Download App

Maths 1 Marks Question

Surface Areas and Volumes · Bihar Board (BSEB) STD 9 (BSEB - English Medium). 27 questions.

Questions

1 Marks Question

Take a timed test

27 questions · self-marked practice — reveal the answer and mark yourself.

Question 11 Mark
How many litres of milk can a hemispherical bowl of diameter $10.5\ cm$ hold$?$
Answer
$\text { Diameter }=10.5 \mathrm{~cm} .$
$\therefore \text { Radius }(r)=\frac{10.5}{2}=5.25 \mathrm{~cm}$
$\therefore \text { Amount of milk }=\frac{2}{3} \pi r^3$
$=\frac{2}{3} \times \frac{22}{7} \times(5.25)^3 \mathrm{~cm}^3$
$=303 \mathrm{~cm}^3=0.303 \mathrm{I}$
View full question & answer→
Question 21 Mark
Find the amount of water displaced by a solid spherical ball of diameter $0.21 m.$
Answer
Diameter $= 0.21 m$
$\therefore $ Radius $(r) = {0.21}\over2m = 0.105 m$
$\therefore $ Amount of water displaced $= {4\over3}\pi r^3$
$={4\over3}\times{22\over7}\times(0.105)^3=0.004851\ m^3$
View full question & answer→
Question 31 Mark
Find the amount of water displaced by a solid spherical ball of diameter $28\ cm.$
Answer
Diameter $= 28\ cm$
$\therefore $ Radius $(r) =\frac{28}{2} cm = 14 \ cm$
$\therefore $ Amount of water displaced $ = {4\over3}\pi r^3$
$={4\over3}\times{22\over7}\times(14)^3={34496\over3}\ cm^3=11498{2\over3}\ cm^3$
View full question & answer→
Question 41 Mark
Find the volume of a sphere whose radius is $0.63\ m.$
Answer
$r = 0.63\ m$
$\therefore $ volume =${4\over3}\pi r^3$
$={4\over3}\times{22\over7}\times(0.63)^3=1.05\ m^3$
View full question & answer→
Question 51 Mark
Find the volume of a sphere whose radius is $7 \ cm$
Answer
$r = 7\ cm$
$∴$ Volume = ${4\over3}\pi r^3$
$={4\over3}\times{22\over7}\times(7)^3={4312\over3}\ cm^3=1437{1\over3}\ cm^3$
View full question & answer→
Question 61 Mark
A capsule of medicine is in the shape of a sphere of diameter $3.5\ mm .$ How much medicine (in $\mathrm{mm}^3$ ) is needed to fill this capsule?
Answer
Diameter of the capsule $=3.5 \mathrm{~mm}$
$\therefore$ Radius of the capsule $(\mathrm{r})=\frac{3.5}{2} \mathrm{~mm}=1.75 \mathrm{~mm}$
$\therefore$ Capacity of the capsule $=\frac{4}{3} \pi r^3$
$=\frac{4}{3} \times \frac{22}{7} \times(1.75)^3 \mathrm{~mm}^3$
$=22.46 \mathrm{~mm}^3$
$\therefore 22.46 \mathrm{~mm}^3$ of medicine is needed to fill this capsule.
View full question & answer→
Question 71 Mark
Find the volume of the right circular cone with radius $3.5$ and height $12\ cm$
Answer
$\mathrm{r}=3.5 \mathrm{~cm}, \mathrm{~h}=12 \mathrm{~cm} .$
$\therefore \text { Volume of the right circular cone }=\frac{1}{3} \pi r^2 h$
$=\frac{1}{3} \times \frac{22}{7} \times(3.5)^2 \times 12=154 \mathrm{~cm}^3$
View full question & answer→
Question 81 Mark
Find the volume of the right circular cone with radius $6\ cm$ and height $7\ cm$
Answer
$\mathrm{r}=6 \mathrm{~cm}, \mathrm{~h}=7 \mathrm{~cm}$
$\therefore$ Volume of the right circular cone $=\frac{1}{3} \pi r^2 h$
$=\frac{1}{3} \times \frac{22}{7} \times(6)^2 \times 7=264 \mathrm{~cm}^3$
View full question & answer→
Question 91 Mark
A hemispherical bowl is made of steel, $0.25\ cm$ thick. The inner radius of the bowl is $5\ cm.$ Find the outer curved surface area of the bowl.
 
Answer
Inner radius of bowl $\left( r \right) = 5 \ cm$
Thickness of steel $\left( t \right)= 0.25\ cm$
$\therefore $ Outer radius of bowl $(R) = r+t = 5 +0.25 = 5.25\ cm$
$\therefore $ Outer curved surface area of bowl $ = 2\pi {{\text{R}}^{2}}= 2\times \frac{22}{7}\times 5.25\times 5.25$
$= 2\times \frac{22}{7}\times \frac{21}{4}\times \frac{21}{4}$
$= \frac{693}{4}$
$=\text{ }173.25\text{ }c{{m}^{2}}$
View full question & answer→
Question 101 Mark
Find the total surface area of a hemisphere of radius $10\ cm$
Answer
$\mathrm{r}=10 \mathrm{~cm}$
Total surface area of the hemisphere $=3 \pi r^2$
$=3 \times 3.14 \times(10)^2=942 \mathrm{~cm}^2$
View full question & answer→
Question 111 Mark
Find the surface area of a sphere of diameter $3.5\ cm.$
Answer
Diameter $=3.5 \mathrm{~cm}$
$\therefore$ Radius $(r)=\frac{3.5}{2} \mathrm{~cm}=1.75 \mathrm{~cm}$
Surface area of sphere $=4 \pi r^2$
$=4 \times \frac{22}{7} \times(1.75)^2=38.5 \mathrm{~cm}^2$
View full question & answer→
Question 121 Mark
Find the surface area of a sphere of diameter $21\ cm.$
Answer
$\text { Diameter }=21 \mathrm{~cm}$
$\therefore \text { Radius }(\mathrm{r})=\frac{21}{2} \mathrm{~cm}$
$\text { Surface area of sphere }=4 \pi r^2$
$=4 \times \frac{22}{7} \times\left(\frac{21}{2}\right)^2$
$=4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}$
$=1386 \mathrm{~cm}^2$
View full question & answer→
Question 131 Mark
Find the surface area of a sphere of diameter $14\ cm.$
Answer
Diameter $=14 \mathrm{~cm}$
$\therefore$ Radius $(\mathrm{r})=\frac{14}{2} \mathrm{~cm}=7 \mathrm{~cm}$
Surface area of sphere $=4 \pi r^2$
$=4 \times \frac{22}{7} \times(7)^2=616 \mathrm{~cm}^2$
View full question & answer→
Question 141 Mark
Find the surface area of a sphere of radius $14\ cm$
Answer
$\mathrm{r}=14 \mathrm{~cm}$
Surface area of a sphere $=4 \pi r^2$
$=4 \times \frac{22}{7} \times(14)^2$
$=4 \times 22 \times 17 \times 2$
$=2464 \mathrm{~cm}^2$
View full question & answer→
Question 151 Mark
Find the surface area of a sphere of radius $5.6\ cm$
Answer
$r = 5.6\ cm$
Surface area of a sphere = $4\pi {r^2}$
$ = 4 \times \frac{{22}}{7} \times {(5.6)^2} = 394.24\;c{m^2}$
View full question & answer→
Question 161 Mark
Find the surface area of a sphere of radius $10.5\ cm$
Answer
$\mathrm{r}=10.5 \mathrm{~cm}$
Surface area of a sphere $=4 \pi r^2$
$=4 \times \frac{22}{7} \times(10.5)^2$
$=4 \times 22 \times 10.5 \times 1.5$
$=1386 \mathrm{~cm}^2$
View full question & answer→
Question 171 Mark
Curved surface area of a cone is $308 \mathrm{~cm}^2$ and its slant height is $14\ cm .$ Find total surface area of the cone.
Answer
Total surface area of cone $=$ Curved surface Area of cone $+$ Area of base
$=\pi r l+\pi r^2$
$=\left[308+\frac{22}{7} \times(7)^2\right] \mathrm{cm}^2$
$=462 \mathrm{~cm}^2$
Thus, the total surface area of the cone is $462 \mathrm{~cm}^2$.
View full question & answer→
Question 181 Mark
Curved surface area of a cone is $308 \mathrm{~cm}^2$ and its slant height is $14\ cm.$ Find the radius of the base
Answer
Slant height of cone $= 14\ cm$
Let radius of circular end of cone be $r.$
Curved surface area of cone $= \pi rl$
$308 \mathrm{~cm}^2 = (\frac { 22 } { 7 } \times r \times 14) \ cm$
$\Rightarrow r = \left( \frac { 308 } { 44 } \right)\ cm = 7\ cm$
Thus, the radius of circular end of the cone is $7\ cm.$
View full question & answer→
Question 191 Mark
The diameter of the base of a cone is $10.5\ cm$ and its slant height is $10\ cm.$ Find its curved surface area.
Answer
Diameter $=10.5 \mathrm{~cm}$
$\Rightarrow$ Radius $(\mathrm{r})=\frac{10.5}{2}=\frac{21}{4} \mathrm{~cm}$
Slant height of cone $(l)=10 \mathrm{~cm}$
Now we have, Curved surface area of cone $=\pi r l=\frac{22}{7} \times \frac{21}{4} \times 10$
$=165 \mathrm{~cm}^2$
View full question & answer→
Question 201 Mark
Find the surface area and total surface area of a hemisphere of radius $21\ cm.$
Answer
We know that the surface area $S$ and total surfaces area $S_1$ of a hemisphere of radius $r$ are given by
$\mathrm{S}=2 \pi r^2$ and, $\mathrm{S}_1=3 \pi r^2$ respectively.
Here, $r=21 \mathrm{~cm}$
$\therefore \mathrm{S}=2 \times \frac{22}{7} \times 21 \times 21 \mathrm{~cm}^2 \text { and }, \mathrm{S}_1=3 \times \frac{22}{7} \times 21 \times 21 \mathrm{~cm}^2$
$\Rightarrow \mathrm{~S}=2772 \mathrm{~cm}^2 \text { and, } \mathrm{S}_1=4158 \mathrm{~cm}^2$
View full question & answer→
Question 211 Mark
Find the surface area of a sphere of radius $7\ cm$
Answer
We known that the surface area $S$ of a sphere of radius $r$ is given by
$\mathrm{S}=4 \pi r^2$
Here, $r=7 \mathrm{~cm}$
$\therefore S=4 \times \frac{22}{7} \times 7 \times 7 \mathrm{~cm}^2=616 \mathrm{~cm}^2$
View full question & answer→
Question 221 Mark
Find the curved surface area of a right circular cone whose slant height is $10\ cm$ and base radius is $7\ cm$
Answer
We know that,
Curved surface area of cone $=\pi r l$
$= \frac {22}7 \times 7 \times 10\ cm^2$
$=220 \ cm^2$
View full question & answer→
Question 231 Mark
Savitri had to make a model of a cylindrical Kaleidoscope for her science project. She wanted to use chart paper to make the curved surface of the Kaleidoscope. What should be the area of chart paper required by her, if she wanted to make a Kaleidoscope of length $25 \ cm$ with a $3.5\  cm$ radius? You may take $\pi=\frac{22}{7}$.
Answer
We have given that,
$r =$ Radius of the base of the cylindrical Kaleidoscope $= 3.5\ cm$
$h =$ Height (length) of Kaleidscope $= 25 \ cm$
Therefore, Area of the chart paper required = Curved surface area of the Kaleidoscope
$= 2\pi rh = 2  \times \frac{22}{7} \times  3.5 \times 25\ cm^2 = 550\ cm^2$
View full question & answer→
Question 241 Mark
A hemi spherical bowl has a radius of $3.5\ cm.$ What would be the volume of water it would contain$?$
Answer
The volume of water the bowl contain  $= \frac{2}{3}\pi {{r}^{3}}$
Radius $= 3.5\ cm$
Then volume $= \frac{2}{3}\times \frac{22}{7}\times {{\left( 3.5 \right)}^{3}}$
$=\frac{2}{3}\times \frac{22}{7}\times 3.5\times 3.5\times 3.5$
$=\frac{2}{3}\times \frac{22}{7}\times \frac{35}{10}\times \frac{35}{10}\times \frac{35}{10}$
$=89.8c{{m}^{3}}$
View full question & answer→
Question 251 Mark
Find the volume of a sphere of radius $11.2\ cm.$
Answer
We know that volume of sphere $=\frac{4}{3} \pi r^3$
$=\frac{4}{3} \times \frac{22}{7} \times 11.2 \times 11.2 \times 11.2 \mathrm{~cm}^3=5887.32 \mathrm{~cm}^3$
View full question & answer→
Question 261 Mark
The height and the slant height of a cone are $21\ cm$ and $28\ cm$ respectively. Find the volume of the cone.
Answer
We know that, $l^2=r^2+h^2$
Therefore, we have
$r=\sqrt{l^2-h^2}=\sqrt{28^2-21^2} \mathrm{~cm}=7 \sqrt{7} \mathrm{~cm}$
So, volume of the cone $=\frac{1}{3} \pi r^2 h=\frac{1}{3} \times \frac{22}{7} \times 7 \sqrt{7} \times 7 \sqrt{7} \times 21 \mathrm{~cm}^3$
$=7546 \mathrm{~cm}^3$
View full question & answer→
Question 271 Mark
A child playing with building blocks, which are of the shape of the cubes, has built a structure as shown in the given figure. If the edge of each cube is $3\ cm,$ find the volume of the structure built by the child.
Answer
$\text { Volume of each cube }=\text { edge } \times \text { edge } \times \text { edge }$
$=3 \times 3 \times 3 \mathrm{~cm}^3=27 \mathrm{~cm}^3$
$\text { Number of cubes in the structure }=15$
$\text { Therefore, volume of the structure }=27 \times 15 \mathrm{~cm}^3=405 \mathrm{~cm}^3$
View full question & answer→
Generate Paper FreeAll Surface Areas and Volumes topics
1 Marks Question - Maths STD 9 Questions - Vidyadip