ML Aggarwal Solution Class 10 Chapter 17 Mensuration MCQs

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 Q30 Q31 Q32

 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)