ML Aggarwal Solution Class 10 Chapter 17 Mensuration Test
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
$\therefore$ 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
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
$\therefore$ 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
$=\frac{1}{3} \times 156 \times 8$
$=\frac{1248}{3} \mathrm{~cm}^{3}=416 \mathrm{~cm}^{3}$
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
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
$\Rightarrow \frac{2}{3} \times \frac{22}{7} \times r^{3}=\frac{4851}{2}$
$\Rightarrow r^{3}=\frac{4851 \times 3 \times 7}{2 \times 2 \times 22}=\frac{9261}{8}=\left(\frac{21}{2}\right)^{3}$
$\therefore r=\frac{21}{2} \mathrm{~cm}$
∴ 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
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
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
$\therefore$ 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
$=\frac{1}{3}(3 \cdot 14)(5 x)^{2} \times 12 x$
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
$=\frac{1}{3} \pi\left(\frac{3 x}{2}\right)^{2} \times h=\frac{1}{3} \pi \frac{9}{4} x^{2} h=\frac{3}{4} \pi x^{2} h$
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
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
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
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
$\therefore$ 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
$=\pi \times 4.41 \times 2.8 \mathrm{~cm}^{3}=12.348 \pi \mathrm{cm}^{3}$
$\therefore$ 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
Diameter of a lead shot = 4.2 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
Question 19
Question 20
$\Rightarrow(4)^{2} h=144 \Rightarrow 16 h=144$
$\therefore h=\frac{144}{16}=9$
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
$\therefore$ 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?
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
$=\frac{4}{3} \times \pi \times(6)^{3} \mathrm{~cm}^{3}=288 \pi \mathrm{cm}^{3}$
$\therefore$ 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
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
$=825-\frac{847}{6}=825-141 \cdot 97 \mathrm{~cm}^{3}$
$=683 \cdot 83 \mathrm{~cm}^{3}$
Comments
Post a Comment