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Maths 3 Marks Question

Volume and Surface Area · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 39 questions.

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Question 13 Marks
The lateral surface area of a cube is 900cm2. Find its volume.
Answer
Suppose that the side of cube is x cm.
Lateral surface area of the cube = 900cm2
Then 900 = 4x2
$\Rightarrow\text{x}^2=\frac{900}{4}=225$
$\Rightarrow\text{x}=\sqrt{225}=15$
i.e., the side of the cube is 15cm.
$\therefore$ Volume of the given cube = x3cm3= 153cm3 = 3375cm3
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Question 23 Marks
In a water heating system, there is a cylindrical pipe of length 28m and diameter 5cm. Find the total radiating surface in the system.
Answer
Diameter of a cylindrical pipe = 5cm
⇒ Radius (r) of a cylindrical pipe, = 2.5cm
Height (h) of a cylindrical pipe = 28m = 2800cm
Now,
Total radiating surface in the system
= Curved surface area of the cylindrical pipe
$=2\pi\text{rh}$
$=\Big(2\times\frac{22}{7}\times2.5\times2800\Big)\text{cm}^2$
$=44000\text{cm}^2$
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Question 33 Marks
How many litres of water flows out of a pipe having an area of cross section of 5cm2 in 1 minute, if the speed of water in the pipe is 30cm/sec?
Answer
Speed of water = 30cm/sec
$\therefore$ Volume of water that flows out of the pipe in one second
= Area of cross section × Length of water flows in one second
= (5 × 30)cm3
= 150cm3
Hence, volume of water that flows out of the pipe in 1 minute
= (150 × 60)cm3
= 9000cm3
= 9 litres
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Question 43 Marks
A spherical ball of radius 3cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5cm and 2cm. Find the radius of the third ball.
Answer
Radius of the original spherical ball = 3cm
Suppose that the radius of third ball is r cm.
Then,
Volume of the original spherical ball = Volume of the three spherical balls
$\Rightarrow\frac{4}{3}\pi\times3^3=\frac{4}{3}\pi\times1.5^3\\+\frac{4}{3}\pi\times2^3+\frac{4}{3}\pi\times\text{r}^3$
$\Rightarrow27=3.375+8+\text{r}^3$
$\Rightarrow\text{r}^3=27-11.375=15.625$
$\Rightarrow\text{r}=2.5\text{cm}$
$\therefore$ The radius of the third ball is 2.5cm.
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Question 53 Marks
The lateral surface area of a cylinder is 94.2cm2 and its height is 5cm $\big(\text{Take}\ \pi=3.14\big)$. Find:
  1. The radius of its base.
  2. Its volume.
Answer
Lateral surface area of a cylinder = 94.2cm2

Height (h) of a cylinder = 5cm

  1. Lateral surface area of cylinder $=2\pi\text{rh}$

$\Rightarrow94.2=2\times3.14\times\text{r}\times5$

$\Rightarrow\text{r}=\frac{94.2}{2\times3.14\times5}=3\text{cm}$

  1. Volume of a cylinder $=\pi\text{r}^2\text{h}$

$=(3.14\times3\times3\times5)\text{cm}^3$

$=141.3\text{cm}^3$

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Question 63 Marks
From a solid right circular cylinder with height 10cm and radius of the base 6cm, a right circular cone of the same height and base is removed. Find the volume of the remaining solid. $\big(\text{Take}\ \pi=3.14\big).$
Answer
Here, height (h) = 10cm and radius = 6cm
$\therefore$ Volume of the remaining solid $=(\pi\text{r}^2\text{h})-\Big(\frac{1}{3}\pi\text{r}^2\text{h}\Big)$
$=\big(\pi\times6\times6\times10\big)\text{cm}^3\\-\Big(\frac{1}{3}\pi\times6\times6\times10\Big)\text{cm}^3$
$=\frac{2}{3}\pi\times6\times6\times10\text{cm}^3$
$=\Big(\frac{2}{3}\times3.14\times360\Big)\text{cm}^3=753.6\text{cm}^3$
$\therefore$ Volume of the remaining solid 753.6cm3.
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Question 73 Marks
The diameter of the moon is approximately aon fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?
Answer
Let the radius of the moon and earth be r units and R units, respectively.
$\therefore2\text{r}=\frac{1}{4}\times2\text{R}$ (Given)
$​​\Rightarrow\text{r}=\frac{\text{R}}{4}\ ...(1)$
$\therefore\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{moon}}{\text{Volume}\ \text{of}\ \text{the}\ \text{earth}}=\frac{\frac{4}{3}\pi\text{r}^3}{\frac{4}{3}\pi\text{R}^3}$
$=\frac{\Big(\frac{\text{R}}{4}\Big)^3}{\text{R}^3}$
$=\frac{1}{64}$ [Using (1)]
Thus, the volume of the moon is $\frac{1}{64}$ of the volume of the earth.
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Question 83 Marks
A hemispherical bowl is made of steel 0.25cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface area of the bowl. $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
Answer
Inner radius of the bowl, r = 5cm
Let the outer radius of the bowl be R cm.
Thickness of the bowl = 0.25cm (Given)
$\therefore$ R - r = 0.25cm
⇒ R = 0.25 + r = 0.25 + 5 = 5.25cm
$\therefore$ Outer curved surface area of the bowl $=2\pi\text{r}^2$
$=2\times\frac{22}{7}\times(5.25)^2=173.25\text{cm}^2$
Thus, the outer curved surface area of the bowl is 173.25cm2.
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Question 93 Marks
A circus tent is cylindrical to a height of 3 metres and conical above it. If its diameter is 105m and the slant height of the conical portion is 53m, calculate the length of the canvas 5m wide to make the required tent. $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$
Answer
Radius of the cylinder, $\text{R}=\Big(\frac{105}{2}\Big)\text{m}$ and its height, H = 3m
Slant height (l) = 53m
$\therefore$ area of canvas $=(2\pi\text{RH}+\pi\text{Rl})$
$=\bigg[\Big(2\times\frac{22}{7}\times\frac{105}{2}\times3\Big)+\Big(\frac{22}{7}\times\frac{105}{2}\times53\Big)\bigg]\text{m}^2$
$=(990 + 8745)\text{m}^2$
$=9735\text{m}^2$
$\therefore$ length of canvas $=\Big(\frac{\text{Area}\ \text{of}\ \text{canvas}}{\text{Width}\ \text{of}\ \text{canvas}}\Big)\text{m}$
$=\Big(\frac{9735}{5}\Big)=1947\text{m}.$
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Question 103 Marks
A cylindrical bucket with base radius 15cm is filled with water up to a height of 20cm. A heavy iron spherical ball of radius 9cm is dropped into the bucket to submerge completely in the water. Find the increase in the level of water.
Answer
Let h cm be the increase in the level of water.
Radius of the cylinder bucket = 15cm
Height up to which water is being filled = 20cm
Radius of the spherical ball = 9cm
Now, volume of the sphere = increased in volume of the cylinder
$\Rightarrow\frac{4}{3}\pi\times9^3=\pi\times15^2\times\text{h}$
$\Rightarrow\text{h}=\frac{4\times729}{3\times15\times15}=4.32\text{cm}$
$\therefore$ The increase in the level of water is 4.32cm.
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Question 113 Marks
Find the volume of a sphere whose surface area is 154cm2. $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
Answer
Let the radius of the sphere be r cm.
Surface area of the sphere = 154cm2
$\therefore4\pi\text{r}^2=154$
$\Rightarrow4\times\frac{22}{7}\times\text{r}^2=154$
$\Rightarrow\text{r}=\sqrt{\frac{154\times7}{4\times22}}=\sqrt{1225}$
$\Rightarrow\text{r}=3.5\text{cm}$
$\therefore$ Volume of the sphere 
$=\frac{4}{3}\pi\text{r}^3=\frac{4}{3}\times\frac{22}{7}\times(3.5)^3\approx179.67\text{m}^3$
Thus , the volume of the sphere is approximately 179.67m3.
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Question 123 Marks
Find the length of 13.2kg of copper wire of diameter 4mm, when 1 cubic centimetre of copper weighs 8.4g.
Answer
Let the length of the wire = 'h' meters
Then,
Volume of the wire × 8.4 = (13.2 × 1000)g
$\Rightarrow\frac{22}{7}\times\Big(\frac{2}{10}\Big)^2\times\text{h}\times8.4=13200$
$\Rightarrow22\times\Big(\frac{1}{5}\Big)^2\times\text{h}\times1.2=13200$
$\Rightarrow\text{h}=\frac{13200\times5\times5}{22\times1.2}=12500\text{cm}$
$\Rightarrow\text{h}=125\text{m}$
Thus, the length of the wire is 125m.
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Question 133 Marks
A joker's cap is in the form of a right circular cone of base radius 7cm and height 24cm. Find the area of the sheet required to make 10 such caps. $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$
Answer
Radius of a conical cap, r = 7cm
Height of a conical cap, h = 24cm
$\therefore$ Slant height of a conical cap,
$\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$\text{l}=\sqrt{7^2+24^2}$
$\text{l}=\sqrt{49+576}$
$\text{l}=\sqrt{625}$
$\text{l}=25\text{cm}$
Now,
Curved surface area of 1 conical cap $=\pi\text{rl}$
$=\Big(\frac{22}{7}\times7\times25\Big)\text{cm}^2$
$=550\text{cm}^2$
$\therefore$ Curved surface of 10 such conical caps = 10 × 550 = 5500cm2
Thus, 5500cm2 sheet will be required to make 10 caps.
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Question 143 Marks
The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank if its length and depth are respectively10m and 2.5m. (Given, 1000 litres = 1m3.)
Answer
Capacity of the tank $50000\text{L}=\frac{50000}{1000}=50\text{m}^3$ (1000L = 1m3)
Length of the tank = 10m
Height (or depth) of the tank = 2.5m
Now,
Volume of the cuboidal tank = Length × Breadth × Height
$\therefore$ Breadth of the tank $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{tank}}{\text{Length}\times\text{Height}}$
$=\frac{50}{10\times2.5}=\frac{50}{25}=2\text{m}$
Thus, the breadth of the tank is 2m.
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Question 153 Marks
A cylindrical container with diameter of base 56cm contains sufficient water to submerge a rectangular solid of iron with dimensions (32cm × 22cm × 14cm). Find the rise in the level of water when the solid is completely submerged.
Answer
Let the rise in the level of water = h cm
Then,
Volume of the cylinder of height h and base radius 28cm
= Volume of rectangular iron solid
$\Rightarrow\frac{22}{7}\times28\times28\times\text{h}=32\times32\times14$
$\Rightarrow22\times28\times4\times\text{h}=32\times32\times14$
$\Rightarrow\text{h}=\frac{32\times22\times14}{22\times28\times4}=4\text{cm}$
Thus, the rise in the level of water is 4cm.
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Question 163 Marks
Find the volume and surface area of a sphere whose radius is: $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
3.5cm
Answer
Radius of the sphere = 3.5cm
Now, volume $=\frac{4}{3}\pi\text{r}^3$
$=\frac{4}{3}\times\frac{22}{7}\times3.5\times3.5\times3.5$
$=179.67\text{cm}^3$
$\therefore$ Surface area $=4\pi\text{r}^2$
$=4\times\frac{22}{7}\times3.5\times3.5$
$=154\text{cm}^2$
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Question 173 Marks
Find the surface area of a sphere whose volume is 606.375m3. $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
Answer
Volume of the sphere = 606.375m3
Then, $\frac{4}{3}\pi\text{r}^3=606.375$
$\Rightarrow\text{r}^3=\frac{606.375\times3\times7}{4\times22}=144.703$
$\Rightarrow\text{r}=5.25\text{m}$
$\therefore$ Surface area $=4\pi\text{r}^2$
$=4\times\frac{22}{7}\times5.25\times5.25$
$=346.5\text{m}^2$
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Question 183 Marks
The diameter of a sphere is 6cm. It is melted and drawn into a wire of diameter 2mm. Find the length of the wire.
Answer
Let the length of the wire be h cm.
Radius of the wire, r = 1mm = 0.1cm
Radius of the sphere, R = 3cm
Now, volume of the sphere = Volume of the cylindrical wire
$\Rightarrow\frac{4}{3}\pi\text{R}^3=\pi\text{r}^2\text{h}$
$\Rightarrow4\times3^2=(0.1)^2\times\text{h}$
$\Rightarrow\text{h}=\frac{4\times9}{0.1\times0.1}=3600\text{cm}=36\text{m}$
$\therefore$ Length of the wire = 36m
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Question 193 Marks
The slant height and base diameter of a conical tomb are 25 and 14 m respectively. Find the cost of whitewashing its curved surface at the rate of ₹ 12 per m2. $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$
Answer
Radius of a cone, r = 7m
Slant height of a cone, l = 25cm
Curved surface area of a cone $=\pi\text{rl}$
$=\Big(\frac{22}{7}\times7\times25\Big)\text{m}^2$
$=550\text{m}^2$
Cost of whitewashing = ₹ 12 per m2
⇒ Cost of whitewashing 550m2 area = ₹ (12 × 550) = ₹ 6600
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Question 203 Marks
A hemisphere of lead of radius 9cm is cast into a right circular cone of height 72cm. Find the radius of the base of the cone.
Answer
Radius of the hemisphere = 9cm
Height of the right circular cone = 72cm
Suppose that the radius of the base of the cone is r cm.
Volume of the hemisphere = thereforeolume of the cone
$\Rightarrow\frac{2}{3}\pi\times9^3=\frac{1}{3}\pi\times\text{r}^2\times72$
$\Rightarrow\text{r}^2=\frac{2\times9\times9\times9}{72}=\frac{81}{4}$
$\Rightarrow\text{r}=\frac{9}{2}=4.5\text{cm}$
$\therefore$ The radius of the base of the cone is 4.5cm.
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Question 213 Marks
The diameter of a copper sphere is 18cm. It is melted and drawn into a long wire of uniform cross section. If the length of the wire is 108m, find its diameter.
Answer
Radius of the spheres, R = 9cm

Length of the wire, h = 108m = 10800cm

Volume of the sphere = Volume of the wire

Suppose that r cm is the radius of the wire.

Then $\frac{4}{3}\pi\text{R}^3=\pi\text{r}^2\text{h}$

$\Rightarrow\frac{4}{3}\times9^3=\text{r}^2\times10800$

$\Rightarrow\text{r}^2=\frac{4\times729}{3\times10800}=\frac{4\times81}{3\times1200}=\frac{9}{100}$

$\Rightarrow\text{r}=\frac{3}{10}=0.3\text{cm}$

$\therefore$ Diameter of the wire = 0.6cm

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Question 223 Marks
Find the volume and surface area of a sphere whose radius is: $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
4.2cm
Answer
Radius of the sphere = 4.2cm

Now, volume $=\frac{4}{3}\pi\text{r}^3$

$=\frac{4}{3}\times\frac{22}{7}\times4.2\times4.2\times4.2$

$=310.46\text{cm}^3$

$\therefore$ Surface area $=4\pi\text{r}^2$

$=4\times\frac{22}{7}\times4.2\times4.2$

$=221.76\text{cm}^2$

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Question 233 Marks
A solid sphere of radius 3cm is melted and then cast into smaller spherical balls, each of diameter 0.6cm. Find the number of small balls thus obtained. $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
Answer
Radius of the solid sphere = 3cm
Volume of the solid sphere $=\frac{4}{3}\pi\text{r}^3$
$=\frac{4}{3}\times\frac{22}{7}\times3^3\text{cm}^3$
Diameter of the spherical ball = 0.6
Radius of the spherical ball = 0.3cm
Volume of the spherical ball $=\frac{4}{3}\times\frac{22}{7}\times(0.3)^3\text{cm}^3$
Now, number of small spherical balls $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{sphere}}{\text{Volume}\ \text{of}\ \text{the}\ \text{spherical}\ \text{ball}}$
$=\frac{\frac{4}{3}\pi\times27}{\frac{4}{3}\pi\times(0.3)^3}$
$=1000$
$\therefore$ The number of small balls thus obtained is 1000.
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Question 243 Marks
A sphere of diameter 15.6cm is melted and cast into a right circular cone of height 31.2cm. Find the diameter of the base of the cone.
Answer
Radius of the sphere, r = 7.8cm
Height of the cone, h = 31.2cm
Suppose that R cm is the radius of the cone.
Now, volume of the sphere = Volume of the cone
$\Rightarrow\frac{4}{3}\pi\text{r}^3=\pi\text{R}^2\text{h}$
$\Rightarrow4\times(7.8)^3=\text{R}^2\times31.2$
$\Rightarrow\text{R}^2=\frac{4\times7.8\times7.8\times7.8}{31.2}=60.84$
$\Rightarrow\text{R}=7.8\text{cm}$
$\therefore$ The diameter of the cone is 15.6cm.
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Question 253 Marks
How many bricks will be required to construct a wall 8m long, 6m high and 22.5cm thick if each brick measures (25cm × 11.25cm × 6cm)?
Answer
Length of the wall = 8m = 800cm
Breadth of the wall = 22.5cm
Height of the wall = 6m = 600cm
i.e., volume of wall = 800 × 22.5 × 600cm3 = 10800000cm3
Length of the brick = 25cm
Breadth of the brick = 11.25cm
Height of the brick = 6cm
i.e., volume of one brick = 25 × 11.25 × 6 = 1687.5 cm3
$\therefore$ Number of bricks $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{wall}}{\text{Volume}\ \text{of}\ \text{the}\ \text{brick}}$
$=\frac{10800000}{1687.5}=6400$
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Question 263 Marks
If V is the volume of a cuboid of dimensions a, b, c and S is its surface area then prove that $\frac{1}{\text{V}}=\frac{2}{\text{S}}\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}\Big).$
Answer
Let the length, breadth and height of the cuboid be a, b and c respectively.
$\therefore$ Surface area of the cuboid, S = 2(ab + bc + ca)
Volume of the cuboid, V = abc
Now,
$\frac{\text{S}}{\text{V}}=\frac{2(\text{ab}+\text{bc}+\text{ca})}{\text{abc}}$
$\Rightarrow\frac{\text{S}}{\text{V}}=2\Big(\frac{\text{ab}}{\text{abc}}+\frac{\text{bc}}{\text{abc}}+\frac{\text{ca}}{\text{abc}}\Big)$
$\Rightarrow\frac{\text{S}}{\text{V}}=2\Big(\frac{1}{\text{c}}+\frac{1}{\text{a}}+\frac{1}{\text{b}}\Big)$
$\Rightarrow\frac{1}{\text{V}}=\frac{2}{\text{S}}\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}\Big)$
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Question 273 Marks
Find the cost of sinking a tube-well 280m deep, having a diameter 3m at the rate of ₹ 15 per cubic metre. Find also the cost of cementing its inner curved surface at ₹ 10 per square metre.
Answer
Radius, r = 1.5m
Height, h = 280m
$\therefore$ Volume of the tubewell $=\pi\text{r}^2\text{h}$
$=\Big(\frac{22}{7}\times1.5\times1.5\times280\Big)\text{cm}^3$
$=\Big(22\times\frac{15}{10}\times\frac{15}{10}\times40\Big)\text{cm}^3$
$=1980\text{cm}^3$
⇒ Cost of sinking the tubewell = ₹ (15 × 1980) = ₹ 29700
Curved surface area of tubewell $=2\pi\text{rh}$
$=\Big(2\times\frac{22}{7}\times1.5\times280\Big)\text{cm}^2$
$=2640\text{cm}^2$
⇒ Cost of cementing = ₹ (10 × 2640) = ₹ 26400
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Question 283 Marks
A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-planting it on the inside at the rate of ₹ 32 per 100cm2. $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
Answer
Inner radius of the bowl, $\text{r}=\frac{10.5}{2}=5.25\text{cm}$
$\therefore$ Inner curved surface area of the bowl $=2\pi\text{r}^2$
$=2\times\frac{22}{7}\times(5.25)^2$
$=173.25\text{cm}^2$
Rate of tin-planting = ₹ 32 per 100cm2
$\therefore$ Cost of tin-planting the bowl on the inside
= Inner curved surface area of the bowl × Rate of tin-planting
$=173.25\times\frac{32}{100}$
$=₹\ 55.44$
Thus, the cost of tin-planting the bowl on the inside is ₹ 55.44.
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Question 293 Marks
Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of the hemisphere?
Answer
Let the radius of the solid hemisphere be r units.
Numerical value of surface area of the solid hemisphere $=3\pi\text{r}^2$
Numerical value of volume of the solid hemisphere $=\frac{2}{3}\pi\text{r}^3$
It is given that volume and surface area of the solid hemisphere are numerically equal.
$\therefore\frac{2}{3}\pi\text{r}^3=3\pi\text{r}^2$
$\Rightarrow2\text{r}=9\ \text{units}$
Thus, the diameter of the hemisphere is 9 units.
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Question 303 Marks
A hemispherical bowl is made of steel 0.5cm thick. The inside radius of the bowl is 4cm. Find the volume of steel used in making the bowl. $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
Answer
Internal radius of the hemispherical bowl = 4cm
Thickness of a the bowl = 0.5cm
Now, external radius of the bowl = (4 + 0.5)cm = 4.5cm
Now, volume of steel used in making the bowl = Volume of the shell
$=\frac{2}{3}\pi(4.5^3-4^3)$
$=\frac{2}{3}\times\frac{22}{7}\times(91.125-64)$
$=\frac{2}{3}\times\frac{22}{7}\times27.125$
$=56.83\text{cm}^3$
$\therefore$ 56.83cm3 of steel is used in making the bowl.
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Question 313 Marks
The total surface area of a cube is 1176cm2. Find its volume.
Answer
Suppose that the side of cube is x cm.
Total surface area of cube = 1176sq cm
Then 1176 = 6x2
$\Rightarrow\text{x}^2=\frac{1176}{6}=196$
$\Rightarrow\text{x}=\sqrt{196}=14$
i.e., the side of the cube is 14cm.
$\therefore$ Volume of the cube = x3 = 143cm3 = 2744cm3
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Question 323 Marks
A hollow spherical shell is made of density 4.5g per cm3. If its internal and external radii are 8cm and 9cm respectively, find the weight of the shell.
Answer
Internal radius of the hollow spherical shell, r = 8cm
External radius of the hollow spherical shell, R = 9cm
Volume of the shell $=\frac{4}{3}\pi(\text{R}^3-\text{r}^3)$
$=\frac{4}{3}\pi(9^3-8^3)$
$=\frac{4}{3}\times\frac{22}{7}\times(729-512)$
$=\frac{4\times22\times217}{21}$
$=\frac{88\times31}{3}$
$=\frac{2728}{3}\text{cm}^3$
Weight of the shell = Volume of the shell × density per cubic cm
$=\frac{2728}{3}\times4.5\approx4092\text{kg}=4.092\text{kg}$
$\therefore$ Weight of the shell = 4.092kg
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Question 333 Marks
Find the volume and surface area of a sphere whose radius is: $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
5m
Answer
Radius of the sphere = 5m
Now, volume $=\frac{4}{3}\pi\text{r}^3$
$=\frac{4}{3}\times\frac{22}{7}\times5^3$
$=523.81\text{cm}^3$
$\therefore$ Surface area $=4\pi\text{r}^2$
$=4\times\frac{22}{7}\times5^2$
$=314.29\text{cm}^2$
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Question 343 Marks
How many lead shots, each 3mm in diameter, can be made from a cuboid with dimensions. (12cm × 11cm × 9cm)? $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
Answer
Here, l = 12cm, b = 11cm and h = 9cm
Volume of the cuboid = l × b × h
= 12 × 11 × 9
= 1188cm3
Radius of one lead shot $=3\text{mm}=\frac{0.3}{2}\text{cm}$
Volume of one lead shot $=\frac{4}{3}\times\frac{22}{7}\times\Big(\frac{0.3}{2}\Big)^3$
$=\frac{11\times9}{7000}$
$=0.014\text{cm}^3$
$\therefore$ Number of lead shots $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{cuboid}}{\text{Volume}\ \text{of}\ \text{one}\ \text{lead}\ \text{shot}}$
$=\frac{1188}{0.014}$
$=84857.14\approx84857$
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Question 353 Marks
How many spheres 12cm in diameter can be made from a metallic cylinder of diameter 8cm and height 90cm?
Answer
Diameter of each sphere, d = 12cm
Radius of each sphere, r = 6cm
Volume of each sphere $=\frac{4}{3}\pi\text{r}^3$
$=\frac{4}{3}\pi(6)^3\text{cm}^3$
Diameter of base of the cylinder, D = 8cm
Radius of base of cylinder, R = 4cm
Height of the cylinder, h = 90cm
Volume of the cylinder $=\pi\text{R}^2\text{h}$
$=\pi(4)^2\times90\text{cm}^3$
Number of spheres $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{cylinder}}{\text{Volume}\ \text{of}\ \text{the}\ \text{sphere}}$
$=\frac{\pi\text{R}^2\text{h}}{\frac{4}{3}\pi\text{r}^3}$
$=\frac{4^2\times90\times3}{4\times6^3}$
$=\frac{12\times90}{216}$
$=5$
$\therefore$ Five spheres can be made.
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Question 363 Marks
How many lead balls, each of radius 1cm, can be made from a sphere of radius 8cm? $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
Answer
Radius of the sphere = 8cm
Volume of the sphere $=\frac{4}{3}\pi\text{r}^3$
$=\frac{4}{3}\times\frac{22}{7}\times8^3$
$=2145.52\text{cm}^2$
Radius of one lead ball = 1cm
Volume of one lead ball $=\frac{4}{3}\times\frac{22}{7}\times1^3=4.19\text{cm}^3$
$\therefore$ Number of lead balls $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{sphere}}{\text{Volume}\ \text{of}\ \text{one}\ \text{lead}\ \text{ball}}$
$=\frac{2145.52}{4.19}$
$=512.05\approx512$
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Question 373 Marks
The curved surface area of a right circular cylinder is 4.4m2. If the radius of its base is 0.7m, find its:
  1. Height.
  2. Volume.
Answer
Curved surface area of a cylinder = 4.4m2

Radius (r) of a cylinder = 0.7m

  1. Curved surface area of cylinder $=2\pi\text{rh}$

$\Rightarrow4.4=2\times\frac{22}{7}\times0.7\times\text{h}$

$\Rightarrow4.4=2\times22\times0.1\times\text{h}$

$\Rightarrow\text{h}=\frac{4.4}{2\times22\times0.1}=1\text{m}$

  1. Volume of a cylinder $=\pi\text{r}^2\text{h}$

$=\Big(\frac{22}{7}\times0.7\times0.7\times1\Big)\text{m}^3$

$=1.54\text{m}^3$

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Question 383 Marks
A metallic sphere of radius 10.5cm is melted and then recast into smaller cones, each of radius 3.5cm and height 3cm. How many cones are obtained?
Answer
Radius of the metallic sphere, r1 = 10.5cm
Volume of the sphere $=\frac{4}{3}\pi\text{r}^3$
$=\frac{4}{3}\pi(10.5)^3\text{cm}^3$
Radius of each smaller cone, r2 = 3.05cm
Height of each smaller cone = 3cm
Volume of each smaller cone $=\frac{1}{3}\pi\text{r}_2\text{h}$
$=\frac{1}{3}\pi(3.05)^2\times3\text{cm}^3$
Number of cones obtained $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{sphere}}{\text{Volume}\ \text{of}\ \text{each}\ \text{smaller}\ \text{cone}}$
$=\frac{\frac{4}{3}\pi\text{r}^3}{\frac{1}{3}\pi\text{r}_2^2\text{h}}$
$=\frac{4\times10.5\times10.5\times10.5}{3.5\times3.5\times3}$
$=126.006\approx126$
$\therefore$ 126 cones are obtained.
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Question 393 Marks
A spherical cannonball 28cm in diameter is melted and cast into a right circular cone mould, whose base is 35cm in diameter. Find the height of the cone.
Answer
 Radius of the spherical cannonball, R = 14cm

Radius of the base of the cone, r = 17.5cm

Let h cm be the height of the cone.

Now, volume of the sphere = Volume of the cone

$\Rightarrow\frac{4}{3}\pi\text{R}^3=\pi\text{r}^2\text{h}$

$\Rightarrow4\times14^3=(17.5)^2\times\text{h}$

$\Rightarrow\text{h}=\frac{4\times14\times14\times14}{17.5\times17.5}=35.84\text{cm}$

$\therefore$ The height of the cone is 35.84cm. 

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