ML Aggarwal Solution Class 10 Chapter 17 Mensuration MCQs

 MCQs

Question 1

In a cylinder, if radius is halved and height is doubled then the volume will be

(a) same

(b) doubled

(c) halved

(d) four times

Sol :

Let radius of cylinder = r

and height = h

then volume = πr²h

If the radius is halved and the height is doubled

Then volume $=\pi\left(\frac{r}{2}\right)^{2} \times 2 h$

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

which is half

Ans (c)

Question 2

In a cylinder, if the radius is doubled and height is halved then its curved surface area will be

(a) halved

(b) doubled

(c) same

(d) four times

Sol :

Let radius of a cylinder = r

and height = h

Then curved surface area = 2πrh

Now if radius is doubled and height is halved,

then curved surface area $=2 \pi \frac{r}{2} \times 2 h=2 \pi r h$ which is same (c)

Question 3

If a well of diameter 8 m has been dug to the depth of 14 m, then the volume of the earth dug out is

(a) 352 $m^{3}$

(b) 704 $m^{3}$

(c) 1408 $m^{3}$

(d) 2816 $m^{3}$

Sol :

Diameter of a well = 8 m

Radius $(r)=\frac{8}{2}=4 m$

Depth (h) = 14 m

Volume of the earth dug put = $\pi r^{2} h$

$=\frac{22}{7} \times 4 \times 4 \times 14 \mathrm{~m}^{3}$

$=704 \mathrm{~m}^{3}$

Ans (b)

Question 4

If two cylinders of the same lateral surface have their radii in the ratio 4 : 9, then the ratio of their heights is

(a) 2 : 3

(b) 3 : 2

(c) 4 : 9

(d) 9 : 4

Sol :

Ratio in two cylinder having same lateral surface area their radii is 4 : 9

Let r1 be the radius of the first and $r_2$ be the second cylinder

and $h_1, h_2$ and their heights

Let $r_{1}=4 x$ and $r_{2}=9 x$

$\therefore 2 \pi r_{1} h_{1}=2 \pi r_{2} h_{2}$

$=2 \pi 4 x \times h_{1}=2 \times \pi \times 9 x h_{2}$

$\frac{h_{1}}{h_{2}}=\frac{9 x}{4 x}=9: 4$

Ratio in their heights = 9 : 4 

Ans (d)

Question 5

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their volumes is

(a) 10 : 17

(b) 20 : 27

(c) 17 : 27

(d) 20 : 37

Sol :

Radii of two cylinder are in the ratio = 2 : 3

Let radius $(r_1)$ = 2x

and radius $(r_2)$ = 3x

Ratio in their height = 5 : 3

Let height of the first cylinder = 5y

and of second = 3y

Now, volume of the first cylinder

$\pi r_{1}^{2} h=\pi(2 x)^{2} \times 5 h=20 \pi x^{2} y$

and volume of second $=\pi(3 x)^{2} \times 3 y$

$=\pi \times 27 x^{2} y$

$\therefore$ Ratio is $20 \pi x^{2} y: 27 \pi x^{2} y$

$=20: 27$

Ans (b)

Question 6

The total surface area of a cone whose radius is $\frac{r}{2}$ and slant height 2l is

(a) 2πr (l + r)

(b) $\pi r\left(l+\frac{r}{4}\right)$

(c) πr(l + r)

(d) 2πrl

Sol :

Radius of a cone $=\frac{r}{2}$

and slant height = 2l

total surface area of a cone

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

$=\pi \frac{r}{2} \times 2 l+\pi\left(\frac{r}{2}\right)^{2}$

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

$=\pi r\left(l+\frac{r}{4}\right)$

Ans (b)

Question 7

If the diameter of the base of cone is 10 cm and its height is 12 cm, then its curved surface area is

(a) 60π $cm^2$

(b) 65π $cm^2$

(c) 90π $cm^2$

(d) 120π $cm^2$

Sol :

Diameter of the base of a cone = 10 cm

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

and height (h) = 12 cm

$l=\sqrt{r^{2}+h^{2}}=\sqrt{5^{2}+12^{2}}$

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

Curved surface area 

$=\pi r l=\pi \times 5 \times 13=65 \pi \mathrm{cm}^{2}$

Ans (b)

Question 8

If the diameter of the base of a cone is 12 cm and height is 20 cm, then its volume is ,

(a) 240π $cm^3$

(b) 480π $cm^3$

(c) 720π $cm^3$

(d) 960π $cm^3$

Sol :

Diameter of the base of a cone = 12 cm

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

and height (h) = 20 cm

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

$=\frac{1}{3} \pi \times 6 \times 6 \times 20 \mathrm{~cm}^{3}$

$=240 \pi \mathrm{cm}^{3}$

Ans (a)

Question 9

If the radius of a sphere is 2r, then its volume will be

(a) $\frac{4}{3} \pi r^{3}$

(b) $4 \pi r^{3}$

(c) $\frac{8 \pi r^{3}}{3}$

(d) $\frac{32 \pi r^{3}}{3}$

Sol :

Radius of a sphere = 2r

Volume $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi \times(2 r)^{3}$

$=\frac{4}{3} \pi \times 8 r^{3}=\frac{32 \pi r^{3}}{3}$

Ans (d)

Question 10

If the diameter of a sphere is 16 cm, then its surface area is

(a) 64π $cm^2$

(b) 256π $cm^2$

(c) 192π $cm^2$

(d) 256 $cm^2$

Sol :

Diameter of a sphere = 16 cm

Radius $(r)=\frac{16}{2}=8 \mathrm{~cm}$

Surface area $=4 \pi^{2}=4 \pi \times 8 \times 8 \mathrm{~cm}^{2}=256 \pi \mathrm{cm}^{2}$

Ans (b)

Question 11

If the radius of a hemisphere is 5 cm, then its volume is

(a) $\frac{250}{3} \pi \quad \mathrm{cm}^{3}$

(b) $\frac{500}{3} \pi \quad c m^{3}$

(c) $75 \pi \mathrm{cm}^{3}$

(d) $\frac{125}{3} \pi \quad c m^{3}$

Sol :

Radius of a hemisphere (r) = 5 cm

Volume $=\frac{2}{3} \pi r^{3}=\frac{2}{3} \pi(5)^{3} \mathrm{~cm}^{3}$

$=\frac{250}{3} \pi \mathrm{cm}^{3}$

Ans (a)

Question 12

If the ratio of the diameters of the two spheres is 3 : 5, then the ratio of their surface areas is

(a) 3 : 5

(b) 5 : 3

(c) 27 : 125

(d) 9 : 25

Sol :

Ratio in the diameters of two spheres = 3 : 5

Let radius of the first sphere = 3x cm

and radius of the second sphere = 5x cm

Ratio in their surface area

$=4 \pi(3 x)^{2}: 4 \pi(5 x)^{2}$

$\Rightarrow 9 x^{2}: 25 x^{2}$

=9: 25

Ans (d)

Question 13

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratio of the surface areas of the balloon in the two cases is

(a) 1 : 4

(b) 1 : 3

(c) 2 : 3

(d) 2 : 1

Sol :

Radius of balloon (hemispherical) in the original position = 6 cm

and in increased position = 12 cm

Ratio in their surface areas

$4 \pi(6)^{2}: 4 \pi(12)^{2}$

$=6^{2}: 12^{2}=36: 144$

=1: 4

Ans (a)

Question 14

The shape of a Gilli, in the game of Gilli- danda, is a combination of

(a) two cylinders

(b) a cone and a cylinders

(c) two cones and a cylinder

(d) two cylinders and a cone

Sol :

The shape of a Gilli is the combination of

two cones and a cylinder (as shown in the figure). 

Ans (c)

Question 15

If two solid hemisphere of same base radius r are joined together along with their bases, then the curved surface of this new solid is

(a) $4πr^2$

(b) $6πr^2$

(c) $3πr^2$

(d) $8πr^2$

Sol :

Radius of two solid hemispheres = r

These are joined together along with the bases

Curved surface area $= 2π^2 × 2 = 4πr^2$

Ans (a)

Question 16

During conversion of a solid from one shape to another, the volume of the new shape will

(a) increase

(b) decrease

(c) remain unaltered

(d) be doubled

Sol :

During the conversion of a solid into another, the volume of the new shaper will be the same.

i.e. remain unaltered

Ans (c)

Question 17

If a solid of one shape is converted to another, then the surface area of the new solid

(a) remains same

(b) increases

(c) decreases

(d) can’t say

Sol :

If a solid of one shape has conversed into another then

the surface area of the new solid will same or not same

i.e. can’t say.

Ans (d)

Question 18

If a marble of radius 2.1 cm is put into a cylindrical cup full of water of radius 5 cm and height 6 cm, then the volume of water that flows out of the cylindrical cup is

(a) 38.8 $cm^3$

(b) 55.4 $cm^3$

(c) 19.4 $cm^3$

(d) 471.4 $cm^3$

Sol :

Radius of a marble = 2.1 cm

Volume of marble $=\frac{4}{3} \pi r^{3}  \mathrm{~cm}^{3}$

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

$=38.88 \mathrm{~cm}^{3}$

Ans (a)

Question 19

The volume of the largest right circular cone that can be carved out from a cube of edge 4.2 cm is

(a) 9.7 $cm^3$

(b) 77.6 $cm^3$

(c) 58.2 $cm^3$

(d) 19.4 $cm^3$

Sol :

Edge of cube = 4.2 cm

Radius of largest cone cut out $=\frac{42.2}{2}=2.1 \mathrm{~cm}$

and height = 4.2 cm

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

$=\frac{1}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 4.2 \mathrm{~cm}^{3}$

= 19.404

= 19.4 $cm^3$

Ans (d)

Question 20

The volume of the greatest sphere cut off from a circular cylindrical wood of base radius 1 cm and height 6 cm is

(a) 288 π $cm^3$

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

(c) 6 π $cm^3$

(d) 4 π $cm^3$

Sol :

Radius of cylinder (r) = 1 cm

Height (h) = 6 cm

The largest sphere that can be cut off from the cylinder of radius 1 cm

$\therefore$ Volume $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi(1)^{3}$

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

Ans (b)

Question 21

The volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is

(a) 3 : 4

(b) 4 : 3

(c) 9 : 16

(d) 16 : 9

Sol :

Ratio in volumes of two spheres = 64 : 27

Ratio in their radii $=\frac{r_{1}^{3}}{r_{2}^{3}}=\frac{64}{27}$

$=\left(\frac{r_{1}}{r_{2}}\right)^{3}=\left(\frac{4}{3}\right)^{3}$

$\Rightarrow \frac{r_{1}}{r_{2}}=\frac{4}{3}$

$\therefore$ Ratio in their surface area

$=\frac{4 \pi r_{1}^{2}}{4 \pi r_{2}^{2}}=\frac{r_{1}^{2}}{r_{2}^{2}}=\left(\frac{4}{3}\right)^{2}=\frac{16}{9}$

$\therefore$ Ratio is 16: 9

Ans (d)

Question 22

If a cone, a hemisphere and a cylinder have equal bases and have same height, then the ratio of their volumes is

(a) 1 : 3 : 2

(b) 2 : 3 : 1

(c) 2 : 1 : 3

(d) 1 : 2 : 3

Sol :

If a cone, a hemisphere and a cylinder have equal bases = r (say)

and height = h in each case and r = h

Ratio in their volumes $=\frac{1}{3} \pi r^{2} h: \frac{2}{3} \pi r^{3}: \pi r^{2} h$

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

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

$=\frac{1}{3}: \frac{2}{3}: 1=1: 2: 3$

Ans (d)

Question 23

If a sphere and a cube have equal surface areas, then the ratio of the diameter of the sphere to the edge of the cube is

(a) 1 : 2

(b) 2 : 1

(c) √π : √6

(d) √6 : √π

Sol :

A sphere and a cube have equal surface area

Let a be the edge of a cube and r be the radius of the sphere, then

$4 \pi r^{2}=6 a^{2}$

$\Rightarrow \pi(2 r)^{2}: 6 a^{2} \quad(\because d=2 r)$

$\Rightarrow \frac{d^{2}}{a^{2}}=\frac{6}{\pi}$

$\Rightarrow \frac{d}{a}=\frac{\sqrt{6}}{\sqrt{\pi}}$

Radii $d: a=\sqrt{6}: \sqrt{\pi}$

Ans (d)

Question 24

A solid piece of iron in the form of a cuboid of dimensions 49 cm x 33 cm x 24 cm is moulded to form a sphere. The radius of the sphere is

(a) 21 cm

(b) 23 cm

(c) 25 cm

(d) 19 cm

Sol :

Dimension of a cuboid = 49 cm × 33 cm × 24 cm

Volume of a cuboid = 49 × 33 × 24 $cm^3$

⇒ Volume of sphere = Volume of a cuboid

Volume of a sphere = 49 × 33 × 24 $cm^3$

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

$=\left(\frac{49 \times 33 \times 24 \times 3 \times 7}{4 \times 22}\right)^{\frac{1}{3}}$

$=(49 \times 7 \times 3 \times 3 \times 3)^{\frac{1}{3}}$

$=(7 \times 7 \times 7 \times 3 \times 3 \times 3)^{\frac{1}{3}}$

$=7 \times 3=21 \mathrm{~cm}$

Ans (a)

Question 25

If a solid right circular cone of height 24 cm and base radius 6 cm is melted and recast in the shape of a sphere, then the radius of the sphere is

(a) 4 cm

(b) 6 cm

(c) 8 cm

(d) 12 cm

Sol :

Height of a circular cone (h) = 24 cm

and radius (r) = 6 cm

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

$=\frac{1}{3} \pi \times 6 \times 6 \times 24 \mathrm{~cm}^{3}$

Volume of sphere=Volume of a cone

Now volume of sphere $=\frac{1}{3} \pi \times 36 \times 24 \mathrm{~cm}^{3}$

Let r be in radius of sphere

Then $\frac{4}{3} \pi r^{3}=\frac{1}{3} \pi \times 36 \times 24$

$4 r^{3}=36 \times 24 \Rightarrow r^{3}=\frac{36 \times 24}{4}$

$r^{3}=3 \times 3 \times 3 \times 2 \times 2 \times 2=3^{3} \times 2^{3}$

$\therefore r=3 \times 2^{\circ}=6 \mathrm{~cm}$

Ans (b)

Question 26

If a solid circular cylinder of iron whose diameter is 15 cm and height 10 cm is melted and recasted into a sphere, then the radius of the sphere is

(a) 15 cm

(b) 10 cm

(c) 7.5 cm

(d) 5 cm

Sol :

Diameter of a cylinder = 15 cm

Radius $=\frac{15}{2} \mathrm{~cm}$

and height = 10 cm

$\therefore$ Volume $=\pi r^{2} h=\pi \times \frac{15}{2} \times \frac{15}{2} \times 10 \mathrm{~cm}^{3}$

$=\frac{1125 \pi}{2} \mathrm{~cm}^{3}$

$\therefore$ Volume of sphere $=\frac{1125 \pi}{2} \mathrm{~cm}^{3}$

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

$=\left(\frac{1125 \pi \times 3}{2 \times 4 \pi}\right)^{\frac{1}{3}}=\left(\frac{3375}{8}\right)^{\frac{1}{3}}$

$=\left(\frac{1125 \pi \times 3}{2 \times 4 \pi}\right)^{\frac{1}{3}}=\left(\frac{3375}{8}\right)^{\frac{1}{3}}$

$\begin{array}{l|l}3 & 3375 \\\hline 3 & 1125 \\\hline 3 & 375 \\\hline 5 & 125 \\\hline 5 & 25 \\\hline 5 & 5 \\\hline & 1\end{array}$

$=\left(\frac{5^{3} \times 3^{3}}{2^{3}}\right)^{\frac{1}{3}}=\frac{5 \times 3}{2} \mathrm{~cm}$

$=\frac{15}{2}=7.5 \mathrm{~cm}$

Ans (c)

Question 27

The number of balls of radius 1 cm that can be made from a sphere of radius 10 cm is

(a) 100

(b) 1000

(c) 10000

(d) 100000

Sol :

Radius of sphere (R) = 10 cm

Volume of sphere $=\frac{4}{3} \pi R^{3}=\frac{4}{3} \pi(10)^{3} \mathrm{~cm}^{3}$

$=\frac{4}{3} \pi \times 1000 \mathrm{~cm}^{3}$

and radius of one ball $(r)=1 \mathrm{~cm}$

Volume of one ball $=\frac{4}{3} \pi(1)^{3} \mathrm{~cm}^{3}=\frac{4}{3} \pi \mathrm{cm}^{3}$

$\therefore$ Number of

$\frac{4 \pi \times 1000 \times 3}{3 \times 4 \times \pi}=1000$

Ans (b)

Question 28

A metallic spherical shell of internal and external diameters 4 cm and 8 cm, respectively is melted and recast into the form of a cone of base diameter 8 cm. The height of the cone is

(a) 12 cm

(b) 14 cm

(c) 15 cm

(d) 18 cm

Sol :

The internal diameter of the metallic shell = 4 cm

and external diameter = 8 cm

$\therefore$ Internal radius $(r)=\frac{4}{2}=2 \mathrm{~cm}$

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

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

$=\frac{4}{3} \times \pi(64-8) \mathrm{cm}^{3}$

$=\frac{4}{3} \pi \times 56 \mathrm{~cm}^{3}$

Diameter of cone $=8 \mathrm{~cm}$

$\therefore$ Radius of cone $=\frac{8}{2}=4 \mathrm{~cm}$

$\Rightarrow$ Height $=\frac{\text { Volume }}{\frac{1}{3} \pi r^{2}}$

$=\frac{4 \pi \times 56 \times 3}{3 \times 1 \times \pi \times 4 \times 4}$

=14 cm

Ans (b)

Question 29

A cubical ice cream brick of edge 22 cm is to be distributed among some children by filling ice cream cones of radius 2 cm and height 7 cm up to its brim. The number of children who will get the ice cream cones is

(a) 163

(b) 263

(c) 363

(d) 463

Sol :

Edge of a cubical icecream brick = 22 cm

Volume = $a^3 = (22)^3$ = 10648 $cm^3$

Radius (r) of ice cream cone (r) = 2 cm

and height (h) = 7 cm

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

$=\frac{1}{3} \times \frac{22}{7} \times 2 \times 2 \times 7 \mathrm{~cm}^{3}=\frac{88}{3} \mathrm{~cm}^{3}$

$\therefore$ Number of cones $=\frac{10648 \times 3}{88}=363$

Ans (c)

Question 30

Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is

(a) 4 cm

(b) 3 cm

(b) 2 cm

(d) 6 cm

Sol :

Diameter of cylinder = 2 cm

Radius $=\frac{2}{2}=1 \mathrm{~cm}$

and height = 16 cm

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

$=\frac{22}{7} \times 1 \times 1 \times 16=\frac{352}{7} \mathrm{~cm}^{3}$

$\therefore$ Volume of 12 solid spheres so formed

$=\frac{352}{7} \mathrm{~cm}^{3}$

$\therefore$ Volume of each sphere $=\frac{352}{7 \times 12}=\frac{352}{84} \mathrm{~cm}^{3}$

$\therefore$ Radius of each sphere $=\left(\frac{352 \times 3 \times 7}{84 \times 4 \times 22}\right)^{\frac{1}{3}}$

$=(1)^{\frac{1}{3}}=1 \mathrm{~cm}$

$\therefore$ Diameter $=2 \times 1=2 \mathrm{~cm}$

Ans (c)

Question 31

A hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that $\frac{1}{8}$ space of the cube remains unfilled. Then the number of marbles that the cube can accommodate is

(a) 142296

(b) 142396

(c) 142496

(d) 142596

Sol :

Internal edge of a hollow cube = 22 cm

Volume $=(\text { side })^{3}=(22)^{3}=22 \times 22 \times 22 \mathrm{~cm}^{3}=10648 \mathrm{~cm}^{3}$

Diameter of spherical marble $=0.5 \mathrm{~cm}=\frac{1}{2}$

$\therefore$ Radius $=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4} \mathrm{~cm}$

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

$=\frac{4}{3} \times \frac{22}{7} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \mathrm{~cm}^{3}$

$=\frac{11}{168} \mathrm{~cm}^{3}$

Space left unfilled $=10648 \times \frac{1}{8} \mathrm{~cm}^{3}$

$=1331 \mathrm{~cm}^{3}$

$\therefore$ Remaining volume for marbles

$=10648-1331=9317 \mathrm{~cm}^{3}$

$\therefore$ Number of marble to accommodate

$=9317 \div \frac{11}{168}=\frac{9317 \times 168}{11}$

=142296

Ans (a)

Question 32

In the given figure, the bottom of the glass has a hemispherical raised portion. If the glass is filled with orange juice, the quantity of juice which a person will get is

(a) 135 π $cm^3$

(b) 117 π $cm^3$

(c) 99 π $cm^3$

(d) 36 π $cm^3$

Sol :

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

and height (h)= 15 cm

$\therefore$ Volume of the glass $=\pi r^{2} h-\frac{2}{3} \pi r^{3}$

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

$=\pi \times 3\left(3 \times 15-\frac{2}{3} \times 9\right)$

$=3 \pi(45-6) \mathrm{cm}^{3}$

$=3 \pi \times 39=117 \pi \mathrm{cm}^{3}$

Ans (b)