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MATHS 2 Marks Questions

Volume and Surface Area · Rajasthan Board (RBSE) STD 9 (Rajasthan - English Medium). 24 questions.

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Question 12 Marks
The volume of a cube is $512\ cm^3.$ Find its surface area.
Answer
Suppose that the side of the given cube is $x\ cm.$Volume of the cube $= 512\ cm^3$
Then $512 = x^3$
$\Rightarrow x=\sqrt[3]{512}=8$
i.e., the side of the cube is $8\ cm.$
$\therefore$ Surface area of the cube $= 6x^2\ cm^2= 6 \times 8^2\ cm^2 = 384\ cm^2$
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Question 22 Marks
The pillars of a temple are cylindrically shaped. Each pillar has a circular base of radius 20cm and height 10m. How much concreate mixture would be required to build 14 such pillars?
Answer
Radius (r) of pillar = 20cm $=\frac{20}{100}\text{m}$
Height (h) of pillar = 10m
$\therefore$ Volume of 1 pillar $=\pi\text{r}^2\text{h}$
$=\Big(\frac{22}{7}\times\frac{20}{100}\times\frac{20}{100}\times10\Big)\text{m}^3$
$=\frac{44}{35}\text{m}^3$
⇒ Volume of concreate mixture in 14 pillars $=14\times\frac{44}{35}=17.6\text{m}^3$
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Question 32 Marks
The outer diameter of a spherical shell is 12cm and its inner diameter is 8cm. Find the volume of metal contained in the shell. Also, find its outer surface area. $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
Answer
Outer radius of the spherical shell = 6cm
Inner radius of the spherical shell = 4cm
Volume of metal contained in the shell $=\frac{4}{3}\times\frac{22}{7}(6^3-4^3)$
$=\frac{88}{21}\times(216-64)$
$=\frac{88}{21}\times152$
$=636.95\text{cm}^3$
$\therefore$ Outer surface area $=4\times\frac{22}{7}\times6\times6$
$=452.57\text{cm}^2$
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Question 42 Marks
The surface area of sphere is $(576\pi)\text{cm}^2.$ Find its volume.$\big(\text{Take}\ \pi=\frac{22}{7}\big).$
Answer
Surface area of the sphere $=(576\pi)\text{cm}^2$
Suppose that r cm is the radius of the sphere.
Then $4\pi\text{r}^2=576\pi$
$\Rightarrow\text{r}^2=\frac{576}{4}=144$
$\Rightarrow\text{r}=12\text{cm}$
$\therefore$ Volume of the sphere $=\frac{4}{3}\times\pi\times12\times12\times12\text{cm}^3$
$=2304\text{cm}^3$
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Question 52 Marks
A solid metallic cuboid of dimensions $(9m \times 8m \times 2m)$ is melted and recast into solid cubes of edge $2m.$ Find the number of cubes so formed.
Answer
Volume of the solid metallic cuboid $= 9m \times 8m \times 2m = 144m^3$
Volume of each solid cube $= ($Edge$)^3 = (2)^3 = 8m^3$
$\therefore$ Number of cubes formed $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{solid}\ \text{metallic}\ \text{cuboid}}{\text{Volume}\ \text{of}\ \text{each}\ \text{solid}\ \text{cube}}$
$=\frac{144}{8}=18$
Thus, the number of cubes so formed is $18.$
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Question 62 Marks
Find the total surface area of a cone, if its slant height is 21m and diameter of its base is 24m. $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$
Answer
Radius of a cone, r = 12cm
Slant height of a cone, l = 21cm
Totalsurface area of a cone $=\pi\text{r}(\text{l}+\text{r})$
$=\Big[\frac{22}{7}\times12(21+12)\Big]\text{m}^2$
$=\Big(\frac{22}{7}\times12\times33\Big)\text{m}^2$
$=1244.57\text{m}^2$
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Question 72 Marks
A godown measures $40m \times 25m \times 15m.$ Find the maximum number of wooden crates, each measuring $1.5m \times 1.25m \times 0.5m,$ that can be stored in the godown.
Answer
Volume of the godown $= 40m \times 25m \times 15m = 15000m^3$
Volume of each wooden create $= 1.5m \times 1.25m \times 0.5m = 0.9375m^{3 }$
$\therefore$ Maximum number of wooden creates that can be stored in the godown
$=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{godown}}{\text{Volume}\ \text{of}\ \text{each}\ \text{wooden}\ \text{crate}}$
$=\frac{15000}{0.9375}$
$=16000$
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Question 82 Marks
A cylindeical tub of radius 12cm contains water to a depth of 20cm. A sphrical iron ball is dropped into the tub and thus the leval of water is raised by 6.75cm. What is the radius of the ball.
Answer
Suppose that the radius of the ball is r cm.
Radius of the cylindrical tub = 12cm
Depth of the tub = 20cm
Now, volume of the ball = volume of water raised in the cylinder
$\Rightarrow\frac{4}{3}\pi\text{r}^3=\pi\times12^2\times6.75$
$\Rightarrow\text{r}^3=\frac{144\times6.75\times3}{4}$
$=36\times6.5\times3=729$
$\Rightarrow\text{r}=9\text{cm}$
$\therefore$ The radius of the ball is 9cm.
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Question 92 Marks
Two cones have their heights in the ratio 1 : 3 and the radii of their bases in the ratio 3 : 1. Show that their volumes are in the ratio 3 : 1.
Answer
Let their heights be h and 3h
And, their radii be 3r and r.
Then, $\text{V}_1=\frac{1}{3} \pi(3\text{r})^2\times\text{h}$
And, $\text{V}_2=\frac{1}{3} \pi\text{r}^2\times3\text{h}$
$\Rightarrow\frac{\text{V}_1}{\text{V}_2}=\frac{\frac{1}{3}\pi(3\text{r})^2\times\text{h}}{\frac{1}{3}\pi\text{r}^2\times3\text{h}}=\frac{3}{1}$
$\therefore\text{V}_1:\text{V}_2=3:1$
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Question 102 Marks
The surface areas of two spheres are in the 1 : 4. Find the ratio of their volumes.
Answer
Suppose that the radii of the spheres are r and R.
We have:
$\frac{4\pi\text{r}^2}{4\pi\text{R}^2}=\frac{1}{4}$
$\Rightarrow\frac{\text{r}}{\text{R}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$
Now, ratio of the volumes $=\frac{\frac{4}{3}\pi\text{r}^3}{\frac{4}{3}\pi\text{R}^3}=\Big(\frac{\text{r}}{\text{R}}\Big)^3$
$=\Big(\frac{1}{2}\Big)^3=\frac{1}{8}$
$\therefore$ The ratio of the volumes of the sphere is 1 : 8.
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Question 112 Marks
A hemispherical bowl of internal radius 9cm contains a liquid. This liquid is to be filled into cylindrical shaped small bottles of diameter 3cm and height 4cm How many bottles required to empty the bowl?
Answer
Internal radius of the hemispherical bowl = 9cm
Radius of a cylindrical shaped bottle = 1.5cm
Height of a bottle = 4cm
Number of bottles required to empty the bowl $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{hemispherical}\ \text{bowl}}{\text{Volume}\ \text{of}\ \text{a}\ \text{cylindrical}\ \text{shaped}\ \text{bottle}}$
$=\frac{\frac{2}{3}\pi\times9^3}{\pi\times1.5^2\times4}$
$=\frac{2\times9\times9\times9}{3\times1.5\times1.5\times4}$
$=54$
$\therefore$ 54 bottles are required to empty the bowl.
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Question 122 Marks
A matchbox measures 4$\ cm \times 2.5\ cm \times 1.5\ cm.$ What is the volume of a packet containing $12$ such matchboxes?
Answer
Volume of each matchbox $= 4 \times 2.5 \times 1.5 = 15\ cm^3$
$\therefore$ Volume of $12$ matchbox $= 12 \times 15 = 180\ cm^3$
Thus, the volume of a packet containing $12$ such matchboxes is $180\ cm^3.$
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Question 132 Marks
A patient in a hospital is given soup in a cylindrical bowl of diameter $7\ cm.$ If the bowl is filled with soup to a height of $4\ cm,$ how much soup the hospital has to prepare daily to serve $250$ patients?
Answer
Radius $(r)$ of cylindrical bowl $=\Big(\frac{7}{2}\Big)\ cm=3.5\ cm$
Height $(h)$ up to which the bowl is filled with soup $= 4\ cm$
Volume of soup in $1$ bowl $= pr^2h$
$=\Big(\frac{22}{7}\times(3.5)^2\times4\Big)\ cm^3$
$=154\ cm^3$
Hence, volume of soup in $250$ bowls $= (250 \times 154)cm^3 = 38500\ cm^3 = 38.5$ litres
Thus, the hospital will have to prepare $38.5$ litres of soup daily to serve $250$ patients.
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Question 142 Marks
The volume of a cuboid is $1536m^3.$ Its length is $16m,$ and its breadth and height are in the ratio $3 : 2.$ Find the breadth and height of the cuboid.
Answer
Length of the cuboid $= 16m$
Suposs that the breadth and height of the cuboid are $3x\ m$ and $2x\ m,$ respectively.
Then $1536 = 16 \times 3x \times 2x$
$\Rightarrow 1536 = 16 \times 6x^2$
$\Rightarrow\text{x}^2=\frac{1536}{96}=16$
$\Rightarrow\text{x}=\sqrt{16}=4$
$\therefore$ The breadth and height of the cuboid are $12m$ and $8m,$ respectively.
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Question 152 Marks
In a shower, $5\ cm$ of rain falls. Find the volume of water that falls on $2$ hectares of ground.
Answer
Volume of the water that falls on the ground $=$ area of ground $\times$ depth $= 20000 \times 0.05m^3$
$= 1000m^3$
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Question 162 Marks
Find the curved surface area of a cone with base radius 5.25cm and slant height 10cm. $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$
Answer
Radius of a cone, r = 5.25cm
Slant height of a cone, l = 10cm
Curved surface area of a cone $=\pi\text{rl}$
$=\Big(\frac{22}{7}\times5.25\times10\Big)\text{cm}^2$
$=165\text{cm}^2$
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Question 172 Marks
A right circular cone is 3.6cm high and the radius of its base is 1.6cm. It is melted and recast into a right circular cone having base radius 1.2cm. Find its height.
Answer
Here, height of cone = 3.6cm and radius = 1.6cm
After melting, its radius = 1.2cm
Volume of original cone = Volume of cone after melting
$\therefore\frac{1}{3}\pi\times1.6\times1.6\times3.6$
$=\frac{1}{3}\pi\times1.2\times1.2\times\text{h}$
$\Rightarrow\text{h}=\frac{\frac{1}{3}\pi\times1.6\times1.6\times3.6}{\frac{1}{3}\pi\times1.2\times1.2}=6.4\text{cm}$
$\therefore$ height of new cone = 6.4cm.
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Question 182 Marks
Find the volume, the lateral surface area, the totle surface area and the diagonal of a cube, each of whose edges measures $9m. ($Take $\sqrt{3}=1.73)$)
Answer
Here, $a = 9m$
Volume of the cube $= a^3 = 9^3m^3 = 729m^3$
Lateral surface area of the cube $= 4a^2 = 4 \times 9^2m^2 = 4 \times 81m^2 = 324m^2$
Total surface area of the cube $= 6a^26 \times 9^2m^2 = 6 \times 81m^2 = 486m^2$
$\therefore$ Diagonal of the cube $=\sqrt{3}\text{a}=\sqrt{3}\times9=15.57m$
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Question 192 Marks
How many persons can be accommodated in a dining hall of dimension $(20m \times 16m \times 4.5m),$ assuming that each person requires $5$ cubic metres of air?
Answer
Volume of the dining hall $= (20 \times 16 \times 4.5) m^3$
$= 1440m^3$
Volume of air required by each person $= 5m^3$
$\therefore$ Capacity of the dining hall $=\frac{\text{Volume}\ \text{of}\ \text{dining}\ \text{hall}}{\text{Volume}\ \text{of}\ \text{air}\ \text{required}\ \text{by}\ \text{each}\ \text{each}\ \text{person}}$
$=\frac{1440}{5}=288$ persons
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Question 202 Marks
A conical pit of diameter $3.5m$ is $12m$ deep. What is its capacity in kilolitres? $($Use $\pi=\frac{22}{7}).$
Hint: $1m^3 = 1$ kilolitre
Answer
Radius of a conical pit, $\text{r}=\frac{3.5}{2}m$
Depth of a conical pit, $h = 12m$
Volume of the conical pit $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\Big(\frac{1}{3}\times\frac{22}{7}\times\frac{3.5}{2}\times\frac{3.5}{2}\times12\Big)m^3$
$=38.5m^3$
$=38.5$ kiloletre
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Question 212 Marks
The radii of two spheres are in the ratio 1 : 2. Find the ratio of their surface areas.
Answer
Suppose that the radii are r and 2r.
Now, ratio of the surface areas $=\frac{4\pi\text{r}^2}{4\pi(2\text{r})^2}=\frac{\text{r}^2}{4\text{r}^2}=\frac{1}{4}$
$=1:4$
$\therefore$ The ratio of their surface areas is 1 : 4.
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Question 222 Marks
Three cubes of metal with edges $3\ cm, 4\ cm$ and $5\ cm$ respectively are melted to form a single cube. Find the lateral surface area of the new cube formed.
Answer
Three cubes of metal with edges $3\ cm, 4\ cm$ and $5\ cm$ are melted to form a single cube.
$\therefore$ Volume of the new cube = sum of the volumes the old cubes
$= (3^3 + 4^3 + 5^3)cm^3$
$= (27 + 64 + 125)cm^3$
$= 216\ cm^3$
Suppose the edge of the new cube $= x \ cm$
Then we have:
Then $216 = x^3$
$\Rightarrow\text{x}=\sqrt[3]{216}=6$
$\therefore$ Lateral surface area of the new cube $= 4x^2\ cm^2 = 4 \times 6^2\ cm^2 = 144\ cm^2$
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Question 232 Marks
Find the length of the longest pole that can be put in a room of dimension (10m × 10m × 5m).
Answer
Length of the longest pole = length of the diagonal of the room
$=\sqrt{\text{l}^2+\text{b}^2+\text{h}^2}\text{m}$
$=\sqrt{10^2+10^2+5^2}\text{m}$
$=\sqrt{100+100+25}$
$=\sqrt{225}=15\text{m}$
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Question 242 Marks
A cuboidal water tank is $6m$ long, $5m$ wide and $4.5m$ deep. How many litres of water can it hold? $($Given, $1m^3 = 100$0 litres.$)$
Answer
Volume of water in the tank $=$ Length $\times$ Breadth $\times$ Height $= 6 \times 5 \times 4.5 = 135m^3$
$\therefore$ Volume of water in litres $= 135 \times 1000 = 135000L (1m^3 = 1000L)$
Thus, the water tank can hold $135000L$ of water.
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