ML Aggarwal Solution Class 10 Chapter 17 Mensuration Test

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22 Q23 Q24

 Test

Question 1

A cylindrical container is to be made of tin sheet. The height of the container is 1 m and its diameter is 70 cm. If the container is open at the top and the tin sheet costs Rs 300 per $m^2$, find the cost of the tin for making the container.

Sol :

Height of container opened at the top (h) = 1 m = 100 cm

and diameter = 70 cm

∴Radius (r) $=\frac{70}{2}=35 \mathrm{~cm}$

$\therefore$ Total surface area $=2 \pi \mathrm{rh}+\pi \mathrm{r}^{2}$

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

$=\frac{22}{7} \times 35(2 \times 100+35) \mathrm{cm}^{2}$

$=110(200+35)=110 \times 235 \mathrm{~cm}^{2}$

$=\frac{110 \times 235}{100 \times 100} \mathrm{~m}^{2}=\frac{517}{200} \mathrm{~m}^{2}$

$\therefore$ Area of sheet required $=\frac{517}{200} \mathrm{~m}^{2}$

Cost of $1 \mathrm{~m}^{2}$ sheet $=₹ 300$

$\therefore$ Total cost $=\frac{517}{200} \times 300$

$=₹ \frac{1551}{2}=₹ 775.50$

Question 2

A cylinder of maximum volume is cut out from a wooden cuboid of length 30 cm and cross-section of square of side 14 cm. Find the volume of the cylinder and the volume of wood wasted.

Sol :

Dimensions of the wooden cuboid = 30 cm × 14 cm × 14 cm

Volume $=30 \times 14 \times 14=5880 \mathrm{~cm}^{3}$

Largest size of cylinder cut out of the wooden cuboid will be of diameter $=14 \mathrm{~cm}$ and height $=30 \mathrm{~cm}$

$\therefore$ Radius of cylinder $=\frac{14}{2}=7 \mathrm{~cm}$

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

$=\frac{22}{7} \times 7 \times 7 \times 30 \mathrm{~cm}^{3}=4620 \mathrm{~cm}^{3}$

$\therefore$ Volume of wooden wasted $=5880-4620$

$=1260 \mathrm{~cm}^{3}$

Question 3

Find the volume and the total surface area of a cone having slant height 17 cm and base diameter 30 cm. Take π = 3.14.

Sol :

Slant height of a cone (l) = 17 cm

Diameter of base = 30 cm

Radius $(r)=\frac{30}{2}=15 \mathrm{~cm}$

$\therefore$ Height $(h)=\sqrt{l^{2}-r^{2}}=\sqrt{17^{2}-15^{2}} \mathrm{~cm}$

$=\sqrt{289-225}=\sqrt{64}=8 \mathrm{~cm}$

Now volume $=\frac{1}{3} \pi r^{3} h$

$=\frac{1}{3}(3.14) \times 15 \times 15 \times 8 \mathrm{~cm}^{3}=1884 \mathrm{~cm}^{3}$

and total surface area $=\pi r l+\pi r^{2}$

$=\pi r(l+r)=3.14 \times 15 \times(17+15) \mathrm{cm}^{2}$

$=3.14 \times 15 \times 32 \mathrm{~cm}^{2}=1507.2 \mathrm{~cm}^{2}$

Question 4

Find the volume of a cone given that its height is 8 cm and the area of base 156 $cm^2$

Sol :

Height of a cone = 8 cm

Area of base = 156 cm

$\therefore$ Volume $=\frac{1}{3} \times$ area of base $\times$ height

$=\frac{1}{3} \times 156 \times 8$

$=\frac{1248}{3} \mathrm{~cm}^{3}=416 \mathrm{~cm}^{3}$

Question 5

The circumference of the edge of a hemispherical bowl is 132 cm. Find the capacity of the bowl.

Sol :

Circumference of the edge of bowl = 132 cm

Radius of a hemispherical bowl

$=\frac{132}{2 \pi}=\frac{132 \times 7}{2 \times 22}=21 \mathrm{~cm}$

Now volume of $=\frac{2}{3} \pi r^{3}$

$=\frac{2}{3} \times \frac{22}{7} \times(21)^{3} \mathrm{~cm}^{3}$

$=\frac{2}{3} \times \frac{22}{7} \times 9261 \mathrm{~cm}^{3}$

$=19404 \mathrm{~cm}^{3}$

Question 6

The volume of a hemisphere is $2425 \frac{1}{2} \mathrm{~cm}^{2}$ Find the curved surface area.

Sol :

Volume of a hemisphere $=2425 \frac{1}{2} \mathrm{~cm}^{3}$

$=\frac{4851}{2} \mathrm{~cm}^{3}$

Let radius = r, then

$\frac{2}{3} \pi r^{3}=\frac{4851}{2}$

$\Rightarrow \frac{2}{3} \times \frac{22}{7} \times r^{3}=\frac{4851}{2}$

$\Rightarrow r^{3}=\frac{4851 \times 3 \times 7}{2 \times 2 \times 22}=\frac{9261}{8}=\left(\frac{21}{2}\right)^{3}$

$\therefore r=\frac{21}{2} \mathrm{~cm}$

∴ Curved surface area $=2 \pi r^{2}$

$=2 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \mathrm{~cm}^{2}=693 \mathrm{~cm}^{2}$

Question 7

A solid wooden toy is in the shape of a right circular cone mounted on a hemisphere. If the radius of the hemisphere is 4.2 cm and the total height of the toy is 10.2 cm, find the volume of the toy

Sol :

A wooden solid toy is of a shape of a right circular cone

mounted on a hemisphere.

Radius of hemisphere (r) = 4.2 cm

Total height = 10.2 cm

$\therefore$ Height of conical part $=10.2-4.2=6 \mathrm{~cm}$

Now volume of the toy

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

$=\frac{1}{3} \pi r^{2}(h+2 r)$

Question 8

A medicine capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of the entire capsule is 2 cm. Find the capacity of the capsule.

Sol :

Diameter of cylindrical part = 0.5 cm

Total length of the capsule = 2 cm

Radius $(r)=\frac{0.5}{2}=0.25 \mathrm{~cm}$

and length of cylindrical part $=2-2 \times 0.25$ $=2-0.5=1.5 \mathrm{~cm}$

$\therefore$ Volume of the capsule $=2 \times \frac{2}{3} \pi r^{3}+\pi r^{2} h$

$=\frac{4}{3} \pi r^{3}+\pi r^{2} h$

$=\pi r^{2}\left(\frac{4}{3} r+h\right)$

$=\frac{22}{7} \times 0.25 \times 0.25\left(\frac{4}{3} \times 0.25+1.5\right)$

$=\frac{22}{7} \times 0.0625\left(\frac{4}{3} \times \frac{1}{4}+1.5\right) \mathrm{cm}^{3}$

$=\frac{22}{7} \times \frac{1}{16}\left(\frac{1}{3}+\frac{3}{2}\right) \mathrm{cm}^{3}$

$=\frac{22}{112} \times \frac{11}{6}=\frac{121}{336} \mathrm{~cm}^{3}$

$=0.360 \mathrm{~cm}^{3}=0.36 \mathrm{~cm}^{3}$

Question 9

A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Find the volume and the total surface area of the solid.

Sol :

Radius of cylinder 

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

and height of cylinder $=19-2 \times \frac{7}{2} \mathrm{~cm}$

= 19 – 7 = 12 cm

and radius of hemisphere $=\frac{7}{2} \mathrm{~cm}$

$\therefore$ Total volume of the solid

$=2 \times \frac{2}{3} \pi r^{3}+\pi r^{2} h$

$=\frac{4}{3} \times \frac{22}{7} \times\left(\frac{7}{2}\right)^{3}+\frac{22}{7} \times\left(\frac{7}{2}\right)^{2} \times 12 \mathrm{~cm}^{3}$

$=\frac{539}{3}+462=\frac{539+1386}{3}=\frac{1925}{3} \mathrm{~cm}^{3}$

$=641 \frac{2}{3} \mathrm{~cm}^{3}$

and Total surface area of the solid $=2 \times 2 \pi r^{2}+2 \pi r h$

$=4 \pi r^{2}+2 \pi r h=2 \pi r(2 r+h)$

$=2 \times \frac{22}{7} \times \frac{7}{2}\left(2 \times \frac{7}{2}+12\right)=22(7+12)$

$=22 \times 19 \mathrm{~cm}^{2}=418 \mathrm{~cm}^{2}$

Question 10

The radius and height of a right circular cone are in the ratio 5 : 12. If its volume is 2512 cm , find its slant height. (Take π = 3.14).

Sol :

Let radius of cone (r) = 5x

then height (h) = 12x

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

$=\frac{1}{3}(3 \cdot 14)(5 x)^{2} \times 12 x$

We know, Volume $=2512 \mathrm{~cm}^{3}$

$\Rightarrow \frac{1}{3}(3 \cdot 14) 25 x^{2} \times 12 x=2512$

$\Rightarrow \frac{1}{3} \times 3 \cdot 14 \times 300 x^{3}=2512$

$x^{3}=\frac{2512 \times 3}{3 \cdot 14 \times 300}$

$=\frac{2512 \times 3 \times 100}{314 \times 300}=8=(2)^{3}$

$\therefore x=2$

$\therefore$ Radius of cone $(r)=5 \times 2=10 \mathrm{~cm}$

and height $(h)=12 \times 2=24 \mathrm{~cm}$

Now slant height $(l)=\sqrt{r^{2}+h^{2}}$

$=\sqrt{(10)^{2}+(24)^{2}}$

$=\sqrt{100+576}=\sqrt{676}=26 \mathrm{~cm}$

Question 11

A cone and a cylinder are of the same height. If diameters of their bases are in the ratio 3 : 2, find the ratio of their volumes.

Sol :

Let height of cone and cylinder = h

Diameter of the base of cone = 3x

Diameter of base of cylinder = 2x

$\therefore$ Volume of cone $=\frac{1}{3} \pi\left(r_{1}\right)^{2} h$

$=\frac{1}{3} \pi\left(\frac{3 x}{2}\right)^{2} \times h=\frac{1}{3} \pi \frac{9}{4} x^{2} h=\frac{3}{4} \pi x^{2} h$

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

$=\pi\left(\frac{2 x}{2}\right)^{2} h=\pi x^{2} h$

$\therefore$ Ratio between the two volumes of two

sides $=\frac{3}{4} \pi x^{2} h: \pi x^{2} h$

$=\frac{3}{4}: 1 \Rightarrow 3: 4$

Question 12

A solid cone of base radius 9 cm and height 10 cm is lowered into a cylindrical jar of radius 10 cm, which contains water sufficient to submerge the cone completely. Find the rise in water level in the jar.

Sol :

Radius of the cone (r) = 9 cm

Height of the cone (h) = 10 cm

Volume of water filled in cone

$=\frac{1}{3} \pi r^{2} h=\frac{1}{3} \pi(9)^{2} \times 10 \mathrm{~cm}^{3}$

$=\frac{810}{3} \pi=270 \pi \mathrm{cm}^{3}$

Now radius of the cylindrical jar =10 cm Let h be the height of water in the jar

$\therefore \pi r^{2} h=270 \pi$

$\pi(10)^{2} h=270 \pi $

$\Rightarrow 100 \pi h=270 \pi$

$\Rightarrow h=\frac{270 \pi}{100 \pi}=2 \cdot 7 \mathrm{~cm}$

Question 13

An iron pillar has some part in the form of a right circular cylinder and the remaining in the form of a right circular cone. The radius of the base of each of cone and cylinder is 8 cm. The cylindrical part is 240 cm high and the conical part is 36 cm high. Find the weight of the pillar if one cu. cm of iron weighs 7.8 grams.

Sol :

Radius of the base of cone = 8 cm

and Radius of cylinder=8 cm

Height of cylindrical part $\left(h_{1}\right)=240 \mathrm{~cm}$ and height of conical part $\left(h_{2}\right)=36 \mathrm{~cm}$

Volume of the iron pillar

$=\frac{1}{3} \pi r^{2} h_{2}+\pi r^{2} h_{1}=\pi r^{2}\left(\frac{1}{3} h_{2}+h_{1}\right)$

$=\frac{22}{7} \times 8 \times 8 \cdot\left[\frac{1}{3} \times 36+240\right] \mathrm{cm}^{3}$

$=\frac{1408}{7}\left[\frac{36}{3}+240\right] \mathrm{cm}^{3}$

$=\frac{1408}{7}[252] \mathrm{cm}^{3}$

$=\frac{1408}{7} \times 252=1408 \times 36=50688 \mathrm{~cm}^{3}$

Weight of $1 \mathrm{~cm}^{3}=7 \cdot 8 \mathrm{gm}$

Total weight of the pillar $=50688 \times 7 \cdot 8 \mathrm{gm}=395366 \cdot 4 \mathrm{gm}$

$=395 \cdot 3664 \mathrm{~kg}$

Question 14

A circus tent is made of canvas and is in the form of right circular cylinder and a right circular cone above it. The diameter and height of the cylindrical part of the tent are 126 m and 5 m respectively. The total height of the tent is 21 m. Find the total cost of the tent if the canvas used costs Rs 36 per square metre.

Sol :

Diameter of the cylindrical part = 126 m

Radius $(r)=\frac{126}{2}=63 m$

Height of cylindrical part =5 m 

Total height of the tent =21 m

$\therefore$ Height of conical portion $=21-5=16 \mathrm{~m}$

$\therefore$ Slant height of the conical portion

$=\sqrt{r^{2}+h^{2}}=\sqrt{63^{2}+16^{2}}$

$=\sqrt{3969+256}=\sqrt{4225} \mathrm{~m}=65 \mathrm{~m}$

$\therefore$ Surface area of the tent

$=2 \pi r h+\pi r l$

$=\pi r(2 h+l)=\frac{22}{7} \times 63(2 \times 5+65)$

$=198 \times(10+65)=198 \times 75 \mathrm{~m}^{2}$

$=14850 \mathrm{~m}^{2}$

Cost of one 1 sq. m cloth= 36

∴Total cost=14850×36

=534600

Question 15

The entire surface of a solid cone of base radius 3 cm and height 4 cm is equal to the entire surface of a solid right circular cylinder of diameter 4 cm. Find the ratio of their

(i) curved surfaces

(ii) volumes.

Sol :

Radius of the base of a cone (r) = 3 cm

Height (h) = 4 cm

$l=\sqrt{r^{2}+h^{2}}=\sqrt{3^{2}+4^{2}}$

$=\sqrt{9+16}=\sqrt{25}=5 \mathrm{~cm}$

$\therefore$ Total surface $=\pi r l+\pi r^{2}$

$=\pi r(l+r)=\frac{22}{7} \times 3(5+3) \mathrm{cm}^{2}$

$=\frac{66}{7} \times 8=\frac{528}{7} \mathrm{~cm}^{2}$

Diameter of cylinder =4 cm

$\therefore$ Radius $\left(r_{1}\right)=\frac{4}{2}=2 \mathrm{~cm}$

Total surface area $=\frac{528}{7} \mathrm{~cm}^{2}$

Let h be the height, then

$\therefore 2 \pi r_{1} h_{1}+2 \pi r^{2}=\frac{528}{7}$

$\Rightarrow 2 \pi r\left(h_{1}+r\right)=\frac{528}{7}$

$\Rightarrow 2 \times \frac{22}{7} \times 2\left(h_{1}+2\right)=\frac{528}{7}$

$\Rightarrow h_{1}+2=\frac{528}{7} \times \frac{7}{2 \times 22 \times 2}$

$\Rightarrow h_{1}+2=6$

$h_{1}=6-2=4 \mathrm{~cm}$

(i) Ratio between curved surface of cone and cylinder

$=\pi r l: 2 \pi r_{1} h_{1}$

$=\pi \times 3 \times 5: 2 \times \pi \times 2 \times 4$

=15: 16

(ii) Ratio between their volumes 

$=\frac{1}{3} \pi r^{2} h: \pi r_{1}^{2} h_{1}$

$=\frac{1}{3} \pi \times 3 \times 3 \times 4: \pi \times 2 \times 2 \times 4$

=3 : 4

Question 16

A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. Find the radius of the sphere.

Sol :

Radius of base of a cone (r) = 2. 1 cm

and height (h) = 8.4 cm

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

$=\pi \times 4.41 \times 2.8 \mathrm{~cm}^{3}=12.348 \pi \mathrm{cm}^{3}$

$\therefore$ Volume of sphere $=12.348 \pi \mathrm{cm}^{3}$

Radius $=\left[\frac{\text { Volume }}{\frac{4}{3} \pi}\right]^{\frac{1}{3}}$

$=\left[\frac{12.348 \pi \times 3}{4 \times \pi}\right]^{\frac{1}{3}}=(9.261)^{\frac{1}{3}}$

$=(2.1 \times 2.1 \times 2.1)^{\frac{1}{3}}$

=2.1 cm

Question 17

How many lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimensions 66 cm, 42 cm and 21 cm.

Sol :

Dimensions of a solid rectangular lead piece

= 66 cm × 42 cm × 21 cm

$\therefore$ Volume $=66 \times 42 \times 21 \mathrm{~cm}^{3}$

Diameter of a lead shot = 4.2 cm

$\therefore$ Radius $(r)=\frac{4.2}{2}=2.1 \mathrm{~cm}$

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

$=\frac{4}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1 \mathrm{~cm}^{3}$

$=88 \times 0: 1 \times 4.41=38.808 \mathrm{~cm}^{3}$

Number of shots $=\frac{66 \times 42 \times 21}{38.808}$

$=\frac{66 \times 42 \times 21 \times 1000}{38808}=1500$

Question 18

Find the least number of coins of diameter 2.5 cm and height 3 mm which are to be melted to form a solid cylinder of radius 3 cm and height 5 cm.

Sol :

Radius of a cylinder (r) = 3 cm

Height (h) = 5 cm

$\therefore$ Volume $=\pi r^{2} h=\pi \times 3 \times 3 \times 5=45 \pi \mathrm{cm}^{2}$

Diameter of a coins $=2.5 \mathrm{~cm}$

$\therefore$ Radius $\left(r_{1}\right)=\frac{2.5}{2}=1.25 \mathrm{~cm}$

and height $\left(h_{1}\right)=3 \mathrm{~mm}=\frac{3}{10} \mathrm{~cm}$

$\therefore$ Volume of a coin $=\pi r_{1}^{2} h$

$=\pi \times 1.25 \times 1.25 \times \frac{3}{10} \mathrm{~cm}^{3}$

$=0.46875 \pi \mathrm{cm}^{3}$

$\therefore$ Number of coins required

$=\frac{45 \pi}{0.46875 \pi}=\frac{45}{0.46875}=96 \mathrm{coins}$

Question 19

A hemisphere of lead of radius 8 cm is cast into a right circular cone of base radius 6 cm. Determine the height of the cone correct to 2 places of decimal.

Sol :

Radius of hemisphere = 8 cm

Volume $=\frac{2}{3} \pi r^{3} c m^{3}$

$=\frac{2}{3} \pi(8)^{3} \mathrm{~cm}^{3}=\frac{2}{3} \pi 512 \mathrm{~cm}^{3}$

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

$\therefore$ Volume of right circular cone $=\frac{1024}{3} \pi \mathrm{cm}^{3}$

Radius =6 cm

Let h be the height of the cone

$\therefore \frac{1}{3} \pi r^{2} h=\frac{1024}{3} \pi$

$\Rightarrow \frac{1}{3} \pi(6)^{2} h=\frac{1024}{3} \pi$

$\Rightarrow 12 \pi h=\frac{1024}{3} \pi$

$\Rightarrow h=\frac{1024 \pi}{3 \times 12 \pi}=\frac{256}{9}=28 \cdot 44 \mathrm{~cm}$

Question 20

A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of the water in the cylinder.

Sol :

Radius of hemispherical bowl = 6 cm

∴ Volume of the water in the bowl

$=\frac{2}{3} \pi r^{3}=\frac{2}{3} \pi(6)^{3} \mathrm{~cm}^{3}=144 \pi \mathrm{cm}^{3}$

$\therefore$ Volume of water in the cylinder $=144 \pi \mathrm{cm}^{3}$

Radius of the cylinder =4 cm 

Let h be the height of water

$\therefore \pi r^{2} h=144 \pi$

$\Rightarrow(4)^{2} h=144 \Rightarrow 16 h=144$

$\therefore h=\frac{144}{16}=9$

Hence height of water in the cylinder =9 cm

Question 21

The diameter of a metallic sphere is 42 cm. It is melted and drawn into a cylindrical wire of 28 cm diameter. Find the length of the wire.

Sol :

Diameter of sphere = 42 cm

Radius of sphere $=\frac{42}{2}=21 \mathrm{~cm}$

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

$=\frac{4}{3} \pi(21)^{3} \mathrm{em}^{3}=12348 \pi \mathrm{cnt}^{3}$

Now volume of the wire drawn $=12348 \pi \mathrm{cm}^{3}$

and diameter =28 cm

$\therefore$ Radius $=\frac{28}{2}=14 \mathrm{~cm}$

Let length of wire $=h \mathrm{~cm}$

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

$=\pi(14)^{2} h \mathrm{~cm}^{2}=196 \pi h \mathrm{~cm}^{2}$

$\therefore 196 \pi h=12348 \pi$

$h=\frac{12348 \pi}{196 \pi}=\frac{12348}{196}=63 \mathrm{~cm}$

Question 22

A sphere of diameter 6 cm is dropped into a right circular cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel?

Sol :

Radius of sphere $=\frac{6}{2}=3 \mathrm{~cm}$

Figure to be added

Radius of cylinder $=\frac{12}{2}=6 \mathrm{~cm}$

Let height of water raised =h cm

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

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

and volume of water in the cylinder

$=36 \pi \mathrm{cm}^{3}$

$\therefore \pi r^{2} h=36 \pi \Rightarrow(6)^{2} h=36$

$\Rightarrow 36 h=36 \Rightarrow h=1$

$\therefore$ Height of raised water $=1 \mathrm{~cm}$

Question 23

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 5 cm and its height is 32 cm, find the uniform thickness of the cylinder.

Sol :

Radius of solid sphere = 6 cm

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

$=\frac{4}{3} \times \pi \times(6)^{3} \mathrm{~cm}^{3}=288 \pi \mathrm{cm}^{3}$

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

External radius of cylinder R=5 cm

and height (h)=32 cm

Let r be the inner radius

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

$\therefore \pi\left(\mathrm{R}^{2}-r^{2}\right) h=288 \pi$

$\left[(5)^{2}-r^{2}\right] \times 32=288 \Rightarrow 25-r^{2}=\frac{288}{32}$

$25-r^{2}=9 \Rightarrow r^{2}=25-9=16=(4)^{2}$

$\therefore r=4$

$\therefore$ Thickness of hollow cylinder $=\mathrm{R}-r$

=5-4=1 cm

Question 24

A solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is 3.5 cm and the height of the cone is 4 cm. The solid is placed in a cylindrical vessel, full of water, in such a Way that the whole solid is submerged in water. If the radius of the cylindrical vessel is 5 cm and its height is 10.5 cm, find the volume of water left in the cylindrical vessel.

Sol :

Radius of hemisphere (r) = 3.5 cm

Height of cone (h1) = 4 cm

Radius of cylindrical vessel $=5 \mathrm{~cm}$

and height =10.5 cm

Volume of solid $=\frac{2}{3} \pi r^{3}+\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \pi r^{2}\left(2 r+h_{1}\right)$

$=\frac{1}{3} \times \frac{22}{7} \times(3 \cdot 5)^{2}[2 \times 3 \cdot 5+4] \mathrm{cm}^{2}$

$=\frac{1}{3} \times \frac{22}{7} \times \frac{12 \cdot 25}{1}[7+4] \mathrm{cm}^{2}$

$=\frac{1}{3} \times \frac{22}{7} \times \frac{1225}{100} \times 11=\frac{847}{6} \mathrm{~cm}^{3}$

Radius of cylinder $=5 \mathrm{~cm}$

Height of cylinder $=10 \cdot 5 \mathrm{~cm}$

$\therefore$ Volume of cylinder which is full of water

$=\pi r^{2} h=\frac{22}{7} \times 5 \times 5 \times 10 \cdot 5 \mathrm{~cm}^{3}$

$=22 \times 25 \times 1 \cdot 5 \mathrm{~cm}^{3}=825 \mathrm{~cm}^{3}$

$\therefore$ Volume of water left in the cylinder

$=825-\frac{847}{6}=825-141 \cdot 97 \mathrm{~cm}^{3}$

$=683 \cdot 83 \mathrm{~cm}^{3}$

ML Aggarwal Solution Class 10 Chapter 17 Mensuration Exercise 17.5

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29

 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 $=\frac{89.83 \times 7}{22 \times 3.5 \times 3.5}$

$=2.32 \mathrm{~cm}$

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

ML Aggarwal Solution Class 10 Chapter 17 Mensuration Exercise 17.4

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21

 Exercise 17.4

Question 1

The adjoining figure shows a cuboidal block of wood through which a circular cylinderical hole of the biggest size is drilled. Find the volume of the wood left in the block.

Sol :

Diameter of the biggest hole = 30 cm.

Radius $(r)=\frac{30}{2}=15 \mathrm{~cm}$

and height (h) = 70 cm.

$\therefore$ Volume of the hole made $=\pi r^{2} h$

$=\frac{22}{7} \times 15 \times 15 \times 70 \mathrm{~cm}^{3}=49500 \mathrm{~cm}^{3}$

Total volume of the $\log =70 \times 30 \times 30 \mathrm{~cm}^{3}$

$=63000 \mathrm{~cm}^{3}$

$\therefore$ Volume of the wood left $=63000-49500$

$=13500 \mathrm{~cm}^{3}$

Question 2

The given figure shows a solid trophy made of shining glass. If one cubic centimetre of glass costs Rs 0.75, find the cost of the glass for making the trophy.

Sol :

Edge of cubical part = 28 cm

and radius of cylindrical part (r) $=\frac{28}{2}=14 \mathrm{~cm}$

Height $(h)=28 \mathrm{~cm}$

$\therefore$ Total volume of the trophy $=(\mathrm{Edg} \mathrm{e})^{3}+\pi r^{2} h$

$=(28)^{3}+\frac{22}{7} \times 14 \times 14 \times 28 \mathrm{~cm}^{3}$

$=(21952+17248) \mathrm{cm}^{3}=39200 \mathrm{~cm}^{3}$

Rate of cost of glass $=₹ 0.75$ per $\mathrm{cm}^{3}$

$\therefore$ Total $\operatorname{cost}=₹ 39200 \times 0.75=₹ 29400$

Question 3

From a cube of edge 14 cm, a cone of maximum size is carved out. Find the volume of the remaining material.

Sol :

Edge of a cube = 14 cm

Volume = (side)³ = (14)³ = 14 × 14 × 14 cm³ = 2744 cm³

Now diameter of the cone cut out from it = 14 cm

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

and height $=14 \mathrm{~cm}$

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

$=\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 14$

$=\frac{2156}{3}=718 \frac{2}{3} \mathrm{~cm}^{3}$

$\therefore$ Volume of the remaining portion

$=2744-718 \frac{2}{3}=2025 \frac{1}{3} \mathrm{~cm}^{3}$

Question 4

A cone of maximum volume is curved out of a block of wood of size 20 cm x 10 cm x 10 cm. Find the volume of the remaining wood.

Sol :

Size of wooden block = 20 cm × 10 cm × 10 cm

Maximum diameter of the cone = 10 cm

and height (h) = 20 cm

$\therefore$ Radius $(r)=\frac{10}{2}=5 \mathrm{~cm}$

Now volume of block $=20 \times 10 \times 10$

$=2000 \mathrm{~cm}^{3}$

and volume of cone $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 20 \mathrm{~cm}^{3}=\frac{11000}{21} \mathrm{~cm}^{3}$

$\therefore$ Volume of the remaining portion

$=2000-\frac{11000}{21}$

$=\frac{42000-11000}{21}=\frac{31000}{21}=1476.19 \mathrm{~cm}^{3}$

$=1476 \frac{2}{3} \mathrm{~cm}^{3}($ approx. $)$

Question 5

16 glass spheres each of radius 2 cm are packed in a cuboidal box of internal dimensions 16 cm x 8 cm x 8 cm and then the box is filled with water. Find the volume of the water filled in the box.

Sol :

Given

Radius of each glass sphere = 2 cm

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

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

Volume of 16 glass spheres $=\frac{704}{21} \times 16 \mathrm{~cm}^{3}$

$=\frac{11264}{21} \mathrm{~cm}^{3}=536.38 \mathrm{~cm}^{3}$

Volume of box $=16 \times 8 \times 8=1024 \mathrm{~cm}^{3}$

$\therefore$ Volume of remaining space for water

$=1024-536.38=487.62 \mathrm{~cm}^{3}$

Question 6

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depression is 0.5 cm and the depth is 1.4 cm. Find the volume of the wood in the entire stand, correct to 2 decimal places.

Sol :

Dimensions of cuboid = 15 cm × 10 cm × 3.5 cm

and radius of each conical depression (r) = 0.5 cm

and depth (h) = 1.4 cm

Volume of cuboid = l × b × h

$=15 \times 10 \times 3.5 \mathrm{~cm}^{3}=525 \mathrm{~cm}^{2}$

and volume 4 conical depressions

$=4 \times \frac{1}{3} \pi r^{2} h=\frac{4}{3} \times \frac{22}{7} \times(0.5)^{2} \times 1.4 \mathrm{~cm}^{2}$

$=\frac{88}{21} \times 0.25 \times 1.4 \mathrm{~cm}^{3}$

$=\frac{30.8}{21} \mathrm{~cm}^{2}=1.467 \mathrm{~cm}^{3}$

$\therefore$ Volume of wood in the stand

$=525-1.467=523.533 \mathrm{~cm}^{3}$

Question 7

A cuboidal block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter that the hemisphere can have? Also, find the surface area of the solid.

Sol :

Side of cuboidal = 7 cm

Diameter of hemisphere = 7 cm

and radius $(r)=\frac{7}{2} \mathrm{~cm}$

$\therefore$ Surface area of the total solid $=$

$5 a^{2}+2 \pi r^{2}=5(7)^{2}+2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \mathrm{~cm}^{2}$

$=245+77=322 \mathrm{~cm}^{2}$

Question 8

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder (as shown in the given figure). If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the total surface area of the article.

Sol :

Height of the cylinder = 10 cm

and radius of the base = 3.5 cm

Total surface area

= Curved surface area of cylinder + 2 × Curved surface area of hemisphere

$=2 \pi r h+2 \times 2 \pi r^{2}$

$=2 \times \frac{22}{7} \times 3.5 \times 10+2 \times 2 \times \frac{22}{7} \times 3.5 \times 3.5 \mathrm{~cm}^{2}$

$=220+154=374 \mathrm{~cm}^{2}$

Question 9

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. If the total height of the toy is 15.5 cm, find the total surface area of the toy.

Sol :

Total height of the toy = 15.5 cm

Radius of the base of the conical part (r) = 3.5 cm

∴Height of conical part (h)=15.5-3.5=12 cm

$\therefore$ Slant height $(l)=\sqrt{r^{2}+h^{2}}$

$=\sqrt{(3.5)^{2}+12^{2}}=\sqrt{12.25+144}$

$=\sqrt{156.25 \mathrm{~cm}}=12.5 \mathrm{~cm}$

Total surface area of the toy $=\pi r l+2 \pi r^{2}$

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

$=\frac{22}{7} \times 3.5(12.5+2 \times 3.5) \mathrm{cm}^{2}$

$=11(12.5+7) \mathrm{cm}^{2}=11 \times 19.5 \mathrm{~cm}^{2}$

$=214.5 \mathrm{~cm}^{2}$

Question 10

A circus tent is in the shape of a cylinder surmounted by a cone. The diameter of the cylindrical portion is 24 m and its height is 11 m. If the vertex of the cone is 16 m above the ground, find the area of the canvas used to make the tent.

Sol :

Radius of base of cylindrical portion of tent (r)

$=\frac{24}{2}=12 \mathrm{~m}$

Height of cylindrical portion (h)=11 m

Height of conical part (l)=16-11=5 m 

Radius of conical part =12 m

$\therefore$ Slant height $(l)=\sqrt{r^{2}+h_{2}^{2}}$

$=\sqrt{(12)^{2}+(5)^{2}}$

$=\sqrt{144+25}=\sqrt{169}=13^{\circ} \mathrm{m}$

Now surface area of tent $=\pi r l+2 \pi r h$

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

$=\frac{22}{7} \times 12(13+2 \times 11) \mathrm{m}^{2}$

$=\frac{22}{7} \times 12 \times 35 \mathrm{~m}^{2}=1320 \mathrm{~m}^{2}$

Question 11

An exhibition tent is in the form of a cylinder surmounted by a cone. The height of the tent above the ground is 85 m and the height of the cylindrical part is 50 m. If the diameter of the base is 168 m, find the quantity of canvas required to make the tent. Allow 20% extra for folds and stitching. Give your answer to the nearest m².

Sol :

Total height of the tent = 85 m

Height of cylindrical part $(h_1)$ = 50 m

$\therefore$ Height of conical part $\left(h_{2}\right)=85-50$ 

$=35 \mathrm{~m}$

Diameter of base=168 m

$\therefore$ Radius $=\frac{168}{2}=84 m$

Slant height $(l)=\sqrt{r^{2}+h_{2}^{2}}$

$=\sqrt{(84)^{2}+(35)^{2}}=\sqrt{7056+1225}$

$=\sqrt{8281}=91 \mathrm{~m}$

Now surface area of the tent

$=\pi r l+2 \pi r h=\pi r\left(l+2 h_{1}\right)$

$=\frac{22}{7} \times 84(91+2 \times 50)$

$=\frac{22}{7} \times 84 \times 191 m^{2}=50424 m^{2}$

Extra canvas for fold and stiching @) $20 \%$

$=50424 \times \frac{20}{100}=10084 \cdot 8 m^{2}$

$\therefore$ Total canvas required $=50424+10084 \cdot 8 m^{2}$

$=60508 \cdot 8 m^{2}=60509 m^{2}$

Question 12

From a solid cylinder of height 30 cm and radius 7 cm, a conical cavity of height 24 cm and of base radius 7 cm is drilled out. Find the volume and the total surface of the remaining solid.

Sol :

Radius of solid cylinder $(r_1)$ = 7 cm

Height $\left(h_{1}\right)=30 \mathrm{~cm}$

Radius of cone $\left(r_{2}\right)=7 \mathrm{~cm}$ 

and height $\left(h_{2}\right)=24 \mathrm{~cm}$

Volume of cylinder $=\pi r^{2} h_{1}$

and volume of cone $=\frac{1}{3} \pi r^{2} h_{2}$

$\therefore$ Volume of remaining part

$=\pi r^{2} h_{1}-\frac{1}{3} \pi r^{2} h_{2}=\pi r^{2}\left(h_{1}-\frac{1}{3} h_{2}\right)$

$=\frac{22}{7} \times 7 \times 7\left(30-\frac{1}{3} \times 24\right)$

$=154(30-8) \mathrm{cm}^{3}=154 \times 22=3388 \mathrm{~cm}^{3}$

$l=\sqrt{r^{2}+h_{2}^{2}}=\sqrt{(7)^{2}+(24)^{2}}$

$=\sqrt{49+576}=\sqrt{625}=25 \mathrm{~cm}$

$\therefore$ Surface area of the remaining part $=2 \pi r h_{1}+\pi r^{2}+\pi r l=\pi r\left(2 h_{1}+r+l\right)$

$=\frac{22}{7} \times 7(2 \times 30+7+25)$

$=22(60+7+25)=22 \times 92=2024 \mathrm{~cm}^{2}$

Question 13

The given figure shows a wooden toy rocket which is in the shape of a circular cone mounted on a circular cylinder. The total height of the rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted green and the cylindrical portion red, find the area of the rocket painted with each of these colors. Also, find the volume of the wood in the rocket. Use π = 3.14 and give answers correct to 2 decimal places.

Sol :

In the given figure,

The total height of the toy rocket = 26 cm

Diameter of cylindrical portion = 3 cm

$\therefore$ Radius $\left(r_{1}\right)=\frac{3}{2} \mathrm{~cm}=1.5 \mathrm{~cm}$

and height $\left(h_{1}\right)=25-6 \mathrm{~cm}=20 \mathrm{~cm}$

Diameter of conical portion $=5 \mathrm{~cm}$

$\therefore$ Radius $\left(r_{2}\right)=\frac{5}{2} \mathrm{~cm}=2.5 \mathrm{~cm}$

and height $\left(h_{2}\right)=6 \mathrm{~cm}$

$\therefore$ Slant height $(l)=\sqrt{r^{2}+h^{2}}=\sqrt{(2.5)^{2}+(6)^{2}}$

$=\sqrt{6.25+36}=\sqrt{42.25}=6.5 \mathrm{~cm}$

Now curved surface area of conical portion $=\pi r_{2} l=3.14 \times 2.5 \times 6.5 \mathrm{~cm}^{2}=51.025 \mathrm{~cm}^{2}$

and curved surface area of cylindrical portion $=2 \pi r_{1} h_{1}$

$=2 \times 3.14 \times 1.5 \times 20 \mathrm{~cm}^{2}=188.4 \mathrm{~cm}^{2}$

$\therefore$ Total surface area of cylindrical portion

$=2 \pi r_{1} h_{1}+\pi r_{1}^{2}$

$=188.4+3.14 \times(1.5)^{2} \mathrm{~cm}^{3}$

$=188.4+3.14 \times 2.25$

$=188.4+7.065=195.465 \mathrm{~cm}^{3}$

$=195.47 \mathrm{~cm}^{3}$

and Total surface area of conical portion 

$=\pi r_{2} l+\pi r_{2}^{2}-\pi r_{1}^{2}$

$=51.025+3.14 \times(2.5)^{2}-7.065$

$=51.025+3.14 \times 6.25-7.065$

$=51.025+19.625-7.065 \mathrm{~cm}^{2}$

$=70.65-7.065=63.585 \mathrm{~cm}^{2}$

$=63.59 \mathrm{~cm}^{2}$

(ii) Total volume of the toy

$=\pi r_{1}^{2} h_{1}+\frac{1}{3} \pi r_{2}^{2} h_{2}$

$=3.14 \times(1.5)^{2} \times 20+\frac{1}{3} \times 3,14 \times(2.5)^{2} \times 6$

$=3.14 \times 2.25 \times 20+\frac{1}{3} \times 3.14 \times 6.25 \times 6$

$=141.3+39.25=180.55 \mathrm{~cm}^{3}$

Question 14

The given figure shows a hemisphere of radius 5 cm surmounted by a right circular cone of base radius 5 cm. Find the volume of the solid if the height of the cone is 7 cm. Give your answer correct to two places of decimal.

Sol :

Height of the conical part (h) = 7 cm

and radius of the base (r) = 5 cm

$\therefore$ Total volume $=\frac{1}{3} \pi r^{2} h+\frac{2}{3} \pi r^{2}$

$=\pi r^{2}\left[\frac{1}{3} h+\frac{2}{3} r\right]$

$=\frac{22}{7} \times 5 \times 5\left[\frac{1}{3} \times 7+\frac{2}{3} \times 5\right] \mathrm{cm}^{3}$

$=\frac{550}{7}\left[\frac{7}{3}+\frac{10}{3}\right]=\frac{550}{7} \times \frac{17}{3} \mathrm{~cm}^{3}$

$=\frac{9350}{21} \mathrm{~cm}^{3}=445.238 \mathrm{~cm}^{3}$

$=445.24 \mathrm{~cm}^{3}$ (approx.)

Question 15

A buoy is made in the form of a hemisphere surmounted by a right cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 metres and its volume is $\frac{2}{3}$ of the hemisphere. Calculate the height of the cone and the surface area of the buoy correct to 2 places of decimal

Sol :

Radius of base of hemisphere $=\frac{7}{2} m$

$\therefore $ Volume of hemisphere $=\frac{2}{3} \pi r^{3}$

$=\frac{2}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} m^{3}=\frac{539}{6} m^{3}$

$\therefore$ Volume of cone $=\frac{539}{6} \times \frac{2}{3}=\frac{539}{9} m^{3}$

Let height of the cone =h

$\therefore  \frac{1}{3} \pi r^{2} h=\frac{539}{9}$

$\Rightarrow \frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} h=\frac{539}{9}$

$\Rightarrow \frac{77}{6} h=\frac{539}{9} $

$\Rightarrow h=\frac{539}{9} \times \frac{6}{77}=\frac{14}{3} m$

$\Rightarrow h=4 \cdot 67 \mathrm{~m}$

$\therefore$ Height of cone $=4 \cdot 67 \mathrm{~m}$

$l=\sqrt{r^{2}+h^{2}}=\sqrt{\left(\frac{7}{2}\right)^{2}+\left(\frac{14}{3}\right)^{2}}$

$=\sqrt{\frac{49}{4}+\frac{196}{9}}$

$=\sqrt{\frac{441+784}{36}}$

$=\sqrt{\frac{1225}{36}}=\frac{35}{6} m$

Now surface area of the buoy

$=2 \pi r^{2}+\pi r l$

$=2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}+\frac{22}{7} \times \frac{7}{2} \times \frac{35}{6}$

$=\frac{77}{1}+\frac{385}{6}=\frac{462+385}{6} \mathrm{~m}^{2}$

$=\frac{847}{6}=141 \cdot 17 m^{2}$

Question 16

A circular hall (big room) has a hemispherical roof. The greatest height is equal to the inner diameter, find the area of the floor, given that the capacity of the hall is 48510 m³.

Sol :

Let h be the greatest height

and r be the radius of the base

Then 2r = h + r ⇒ h = r

Volume of the hall $=\frac{2}{3} \pi r^{3}+\pi r^{2} h$

$=\frac{2}{3} \pi r^{3}+\pi r^{2} \times(r)$

$=\frac{2}{3} \pi r^{3}+\pi r^{3}=\frac{5}{3} \pi r^{3}$

$\therefore \frac{5}{3} \pi r^{3}=48510$

$\Rightarrow \frac{5}{3} \times \frac{22}{7} r^{3}=48510$

$\Rightarrow r^{3}=\frac{48510 \times 3 \times 7}{5 \times 22}$

$\Rightarrow r^{3}=\frac{46305}{5}=9261=(21)^{3}$

$\therefore r=21 m$

Area of floor $=\pi r^{2}$

$=\frac{22}{7} \times 21 \times 21=1386 \mathrm{~m}^{2}$

Question 17

A building is in the form of a cylinder surmounted by a hemisphere valted dome and contains $41 \frac{19}{21} m^3$ of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building

Sol :

Volume of air in dome = $41 \frac{19}{21} \mathrm{~m}^{3}$

$=\frac{880}{21} m^{3}$

Let radius of the dome = r m

Then height (h) = r m

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

$=\pi r^{2} \times r+\frac{2}{3} \pi r^{3}$

$=\pi r^{3}+\frac{2}{3} \pi r^{3}=\frac{5}{3} \pi r^{3}$

$\therefore \frac{5}{3} \times \frac{22}{7} r^{3}=\frac{880}{21}$

$ \Rightarrow r^{3}=\frac{880}{21} \times \frac{21}{110}$

$r^{3}=\frac{88}{11}=8=(2)^{3}$

$\therefore r=2 \mathrm{~m}$

Height of the building $=2 r=2 \times 2=4 \mathrm{~m}$

Question 18

A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and the height of the cylinder are 6 cm and 12 cm respectively. If the slant height of the conical portion is 5 cm, find the total surface area and the volume of the rocket. (Use π = 3.14).

Sol :

Height of the cylindrical part (A) = 12 cm

Diameter = 6 cm

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

Slant height of the conical part (l) = 5 cm

(i) ∴Total surface area of the rocket so formed

$=\pi r l+2 \pi r h+\pi r^{2}$

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

$=3.14 \times 3[5+2 \times 12+3] \mathrm{cm}^{2}$

$=9.42 \times[5+24+3] \mathrm{cm}^{2}$

$=9.42 \times 32=301.44 \mathrm{~cm}^{2}$

(ii) Volume $=\frac{1}{3} \pi r^{2} h_{1}+\pi r^{2} h$

$=\pi r^{2}\left(\frac{1}{3} h_{1}+h\right)$

$=3.14 \times 3 \times 3\left[\frac{1}{3} \sqrt{l^{2}-r^{2}}+12\right] \mathrm{cm}^{3}$

$=28.26 \times\left[\frac{1}{3} \sqrt{5^{2}-3^{2}}+12\right] \mathrm{cm}^{3}$

$=28.26 \times\left[\frac{1}{3} \sqrt{16}+12\right] \mathrm{cm}^{3}$

$=28.26 \times\left(\frac{4}{3}+12\right)=28.26 \times \frac{40}{3} \mathrm{~cm}^{3}$

$=9.42 \times 40=376.80=376.8 \mathrm{~cm}^{3}$

Question 19

A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. Their common diameter is 3.5 cm and the height of the cylindrical and conical portions are 10 cm and 6 cm respectively. Find the volume of the solid. (Take π = 3.14)

Sol :

Diameter = 3.5 cm

Radius $(r)=\frac{3.5}{2}=1.75 \mathrm{~cm}$

Height of cylindrical part $(h_1)$ = 10 cm

and height of conical part $(h_2)$ = 6 cm

$\therefore$ Total volume of the so formed solid

$=\frac{1}{3} \pi r^{2} h_{1}+\pi r^{2} h+\frac{2}{3} \pi r^{3}$

$=\pi r^{2}\left[\frac{1}{3} h_{1}+h+\frac{2}{3} r\right]$

$=3.14 \times 1.75 \times 1.75\left[\frac{1}{3} \times 6+10+\frac{2}{3} \times 1.75\right] \mathrm{cm}^{3}$

$=9.61625\left[2+10+\frac{3.5}{3}\right] \mathrm{cm}^{3}$

$=9.61625[12+1.167] \mathrm{cm}^{3}$

$=9.61625 \times 13.167 \mathrm{~cm}^{3}=126.617 \mathrm{~cm}^{3}$

$=126.61 \mathrm{~cm}^{3}$

Question 20

A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The height and radius of the cylindrical part are 13 cm and 5 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Calculate the surface area of the toy if the height of the conical part is 12 cm.

Sol :

Height of cylindrical part = 13 cm

Radius = 5 cm

Radius of cone (r) = 5 cm

Height of cone (h) = 12 cm

$\therefore$ slant height $(l)=\sqrt{r^{2}+h^{2}}$

$=\sqrt{(5)^{2}+(12)^{2}}$

$=\sqrt{25+144}=\sqrt{169}=13 \mathrm{~cm}$

Now surface area of the toy $=\pi r l+2 \pi r h+2 \pi r^{2}=\pi r(l+2 h+2 r)$

$=\frac{22}{7} \times 5[13+2 \times 13+2 \times 5] \mathrm{cm}^{2}$

$=\frac{110}{7}(13+26+10) \mathrm{cm}^{2}$

$=\frac{110}{7} \times 49=770 \mathrm{~cm}^{2}$

Question 21

The given figure shows a model of a solid consisting of a cylinder surmounted by a hemisphere at one end. If the model is drawn to a scale of 1 : 200, find

(i) the total surface area of the solid in π m².

(ii) the volume of the solid in π litres.

Sol :

(i) In the given figure,

Height of cylindrical portion (h) = 8 cm

Radius (r) = 3 cm

Scale = 1 : 200

Total surface area $=2 \pi r^{2}+2 \pi r h$

$=2 \pi r[r+h] \mathrm{cm}^{2}$

$=2 \times \pi \times 3[3+8] \mathrm{cm}^{2}$

$=6 \pi \times 11=66 \pi \mathrm{cm}^{2}$

$\therefore$ Surface area of the solid $=66 \pi \times \frac{(200)^{2}}{1} \mathrm{~cm}^{2}$

$=66 \pi \times 40000 \mathrm{~cm}^{2}$

$=\frac{66 \times 40000}{100 \times 100} \pi=264 \pi \mathrm{m}^{2}$

(ii) and volume in π litres (of model)

$=\frac{2}{3} \pi r^{3}+\pi r^{2} h$

$=\pi r^{2}\left(\frac{2}{3} r+h\right) \mathrm{cm}^{3}$

$=\pi \times(3)^{2}\left(\frac{2}{3} \times 3+8\right) \mathrm{cm}^{3}$

$=9 \pi(2+8)=90 \pi \mathrm{cm}^{3}$

Scale =1: 200

$\therefore$ Capacity $=90 \pi \times(200)^{3} \mathrm{~cm}^{3}$

$=90 \pi \times 8000000 \mathrm{~cm}^{3}$

$=720000000 \pi \mathrm{cm}^{3}$

$=\frac{720000000 \pi}{100 \times 100 \times 100}=720 \pi \mathrm{m}^{3}$

$=720 \pi \times 1000$ litres $=720000 \pi$ litres