# ICSE Class 10 Maths Mensuration PYQs with Solution

Mathematics is a subject that holds great significance in our lives. It not only develops problem-solving skills but also enhances logical reasoning. ICSE (Indian Certificate of Secondary Education) is a renowned educational board in India that follows a comprehensive curriculum, providing students with a strong foundation in various subjects. One such subject is Mathematics, which is a core component of the ICSE syllabus. In this post, we provide you ICSE Class 10 Maths Mensuration PYQs with Solution.

In the ICSE Class 10 Mathematics syllabus, Mensuration is an important topic that plays a crucial role in understanding the measurement of various geometric shapes and their properties. It deals with the calculation of length, area, and volume of different figures, enabling students to solve real-life problems related to shape and space.

## Importance of Mensuration in Class 10 Maths

Mensuration holds immense importance in the Class 10 Maths curriculum as it lays the foundation for advanced topics in higher classes. It provides students with the necessary tools to calculate the dimensions of objects, understand the relationship between different measurements, and apply these concepts to solve complex problems.

By mastering Mensuration, students can develop spatial visualization skills, which are valuable not only in the field of Mathematics but also in various other disciplines such as architecture, engineering, and design. Moreover, it helps students in everyday life situations like measuring the area of a room, estimating the volume of a container, or understanding the cost of materials required for construction.

## ICSE Class 10 Maths Mensuration PYQs with Solution

Now, let’s delve into some ICSE Class 10 Maths Mensuration PYQs with Solution from the year 2010 to 2023. These questions will help you practice your understanding of Mensuration concepts and strengthen your problem-solving skills.

Q1. A solid is in the shape of a hemisphere of radius 7 cm, surmounted by a cone of height 4 cm. The solid is immersed completely in a cylindrical container filled with water to a certain height. If the radius of the cylinder is 14 cm, find the rise in the water level. [2023]

Solution: 1.5 cm

Step-by-step Explanation:

$Volume\;of\;the\;solid\;=\;Volume\;of\;cone+Volume\;of\;hemisphere\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\;\frac13\mathrm{πr}^2\mathrm h+\frac23\mathrm{πr}^3\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac13\mathrm{πr}^2\;(\mathrm h+2\mathrm r)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\;\frac13\times\frac{22}7\times7^2\;(4+14)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac13\times\frac{22}7\times49\times18\;\\\mathrm{Water}\;\mathrm{will}\;\mathrm{raise}\;\mathrm{in}\;\mathrm{the}\;\mathrm{shape}\;\mathrm{of}\;\mathrm{cylinder}.\\\mathrm{Volume}\;\mathrm{of}\;\mathrm{the}\;\mathrm{water}\;\mathrm{raised}\;=\;\mathrm{volume}\;\mathrm{of}\;\mathrm{the}\;\mathrm{solid}\\\mathrm{πR}^2\mathrm H\;=\;\frac13\times\frac{22}7\times49\times18\;\\\Rightarrow\frac{22}7\times14\times14\times\mathrm H=\frac13\times\frac{22}7\times49\times18\;\\\Rightarrow14\times14\times\mathrm H=49\times6\\\Rightarrow\mathrm H=\frac{49\times6}{14\times14}\\\Rightarrow\mathrm H=\frac32\\\Rightarrow\mathrm H=1.5\;\mathrm{cm}\\\mathrm{Height}\;\mathrm{of}\;\mathrm{water}\;\mathrm{raised}\;\mathrm{is}\;1.5\;\mathrm{cm}.$

You can watch video solution of ICSE Class 10 Maths Mensuration PYQs with Solution HERE.

Q2. Volume of a cylinder of height 3 cm is 48 π. Radius of the cylinder is:
(a) 48 cm

(b) 16 cm
(c) 4 cm
(d) 24 cm [2023]

Solution: (c) 4 cm

Step-by-step solution:

$Volume\;of\;the\;cylinder=\mathrm{πr}^2\mathrm h\\\mathrm{πr}^2\times3=48\mathrm\pi\\\mathrm r^2=\frac{48}3\\\mathrm r^2\;=16\\\mathrm r\;=\sqrt{16}\\\mathrm r=\;4\;\mathrm{cm}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$

Q3. A solid cone of radius 5 cm and height 9 cm is melted and made into small cylinders of radius of 0.5 cm and height 1.5 cm. find the number of cylinders so formed. [2022 Semester-2]

Solution: 200

Step-by-step Explanation:

$\mathrm{Let}\;\mathrm{the}\;\mathrm{number}\;\mathrm{of}\;\mathrm{small}\;\mathrm{cylinders}\;\mathrm{be}\;\mathrm n.\\\mathrm{volume}\;\mathrm{of}\;\mathrm{all}\;\mathrm{the}\;\mathrm{small}\;\mathrm{cylinders}=\mathrm{Volume}\;\mathrm{of}\;\mathrm{cone}\\\mathrm n\times\mathrm{πr}^2\mathrm h\;=\frac13\mathrm{πR}^2\mathrm H\\\mathrm n\times{(0.5)}^2\times1.5\;=\frac13\times5^2\times9\\\mathrm n=\frac{25\times9}{3\times0.25\times1.5}\\\mathrm n=\frac{25\times9\times1000}{3\times25\times15}\\\mathrm n=200\\\mathrm{Total}\;\mathrm{number}\;\mathrm{of}\;\mathrm{small}\;\mathrm{cylinders}\;\mathrm{formed}\;\mathrm{is}\;200.\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$

Q4. The radius of a roller 100 cm long is 14 cm. The curved surface area of the roller is: (Take π = 22/7)

(a) 13200 cm2
(b) 15400 cm2
(c) 4400 cm2
(d) 8800 cm2 [2022 Semester-2]

Solution: (d) 8800 cm2

Step-by-step explanation:

$\mathrm{curved}\;\mathrm{surface}\;\mathrm{area}\;\mathrm{of}\;\mathrm{the}\;\mathrm{roller}=\mathrm{curved}\;\mathrm{surface}\;\mathrm{area}\;\mathrm{of}\;\mathrm{the}\;\mathrm{cylinder}\\=\;2\mathrm{πrh}\\=2\times\frac{22}7\times14\times100\\=8800\;\mathrm{cm}^2\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$

Q5. A solid wooden cylinder is of radius 6cm and height 16cm. Two cones each of radius 2cm and height 6cm are drilled out of the cylinder. Find the volume of the remaining solid. [Take π = 22/7]

[2022 Semester-2]

Solution:
1760 cm3

Step-by-step Explanation:

$Volume\;of\;the\;remaining\;solid\\=\;Volume\;of\;Cylinder-Volume\;of\;2\;cones\\=\;\mathrm{πR}^2\mathrm H\;-\;2(\frac13\mathrm{πr}^2\mathrm h)\\=\mathrm\pi(\mathrm R^2\mathrm H\;-\;\frac23\mathrm r^2\mathrm h)\\=\;\frac{22}7\;(6\times6\times16\;-\frac23\times2\times2\times6)\\=\;\frac{22}7\;(576\;-16)\\=\;\frac{22}7\times560\\=\;22\times80\\=\;1760\;\mathrm{cm}^3$

Q6. The volume of a conical tent is 462 m³ and the area of the base is 154 m². The height of the cone is:
(a) 15 m
(b) 12 m
(c) 9 m
(d) 24 m [2022 Semester-2]

Solution: (c) 9 m

Step-by-step Explanation:

$Area\;of\;the\;base\;of\;the\;cone=\;\mathrm{πr}^2\\\Rightarrow\mathrm{πr}^2\;=154\\\Rightarrow\mathrm r^2\;=\;154\times\frac7{22}\\\Rightarrow\mathrm r\;=\;\sqrt{7\times7}\\\Rightarrow\mathrm r\;=7\\\\\mathrm{Volume}\;\mathrm{of}\;\mathrm{conical}\;\mathrm{tent}\;=\;462\;\mathrm m^3\\\frac13\mathrm{πr}^2\mathrm h\;=\;462\\\Rightarrow\frac13\times\frac{22}7\times7\times7\times\mathrm h\;=\;462\\\Rightarrow\frac{22\times7\times\mathrm h}3\;=\;462\\\Rightarrow\mathrm h\;=\;\frac{462\times3}{22\times7}\\\Rightarrow\mathrm h=\;9\\\mathrm{Hence},\;\mathrm{height}\;\mathrm{of}\;\mathrm{the}\;\mathrm{cone}\;\mathrm{is}\;9\;\mathrm m.$

Q7. From a solid wooden cylinder of height 28 cm and diameter 6 cm, two conical cavities are hollowed out. The diameters of the cones are also of 6 cm and height 10.5 cm. Taking π = 22/7 find the volume of the remaining solid. [2020]

Solution: 594 cm3

Step-by-step Explanation:

$Volume\;of\;the\;remaining\;solid\;\\=\;Volume\;of\;Cylinder\;-\;Volume\;of\;2\;cones\\=\;\mathrm{πr}^2\mathrm H\;-\;2\times\frac13\mathrm{πr}^2\mathrm h\\=\;\mathrm{πr}^2\;\left(\mathrm H\;-\;\frac23\mathrm h\right)\\=\;\frac{22}{7\;}\times3\times3\;\left(28\;-\;\frac23\times10.5\right)\\=\;\frac{198}7\;\left(28\;-\;7\right)\\=\;\frac{198}7\times21\\=594\\\mathrm{Hence},\;\mathrm{volume}\;\mathrm{of}\;\mathrm{the}\;\mathrm{remaining}\;\mathrm{solid}\;\mathrm{is}\;594\;\mathrm{cm}^3.$

Q8. A solid spherical ball of radius 6 cm is melted and recast into 64 identical spherical marbles. Find the radius of each marble. [2020]

Solution: 1.5 cm

Step-by-step Explanation:

$Volume\;of\;64\;identical\;spheres\;=\;Volume\;of\;big\;sphere\\64\times\frac43\mathrm{πr}^3\;=\;\frac43\mathrm{πR}^3\\\Rightarrow64\times\mathrm r^3\;=\;6\times6\times6\\\Rightarrow\mathrm r^3\;=\;\frac{6\times6\times6}{64}\\\Rightarrow\mathrm r\;=\sqrt[3]{\frac{6\times6\times6}{64}}\\\Rightarrow\mathrm r=\;\frac64\\\Rightarrow\mathrm r=1.5\\\mathrm{Hence},\;\mathrm{radius}\;\mathrm{of}\;\mathrm{each}\;\mathrm{identical}\;\mathrm{hemisphere}\;\mathrm{is}\;1.5\;\mathrm{cm}.$

Q9. A solid metallic sphere of radius 6 cm is melted and made into a solid cylinder of height 32 cm. Find the : [4]

(ii) curved surface area of the cylinder [Take π = 3.1] [2019]

Solution: (i) 3 cm (ii) 595.2 cm2

Step-by-step Explanation:

$Volume\;of\;solid\;cylinder\;=\;Volume\;of\;sphere\\\mathrm{πr}^2\;h\;=\;\frac43\mathrm{πR}^3\\\Rightarrow\mathrm r^2\times32\;=\;\frac43\times6\times6\times6\\\mathrm r^2=\;\frac{4\times6\times6\times6}{3\times32}\\\Rightarrow\mathrm r^2=3\times3\\\Rightarrow\mathrm r=\;3\\\\\mathrm{Curved}\;\mathrm{surface}\;\mathrm{area}\;\mathrm{of}\;\mathrm{the}\;\mathrm{cylinder}\\=\;2\mathrm{πrh}\\=\;2\times3.1\times3\times32\\=\;595.2\;\mathrm{cm}^2\\$

Q10. A hemispherical and conical hole is scooped out of a solid wooden cylinder. Find the volume of the remaining solid where the measurements are as follows: [4] [2019]

The height of the solid cylinder is 7 cm, radius of each of hemisphere, cone and cylinder is 3 cm. Height of cone is 3 cm. Give your answer correct to the nearest whole number. [Take π = 22/7]

Solution: 113 cm3

Step-by-step Explanation:

$Volume\;of\;the\;remaining\;solid\;=\\Volume\;of\;cylinder-\left(volume\;of\;hemisphere+volume\;of\;cone\right)\\=\mathrm{πr}^2\mathrm H-\left(\frac23\mathrm{πr}^3+\frac13\mathrm{πr}^2\mathrm h\right)\\=\;\mathrm{πr}^2\;\left(\mathrm H-\frac23\mathrm r-\frac13\mathrm h\right)\\=\frac{22}7\times3\times3\;\left(7-\frac23\times3-\frac13\times3\right)\\=\;\frac{198}7\left(7-2-1\right)\\=\frac{198}7\times4\\=\frac{792}7\\=113.14\\=\;113\;\mathrm{cm}^3\\$

Q11. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. Find the:
(ii) volume of cylinder, (use π = 22/7) [3] [2018]

Solution: (i) 21 cm (ii) 34650 cm3

Step-by-step Explanation:

$(i)\;Circumference\;of\;the\;base\;of\;the\;cylinder\;=\;132\;cm\\\Rightarrow2\mathrm{πr}\;=\;132\\\Rightarrow r=\;\frac{132\times7}{22\times2}\\\Rightarrow r=21\\Hence,\;radius\;of\;the\;cylinder\;is\;21\;cm.\\(ii)\;Volume\;of\;the\;cylinder\;=\;\mathrm{πr}^2\mathrm h\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\;\frac{22}7\times21\times21\times25\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\;22\times3\times21\times25\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\;34650\;cm^3\\Hence,\;volume\;of\;the\;cylinder\;is\;34650\;cm^3.\\\\$

Q12. The following figure represents a solid consisting of a right circular cylinder with a hemisphere at one end and a cone at the other. Their common radius is 7. The height of the cylinder and cone are each of 4 cm. Find the volume of the solid. [4] [2018]

Solution: 1540 cm3

Step-by-step Explanation:

$Volume\;of\;the\;solid=\\Volume\;of\;cone+volume\;of\;cylinder+volume\;of\;hemisphere\\=\frac13\mathrm{πr}^2\mathrm h+\mathrm{πr}^2\mathrm h+\frac23\mathrm{πr}^3\\=\;\mathrm{πr}^2\;\left(\frac13\mathrm h\;+\;\mathrm h\;+\;\frac23\mathrm r\right)\\=\;\frac{22}7\times7\times7\left(\frac13\times4\;+\;4\;+\;\frac23\times7\right)\\=\;154\;\left(\frac43\;+\;4\;+\;\frac{14}3\right)\\=154\;\left(\frac{4+12+14}3\right)\\=\;154\times\frac{30}3\\=\;1540\;\mathrm{cm}^3\\\\$

Q13. A conical tent is to accommodate 77 persons. Each person must have 16 m3 of air to breathe. Given the radius of the tent as 7 m, find the height of the tent and also its curved surface area. [2017]

Solution: 24 m ; 550 m2

Step-by-step Explanation:

$In\;a\;conical\;tent,\\1\;person\;can\;breathe\;16\;m^3\;of\;air.\\Hence,\;amount\;of\;air\;taken\;by\;77\;persons=\;16\times77=1232\;m^3\\Hence,\;volume\;of\;the\;conical\;tent\;=\;1232\;m^3\\\frac13\mathrm{πr}^2\mathrm h=\;1232\\\Rightarrow\mathrm h=\frac{1232\times3\times7}{22\times7\times7}\\\Rightarrow\mathrm h=\;24\\\mathrm{Therefore},\;\mathrm{height}\;\mathrm{of}\;\mathrm{the}\;\mathrm{cone}\;\mathrm{is}\;24\;\mathrm m\\\mathrm{slant}\;\mathrm{height}\;(\mathrm l)\;=\;\sqrt{\mathrm h^2+\mathrm r^2}\\\mathrm l=\sqrt{24^2+7^2}\\\mathrm l=\;\sqrt{625}\\\mathrm l=\;25\\\mathrm{curved}\;\mathrm{surface}\;\mathrm{area}\;\mathrm{of}\;\mathrm{the}\;\mathrm{cone}=\;\mathrm{πrl}\\=\frac{22}7\times7\times25\\=550\;\mathrm m^2\\\mathrm{Hence},\;\mathrm{curved}\;\mathrm{surface}\;\mathrm{area}\;\mathrm{of}\;\mathrm{the}\;\mathrm{conical}\;\mathrm{tent}\;\mathrm{is}\;550\;\mathrm m^2.\\\\$

Q14. A certain number of metallic cones, each of radius 2 cm and height 3 cm are melted and recast into a solid sphere of radius 6 cm. Find the number of cones. [3] [2016]

Solution: 72

Step-by-step Explanation:

$Let\;the\;number\;of\;cones\;melted\;be\;x.\\Hence,\;volume\;of\;x\;cones\;=\;volume\;of\;sphere\\\Rightarrow x\times\frac13\mathrm{πr}^2\mathrm h\;=\;\frac43\mathrm{πR}^3\\\Rightarrow x\times\frac13\times2\times2\times3\;=\;\frac43\times6\times6\times6\\\Rightarrow x\times4=8\times36\\\Rightarrow x=\;\frac{8\times36}4\\\Rightarrow x=\;72\\Hence\;total\;number\;of\;cones\;melted\;is\;72.\\$

Q15. Two solid spheres of radii 2 cm and 4 cm are melted and recast into a cone of height 8 cm. Find the radius of the cone so formed. [3] [2015]

Solution: 6 cm

Step-by-step Explanation:

$Volume\;of\;cone=\;Volume\;of\;1st\;sphere+volume\;of\;2nd\;sphere\\\frac13\mathrm{πR}^2\mathrm H=\frac43\mathrm\pi{({\mathrm r}_1)}^3+\frac43\mathrm\pi{({\mathrm r}_2)}^3\\\Rightarrow\frac13\mathrm{πR}^2\mathrm H=\frac43\mathrm\pi\left({({\mathrm r}_1)}^3+{({\mathrm r}_2)}^3\right)\\\Rightarrow\mathrm R^2\mathrm H=4\left({({\mathrm r}_1)}^3+{({\mathrm r}_2)}^3\right)\\\Rightarrow\mathrm R^2\times8=4\left(2^3+4^3\right)\\\Rightarrow\mathrm R^2\times8=\;4\times72\\\Rightarrow\mathrm R^2=4\times9\\\Rightarrow\mathrm R=\;\sqrt{36}\\\Rightarrow\;\mathrm R=\;6\\\mathrm{Hence},\;\mathrm{radius}\;\mathrm{of}\;\mathrm{the}\;\mathrm{cone}\;\mathrm{is}\;6\;\mathrm{cm}.\\$

Q16. The surface area of a solid metallic sphere is 2464 cm2. It is melted and recast into solid right circular cones of radius 3.5 cm and height 7 cm. Calculate :
(i) the radius of the sphere.
(ii) the number of cones recast. (Take π = 22/7) [4] [2014]

Solution: (i) 14 cm (ii) 128

Step-by-step Explanation:

$(i)\;Surface\;area\;of\;solid\;metallic\;sphere=2464\;cm^2\\4\mathrm{πr}^2=\;2464\\\Rightarrow\mathrm r^2=\frac{2464\times7}{22\times4}\\\Rightarrow\mathrm r^2=28\times7\\\Rightarrow\mathrm r=\sqrt{28\times7}\\\Rightarrow\mathrm r=14\\\mathrm{Hence},\;\mathrm{radius}\;\mathrm{of}\;\mathrm{the}\;\mathrm{sphere}\;\mathrm{is}\;14\;\mathrm{cm}.\\(\mathrm{ii})\;\mathrm{let}\;\mathrm{the}\;\mathrm{number}\;\mathrm{of}\;\mathrm{cones}\;\mathrm{recast}\;\mathrm{by}\;\mathrm{the}\;\mathrm{sphere}\;\mathrm{be}\;\mathrm x.\\\mathrm{So},\;\mathrm{volume}\;\mathrm{of}\;\mathrm{the}\;\mathrm{cones}=\;\mathrm{volume}\;\mathrm{of}\;\mathrm{sphere}\\\mathrm x\times\frac13\mathrm{πr}^2\mathrm h\;=\;\frac43\mathrm{πR}^3\\\Rightarrow\mathrm x\times\left(3.5\right)^2\times7=4\times14\times14\times14\\\Rightarrow\mathrm x=\frac{4\times14\times14\times14}{3.5\times3.5\times7}\\\Rightarrow\mathrm x=\;128\\\mathrm{Hence}\;,\;128\;\mathrm{cones}\;\mathrm{are}\;\mathrm{recast}.\\$

Q17. A solid sphere of radius 15 cm is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate the number of cones recast. [2013]

Solution: 270

Step-by-step Explanation:

$\mathrm{let}\;\mathrm{the}\;\mathrm{number}\;\mathrm{of}\;\mathrm{cones}\;\mathrm{recast}\;\mathrm{by}\;\mathrm{the}\;\mathrm{sphere}\;\mathrm{be}\;\mathrm x.\\\mathrm{So},\;\mathrm{volume}\;\mathrm{of}\;\mathrm{the}\;\mathrm{cones}=\;\mathrm{volume}\;\mathrm{of}\;\mathrm{sphere}\\\mathrm x\times\frac13\mathrm{πr}^2\mathrm h\;=\;\frac43\mathrm{πR}^3\\\Rightarrow\mathrm x\times\left(2.5\right)^2\times8=4\times15\times15\times15\\\Rightarrow\mathrm x=\frac{4\times15\times15\times15}{2.5\times2.5\times8}\\\Rightarrow\mathrm x=\;270\\\mathrm{Hence}\;,\;270\;\mathrm{cones}\;\mathrm{are}\;\mathrm{recast}.$

ICSE Class 10 Maths All chapters Previous Years Questions with Solution

Q18. A hollow sphere of internal and external radii 6 cm and 8 cm respectively is melted and recast into small cones of base radius 2 cm and height 8 cm. Find the number of cones. [3] [2012]

Solution: 37

Step-by-step Explanation:

$\mathrm{let}\;\mathrm{the}\;\mathrm{number}\;\mathrm{of}\;\mathrm{cones}\;\mathrm{recast}\;\mathrm{by}\;\mathrm{the}\;\mathrm{hollow}\;\mathrm{sphere}\;\mathrm{be}\;\mathrm x.\\\mathrm{So},\;\mathrm{volume}\;\mathrm{of}\;\mathrm{the}\;\mathrm{cones}=\;\mathrm{volume}\;\mathrm{of}\;\mathrm{hollow}\;\mathrm{sphere}\\\mathrm x\times\frac13\mathrm{πr}^2\mathrm h\;=\;\frac43\mathrm\pi\;\left(\mathrm R^3-\mathrm r^3\right)\\\Rightarrow\mathrm x\times2\times2\times8=4\times\left(8^3-6^3\right)\\\Rightarrow\mathrm x\times32=4\times296\\\Rightarrow\mathrm x=\;\frac{4\times296}{32}\\\Rightarrow\mathrm x=37\\\mathrm{Hence}\;,\;37\;\mathrm{cones}\;\mathrm{are}\;\mathrm{recast}.$

Q19. A solid cone of radius 5 cm and height 8 cm is melted and made into small spheres of radius 0.5 cm. Find the number of spheres formed. [2011]

Solution: 400

Step-by-step Explanation:

$\mathrm{let}\;\mathrm{the}\;\mathrm{number}\;\mathrm{of}\;\mathrm{spheres}\;\mathrm{recast}\;\mathrm{by}\;\mathrm{the}\;\mathrm{cone}\;\mathrm{be}\;\mathrm x.\\\mathrm{So},\;\mathrm{volume}\;\mathrm{of}\;\mathrm{spheres}=\;\mathrm{volume}\;\mathrm{of}\;\mathrm{the}\;\mathrm{cone}\\\mathrm x\times\frac43\mathrm{πr}^3\;=\;\frac13\mathrm{πR}^2\mathrm h\\\Rightarrow\mathrm x\times4\times{(0.5)}^3=5^2\times8\\\Rightarrow\mathrm x=\;\frac{25\times8}{4\times0.5\times0.5\times0.5}\\\Rightarrow\mathrm x=\;400\\\mathrm{Hence}\;,\;400\;\mathrm{spheres}\;\;\mathrm{are}\;\mathrm{formed}.$

Q20. A hemispherical bowl of diameter 7.2 cm is filled completely with chocolate sauce. This sauce is poured into an inverted cone of radius 4.8 cm. Find the height of the cone. [3] [2010]

Solution: 4.05 cm

Step-by-step Explanation:

$Volume\;of\;the\;chocolate\;sauce=\;volume\;of\;the\;hemisphere\\Hence,\;Volume\;of\;the\;chocolate\;sauce=\frac23\mathrm{πr}^3\\\;\;\;\;\;\;\;\;\;\;\;\;=\frac23\times\frac{22}7\times\frac{7.2}2\times\frac{7.2}2\times\frac{7.2}2\\\mathrm{let}\;\mathrm{the}\;\mathrm{height}\;\mathrm{of}\;\mathrm{the}\;\mathrm{sauce}\;\mathrm{in}\;\mathrm{the}\;\mathrm{cone}\;\mathrm{be}\;\mathrm h\;\mathrm{cm}.\\\mathrm{Therefore},\;\frac13\mathrm{πr}^2\mathrm h=\;\frac23\times\frac{22}7\times\frac{7.2}2\times\frac{7.2}2\times\frac{7.2}2\\\Rightarrow\frac13\times\frac{22}7\times4.8\times4.8\times\mathrm h=\;\frac23\times\frac{22}7\times\frac{7.2}2\times\frac{7.2}2\times\frac{7.2}2\\\Rightarrow\mathrm h=\frac{2\times3.6\times3.6\times3.6}{4.8\times4.8}\\\Rightarrow\mathrm h=\;4.05\\\mathrm{height}\;\mathrm{of}\;\mathrm{the}\;\mathrm{chocolate}\;\mathrm{sauce}\;\mathrm{in}\;\mathrm{the}\;\mathrm{cone}\;\mathrm{is}\;4.05\;\mathrm{cm}.$

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ICSE Class 10 Maths All chapters Previous Years Questions with Solution

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