ML Aggarwal Solution Class 10 Chapter 17 Mensuration Exercise 17.5

 Exercise 17.5

Question 1

The diameter of a metallic sphere is 6 cm. The sphere is melted and drawn into a wire of uniform cross-section. If the length of the wire is 36 m, find its radius.

Sol :

Diameter of metallic sphere = 6 cm

Radius $(r)=\frac{6}{2}=3 \mathrm{~cm}$

Volume $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \times \pi \times(3)^{3} \mathrm{~cm}^{3}$

$=\frac{4}{3} \pi \times 3 \times 3 \times 3 \mathrm{~cm}^{3}=36 \pi \mathrm{cm}^{3}$

$\therefore$ Volume of wire $=36 \pi \mathrm{cm}^{3}$

Length of wire (h)=36 m 

Let r be the radius of the wire

$\therefore \pi r^{2} h=36 \pi$

$r^{2} \times 36 \times 100=36$

$r^{2}=\frac{1}{100}=\left(\frac{1}{10}\right)^{2}$

$r=\frac{1}{10} \mathrm{~cm}=1 \mathrm{~mm}$

Question 2

The radius of a sphere is 9 cm. It is melted and drawn into a wire of diameter 2 mm. Find the length of the wire in metres.

Sol :

Radius of sphere = 9 cm

Volume $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi \times(9)^{3} \mathrm{~cm}^{3}$

$=\frac{4}{3} \pi \times 9 \times 9 \times 9 \mathrm{~cm}^{3}=972 \pi \mathrm{cm}^{3}$

Diameter of wire =2 mm

$\therefore$ Radius $(r)=\frac{2}{2}=1 \mathrm{~mm}=\frac{1}{10} \mathrm{~cm}$

Let h be the length of wire then

$\pi r^{2} h=972 \pi$

$\frac{1}{10} \times \frac{1}{10} \times h=972$

$\Rightarrow h=972 \times 10 \times 10$

h = 97200 cm = 972 m

Length of wire = 972 m

Question 3

A solid metallic hemisphere of radius 8 cm is melted and recasted into right circular cone of base radius 6 cm. Determine the height of the cone.

Sol :

Radius of a solid hemisphere (r) = 8 cm

Volume $=\frac{2}{3} \pi r^{3}=\frac{2}{3} \pi \times(8)^{3} \mathrm{~cm}^{3}$

$=\frac{2}{3} \times 512 \pi=\frac{1024}{3} \pi \mathrm{cm}^{3}$

Radius (r)=6 cm

$\therefore$ Height $(h)=\frac{\text { Volume }}{\frac{1}{3} \pi r^{2}}=\frac{1024 \pi \times 3}{3 \times 1 \times \pi \times 6 \times 6}$

$=\frac{256}{9} \mathrm{~cm}=28 \frac{4}{9} \mathrm{~cm}$

Question 4

A rectangular water tank of base 11 m x 6 m contains water upto a height of 5 m. if the water in the tank is transferred to a cylindrical tank of radius 3.5 m, find the height of the water level in the tank.

Sol :

Base of a water tank = 11 m × 6 m

Height of water level in it (h) = 5 m

Volume of water =11 × 6 × 5 = 330 m³

Volume of water in the cylindrical tank

$=330 \mathrm{~m}^{3}$

Radius of its base $=3.5 \mathrm{~m}=\frac{7}{2} \mathrm{~m}$

$\therefore$ Height of water level $=\frac{\text { Volume }}{\pi r^{2}}$

$=\frac{330 \times 7 \times 2 \times 2}{22 \times 7 \times 7} \mathrm{~m}$

$=\frac{60}{7} \mathrm{~m}=8 \frac{4}{7} \mathrm{~m}$

Question 5

The rain water from a roof of dimensions 22 m x 20 m drains into a cylindrical vessel having diameter of base 2 m and height; 3.5 m. If the rain water collected from the roof just fill the cylindrical vessel, then find the rainfall in cm.

Sol :

Dimensions of roof = 22 m × 20 m

Let rainfall = x m

.’. Volume of water = 22 × 20 × x m³

Volume of water in cylinder = 22 × 20 × x m³

Diameter of its base = 2 m

$\therefore$ Radius $=\frac{2}{2}=1 \mathrm{~m}$

and height of water level $=3.5 \mathrm{~m}=\frac{7}{2} \mathrm{~m}$

$\therefore$ Volume of water $=\pi r^{2} h$

$=\frac{22}{7} \times 1 \times 1 \times \frac{7}{2} \mathrm{~m}^{3}=11 \mathrm{~m}^{3}$

$\therefore$ Volume of water in cylinder $=$ Volume of water

$22 \times 20 \times x=11$

$x=\frac{11}{22 \times 20} \mathrm{~m}=\frac{1}{40} \mathrm{~m}=\frac{1}{40} \times 100$

$=\frac{5}{2} \mathrm{~cm}=2.5 \mathrm{~cm}$

$\therefore$ Rainfall $=2.5 \mathrm{~cm}$

Question 6

The volume of a cone is the same as that of the cylinder whose height is 9 cm and diameter 40 cm. Find the radius of the base of the cone if its height is 108 cm.

Sol :

Diameter of a cylinder = 40 cm

Radius $(r)=\frac{40}{2}=20 \mathrm{~cm}$

Height(h) = 9 cm

$\therefore$ Volume $=\pi r^{2} h=\pi \times 20 \times 20 \times 9 \mathrm{~cm}^{3}$

$=3600 \pi \mathrm{cm}^{3}$

Now volume of cone $=3600 \pi \mathrm{cm}^{3}$

Height of cone =108 cm

$\therefore$ Radius $=\sqrt{\frac{\text { Volume } \times 3}{1 \times \pi h}}$

$=\sqrt{\frac{3600 \pi \times 3}{\pi \times 108}}=\sqrt{100}=10 \mathrm{~cm}$

Question 7

Eight metallic spheres, each of radius 2 cm, are melted and cast into a single sphere. Calculate the radius of the new (single) sphere.

Sol :

Radius of each metallic sphere (r) = 2 cm

Volume of one sphere $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \pi \times 2^{3} \mathrm{~cm}^{3}$

$=\frac{32}{3} \pi \mathrm{cm}^{3}$

and volume of 8 such spheres $=\frac{32}{3} \pi \times 8 \mathrm{~cm}^{3}$

$=\frac{256}{3} \pi \mathrm{cm}^{3}$

Now volume of a single sphere $=\frac{256}{3} \pi \mathrm{cm}^{3}$

$\therefore$ Radius of new sphere $=\sqrt[3]{\frac{\text { Volume }}{4} \pi}$

$=\left(\frac{256 \pi}{3} \times \frac{3}{4 \times \pi}\right)^{\frac{1}{3}} \mathrm{~cm}=(64)^{\frac{1}{3}}=4 \mathrm{~cm}$

Question 8

A metallic disc, in the shape of a right circular cylinder, is of height 2.5 mm and base radius 12 cm. Metallic disc is melted and made into a sphere. Calculate the radius of the sphere.

Sol :

Height of disc cylindrical shaped = 2.5 mm

and base radius = 12 cm

Volume of the disc = πr²h

$=\pi \times 12 \times 12 \times \frac{2.5}{10} \mathrm{~cm}$

$=144 \pi \times \frac{25}{100}=36 \pi \mathrm{cm}^{3}$

Now volume of sphere $=36 \pi \mathrm{cm}^{3}$

$\therefore$ Radius $=\left(\frac{\text { Volume }}{\frac{4}{3} \pi}\right)^{\frac{1}{3}}$

$=\left(\frac{36 \pi \times 3}{4 \times \pi}\right)^{\frac{1}{3}}=(27)^{\frac{1}{3}}=3 \mathrm{~cm}$

Question 9

Two spheres of the same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the big sphere.

Sol :

Weight of first sphere = 1 kg

and weight of second sphere = 7 kg

Radius of smaller sphere = 3 cm

Let r be the radius of a larger sphere

Now volume of smaller sphere $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \pi(3)^{3}=36 \pi \mathrm{cm}^{3}$

and volume of layer sphere $=\frac{4}{3} \pi \mathrm{R}^{3}$

Weight of smaller is 1 kg and of bigger is 7 kg Weight of both sphere 

=1+7=8 kg

$\therefore 36 \pi: \frac{4}{3} \pi \mathrm{R}^{3}=1: 8$

$\frac{36 \pi \times 3}{4 \pi \mathrm{R}^{3}}=\frac{1}{8}$

$\frac{27}{R^{3}}=\frac{1}{8} \Rightarrow R^{3}=27 \times 8=(3 \times 2)^{3}$

R = 3 x 2 = 6 cm

Diameter of big sphere = 2 x 6 = 12 cm

Question 10

A hollow copper pipe of inner diameter 6 cm and outer diameter 10 cm is melted and changed into a solid circular cylinder of the same height as that of the pipe. Find the diameter of the solid cylinder.

Sol :

Inner diameter of a hollow pipe = 6 cm

and outer diameter = 10 cm

Inner radius $(r)=\frac{6}{2}=3 c m$

and outer radius $(R)=\frac{10}{2}=5 \mathrm{~cm}$

Let height of the pipe =h cm and r be the solid cylinder

$\therefore$ Volume of pipe $=\pi\left(\mathrm{R}^{2}-r^{2}\right) h$

$=\pi\left(5^{2}-3^{2}\right) h$

$\therefore \pi h(25-9)=\pi R^{2} h$

$16=R^{2}$

$\Rightarrow R=\sqrt{16}=4$

$\therefore$ Diameter of solid cylinder

$=4 \times 2=8 \mathrm{~cm}$

Question 11

A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 4 cm and height is 72 cm, find the uniform thickness of the cylinder.

Sol :

Radius of a solid sphere (r) = 6 cm

Volume $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi \times(6)^{3} \mathrm{~cm}^{3}$

$=\frac{4}{3} \times 216 \pi=288 \pi \mathrm{cm}^{3}$

$\therefore$ Volume of hollow cylinder $=288 \pi \mathrm{cm}^{3}$

External radius $(\mathrm{R})=4 \mathrm{~cm}$ 

Height $(h)=72 \mathrm{~cm}$

Let internal radius $=r,$ then 

Volume $=\pi h\left(\mathrm{R}^{2}-r^{2}\right)$

$\Rightarrow 288 \pi=\pi \times 72\left(4^{2}-r^{2}\right)$

$4=16-r^{2}$

$\Rightarrow r^{2}=16-4=12$

$r=\sqrt{12}=2 \sqrt{3}=2(1.732)=3.464 \mathrm{~cm}$

$\therefore$ Thickness $=\mathrm{R}-r=4-3.464=0.536 \mathrm{~cm}$

Question 12

A hollow metallic cylindrical tube has an internal radius of 3 cm and height 21 cm. The thickness of the metal of the tube is $\frac{1}{2} \mathrm{~cm}$ . The tube is melted and cast into a right circular cone of height 7 cm. Find the radius of the cone correct to one decimal place.

Sol :

Internal radius of a hollow metallic cylindrical tube (r) = 3 cm

and height (h) = 21 cm

Thickness of metal $=\frac{1}{2} \mathrm{~cm}=0.5 \mathrm{~cm}$

$\therefore$ External radius (R)=3+0.5=3.5 cm

$\therefore$ Volume of metallic tube $=\pi h\left(\mathbf{R}^{2}-r^{2}\right)$

$=\pi \times 21\left[(3.5)^{2}-3.0^{2}\right] \mathrm{cm}^{3}$

$=21 \pi[12.25-9]=21 \pi \times 3.25 \mathrm{~cm}^{3}$

$=21 \times 3.25 \pi \mathrm{cm}^{3}$

Now volume of circular cone $=21 \times 3.25 \pi \mathrm{cm}^{3}$

Height of cone =7 cm

$\therefore$ Radius $=\left(\frac{\text { Volume }}{\frac{1}{3} \pi h}\right)^{\frac{1}{2}}$

$=\left(\frac{21 \times 3.25 \pi \times 3}{\pi \times 7}\right)^{\frac{1}{2}}=(29.25)^{\frac{1}{2}}=5.4 \mathrm{~cm}$

Question 13

A hollow sphere of internal and external diameters 4 cm and 8 cm respectively, is melted into a cone of base diameter 8 cm. Find the height of the cone. (2002)

Sol :

Internal diameter of a hollow sphere = 4 cm

and external diameter = 8 cm

Internal radius (r) = 2 cm

and external radius (R) = 4 cm

Volume of hollow sphere

$=\frac{4}{3} \pi\left(\mathrm{R}^{3}-r^{3}\right)$

$=\frac{4}{3} \pi\left(4^{3}-2^{3}\right) \mathrm{cm}^{3}$

$=\frac{4}{3} \pi(64-8)=\frac{56 \times 4}{3} \pi$

$=\frac{224}{3} \pi \mathrm{cm}^{3}$

Diameter of cone =8 cm

Radius =4 cm

Let the height of the cone be h cm

$\therefore$ Volume of the cone $=$ Volume of the metal

$\frac{1}{3} \pi r_{.}^{2} h=\frac{224}{3} \pi \mathrm{cm}^{3}$

$h=\frac{224 \times \pi \times 3}{3 \times \pi \times 4 \times 4}=14 \mathrm{~cm}$

$\therefore$ The height of the cone is $14 \mathrm{~cm}$

Question 14

A well with inner diameter 6 m is dug 22 m deep. Soil taken out of it has been spread evenly all round it to a width of 5 m to form an embankment. Find the height of the embankment.

Sol :

Inner diameter of a well = 6 m

Depth (h) = 22 m

$\therefore$ Radius $(r)=\frac{6}{2}=3 \mathrm{~m}$

$\therefore$ Volume of earth dug out $=\pi r^{2} h$

$=\pi \times 3 \times 3 \times 22 \mathrm{~m}^{3}=198 \pi \mathrm{m}^{3}$

Width of an embankment $=5 \mathrm{~m}$ 

Inner radius $(r)=3 \mathrm{~m}$

and outer radius $(\mathrm{R})=3+5=8 \mathrm{~m}$

Let height of embankment =h m

$\therefore$ Volume of embankment $=\pi\left(\mathrm{R}^{2}-r^{2}\right) \times h$

$=\pi \times h\left(8^{2}-3^{2}\right) \mathrm{m}^{3}$

$=\pi h \times 55 \mathrm{~m}^{3}$

$\therefore 55 \pi h=198 \pi$

$h=\frac{198 \pi}{55 \pi}=3.6 \mathrm{~m}$

Hence height of embankment =3.6 m

Question 15

A cylindrical can of internal diameter 21 cm contains water. A solid sphere whose diameter is 10.5 cm is lowered into the cylindrical can. The sphere is completely immersed in water. Calculate the rise in water level, assuming that no water overflows.

Sol :

Internal diameter of cylindrical can = 21 cm

Radius $(R)=\frac{21}{2} \mathrm{~cm}$

Diameter of a solid sphere =10.5 cm

$\therefore$ Radius $(r)=\frac{10.5}{2}=5.25 \mathrm{~cm}$

$\therefore$ Volume of sphere $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \pi(5.25)^{3} \mathrm{~cm}^{3}$

Let rise of water in cylindrical can =h

$\therefore \pi R^{2} h=\frac{4}{3} \pi(5.25)^{3}$

$\frac{21}{2} \times \frac{21}{2} h=\frac{4}{3} \times 5.25 \times 5.25 \times 5.25$

$\frac{21}{2} \times \frac{21}{2} h=\frac{4}{3} \times \frac{21}{4} \times \frac{21}{4} \times \frac{21}{4}$

$\therefore h=\frac{4}{3} \times \frac{21}{4} \times \frac{21}{4} \times \frac{21}{4} \times \frac{2}{21} \times \frac{2}{21}$

$=\frac{7}{4} m=1.75 \mathrm{~m}$

Question 16

There is water to a height of 14 cm in a cylindrical glass jar of radius 8 cm. Inside the water there is a sphere of diameter 12 cm completely immersed. By what height will the water go down when the sphere is removed?

Sol :

Radius of the cylindrical jar (R) = 8 cm

Height of water level (h) = 14 cm

Volume of water = πR²h

$=\pi \times 8 \times 8 \times 14 \mathrm{~cm}^{3}=896 \pi \mathrm{cm}^{3}$

Diameter of sphere $=12 \mathrm{~cm}$

$\therefore$ Radius $(r)=\frac{12}{2}=6 \mathrm{~cm}$

Volume $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi \times 6 \times 6 \times 6 \mathrm{~cm}^{3}$

$=288 \pi \mathrm{cm}^{3}$

By immersing the sphere in the cylinder water rose up $=288 \pi \mathrm{cm}^{3}$

$\therefore$ Let height of water rose $=h \mathrm{~cm}$

$h=\frac{288 \pi}{\pi \times 8 \times 8}=\frac{9}{2} \mathrm{~cm}$

$\therefore \pi \times 8 \times 8 \times h=288 \pi$

$h=\frac{288 \pi}{\pi \times 8 \times 8}=\frac{9}{2} \mathrm{~cm}$

$\therefore$ Water rise $=\frac{9}{2}=4.5 \mathrm{~cm}$

Question 17

A vessel in the form of an inverted cone is filled with water to the brim. Its height is 20 cm and diameter is 16.8 cm. Two equal solid cones are dropped in it so that they are fully submerged. As a result, one-third of the water in the original cone overflows. What is the volume of each of the solid cone submerged? (2002)

Sol :

Height of conical vessel (h) = 20 cm

and diameter = 16.8 cm

$\therefore$ Radius $(r)=\frac{16.8}{2}=8.4 \mathrm{~cm}$

Volume of water filled in it

$=\frac{1}{3} \pi r^{2} h=\frac{1}{3} \pi \times 8.4 \times 8.4 \times 20 \mathrm{~cm}^{3}$

$=\frac{1}{3} \times \frac{22}{7} \times 8.4 \times 8.4 \times 20 \mathrm{~cm}^{3}$

$=1478.4 \mathrm{~cm}^{3}$

$\therefore \frac{1}{3} \%$ volume of water $=1478.4 \times \frac{1}{3}$

$=492.8 \mathrm{~cm}^{3}$

$\therefore$ Volume of two equal solid cones $=492.8 \mathrm{~cm}^{3}$

and volume of one cone $=\frac{492.8}{2}$

$=246.4 \mathrm{~cm}^{3}$

Question 18

A solid metallic circular cylinder of radius 14 cm and height 12 cm is melted and recast into small cubes of edge 2 cm. How many such cubes can be made from the solid cylinder?

Sol :

Radius of a solid metallic cylindrical (r) = 14 cm

and height (h) = 12 cm

Volume $=\pi r^{2} h=\frac{22}{7} \times 14 \times 14 \times 12 \mathrm{~cm}^{3}$

$=7392 \mathrm{~cm}^{3}$

Volume of one cube $=(2 \mathrm{~cm})^{3}=8 \mathrm{~cm}^{3}$

$\therefore$ Number of cube so formed $=\frac{7392}{8}=924$

Question 19

How many shots each having diameter 3 cm can be made from a cuboidal lead solid of dimensions 9 cm x 11 cm x 12 cm?

Sol :

Diameter of a shot = 3 cm

Radius $(r)=\frac{3}{2} \mathrm{~cm}$

Volume of one shot $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \times \frac{22}{7} \times\left(\frac{3}{2}\right)^{3} \mathrm{~cm}^{3}$

$=\frac{88}{21} \times \frac{27}{8} \mathrm{~cm}^{3}$

Dimensions of a cuboidal lead

$=9 \mathrm{~cm} \times 11 \mathrm{~cm} \times 12 \mathrm{~cm}$

$\therefore$ Volume $=9 \times 11 \times 12=1188 \mathrm{~cm}^{3}$

$\therefore$ Number of shots to be made $=\frac{1188}{\frac{88}{21} \times \frac{27}{8}}$

$=\frac{1188 \times 21 \times 8}{88 \times 27}=84$ shots

Question 20

How many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge measures 44 cm?

Sol :

Diameter of lead shot = 4 cm

Radius $(r)=\frac{4}{2}=2 \mathrm{~cm}$ and

volume $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \times \frac{22}{7} \times 2 \times 2 \times 2 \mathrm{~cm}^{3}$

$=\frac{704}{21} \mathrm{~cm}^{3}$

Edge (side) of a solid cube $=44 \mathrm{~cm}$

$\therefore$ Volume $=(a)^{3}=44 \times 44 \times 44 \mathrm{~cm}^{3}$

Number of lead shots to be made

$=\frac{44 \times 44 \times 44}{\frac{704}{21}}$

$=\frac{44 \times 44 \times 44 \times 21}{704}$

=2541 shots

Question 21

Find the number of metallic circular discs with 1.5 cm base diameter and height 0.2 cm to be melted to form a circular cylinder of height 10 cm and diameter 4.5 cm.

Sol :

Radius of the circular disc (r) = 0.75 cm

Height of circular disc (h) = 0.2 cm

Radius of cylinder (R) = 2.25 cm

Height of cylinder (H) = 10 cm

Now,

Number of metallic circular disc

$=\frac{\text { Volume of cylinder }}{\text { Volume of each circular } \operatorname{disc}}$

$=\frac{\pi \mathrm{R}^{2} \mathrm{H}}{\pi r^{2} h}=\frac{\mathrm{R}^{2} \mathrm{H}}{r^{2} h}$

$=\frac{(2.25)^{2}(10)}{(0.75)^{2}(0.2)}=\frac{2.25 \times 2.25 \times 10}{0.75 \times 0.75 \times 0.2}$

$=\frac{225 \times 2.25 \times 10 \times 100 \times 100 \times 10}{75 \times 75 \times 2 \times 100 \times 100}=450$ shots

Question 22

A solid metal cylinder of radius 14 cm and height 21 cm is melted down and recast into spheres of radius 3.5 cm. Calculate the number of spheres that can be made.

Sol :

Radius of a solid metallic cylinder (r) = 14 cm

and height (h) = 21 cm

Volume of cylinder = πr²h

$=\frac{22}{7} \times 14 \times 14 \times 21 \mathrm{~cm}^{3}=12936 \mathrm{~cm}^{3}$

Radius of sphere $\left(r_{1}\right)=3.5 \mathrm{~cm}=\frac{7}{2} \mathrm{~cm}$

$\therefore$ Volume of one sphere $=\frac{4}{3} \pi r_{1}^{3}$

$=\frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} \mathrm{~cm}^{3}=\frac{539}{3} \mathrm{~cm}^{3}$

$\therefore$ Number of sphere so formed $=\frac{12936 \times 3}{539}$

=72 spheres

Question 23

A metallic sphere of radius 10.5 cm is melted and then recast into small cenes, each of radius 3.5 cm and height 3 cm. Find the number of cones thus obtained. (2005)

Sol :

Radius of a metallic sphere (r) = 10.5 cm

Volume $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \times \pi \times 10.5 \times 10.5 \times 10.5=1543.5 \pi \mathrm{cm}^{3}$

Radius of each cone $\left(r_{1}\right)=3.5 \mathrm{~cm}$

and height (h)=3 cm

$\therefore$ Volume of one cone $=\frac{1}{3} \pi r_{1}^{2} h$

$=\frac{1}{3} \pi \times 3.5 \times 3.5 \times 3 \mathrm{~cm}^{3}=12.25 \pi \mathrm{cm}^{3}$

$\therefore$ Number of cones so formed $=\frac{1543.5 \pi}{12.25 \pi}$

=126 cones

Question 24

A certain number of metallic cones each of radius 2 cm and height 3 cm are melted and recast in a solid sphere of radius 6 cm. Find the number of cones. (2016)

Sol :

Radius of each cone (r) = 2 cm

and height (h) = 3 cm

$\therefore$ Volume of one cone $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \pi \times 2 \times 2 \times 3=4 \pi \mathrm{cm}^{3}$

Radius of a solid sphere $(\mathrm{R})=6 \mathrm{~cm}$

$\therefore$ Volume $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \pi \times 6 \times 6 \times 6 \mathrm{~cm}^{3}$

$=288 \pi \mathrm{cm}^{3}$

$\therefore$ Number of cones required $=\frac{288 \pi}{4 \pi}$

=72 cones

Question 25

A vessel is in the form of an inverted cone. Its height is 11 cm and the radius of its top, which is open, is 2.5 cm. It is filled with water upto the rim. When some lead shots, each of which is a sphere of radius 0.25 cm, are dropped into the vessel, $\frac{2}{5}$ of the water flows out. Find the number of lead shots dropped into the vessel. (2003)

Sol :

Radius of the top of the inverted conical vessel (R) = 2.5 cm

and height (h)= 11 cm

$\therefore$ Volume of the water in the vessel $=\frac{1}{3} \pi \mathrm{R}^{2} h$

$=\frac{1}{3} \pi(2.5)^{2} \times 11 \mathrm{~cm}^{2}$

$=\frac{11}{3} \pi \times 6.25 \mathrm{~cm}^{3}$

spherical shot $=0.25 \mathrm{~cm}=\frac{1}{4} \mathrm{~cm}$

Volume of water flows out

$=\frac{2}{5}$ of $\frac{11}{3} \pi \times 6.25 \mathrm{~cm}^{3}=\frac{137.5}{15} \pi \mathrm{cm}^{3}$

and volume of one shot $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \pi \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \mathrm{~cm}^{3}$

$=\frac{\pi}{48} \mathrm{~cm}^{3}$

$\therefore$ Number of shots required $=\frac{137.5 \pi \times 48}{15 \times \pi}$

=440 shots

Question 26

The surface area of a solid metallic sphere is 616 cm². It is melted and recast into smaller spheres of diameter 3.5 cm. How many such spheres can be obtained? (2007)

Sol :

Surface area of a metallic sphere = 616 cm²

$\therefore$ Radius $(\mathrm{R})=\sqrt{\frac{\text { Surface area }}{4 \pi}}$

$=\sqrt{\frac{616 \times 7}{4 \times 22}} \mathrm{~cm}=\sqrt{49}=7 \mathrm{~cm}$

$\therefore$ Volume $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 \mathrm{~cm}^{3}$

$=\frac{4312}{3} \mathrm{~cm}^{3}$

Diameter of smaller sphere $=3.5 \mathrm{~cm}$

$\therefore$ Radius $(r)=\frac{3.5}{2}=\frac{7}{4} \mathrm{~cm}$

Volume $=\frac{4}{3} \pi r^{3}$

$\therefore$ Number of smaller spheres $=\frac{4312}{3} \div \frac{539}{24}$

$=\frac{4312}{3} \times \frac{24}{539}=64$ spheres

Question 27

The surface area of a solid metallic sphere is 1256 cm². It is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate

(i) the radius of the solid sphere.

(ii) the number of cones recast. (Use π = 3.14).

Sol :

Surface area of a solid metallic sphere = 1256 cm²

$(i) \therefore$ Radius $(r)=\sqrt{\frac{\text { Surface area }}{4 \pi}}$

$=\sqrt{\frac{1256}{4 \times 3.14}} \mathrm{~cm}=\sqrt{\frac{314 \times 100}{314}}$

$=\sqrt{100}=10 \mathrm{~cm}$

Radius of a solid cone $\left(r_{1}\right)=2.5 \mathrm{~cm}$ and height $(h)=8 \mathrm{~cm}$

$\therefore$ Volume $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3}(3.14) \times 2.5 \times 2.5 \times 8 \mathrm{~cm}^{3}=\frac{157}{3} \mathrm{~cm}^{3}$

(ii) Volume of solid sphere $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \times 3.14 \times 10 \times 10 \times 10 \mathrm{~cm}^{3}$

$=\frac{12560}{3} \mathrm{~cm}^{3}$

$\therefore$ Number of cones formed

$=\frac{12560}{3} \div \frac{157}{3}$

$=\frac{12560}{3} \times \frac{3}{157}=80$ cones

Question 28

Water is flowing at the rate of 15 km/h through a pipe of diameter 14 cm into a cuboid pond which is 50 m long and 44 m wide. In what time will the level of water in the pond rise by 21 cm?

Sol :

Speed of water flow = 15 km/h

Diameter of pipe = 14 cm

$\therefore$ Radius $(\mathrm{R})=\frac{14}{2}=7 \mathrm{~cm}=\frac{7}{100} \mathrm{~m}$

and dimension of a cuboid pond

$=50 \mathrm{~m} \times 44 \mathrm{~m}$

Level of water in the pond $=21 \mathrm{~cm}$

$=\frac{21}{100} \mathrm{~m}$

$\therefore$ Volume of water in the pond

$=50 \times 44 \times \frac{21}{100} \mathrm{~m}^{3}=462 \mathrm{~m}^{2}$

$\therefore$ Length of flow of water in the pipe

$=\frac{462}{\pi r^{2}}=\frac{462 \times 7 \times 100 \times 100}{22 \times 7 \times 7}$

$=30000 \mathrm{~m}=\frac{30000}{1000}=30 \mathrm{~km}$

$\therefore$ Time taken $=\frac{\text { Distance }}{\text { Speed }}=\frac{30}{15}=2$ hours

Question 29

A cylindrical can whose base is horizontal and of radius 3.5 cm contains sufficient water so that when a sphere is placed in the can, the water just covers the sphere. Given that the sphere just fits into the can, calculate :

(i) the total surface area of the can in contact with water when the sphere is in it.

(ii) the depth of the water in the can before the sphere was put into the can. Given your answer as proper fractions.

Sol :

Radius of a cylindrical can = 3.5 cm

Radius of the sphere = 3.5 cm

and height of water level in the can = 3.5 × 2 = 7 cm

(i) $\therefore$ Total surface area of the can in touch with water $=2 \pi r h+\pi r^{2}$

$=\pi r(2 h+r)$

$=\frac{22}{7} \times 3.5(2 \times 7+3.5) \mathrm{cm}^{2}$

$=\frac{22}{7} \times 3.5(2 \times 7+3.5) \mathrm{cm}^{2}$

$=11(14+3.5) \mathrm{cm}^{2}=11 \times 17.5 \mathrm{~cm}^{2}$

$=192.5 \mathrm{~cm}^{2}$

(ii) Volume of sphere $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 3.5 \mathrm{~cm}^{3}$

$=\frac{88}{3} \times 6.125=\frac{539}{3} \mathrm{~cm}^{3}=179.67 \mathrm{~cm}^{3}$

and volume of $\mathrm{can}=\pi r^{2} h$

$=\frac{22}{7} \times 3.5 \times 3.5 \times 7 \mathrm{~cm}^{3}=269.5 \mathrm{~cm}^{3}$

$\therefore$ Actual water in the can $=269.5-179.67$

$=89.83 \mathrm{~cm}$

Height of water level $=\frac{89.83 \times 7}{22 \times 3.5 \times 3.5}$

$=2.32 \mathrm{~cm}$

$=2 \frac{1}{3}=\frac{7}{3} \mathrm{~cm}$