9625 × h


h =  = 6 m



Question 9.

A hemispherical pond is filled with 523.908 m3 of water. Find the maximum depth of pond.


Answer:

Given.


Volume of hemisphere = 523.908 m3


Formula used/Theory.


Volume of hemisphere = πr3


As maximum depth is height of pound


But in case of hemisphere radius is height of pound


πr3 =  × r3 = volume


Volume = 523.908 m3


Equating both


 × r3 = 523.908 m3


 × r3 = 523.908 m3


r3 = 


r3 = 250.047 m3


r = ∛(250.047 m3)


r = 6.3 m



Question 10.

A gulab - jamun contain 40% sugar syrup in it. Find how much syrup would be there in 50 gulab - jamuns, each shaped like a cylinder with two hemispherical ends with total length 5 cm and diameter 2.8 cm.


Answer:

Given.


40% of sugar syrup in 1 gulab - jamun


Having length is 5 cm


And diameter is 2.8 cm


Formula used/Theory.


Volume of cylinder = πr2h


Volume of hemisphere = πr3


If gulab - jamun shape like cylinder between 2 hemisphere


Volume of gulab - jamun = volume of cylinder


+ 2 × Volume of hemisphere


Diameter of hemisphere = 2.8 cm


Radius of hemisphere =  = 1.4 cm


Height of cylinder = length of gulab - jamun–2 × radius


= 5 cm–2 × 1.4


= 5 cm – 2.8 cm


= 2.2 cm


Volume of gulab - jamun = volume of cylinder


+ 2 × Volume of hemisphere


= πr2h + 2 × πr3


= πr2[h + r]


 × 1.4 × 1.4 × [2.2 +  × 1.4]


= 22 × 0.2 × 1.4 × [2.2 + 1.86]


= 22 × 0.2 × 1.4 × [4.06]


= 25.0096 cm3


Volume of sugar syrup =  × 25.0096 cm3


Volume of sugar syrup = 10.0038 cm3


Volume of sugar syrup in 50 gulab - jamun = 50 × 10.0038 cm3


= 500.192 cm3


1 cm3 =  litres


500.192 cm3 =  litres


= 0.500192 litres


= 0.5 litres (approx.)



Question 11.

The height and the slant height of a cone are 12 cm and 20 cm respectively. Find its volume. (π = 3.14)


Answer:

Given.


Slant height of cone = 20 cm


Height of cone = 12 cm


Formula used/Theory.


Volume of cone = πr2h


In cone,


The Radius, height and slant height makes a right angled triangle


With hypotenuse as slant height of triangle


∴ By Pythagoras Theorem


Radius2 + Height2 = (slant height) 2


Radius2 + (12 cm) 2 = (20 cm) 2


Radius2 + 144 cm2 = 400 cm2


Radius2 = 400 cm2 – 144 cm2


Radius = √(256 cm2) = 16 cm


Volume of Cone = πr2h


 × 3.14 × 16cm × 16cm × 12cm


= 3.14 × 16cm × 16cm × 4cm


= 3215.36 cm3



Question 12.

Find the total volume of a cone having a hemispherical base. If the radius of the base is 21 cm and height 60 cm.


Answer:

Given.


Radius of cone and hemisphere = 21 cm


Total height of cone = 60 cm


Formula used/Theory.


Volume of Cone = πr2h


Volume of hemisphere = πr3


⇒ As we put Cone on hemisphere


The circle part of both cone and hemisphere will attach


∴ Volume of solid is sum of volume of both cone and hemisphere


Volume of solid = Volume of cone + Volume of Hemisphere


πr2h + πr3


πr2[h + 2r]


As height of hemisphere is equal to radius of hemisphere


Then;


Volume of top =  × 21 × 21 × [60 + 2 × 21]


 × 21 × 21 × [60 + 42]


 × 21 × 21 × [102]


= 22 × 21 × 102


= 47124 cm3



Question 13.

If the slant height of a cone is 18.7 cm and the curved surface area is 602.8 cm2, find the volume of cone. (π = 3.14)


Answer:

Given.


Slant height of cone = 18.7 cm


CSA of cone = 602.8 cm2


Formula used/Theory.


Volume of cone = πr2h


CSA of cone = πrl


In cone,


CSA of cone = 602.8 cm2


πrl = 602.8 cm2


3.14 × r × 18.7cm = 602.8 cm2


r =  = 10.26 cm


In cone,


The Radius, height and slant height makes a right angled triangle


With hypotenuse as slant height of triangle


∴ By Pythagoras Theorem


Radius2 + Height2 = (slant height) 2


Height2 + (10.26 cm) 2 = (18.7 cm) 2


Height2 + 105.26 cm2 = 349.69 cm2


Height2 = 349.69 cm2 – 105.26 cm2


Height = √(244.43 cm2) = 15.63 cm


Volume of Cone = πr2h


 × 3.14 × 10.26cm × 10.26cm × 15.63cm


= 3.14 × 10.26cm × 10.26cm × 5.21cm


= 1722.11 cm3



Question 14.

If the surface area of a spherical ball is 1256 cm2, then find the volume of sphere. (Take π = 3.14)


Answer:

Given.


The surface area of a spherical ball is 1256 cm2


Formula used/Theory.


Area of sphere = 4πr2


Volume of sphere = πr3


Volume of sphere = πr3


 × (4πr2) × r


Area of sphere = 4πr2


= 1256 cm2


Radius of sphere is


r2 =  = 100 cm2


r = √100 cm2


r = 10 cm


∴ Volume of sphere =  × 4πr2 × r


 × 1256 cm2 × 10 cm


= 418.66 cm2 × 10 cm


= 4186.67 cm3




Exercise 14.3
Question 1.

A hemispherical bowl of internal radius 12 cm contains some liquid. This liquid is to be filled into cylindrical bottles of diameter 4 cm and height 6 cm. How many bottles can be filled with this liquid?


Answer:

Given.


Hemispherical bowl of internal radius 12 cm


Cylindrical bottles of diameter 4 cm and height 6 cm


Formula used/Theory.


Volume of Cylinder = πr2h


Volume of hemisphere = πr3


Volume of water will be equal to volume of hemisphere


Volume of hemisphere = πr3


 × π × (12 cm) 3


= 1152 × π cm3


Radius of bottle =  = 2 cm


Volume of each bottle = πr2h


= π × (2cm) 2 × 6 cm


= 24π cm3


**Note we will not put value of π as it will be divided in next step


Number of bottles filled will be dividing the total volume of water by volume of water contained by 1 bottle


Number of bottles =  = 48



Question 2.

A cylindrical container having diameter 16 cm and height 40 cm is full of ice - cream. The ice - cream is to be filled into cones of height 12 cm and diameter 4 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with the ice - cream.


Answer:

Given.


Radius of Hemispherical bowl and cone is  = 2 cm


Cylindrical container of diameter 16 cm and height 40 cm


Formula used/Theory.


Volume of Cylinder = πr2h


Volume of hemisphere = πr3


Volume of Cone = πr2h


Volume of ice - cream will be equal to sum of volume of hemisphere bowl and volume of cone


Volume of hemisphere = πr3


 × π × (2 cm) 3


π cm3


Volume of Cone = πr2h


 × π × (2 cm) 2 × 12 cm


16π cm3


Volume of ice - cream = 16π cm3 + π cm3


 × 16π cm3


Radius of container =  = 8 cm


Volume of container = πr2h


= π × (8cm) 2 × 40 cm


= 2560π cm3


**Note we will not put value of π as it will be divided in next step


Number of ice - cream will be dividing the total volume of ice - cream in container by volume of ice - cream contained by 1 cone


Number of ice - creams = 


= 120


∴ 120 ice - cream can be filled by the container



Question 3.

A cylindrical tank of diameter 3 m and height 7 m is completely filled with groundnut oil. It is to be emptied in 15 tins each of capacity 15 litres. Find the number of such tins required.


Answer:

Given.


Cylindrical tank of diameter 3 m and height 7 m


Tins of capacity 15 litres


Formula used/Theory.


Volume of cylinder = πr2h


1 m3 = 1000 litres


Volume of cylinder tank = πr2h


Radius of tank =  = 1.5 m


⇒ Volume of cylinder =  × (1.5 m) 2 × 7 m


= 22 × 2.25 m3


= 49.5 m3


If 1 m3 = 1000 litres


Then 49.5 m3 = 49.5 × 1000 litres


= 49500 litres


Volume of 1 tin = 15 litres


Number of tins is dividing volume of cylindrical tank by volume of tins


⇒ Number of tins =  = 3300 tins


∴ 3300 tins required to empty the tank



Question 4.

A cylinder of radius 2 cm and height 10 cm is melted into small spherical balls of diameter 1 cm. Find the number of such balls.


Answer:

Given.


Cylinder of radius 2 cm and height 10 cm


spherical balls of diameter 1 cm


Formula used/Theory.


Volume of cylinder = πr2h


Volume of sphere = πr3


Volume of cylinder = πr2h


= π × (2 cm) 2 × 10 cm


= 40π cm3


Radius of sphere =  = 0.5 cm


Volume of sphere = πr3


π (0.5 cm) 3


π cm3


**Note we will not put value of π as it will be divided in next step


Number of spherical balls will be dividing the total volume of cylinder by volume of 1 spherical ball


Number of spherical balls =  = 240


∴ 240 Spherical balls can be made by melting cylinder



Question 5.

A metallic sphere of radius 15 cm is melted and a wire of diameter 1 cm is drawn from it. Find the length of the wire.


Answer:

Given.


Cylindrical wire of diameter 1 cm


Metallic spherical of radius 15 cm


Formula used/Theory.


Volume of cylinder = πr2h


Volume of sphere = πr3


Let the length of wire be x


Volume of cylinder = πr2h


= π × ( cm) 2 × x


π cm2


Volume of sphere = πr3


π (15 cm) 3


= 4500π cm3


**Note we will not put value of π as it will be divided in next step


As we melt the metallic sphere in cylindrical wire the volume of both will be equal


∴ equating both we will get the length of wire


Volume of wire = Volume of metallic sphere


π cm2 = 4500π cm3


x = 4500 × 4 


x = 18000 cm = 180 m


∴ length of wire is 180 m



Question 6.

There are 45 conical heaps of wheat, each of them having diameter 80 cm and height 30 cm. To store the wheat in a cylindrical container of the same radius, what will be the height of cylinder?


Answer:

Given.


Height of cone is 30 cm


Cone and cylinder of diameter 80 cm


Formula used/Theory.


Volume of cylinder = πr2h


Volume of cone = πr2h


Let the Height of container be x


Radius of cylinder and cone is  = 40cm


Volume of container = πr2h


= π × (40 cm)2 × x


= 1600πx cm2


Volume of conical heaps = πr2h


π(40 cm)2 × 30 cm


= 16000π cm3


Volume of 45 conical heaps = 45 × 16000π cm3


= 720000π cm3


**Note we will not put value of π as it will be divided in next step


As we put 45 conical heaps of wheat in cylindrical container volume of 45 conical heaps will be equal to volume of container


∴ equating both we will get the Height of container


Volume of container = Volume of 45 conical heaps


1600π × x cm2 = 720000π cm3


x = 


x = 450 cm


∴ Height of container is 450 cm



Question 7.

A cylindrical bucket, 44 cm high and having radius of base 21 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 33 cm, find the radius and the slant height of the heap.


Answer:

Given.


Height of bucket is 44 cm


Radius of bucket is 21 cm


Height of conical heap is 33 cm


Formula used/Theory.


Volume of cylinder = πr2h


Volume of cone = πr2h


Let the Radius of conical heap be x


Volume of bucket = πr2h


= π × (21 cm)2 × 44 cm


= 19404π cm3


Volume of conical heaps = πr2h


π × x2 × 33 cm


= 11π × x2 cm


**Note we will not put value of π as it will be divided in next step


As we put bucket of sand on ground it will form a conical heap volume of conical heaps will be equal to volume of bucket


∴ equating both we will get the Radius of conical heap


Volume of bucket = Volume of conical heaps


11π × x2 cm = 19404π cm3


x2 = 


x2 = 1764 cm2


x = √ (1764 cm2)


x = 42 cm


∴ Radius of conical heap is 42 cm


In cone;


As the radius , height and slant height makes Right angled triangle where hypotenuse is slant height


Then by Pythagoras theorem


(Slant height)2 = (height)2 + (radius)2


(Slant height)2 = (33 cm)2 + (42 cm)2


(Slant height)2 = 1089 cm2 + 1764 cm2


(Slant height)2 = 2853 cm2


Slant height = √(2853 cm2)


= 53.41 cm




Exercise 14.4
Question 1.

A metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The total vertical height of the bucket is 40 cm and that of cylindrical base is 10 cm, radii of two circular ends are 60 cm and 20 cm. Find the area of the metallic sheet used. Also find the volume of water the bucket can hold. (π = 3.14)


Answer:

Given.


Height of bucket is 40 cm


Height of cylinder is 10 cm


Radii of both circular ends of frustum are 60 cm, 20 cm


Formula used/Theory.


CSA of cylinder = 2πrh


CSA of frustum = π(r + R) × l


Volume of cylinder = πr2h


Volume of frustum = h[R2 + r2 + Rr]


As the bucket is always open from mouth


Metallic sheet require will be sum of CSA of Frustum and CSA of cylinder and Area of base circle


In frustum


L2 = height2 + 


Height of frustum = height of bucket – height of cylinder


= 40 – 10 = 30 cm


L2 = 302 + [60 – 20]2


L2 = 900 cm2 + 1600 cm2 = 2500 cm2


L = √(2500 cm2) = 50cm


⇒ CSA of Frustum = π(60 + 20)50 cm


= 3.14 × 80cm × 50cm


= 12560cm2


⇒ CSA of cylinder = 2πrh


= 2 × 3.14 × 20cm × 10cm


= 1256 cm2


⇒ Area of base = πr2


= 3.14 × 20 × 20


= 1256 cm2


Area of metallic sheet = 12560 cm2 + 1256 cm2 + 1256 cm2


= 15072 cm2


⇒ Volume of Frustum = h[R2 + r2 + Rr]


 × 30cm × [(60 cm)2 + (20 cm)2 + 60 cm × 20 cm]


= 31.4 cm × [3600cm2 + 400cm2 + 1200cm2]


= 31.4 cm × 5200 cm2


= 163280 cm3


1 cm3 = litres


163280 cm3 = litres


= 163.280 litres


∴ Area of metallic sheet used in making bucket is 15072 cm2


Volume of bucket is 163.280 litres



Question 2.

A container, open from the top and made up of a metal sheet is the form of frustum of a cone of height 30 cm with radii 30 cm and 10 cm. Find the cost of the milk which can completely fill container at the rate of Rs. 30 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs. 50 per 100 cm2. (π = 3.14)


Answer:

Given.


Radii of frustum is 10 cm and 30 cm


Height of frustum is 30 cm


Rate of milk is Rs. 30 per litre


Rate metallic sheet is Rs . 50 per 100 cm2


Formula used/Theory.


CSA of frustum = π(r + R) × l


Volume of frustum = h[R2 + r2 + Rr]


⇒ Volume of Frustum = h[R2 + r2 + Rr]


 × 30cm × [(30 cm)2 + (10 cm)2 + 30 cm × 10 cm]


= 31.4 cm × [900cm2 + 100cm2 + 300cm2]


= 31.4 cm × 1300 cm2


= 40820 cm3


1 cm3 = litres


40820 cm3 = litres


= 40.820 litres


As milk is Rs. 30 per litre


For 40.82 litres


Rs. 30 × 40.82


= Rs. 1224.6


In frustum


L2 = height2 + 


L2 = 302 + [30 – 10]2


L2 = 900 cm2 + 400 cm2 = 1300 cm2


L = √(1300 cm2) = 10√13 cm


⇒ CSA of Frustum = π(30 + 10)10√13 cm


= 3.14 × 40cm × 10√13cm


= 1256√13cm2


⇒ Area of base = πr2


= 3.14 × 10 × 10


= 314 cm2


Area of container = 1256√13 + 314


= 314[4√13 + 1]


= 314[4 × 3.6 + 1]


= 314 × 15.4


= 4835.6 cm2


Cost of 100 cm2 metal sheet = Rs. 50


Cost of 1 cm2 metal sheet = Rs. 0.5


Cost of 4835.6 cm2 metal sheet = Rs. 4835.6 × 0.5


= Rs. 2417.8




Exercise 14
Question 1.

A tent is in the shape of cylinder surmounted by a conical top. If the height and the radius of the cylindrical part are 3.5 m and 2 m respectively and the slant height of the top is 3.5 m, find the area of the canvas used for making the tent. Also find the cost of canvas of the tent at the rate of Rs. 1000 per m2.


Answer: 

Given.


Radius of cylinder and cone is 2 m


Height of cylinder is 3.5 m


Slant height of cone is 3.5 m


Formula used/Theory.


CSA of cylinder = 2πrh


CSA of cone = πrl


CSA of cylinder = 2πrh


= 2 ×  × 3.5 m × 2 m


= 44 m2


CSA of cone = πrl


 × 2 m × 3.5 m


= 22 m2


Total Area of cloth = 44 m2 + 22 m2


= 66 m2


Rate of cloth for 1 m2 = Rs. 1000


Rate of cloth for 66 m2 = Rs. 1000 × 66


= Rs. 66000


Cost of canvas is Rs. 66000



Question 2.

A metallic sphere of radius 5.6 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.


Answer:

Given.


Radius of sphere is 5.6 cm


Radius of cylinder is 6 cm


Formula used/Theory.


Volume of cylinder = πr2h


Volume of sphere = πr3


Volume of sphere = πr3


 × π × (5.6 cm)3


Volume of cylinder = πr2h


= π × (6 cm)2 × h


When a metallic sphere is recast in shape of cylinder


Volume of both sphere and cylinder will be equal


∴ comparing both we get height of cylinder


π × (6 cm)2 × h =  × π × (5.6 cm)3


h = 


h = 6.5 cm


∴ Height of cylinder is 6.5 cm



Question 3.

How many spherical balls of radius 2 cm can be made out of a solid cube of lead whose side measures 44 cm?


Answer: 

Given.


Radius of sphere is 2 cm


Side of cube is 44 cm


Formula used/Theory.


Volume of sphere = πr3


Volume of cube = side3


Volume of sphere = πr3


 × (2 cm)3


Volume of cube = side3


= (44 cm)3


Number of spherical balls will be get by dividing Volume of cube by Volume of each Sphere


Number of spheres = 




= 2541



Question 4.

A hemispherical bowl of internal radius 18 cm contains an edible oil to be filled in cylindrical bottles of radius 3 cm and height 9 cm. How many bottles are required to empty the bowl?


Answer:

Given.


Radius of hemisphere is 18 cm


Radius of cylinder is 3 cm


Height of cylinder is 9 cm


Formula used/Theory.


Volume of cylinder = πr2h


Volume of hemisphere = πr3


Volume of cylindrical bottle = πr2h


= π × (3cm)2 × 9 cm


= 81π cm3


Volume of hemispherical bowl = πr3


 × π × (18 cm)3


= 3888 π cm3


Number of cylindrical bottle will be get by dividing Volume of hemispherical bowl by Volume of each Cylindrical bottle


Number of cylindrical bottle = 



= 48 bottles



Question 5.

A hemispherical tank of radius 2.4 m is full of water. It is connected with a pipe which empties it at the rate of 7 litres per second. How much time will it take to empty the tank completely?


Answer:

Given.


A hemispherical tank of radius 2.4 m is full of water


Rate of water flow is 7 litres per second


Formula used/Theory.


Volume of hemisphere = πr3


Volume of hemispherical bowl = πr3


 × (2.4 m)3


= 28.96m3


1m3 = 1000 litres


28.96m3 = 28.964 × 1000 litres


= 28964 litres


Rate of flow is 7 litres in 1 second


Rate of flow is 1 litres in  second


Rate of flow is 28964 litres in  second


Is 4137.8 seconds


In minutes =  = 68.96 minutes



Question 6.

A shuttle cock used for playing badminton has the shape of a frustum of a cone mounted on a hemisphere. The external diameter of the frustum are 5 cm and 2 cm. The height of the entire shuttle cock is 7 cm. Find its external surface area.


Answer:

Given.


The external diameter of the frustum are 5 cm and 2 cm. The height of the entire shuttle cock is 7 cm


Formula used/Theory.


CSA of frustum = π(R + r) × L


CSA of hemisphere = 2πr2


Radii of frustum is and  = 2.5cm and 1cm


As in hemisphere radius and height are same


Height of frustum = height of cock – Radius


= 7 cm – 1 cm


= 6 cm


In frustum


L2 = height2 + 


L2 = (6 cm)2 + [2.5 cm – 1 cm]2


L2 = 36 cm2 + 2.25 cm2 = 38.25 cm2


L = √(38.25 cm2) = 6.18 cm


CSA of hemisphere = 2 × 3.14 × (1 cm)2


= 6.28 cm2


CSA of frustum = 3.14 × [2.5cm + 1cm] × 6.18cm


= 3.14 × 3.5cm × 6.18cm


= 67.92 cm2


External surface area = CSA of hemisphere + CSA of frustum


= 6.28 cm2 + 67.92 cm2


= 74.2 cm2



Question 7.

A fez, the headgear cap used by the trucks is shaped like the frustum of a cone. If its radius on the open side is 12 cm and radius at the upper base is 5 cm and its slant height is 15 cm, find the area of material used for making it. (π = 3.14)


Answer:

Given.


Radii of frustum are 12 cm and 5 cm


Slant height is 15 cm


Formula used/Theory.


CSA of frustum is π[R + r] × l


Area of material = CSA of frustum


= π[R + r] × l


= 3.14[12cm + 5cm] × 15cm


= 3.14 × 17cm × 15cm


= 800.7 cm2



Question 8.

A bucket is in the form of a frustum of a cone with capacity of 12308.8 cm3 of water. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of bucket and the cost of making it at the rate of Rs. 10 per cm2.


Answer: 

Given.


Volume of frustum = 12308.8 cm3


Radii of frustum are 20cm and 12cm


Formula used/Theory.


CSA of frustum = π[R + r]l


Volume of frustum = h[R2 + r2 + Rr]


Volume of frustum = h[R2 + r2 + Rr]


 × h[(20cm)2 + (12cm)2 + 20cm × 12cm]


 × h[400cm2 + 144cm2 + 240cm2]


 × h × 784cm2


 × h × 22 × 112


 × 22 × 112 × h = 12308.8 cm3


h =  = 15 cm


In frustum


(Slant height)2 = (difference in radii )2 + (height)2


(Slant height)2 = (R - r)2 + (height)2


(Slant height)2 = (20cm – 12cm)2 + (15 cm)2


(Slant height)2 = 64 cm2 + 225 cm2


(Slant height)2 = 289 cm2


Slant height = √ (289 cm2)


Slant height = 17 cm


CSA of cone = π[R + r]l


[12cm + 20cm] × 17cm


 × 32cm × 17cm


= 1709.71 cm2


Rate of 1 cm2 = Rs. 10


Rate of 1709.71 cm2 = Rs. 10 × 1709.71


= Rs. 17097.1



Question 9.

The volume of sphere with diameter 1 cm is …….. cm3.
A. π

B. π

C. π 

D. π


Answer:

Volume of sphere = πr3


When diameter of sphere is 1cm


Radius of sphere is  cm


π( cm)3


π cm3


π cm3


Question 10.

The volume of hemisphere with radius 1.2 cm is ……….. cm3.
A. 1.152π

B. 0.96π

C. 2.152π

D. 3.456π


Answer:

Volume of hemisphere = πr3


When radius is 1.2 cm


π(1.2 cm)3


= 1.152π cm3


Question 11.

The volume of sphere is in 4/3 π cm3. Then its diameter is ………. cm.
A. 0.5

B. 1

C. 2

D. 2.5


Answer:

Volume of sphere = πr3


π cm3


π × r3cm3 = π cm3


(r cm)3 = 1 cm3


r = ∛1 cm3


r = 1


if radius = 1 cm then diameter = 2 × radius


= 2 cm


Question 12.

The volume of cone with radius 2 cm and height 6 cm is ……….. cm3.
A. 8π

B. 12π

C. 14π

D. 16π


Answer:

Volume of cone = πr2h


π × (2cm)2 × 6 cm


= 8 π


Question 13.

The diameter of the base of cone is 10 cm and its slant height is 17 cm. Then the curved surface area of the cone is ………. cm2.
A. 85π

B. 170π

C. 95π

D. 88π


Answer:

CSA of cone = πrl


= π × cm × 17cm


= π × 5cm × 17cm


= 85 π cm


Question 14.

The diameter and the height of the cylinder are 14 cm and 10 cm respectively. Then the total surface area is ……….. cm2.
A. 44

B. 308

C. 748

D. 1010


Answer:

Radius of cylinder =  =  = 7cm


TSA of cylinder = 2πr[r + h]


= 2 ×  × 7cm × [7cm + 10cm]


= 44cm × 17cm


= 748 cm2


Question 15.

The ratio of the radii of two cones having equal height is 2 : 3. Then, the ratio of their volumes is ……..
A. 4 : 6

B. 8 : 27

C. 3 : 2

D. 4 : 9


Answer:

If ratio of radii are 2:3


Then;


Let the radii of both cones be 2x and 3x


Volume of cone = πr2h


Volume of 1st cone = π(2r)2h


πr2h


Volume of 2nd cone = π(3r)2h


= 3πr2h


The ratio of their volumes is


 πr2h: 3πr2h


4:9


Question 16.

If the radii of a frustum of a cone are 7 cm and 3 cm and the height is 3 cm, then the curved surface area is …………cm2.
A. 50π

B. 25π

C. 35π

D. 63π


Answer:

In frustum


(Slant height)2 = (difference in radii )2 + (height)2


(Slant height)2 = (R - r)2 + (height)2


(Slant height)2 = (7cm – 3cm)2 + (3 cm)2


(Slant height)2 = 16 cm2 + 9 cm2


(Slant height)2 = 25 cm2


Slant height = √ (25 cm2)


Slant height = 5 cm


CSA of frustum = π[R + r]l


= π[7cm + 3cm] × 5cm


= 50 π cm2


Question 17.

The radii of a frustum of a cone are 5 cm and 9 cm and height is 6 cm, then the volume is ………….. cm3.
A. 320π

B. 151π

C. 302π

D. 98π


Answer:

Volume of frustum = h[R2 + r2 + Rr]


 × 6 cm[(9cm)2 + (5cm)2 + 9cm × 5cm]


= 2π [81cm2 + 25cm2 + 45cm2]


= 2π × 151cm2


= 302 cm2


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