Question 13 Marks
In a water heating system, there is a cylindrical pipe of length $28m$ and diameter $5\ cm.$ Find the total radiating surface in the system.
AnswerDiameter of a cylindrical pipe $= 5\ cm$
$\Rightarrow $ Radius $(r)$ of a cylindrical pipe, $= 2.5\ cm$
Height $(h)$ of a cylindrical pipe $= 28\ m = 2800\ cm$
Now, Total radiating surface in the system $=$ Curved surface area of the cylindrical pipe
$=2\pi\text{rh}$
$=\Big(2\times\frac{22}{7}\times2.5\times2800\Big)\text{cm}^2$
$=44000\text{cm}^2$
View full question & answer→Question 23 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. $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$
AnswerRadius 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 $=\Big(\frac{\text{Area}\ \text{of}\ \text{canvas}}{\text{Width}\ \text{of}\ \text{canvas}}\Big)\text{m}$
$=\Big(\frac{9735}{5}\Big)=1947\text{m}.$
View full question & answer→Question 33 Marks
How many litres of water flows out of a pipe having an area of cross section of $5 cm^2$ in $1$ minute, if the speed of water in the pipe is $30 cm / sec$ ?
AnswerSpeed of water $=30 cm / sec$
$\therefore$ Volume of water that flows out of the pipe in one second
$=$ Area of cross section $\times$ Length of water flows in one second $=(5 \times 30) cm ^3=150 cm^3$
Hence, volume of water that flows out of the pipe in 1 minute $=(150 \times 60) cm ^3=9000 cm^3=9$ litres
View full question & answer→Question 43 Marks
A spherical ball of radius $3\ cm$ is melted and recast into three spherical balls. The radii of two of these balls are $1.5\ cm$ and $2\ cm.$ Find the radius of the third ball.
AnswerRadius of the original spherical ball $= 3\ cm$
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 $2.5\ cm$
View full question & answer→Question 53 Marks
The lateral surface area of a cylinder is $94.2\ cm^2$ and its height is $5\ cm \big($Take$\ \pi=3.14\big)$. Find:
$i.$ The radius of its base.
$ii.$ Its volume.
AnswerLateral surface area of a cylinder $= 94.2\ cm^2$
Height $(h)$ of a cylinder $= 5\ cm$
$i.$ Lateral surface area of cylinder $=2\pi\text{rh}$
$\Rightarrow94.2=2\times3.14\times\text{r}\times5$
$\Rightarrow\text{r}=\frac{94.2}{2\times3.14\times5}=3\text{ cm}$
$ii.$ Volume of a cylinder $=\pi\text{r}^2\text{h}$
$=(3.14\times3\times3\times5)\text{cm}^3$
$=141.3\text{ cm}^3$
View full question & answer→Question 63 Marks
A metallic sphere of radius $10.5\ cm$ is melted and then recast into smaller cones, each of radius $3.5\ cm$ and height $3\ cm$. How many cones are obtained?
AnswerRadius of the metallic sphere, $r_{1}= 10.5cm$
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, $r_2 = 3.05cm$
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.
View full question & answer→Question 73 Marks
From a solid right circular cylinder with height $10\ cm$ and radius of the base $6\ cm$, a right circular cone of the same height and base is removed. Find the volume of the remaining solid. $\big(\text{Take}\ \pi=3.14\big).$
AnswerHere, 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 $753.6cm^3.$
View full question & answer→Question 83 Marks
A cylindrical bucket with base radius $15\ cm$ is filled with water up to a height of $20\ cm$. A heavy iron spherical ball of radius $9\ cm$ is dropped into the bucket to submerge completely in the water. Find the increase in the level of water.
AnswerLet $h \ cm$ be the increase in the level of water.
Radius of the cylinder bucket $= 15\ cm$
Height up to which water is being filled $= 20\ cm$
Radius of the spherical ball $= 9\ cm$
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 $4.32\ cm.$
View full question & answer→Question 93 Marks
A spherical cannonball $28\ cm$ in diameter is melted and cast into a right circular cone mould, whose base is $35\ cm$ in diameter. Find the height of the cone.
AnswerRadius of the spherical cannonball, $R = 14\ cm$
Radius of the base of the cone,$ r = 17.5\ cm$
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 $35.84\ cm.$
View full question & answer→Question 103 Marks
A joker's cap is in the form of a right circular cone of base radius $7\ cm$ and height $24\ cm$. Find the area of the sheet required to make $10$ such caps. $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$
AnswerRadius 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 $= 10 \times 550 = 5500cm^2$
Thus, $5500cm^2$ sheet will be required to make $10$ caps.
View full question & answer→Question 113 Marks
How many lead shots, each $3\ mm$ in diameter, can be made from a cuboid with dimensions. $(12\ cm \times 11\ cm \times 9\ cm)?$ $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
AnswerHere, $l = 12\ cm, b = 11\ cm$ and $h = 9\ cm$
Volume of the cuboid $= l \times b \times h = 12 \times 11 \times 9 = 1188\ cm^3$
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 $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{cuboid}}{\text{Volume}\ \text{of}\ \text{one}\ \text{lead}\ \text{shot}}$
$=\frac{1188}{0.014}$
$=84857.14\approx84857$
View full question & answer→Question 123 Marks
Find the length of $13.2\ kg$ of copper wire of diameter 4mm, when $1$ cubic centimetre of copper weighs $8.4g.$
AnswerLet the length of the wire $= 'h' $ meters Then,
Volume of the wire $\times 8.4 = (13.2 \times 1000)g$
$\Rightarrow\frac{22}{7}\times\Big(\frac{2}{10}\Big)^2\times\text{h}\times8.4=13200$
$\Rightarrow22\times\Big(\frac{1}{5}\Big)^2\times\text{h}\times1.2=13200$
$\Rightarrow\text{h}=\frac{13200\times5\times5}{22\times1.2}=12500\text{cm}$
$\Rightarrow\text{h}=125\text{m}$
Thus, the length of the wire is $125m.$
View full question & answer→Question 133 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?
AnswerLet 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.
View full question & answer→Question 143 Marks
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. $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
AnswerInner radius of the bowl, $r =5 \ cm$
Let the outer radius of the bowl be $R \ cm .$
Thickness of the bowl $=0.25 \ cm$ (Given)
$\therefore R - r =0.25 \ cm$
$ \Rightarrow R =0.25+ r =0.25+5=5.25 \ cm$
$\therefore$ Outer curved surface area of the bowl
$=2 \pi r ^2$ $=2 \times \frac{22}{7} \times(5.25)^2=173.25 \ cm^2$
Thus, the outer curved surface area of the bowl is $173.25 \ cm^2$.
View full question & answer→Question 153 Marks
Find the volume of a sphere whose surface area is $154cm^2$. $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
AnswerLet the radius of the sphere be r cm. Surface area of the sphere $= 154cm^2$
$\therefore4\pi\text{r}^2=154$
$\Rightarrow4\times\frac{22}{7}\times\text{r}^2=154$
$\Rightarrow\text{r}=\sqrt{\frac{154\times7}{4\times22}}=\sqrt{1225}$
$\Rightarrow\text{r}=3.5\text{cm}$
$\therefore$ 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 $179.67m^3.$
View full question & answer→Question 163 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 $2.5\ m.$ (Given, $1000$ litres $= 1m^3.)$
AnswerCapacity of the tank $50000\text{L}=\frac{50000}{1000}=50\text{m}^3(1000L = 1m^3)$
Length of the tank $ = 10m$ Height (or depth) of the tank $= 2.5m$
Now, Volume of the cuboidal tank $=$ Length $\times$ Breadth $\times$ 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 $2\ m.$
View full question & answer→Question 173 Marks
A cylindrical container with diameter of base $56\ cm$ contains sufficient water to submerge a rectangular solid of iron with dimensions $(32\ cm \times 22\ cm \times 14\ cm)$. Find the rise in the level of water when the solid is completely submerged.
AnswerLet the rise in the level of water $= h\ cm$
Then, Volume of the cylinder of height $h$ and base radius $28\ cm =$ 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 $4\ cm.$
View full question & answer→Question 183 Marks
Find the volume and surface area of a sphere whose radius is: $\big(\text{Take}\ \pi=\frac{22}{7}\big). 3.5\ cm$
AnswerRadius of the sphere $= 3.5\ cm$
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 $=4\pi\text{r}^2$
$=4\times\frac{22}{7}\times3.5\times3.5$
$=154\text{cm}^2$
View full question & answer→Question 193 Marks
Find the surface area of a sphere whose volume is $606.375m^3.$ $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
AnswerVolume of the sphere $= 606.375m^3$
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 $=4\pi\text{r}^2$
$=4\times\frac{22}{7}\times5.25\times5.25$
$=346.5\text{m}^2$
View full question & answer→Question 203 Marks
The curved surface area of a right circular cylinder is $4.4\ m^2$. If the radius of its base is$ 0.7\ m$ , find its:
$i.$ Height.
$ii.$ Volume.
AnswerCurved surface area of a cylinder $=4.4\ m^2$ Radius $(r)$ of a cylinder $=0.7\ m$
$i.$ Curved surface area of cylinder $=2 \pi rh$
$\Rightarrow4.4=2\times\frac{22}{7}\times0.7\times\text{h}$
$\Rightarrow4.4=2\times22\times0.1\times\text{h}$
$\Rightarrow\text{h}=\frac{4.4}{2\times22\times0.1}=1\text{ m}$
$ii.$ Volume of a cylinder $=\pi\text{r}^2\text{h}$
$=\Big(\frac{22}{7}\times0.7\times0.7\times1\Big)\text{m}^3$
$=1.54\text{ m}^3$
View full question & answer→Question 213 Marks
The total surface area of a cube is $1176\ cm^2$. Find its volume.
AnswerSuppose that the side of cube is $x \ cm$.
Total surface area of cube $=1176\ sq\ cm$
Then $1176=6 x ^2 \Rightarrow x ^2=\frac{1176}{6}=196$
$\Rightarrow x =\sqrt{196}=14$
i.e., the side of the cube is $14\ cm .$
$\therefore$ Volume of the cube $= x ^3=14^3\ cm^3=2744\ cm^3$
View full question & answer→Question 223 Marks
If $V$ is the volume of a cuboid of dimensions $a, b, c $and S is its surface area then prove that $\frac{1}{\text{V}}=\frac{2}{\text{S}}\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}\Big).$
AnswerLet 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, $\frac{\text{S}}{\text{V}}=\frac{2(\text{ab}+\text{bc}+\text{ca})}{\text{abc}}$
$\Rightarrow\frac{\text{S}}{\text{V}}=2\Big(\frac{\text{ab}}{\text{abc}}+\frac{\text{bc}}{\text{abc}}+\frac{\text{ca}}{\text{abc}}\Big)$
$\Rightarrow\frac{\text{S}}{\text{V}}=2\Big(\frac{1}{\text{c}}+\frac{1}{\text{a}}+\frac{1}{\text{b}}\Big)$
$\Rightarrow\frac{1}{\text{V}}=\frac{2}{\text{S}}\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}\Big)$
View full question & answer→Question 233 Marks
The diameter of a sphere is $6\ cm$. It is melted and drawn into a wire of diameter $2\ mm$. Find the length of the wire.
AnswerLet the length of the wire be $h\ cm.$
Radius of the wire, $r = 1\ mm = 0.1\ cm$
Radius of the sphere, $R = 3\ cm$
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 $= 36\ m$
View full question & answer→Question 243 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 $m ^2$. (Use $\pi=\frac{22}{7}$ ).
AnswerRadius of a cone, $r =7\ m$
Slant height of a cone, $I =25\ cm$
Curved surface area of a cone $=\pi rl$
$=\Big(\frac{22}{7}\times7\times25\Big)\text{m}^2$
$=550\text{m}^2$ Cost of whitewashing = ₹ $12$ per $m^2$
$⇒$ Cost of whitewashing $550m^2$
area $= ₹ (12 \times 550) = ₹\ 6600$
View full question & answer→Question 253 Marks
A hemisphere of lead of radius $9\ cm$ is cast into a right circular cone of height $72\ cm$. Find the radius of the base of the cone.
AnswerRadius of the hemisphere $= 9\ cm$
Height of the right circular cone $= 72\ cm$
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 $4.5\ cm.$
View full question & answer→Question 263 Marks
The diameter of a copper sphere is $18\ cm$. It is melted and drawn into a long wire of uniform cross section. If the length of the wire is $108\ m$, find its diameter.
AnswerRadius of the spheres, $R = 9\ cm$
Length of the wire, $h = 108\ m = 10800\ cm$
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 $= 0.6\ cm$
View full question & answer→Question 273 Marks
Find the volume and surface area of a sphere whose radius is: $\big(\text{Take}\ \pi=\frac{22}{7}\big). 4.2cm$
AnswerRadius 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 $=4\pi\text{r}^2$
$=4\times\frac{22}{7}\times4.2\times4.2$
$=221.76\text{cm}^2$
View full question & answer→Question 283 Marks
A solid sphere of radius $3\ cm$ is melted and then cast into smaller spherical balls, each of diameter $0.6\ cm$. Find the number of small balls thus obtained. $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
AnswerRadius 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 $1000.$
View full question & answer→Question 293 Marks
A sphere of diameter $15.6\ cm$ is melted and cast into a right circular cone of height $31.2\ cm$. Find the diameter of the base of the cone.
AnswerRadius of the sphere,$ r = 7.8\ cm$
Height of the cone, $h = 31.2\ cm$
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 $15.6\ cm$
View full question & answer→Question 303 Marks
How many bricks will be required to construct a wall $8m$ long, $6\ m$ high and $22.5\ cm$ thick if each brick measures $(25cm \times 11.25cm \times 6cm)?$
AnswerLength of the wall $=8 m=800 cm$
Breadth of the wall $=22.5 cm$
Height of the wall $=6 m=600 cm$
i.e., volume of wall $=800 \times 22.5 \times 600 cm^3=10800000 cm^3$
Length of the brick $=25 cm$ Breadth of the brick $=11.25 cm$
Height of the brick $=6 cm$
i.e., volume of one brick $=25 \times 11.25 \times 6=1687.5 cm^3$
$\therefore$ Number of bricks $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{wall}}{\text{Volume}\ \text{of}\ \text{the}\ \text{brick}}$
$=\frac{10800000}{1687.5}=6400$
View full question & answer→Question 313 Marks
Find the cost of sinking a tube-well $280\ m$ deep, having a diameter $3\ m$ at the rate of ₹ $15$ per cubic metre. Find also the cost of cementing its inner curved surface at ₹ $10$ per square metre.
AnswerRadius, $r = 1.5m$
Height, $h = 280m$
$\therefore$ Volume of the tubewell $=\pi\text{r}^2\text{h}$
$=\Big(\frac{22}{7}\times1.5\times1.5\times280\Big)\text{cm}^3$
$=\Big(22\times\frac{15}{10}\times\frac{15}{10}\times40\Big)\text{cm}^3$
$=1980\text{cm}^3$
\Rightarrow Cost of sinking the tubewell $= ₹ (15 \times 1980) = ₹ 29700$
Curved surface area of tubewell $=2\pi\text{rh}$
$=\Big(2\times\frac{22}{7}\times1.5\times280\Big)\text{cm}^2$
$=2640\text{cm}^2$
$\Rightarrow$ Cost of cementing $= ₹ (10 \times 2640) = ₹ 26400$
View full question & answer→Question 323 Marks
How many spheres $12\ cm$ in diameter can be made from a metallic cylinder of diameter $8\ cm$ and height $90\ cm?$
AnswerDiameter 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, $h = 90\ cm$
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.
View full question & answer→Question 333 Marks
A hemispherical bowl made of brass has inner diameter $10.5\ cm$. Find the cost of tin-planting it on the inside at the rate of ₹ $32$ per $100cm^2$. $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
AnswerInner 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 $100cm^2$
$\therefore$ 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 ₹ $55.44.$
View full question & answer→Question 343 Marks
The lateral surface area of a cube is $900cm^2$. Find its volume.
AnswerSuppose that the side of cube is $x cm .$
Lateral surface area of the cube $=900 cm^2$
Then $900=4 x ^2$
$\Rightarrow\text{x}^2=\frac{900}{4}=225$
$\Rightarrow\text{x}=\sqrt{225}=15$
i.e., the side of the cube is $15 cm $.
$\therefore$ Volume of the given cube
$=x^3 cm^3=15^3 cm^3=3375 cm^3$
View full question & answer→Question 353 Marks
Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of the hemisphere?
AnswerLet 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.
View full question & answer→Question 363 Marks
A hemispherical bowl is made of steel $0.5\ cm$ thick. The inside radius of the bowl is $4\ cm$. Find the volume of steel used in making the bowl. $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
AnswerInternal radius of the hemispherical bowl $= 4\ cm$ Thickness of a the bowl $= 0.5\ cm$
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 $=\frac{2}{3}\pi(4.5^3-4^3)$
$=\frac{2}{3}\times\frac{22}{7}\times(91.125-64)$
$=\frac{2}{3}\times\frac{22}{7}\times27.125$
$=56.83\text{cm}^3$
$\therefore 56.83\ cm^3$ of steel is used in making the bowl.
View full question & answer→Question 373 Marks
How many lead balls, each of radius $1\ cm$, can be made from a sphere of radius $8\ cm$? $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
AnswerRadius of the sphere $= 8\ cm$
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 =$ 1\ cm$
Volume of one lead ball $=\frac{4}{3}\times\frac{22}{7}\times1^3=4.19\text{cm}^3$
$\therefore$ Number of lead balls $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{sphere}}{\text{Volume}\ \text{of}\ \text{one}\ \text{lead}\ \text{ball}}$
$=\frac{2145.52}{4.19}$
$=512.05\approx512$
View full question & answer→Question 383 Marks
A hollow spherical shell is made of density $4.5g$ per $cm^3$. If its internal and external radii are $8\ cm$ and $9\ cm $respectively, find the weight of the shell.
AnswerInternal radius of the hollow spherical shell, $r = 8\ cm$
External radius of the hollow spherical shell,
$R = 9\ cm$ 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 $\times$ density per cubic $cm$
$=\frac{2728}{3}\times4.5\approx4092\text{kg}=4.092\text{kg}$
$\therefore$ Weight of the shell $= 4.092kg$
View full question & answer→Question 393 Marks
Find the volume and surface area of a sphere whose radius is: $\big(\text{Take}\ \pi=\frac{22}{7}\big). 5m$
AnswerRadius 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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