Question 14 Marks
It costs $₹ 3300$ to paint the inner curved surface of a cylindrical vessel $10m$ deep at the rate of $₹ 30\ per\ m^2.$ Find the:
- Inner curved surface area of the vessel.
- Inner radius of the base.
- Capacity of the vessel.
Answer
- Cost of painting inner curved surface area of vassel
$=$ Cost of painting per $m^2\ \times$ Inner curved surface of vessel
$\Rightarrow ₹ 3300 = ₹ 30\ \times$ Inner curved surface of vessel
$\Rightarrow$ Inner curved surface of vessel $= 110m^2$
- Let inner radius of the base $= r$
Depth, $h = 10m$
Inner curved surface of vessel $=2\pi\text{rh}$
$\Rightarrow10=2\times\frac{22}{7}\times\text{r}\times10$
$\Rightarrow\text{r}=\frac{110\times7}{2\times22\times10}=1.75\text{m}$
- Capacity of the vessel $=\pi\text{r}^2\text{h}$
$=\Big(\frac{22}{7}\times1.75\times1.75\times10\Big)\text{m}^3$
$=96.25\text{m}^3$ View full question & answer→Question 24 Marks
$1\ cm^3$ of gold is drawn into a wire $0.1\ mm$ in diameter. Find the length of the wire.
Answer$ 1\ cm^3 = 1\ cm \times 1\ cm \times 1\ cm = 0.01m$
Therefore,
Volume of the
gold $= 0.01m \times 0.01m \times 0.01m = 0.000001m^3 ...(1)$
Diameter of the wire drawn $= 0.1mm$
Radius of the wire drawn $=\frac{0.1}{2}\text{mm}=0.05\text{mm}$
$r = 0.00005m ...(2)$
Length of the wire $= h\ m ...(3)$
Volume $pf$ the wire drawn $=$ Volume of the gold
$\Rightarrow\pi\text{r}^2\text{h}=0.000001$
$\Rightarrow\pi\times0.00005\times0.00005\times\text{h}=0.000001 [$from equations $(1), (2)$ and $(3)]$
$\text{h}=\frac{0.00000\times7}{0.00005\times0.00005\times22}$
$\therefore\text{h}=127.27\text{m}$
$\therefore$ the length of the wire is $127.27m.$
View full question & answer→Question 34 Marks
A classroom is $10m$ long, $6.4m$ wide and $5m$ high. If each student be given $1.6m^2$ of the floor area, how many students can be accommodated in the room$?$ How many cubic metres of air would each student get$?$
AnswerLength of the classroom $= 10m$
Breadth of the classroom $= 6.4m$
Height of the classroom $= 5m$
Area of the floor $=$ length $\times $ breadth
$= 10 \times 6.4m^2$
No. of students $=\frac{\text{Area}\ \text{of}\ \text{the}\ \text{floor}}{\text{Area}\ \text{given}\ \text{to}\ \text{student}\ \text{on}\ \text{the}\ \text{floor}}$
$=\frac{640}{16}=40\ \text{students}$
Now, volume of the classroom $= 10 \times 6.4 \times 5m^3$
$\therefore$ Air required by each student $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{room}}{\text{Number}\ \text{of}\ \text{students}}$
$=\frac{10\times6.4\times5}{40}\text{m}^3$
$=8\text{m}^3$
View full question & answer→Question 44 Marks
How many planks of dimensions $(5m \times 25\ cm \times 10\ cm)$ can be stored in a pit which is $20m$ long, $6m$ wide and $80\ cm$ deep?
AnswerNumber of planks $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{pit}\ \text{in}\ \text{cm}^3}{\text{Volume}\ \text{of}\ 1\ \text{plank}\ \text{in}\ \text{cm}^3}$
Volume of one plank $= (l \times b \times h)cm^3$
$= 500 \times 25 \times 10\ cm^3$
$= 125000\ cm^3$
Volume of the pit $= (l \times b \times h)cm^3$
Here, $l = 20m = 2000\ cm, b = 6m = 600\ cm, h = 80\ cm$
i.e., volume of the pit $= 2000 \times 600 \times 80\ cm^3$
$= 96000000\ cm^3$
$\therefore$ Number of planks $=\frac{96000000}{125000}$
$=\frac{96000}{125}=768$
View full question & answer→Question 54 Marks
The capacity of a closed cylindrical vessel of height $1m$ is $15.4$ litres. Find the area of the metal sheet needed to make it.
AnswerVolume of a cylinder $= 15.4$ litres $= 15400\ cm^3$
Height $(h)$ of a cylinder $= 1m = 100\ cm$
Volume of a cylinder $=\pi\text{r}^2\text{h}$
$\Rightarrow15400=\Big(\frac{22}{7}\times\text{r}^2\times100\Big)$
$\Rightarrow\text{r}^2=\frac{15400\times7}{22\times100}=49$
$\Rightarrow\text{r}=7\text{cm}$
Now,
Area of metal sheet needed $=$ Total surface area of cylinder
$=2\pi\text{r}(\text{h}+\text{r})$
$=\Big[2\times\frac{22}{7}\times7(100+7)\Big]\text{cm}^2$
$=\big[2\times22\times107\big]\text{cm}^2$
$=4708\text{cm}^2$
View full question & answer→Question 64 Marks
A box made of sheet metal costs $₹ 6480$ at $₹ 120$ per square metre. If the box is $5m$ long and 3m wide, find its height.
AnswerLength of the box $= 5m$
Breadth of the box $= 3m$
Area of the sheet required $=\frac{\text{Total}\ \text{cost}}{\text{Cost}\ \text{per}\ \text{metre}\ \text{square}}$
Let $h\ m$ be the height of the box.
Then area of the sheet $=$ total surface area of the box
$= 2(lb + lh + bh)m^2$
$=2(5 \times 3 + 5 \times h + 3 \times h)m^2$
$= 2(15 + 8h) = (30 + 16h )m^2$
Now, $30 + 16\text{h}=\frac{6480}{120}$
$\Rightarrow 30 + 16h = 54$
$\Rightarrow 16h = 24$
$\Rightarrow h = 1.5m$
$\therefore$ The height of the box is $1.5m.$
View full question & answer→Question 74 Marks
Each edge of a cube is increased by $50\%$. Find the percentage increase in the surface area of the cube.
AnswerLet the initial edge of the cube be a units.
$\therefore$ Initial surface area of the cube $= 6a^2$ square units
New edge of the cube $= a + 50\%$ of a $=\text{a}+\frac{50}{100}\text{a}=1.5\text{a}\ \text{units}$
$\therefore$ New surface of the cube $= 6(1.5a)^2 = 13.5a^2$ square units
Increase in surface area of the cube $= 13.5a^2 - 6a^2 = 7.5a^2$ square units
$\therefore$ Percentage increase in the surface area of the cube
$=\frac{\text{Increase}\ \text{in}\ \text{surface}\ \text{area}\ \text{of}\ \text{the}\ \text{cube}}{\text{Initial}\ \text{surface}\ \text{area}\ \text{of}\ \text{the}\ \text{cube}}\times100\%$
$=\frac{7.5\text{a}^2}{6\text{a}^2}\times100\%$
$=125\%$
View full question & answer→Question 84 Marks
There are $20$ cylindrical pillars in a building, each having a diameter of $50\ cm$ and height $4m.$ Find the cost of cleaning them at $₹\ 14\ per\ m^2.$
AnswerRadius $(r)$ of $1$ pillar $=\frac{50}{100\times2}=\frac{1}{4}\text{m}$
Height $(h)$ of $1$ pillar $= 4m$
$\therefore$ Lateral surface area of $1$ pillar $=2\pi\text{rh}$
$=\Big(2\times\frac{22}{7}\times\frac{1}{4}\times4\Big)\text{m}^2$
$=\frac{44}{7}\text{m}^2$
$\Rightarrow $ Lateral surface area of $20$ such pillars $=20\times\frac{44}{7}=\frac{880}{7}\text{m}^2$
Cost of cleaning $= ₹ 14/m^2$
$\Rightarrow $ Cost of cleaning $\frac{880}{7}\text{m}^2=₹\ \Big(14\times\frac{880}{7}\Big)$
$=₹\ 1760$
View full question & answer→Question 94 Marks
Water flows at the rate of 10 metres per minute through a cylindrical pipe 5mm in diameter. How long would it take to fill a conical vessel whose diameter art the surface 40cm and depth 24cm?
AnswerDiameter of the pipe = 5mm = 0.5cm
Radius of the pipe $=\frac{0.5}{2}=0.25\text{cm}$
Length of the pipe = 10 metres = 1000cm
Volume that flows in 1 min $=\big[\pi\times(0.25)^2\times1000\big]\text{cm}^3$
$\therefore$ Volume of the conical vessel $=\Big[\frac{1}{3}\pi\times(20)^2\times24\Big]\text{cm}^3$
$\therefore$ Required time $=\Bigg[\frac{\frac{1}{3}\pi\times(20)^2\times24}{\pi\times(0.25)^2\times1000}\Bigg]\text{min}$
$=\Bigg[\frac{\frac{1}{3}\times400\times24}{\pi\times0.0625\times1000}\Bigg]\text{min}$
$=51.2\ \text{min}$
$=51\ \text{min}\ 12\ \text{sec}$
View full question & answer→Question 104 Marks
Find the capacity of a closed rectangular cistern whose length is $8m,$ breadth 6m and depth $2.5m.$ Also, find the area of the iron sheet required to make the cistern.
AnswerLength of the cistern, $l = 8m$
Breadth of the cistern, $b = 6m$
Height $($or depth$)$ of the cistern, $h = 2.5m$
$\therefore$ Capacity of the cistern
$=$ Volume of the cistern
$= l \times b \times h$
$= 8 \times 6 \times 2.5$
$= 120m^3$
Also,
Area of the iron sheet required to make the cistern
$=$ Total surface area of the cistern
$= 2(lb + bh + hl)$
$= 2(8 \times 6 + 6 \times 2.5 + 2.5 \times 8)$
$= 2 \times 83$
$= 166m^2$
View full question & answer→Question 114 Marks
Find the volume, the lateral surface area and the total surface area of the cuboid whose dimensions are:
Length $= 24m,$ breadth $= 25\ cm$ and height $= 6m$
AnswerHere, $l = 24m, b = 25\ cm = 0.25m, h = 6m$
Volume of the cuboid $= l \times b \times h$
$ = (24 \times 0.25 \times 6)m^3$
$= 36m^3$
Total surface area $= 2(lb \times lh \times bh)$
$= 2(24 \times 0.25 + 24 \times 6 + 0.25 \times 6)m^2$
$= 2(6 + 144 + 1.5)m^2$
$= 2 \times 151.5m^2$
$= 303m^2$
Lateral surface area $= 2(l + b) \times h$
$= [2(24 + 0.25) \times 6]m^2$
$= [2 \times 24 \times 6]m^2$
$= 291m^2$
View full question & answer→Question 124 Marks
It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square metres of the sheet are required for the same?
AnswerDiameter of a cylinder = 140cm
⇒ Radius, r = 70cm
Height (h) of a cylinder = 1m = 100cm
Now,
Area of sheet required = Total surface area of cylinder
$=2\pi\text{r}(\text{h}+\text{r})$
$=\Big[2\times\frac{22}{7}\times70(100+70)\Big]\text{cm}^2$
$=\big[2\times22\times10\times170\big]\text{cm}^2$
$=74800\text{cm}^2$
$=7.48\text{m}^2$
View full question & answer→Question 134 Marks
A cylindrical water tank of diameter 1.4m and height 2.1m is being fed by a pipe of diameter 3.5cm through which water flows at the rate of 2m per second. In how much time will the tank be filled?
AnswerSuppose the tank is filled in x minutes. Then,
Volume of the water that flows out through the pipe in x minutes
= Volume of the tank
$\Rightarrow\pi\times\Big(\frac{3.5}{2\times100}\Big)^2\times(2\times60\text{x})=\pi\times(0.7)\times2.1$
$\Rightarrow\Big(\frac{35}{2000}\Big)^2\times120\text{x}=\Big(\frac{7}{10}\Big)^2\times\Big(\frac{21}{10}\Big)$
$\Rightarrow\frac{35}{2000}\times\frac{35}{2000}\times120\text{x}=\frac{7}{10}\times\frac{7}{10}\times\frac{21}{10}$
$\Rightarrow\text{x}=\frac{7\times7\times21\times2000\times2000}{10\times10\times10\times35\times35\times120}$
$\Rightarrow\text{x}=28$
Hence, the tank will be flled in 28 minutes.
View full question & answer→Question 144 Marks
A man uses a piece of canvas having an area of $551m^2,$ to make a conical tent of base radius $7m.$ Assuming that all the stitching margins and wastage incurred while cutting, amount to approximately $1m^2,$ find the volume of the tent that can be made with it. $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$
AnswerRadius of a conical tent, $r = 7m$
Area of $49$ canvas used in making conical tent $= (551 - 1)m^2 = 550m^2$
$\Rightarrow $ Curved surface areaof a conical tent $= 550m^2$
$\Rightarrow\pi\text{rl}=550$
$\Rightarrow\frac{22}{7}\times7\times\text{l}$
$\Rightarrow\text{l}=\frac{550}{22}=25\text{m}=\text{slant}\ \text{height}$
Now, $l^2 = r^2 + h^2$
$\Rightarrow 25^2 = 7^2 + h^2$
$\Rightarrow h^2 = 25^2 - 7^2 = 625 - 49 = 576$
$\Rightarrow h = 24m =$ height
$\therefore$ Volume of conical tent $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\Big(\frac{1}{3}\times\frac{22}{7}\times7\times7\times24\Big)\text{m}^3$
$=1232\text{m}^3$
View full question & answer→Question 154 Marks
The sum of length, breadth and depth of a cuboid is $19\ cm$ and the length of its diagonal is $11\ cm$. The surface area of the cuboid.
AnswerLet the length and height $($or depth$)$ of the cuboid be $l \ cm, b \ cm$ and $h \ cm,$ respectively.
$\therefore l + b + h = 19 ....(1)$
Also,
Length of the diagonal $= 11\ cm$
$\Rightarrow\sqrt{\text{l}^2+\text{b}^2+\text{h}^2}=11$
$\Rightarrow\text{l}^2+\text{b}^2+\text{h}^2=121\ ...(2)$
Squaring $(1)$ we get
$(l + b + h)^2 = 19^2$
$\Rightarrow l^2 + b^2 + h^2 + 2(lb + bh + hl) = 361$
$\Rightarrow 121 + 2(lb + bh + hl) = 361 [$Using $(2)]$
$\Rightarrow 2(lb + bh + hl) = 361 - 121 = 240\ cm^2$
Thus, the surface area of the cuboid is $240\ cm^2.$
View full question & answer→Question 164 Marks
A wall $15m$ long, $30\ cm$ wide and $4m$ high is made of bricks, each measuring $(22\ cm \times 12.5\ cm \times 7.5\ cm)$. If $\frac{1}{12}$ of the total volume of the wall consists of mortar, how many bricks are there in the wall?
AnswerLength of the wall $= 15m = 1500\ cm$
Breadth of the wall $= 30\ cm$
Height of the wall $= 4m = 400\ cm$
Volume of wall $= 1500 \times 30 \times 400\ cm^{3 }$
$= 18000000\ cm^3$
Now, volume of each brick $= 22 \times 12.5 \times 7.5\ cm^3$
$= 2062.5\ cm^3$
Also, volume of the morter $=\frac{1}{12}\times$ Volume of the wall
$=\frac{18000000}{12}$
$=1500000\ \text{cm}^3$
Total volume of the bricks in the wall $=$ Volume of the wall $-$ Volume of the mortar
$= (18000000 - 1500000)\ cm^3 $
$= 16500000\ cm^3$
$\therefore$ Number of briks $=\frac{\text{Volume}\ \text{of}\ \text{bricks}}{\text{Volume}\ \text{of}\ \text{one}\ \text{brick}}$
$=\frac{16500000}{2062.5}=8000$ bricks
View full question & answer→Question 174 Marks
The curved surface area of a cone is $308\ cm^2$ and its slant height is $14\ cm.$ Find the radius of the base and total surface area of the cone. $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$
AnswerLet $r$ be the radius of a cone.
Slant height of a cone, $l = 14\ cm$
Curved surface area of a cone $= 308\ cm^2$
$\Rightarrow\pi\text{rl}=308$
$\Rightarrow\frac{22}{7}\times\text{r}\times14=308$
$\Rightarrow22\times\text{r}\times2=308$
$\Rightarrow\text{r}=\frac{308}{22\times2}=7\text{cm}$
Total surface area of a cone $=\pi\text{r}(\text{l}+\text{r})$
$=\Big[\frac{22}{7}\times7(14+7)\Big]\text{cm}^2$
$=(22\times21)\text{cm}^2$
$=462\text{cm}^2$
View full question & answer→Question 184 Marks
The dimension of a room are $(9m \times 8m \times 6.5m).$ It has one door of dimension $(2m \times 1.5m)$ and two window, each of dimension $(1.5 \times 1m).$ Find the cost of whitewashing the walls at $₹ 25$ per square metre.
AnswerLength of the room, $l = 9m$
Breadth of the room, $b = 8m$
Height of the room, $h = 6.5m$
Now,
Area of the walls to be whitewashed
$=$ Curved surface area of the room $-$ Area of the door $-\ 2\ \times $ Area of each window
$= 2h(l + b) - 2m \times 1.5m - 2 \times 1.5m \times 1m$
$= 2 \times 6.5 \times (9 + 8) - 3 - 3$
$= 221 - 6$
$= 215m^2$
$\therefore$ Cost of whitewashing the walls at $₹ 25$ per square metre
$=$ Area of the walls to be whitewashed $\times\ ₹ 25$ per square metre
$= 215 \times 25$
$= ₹ 5,375$
View full question & answer→Question 194 Marks
How many metres of cloth, 2.5m wide, will be required to make a conical tent whose base radius is 7m and height 24m? $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$
AnswerHere, radius = 7m and height (h) = 24m
$\therefore$ slant height $\text{l}=\sqrt{\text{h}^2+\text{r}^2}$
$=\sqrt{(24)^2+(7)^2}$
$\text{l}=\sqrt{576+49}$
$\text{l}=\sqrt{625}$
$\text{l}=25\text{m}$
Now, area of cloth $=\pi\text{rl}$
$=\Big(\frac{22}{7}\times7\times25\Big)\text{m}^2$
$=550\text{m}^2$
$\therefore$ length of cloth $=\frac{\text{Area}\ \text{of}\ \text{cloth}}{\text{Width}\ \text{of}\ \text{cloth}}$
$=\Big(\frac{550}{2.5}\Big)\text{m}$
$=220\text{m}$
$\therefore$ Length of cloth required to make conical tent = 220m.
View full question & answer→Question 204 Marks
Water in a canal, $30\ dm$ wide and $12\ dm$ deep, is flowing with a velocity of $20\ km$ per hour. How much area will it irrigate, if $9\ cm$ of standing water is desired?
AnswerWidth of the canal $= 30\ dm = 3m (1m = 10\ dm)$
Depth of the canal $= 12\ dm = 1.2m$
Speed of the water flow $= 20\ km/h = 20000\ m/h$
$\therefore$ Volume of water flowing out of the canal in $1h = 3 \times 1.2 \times 20000 = 72000m^3$
Height of standing water on field $- 9\ cm - 0.09m (1m - 100\ cm)$
Assume that water flows of the canal for $1h.$ Then,
Area of the field irrigated
$=\frac{\text{Volume}\ \text{of}\ \text{water}\ \text{flowing}\ \text{out}\ \text{of}\ \text{the}\ \text{canal}}{\text{Height}\ \text{of}\ \text{standing}\ \text{water}\ \text{on}\ \text{the}\ \text{field}}$
$=\frac{72000}{0.09}$
$=800000\text{m}^2$
$=\frac{800000}{10000}$ $\big(1$ hectare $=1000\text{m}^2\big)$
$=80$ hectare
Thus, the area of the field irrigated is $80$ hectares.
Disclaimer: In this question time is not given, so the question is solved assuming that the water flows out of the canal for i hour.
View full question & answer→Question 214 Marks
The curved surface area of a cylinder is $1210\ cm^2$ and its diameter is $20\ cm.$ Find its height and volume.
AnswerHere, curved surface area $= 1210\ cm^2$
Diameter $= 20\ cm $
$\Rightarrow\text{radius}=\frac{20}{2}=10\text{cm}$
$\therefore$ Curved surface area of the cylinder $=2\pi\text{rh}$
$\Rightarrow1210=2\times\frac{22}{7}\times10\times\text{h}$
$\Rightarrow\text{h}=\Big(\frac{1210\times7}{2\times22\times10}\Big)\text{cm}=19.25\text{cm}$
$\therefore\text{Height}=19.25\text{cm}$
$\therefore$ Volume of the cylinder $=(\pi\text{r}^2\text{h})$
$=\Big(\frac{22}{7}\times10^2\times19.25\Big)\text{cm}^3$
$=\Big(\frac{22}{7}\times10\times10\times19.25\Big)\text{cm}^3$
$=6050\text{cm}^3$
$\therefore$ Volume of the cylinder $= 6050\ cm^3.$
View full question & answer→Question 224 Marks
Water flows at the rate of 10 metres per minute through a cylindrical pipe 5mm in diameter. How long would it take to fill a conical vessel whose diameter art the surface 40cm and depth 24cm?
AnswerDiameter of the pipe = 5mm = 0.5cm
Radius of the pipe $=\frac{0.5}{2}=0.25\text{cm}$
Length of the pipe = 10 metres = 1000cm
Volume that flows in 1 min $=\big[\pi\times(0.25)^2\times1000\big]\text{cm}^3$
$\therefore$ Volume of the conical vessel $=\Big[\frac{1}{3}\pi\times(20)^2\times24\Big]\text{cm}^3$
$\therefore$ Required time $=\Bigg[\frac{\frac{1}{3}\pi\times(20)^2\times24}{\pi\times(0.25)^2\times1000}\Bigg]\text{min}$
$=\Bigg[\frac{\frac{1}{3}\times400\times24}{\pi\times0.0625\times1000}\Bigg]\text{min}$
$=51.2\ \text{min}$
$=51\ \text{min}\ 12\ \text{sec}$
View full question & answer→Question 234 Marks
Find the volume, the lateral surface area and the total surface area of the cuboid whose dimensions are :
Length $= 26m,$ breadth $= 14m$ and height $= 6.5m$
AnswerHere$, l = 26m, b = 14m, h = 6.5m$
Volume of the cuboid $= l \times b \times h$
$= (26 \times 14 \times 6.5)m^3$
$= 2366m^3$
Total surface area $= 2(lb \times lh \times bh)$
$= 2(26 \times 14 + 26 \times 6.5 + 14 \times 6.5)m^2$
$= 2(364 + 169 + 91)m^2$
$= 2 \times 624m^2$
$= 1248m^2$
Lateral surface area $= 2(l + b) \times h$
$= [2(26 + 14) \times 6.5]m^2$
$= [2 \times 40 \times 6.5]m^2$
$= 520m^2$
View full question & answer→Question 244 Marks
Find the volume, the lateral surface area and the total surface area of the cuboid whose dimensions are:
Length $= 15m,$ breadth $= 6m$ and height $= 5dm$
AnswerHere$, l = 15m, b = 64m, h = 5dm = 0.5$
Volume of the cuboid $= l \times b \times h$
$= (15 \times 6 \times 0.5)m^3$
$= 45m^3$
Total surface area $= 2(lb \times lh \times bh)$
$= 2(15 \times 6 + 15 \times 0.5 + 6 \times 0.5)m^2$
$= 2(90 + 7.5 + 3)m^2$
$= 2 \times 100m^2$
$= 201m^2$
Lateral surface area $= 2(l + b) \times h$
$= [2(15 + 6) \times 0.5]m^2$
$= [2 \times 21 \times 0.5]m^2$
$= 21m^2$
View full question & answer→Question 254 Marks
Find the volume, curved surface area and the total surface area of a cone whose height and slant height are 6cm and 10cm respectively. $\big(\text{Take}\ \pi=3.14\big)$
AnswerHere, height (h) = 6cm and slant height (l) = 10cm
$\therefore$ radius (r) $=\sqrt{\text{l}^2-\text{h}^2}$
$\text{r}=\sqrt{10^2-6^2}$
$\text{r}=\sqrt{100-36}$
$\text{r}=\sqrt{64}$
$\text{r}=8\text{cm}$
$\therefore$ Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\Big(\frac{1}{3}\times3.14\times8\times8\times8\Big)\text{cm}^3$
$=401.92\text{cm}^3$
$\therefore$ Curved surface area $=\pi\text{rl}$
$=\big(3.14\times8\times10\big)\text{cm}^2$
$=251.2\text{cm}^2$
$\therefore$ Total surface area $=\pi\text{r}(\text{l}+\text{r})$
$=\pi\text{r}(10+8)$
$=(3.14\times8\times18\big)\text{cm})^2$
$=452.16\text{cm}^2$
View full question & answer→Question 264 Marks
The inner diameter of a cylindrical wooden pipe is $24cm$ and its outer diameter is $28cm.$ The length of the pipe is $35cm.$ Find the mass of the pipe, if $1\ cm^3$ of wood has a mass of $0.6g.$
AnswerInternal diameter of a cylinder = 24cm
⇒ Internal radius of a cylinder, r = 12cm
External diameter of a cylinder = 28cm
⇒ External radius of a cylinder, R = 14cm
Length of the pipe, i.e. height, h = 35cm
Now,
Volume of pipe = Volume of cylinder $=\pi(\text{R}^2-\text{r}^2)\text{h}$
$=\Big[\frac{22}{7}(14^2-12^2)\times35\Big]\text{cm}^3$
$=\big[22\times(196-144)\times5\big]\text{cm}^3$
$=(22\times52\times5)\text{cm}^3$
$=5720\text{cm}^3$
Given, 1cm3 = 0.6g
$\therefore$ Mass of pipe = (5720 × 0.6)g = 3432g = 3.432kg
View full question & answer→Question 274 Marks
The barrel of a fountain pen, cylindrical in shape, is 7cm long and 5mm in diameter. A full barrel of ink in the pen will be used up on writing 330 words on an average. How many words would use up a bottle of ink containing one fifth of a litre?
AnswerLength = 7cm = (height)
Diameter = 5mm $\Rightarrow\text{radius}=\Big(\frac{5}{2}\Big)\text{mm}=2.5\text{mm}$
$=0.25\text{cm}$
$\therefore$ Volume of the barrel $=\pi\text{r}^2\text{h}$
$=\Big(\frac{22}{7}\times0.25\times0.25\times7\Big)\text{cm}^3$
$=\frac{11}{8}\text{cm}^3$
$\frac{11}{8}\text{cm}^3$ isused for writing 330 words.
So, $\Big(\frac{1}{5}\times1000\Big)\text{cm}^3$ will be used for writing
$\Big(330\times\frac{8}{11}\times\frac{1}{5}\times1000\Big)\text{words}$
$=48000\ \text{words}$
View full question & answer→Question 284 Marks
A bus stop is barricated from the remaining part of the road by using $50$ hollow cones made of recycled cardboard. Each one has a base diameter of $40\ cm$ and height $1m$. If the outer side of each of the cones is to be painted and the cost of painting is $₹\ 25$ per $m^2,$ what will be the cost of painting all these cones?
$\big(\text{Use}\ \pi=3.14$ and $\sqrt{1.04}=1.02\big)$
AnswerRadius of a cone$, r = 20\ cm =\frac{20}{100}\text{m}=\frac{1}{5}\text{m}$
Height of a cone$, h = 1m$
$\therefore$ Slant height of a cone, $\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$\text{l}=\sqrt{\Big(\frac{1}{5}\Big)^2+1}$
$\text{l}=\sqrt{\frac{26}{25}}$
$\text{l}=\sqrt{1.04}$
$\text{l}=1.02$
Curved surface area of a cone $=\pi\text{rl}$
$=\Big(3.14\times\frac{1}{5}\times1.02\Big)\text{m}^2$
$=\frac{3.2028}{5}\text{m}^2$
$\Rightarrow$ Curved surface area of $50$ cones $=\Big(50\times\frac{3.2028}{5}\Big)\text{m}^2$
$=32.028\text{m}^2$
Cost of painting $= ₹ 25$ per $m^2$
$\Rightarrow$ Cost of painting $32.028m^2$ area $= ₹ (25 \times 32.028) = ₹ 800.70$
View full question & answer→Question 294 Marks
The volume of a sphere is $38808\ cm^3$. Find its radius and hence its surface area. $($Take pi$ =\frac{22}{7}).$
AnswerVolume of the sphere $= 38808\ cm^3$
Suppose that $r \ cm$ is the radius of the given sphere.
$\therefore\frac{4}{3}\pi\text{r}^3=38808$
$\Rightarrow\text{r}^3=\frac{38808\times3\times7}{4\times22}=9261$
$\Rightarrow\text{r}=\sqrt[3]{9261}=21\text{ cm}$
$\therefore$ Surface area of the sphere $=4\pi\text{r}^2$
$=4\times\frac{22}{7}\times21\times21$
$=5544\text{ cm}^2$
View full question & answer→Question 304 Marks
The surface area of a cuboid is $758\ cm^2.$ Its length and breadth are $14\ cm$ and $11\ cm$ respectively. Find its height.
AnswerLength of the cuboid $= 14\ cm$
Breadth of the cuboid $= 11\ cm$
Let the height of the cuboid be $x\ cm.$
Surface area of the cuboid $= 758\ cm^2$
Then $758 = 2(14 \times 11 + 14 \times x + 11 \times x)$
$\Rightarrow 758 = 2(154 + 14x + 11x)$
$\Rightarrow 758 = 2(154 + 25x)$
$\Rightarrow 758 = 308 + 50x$
$\Rightarrow 50x = 758 - 308 = 450$
$\Rightarrow\text{x}=\frac{450}{50}=9$
$\therefore$ The height of the cuboid is $9\ cm.$
View full question & answer→Question 314 Marks
Find the volume, the lateral surface area and the total surface area of the cuboid whose dimensions are:
Length $= 12\ cm,$ breadth $= 8\ cm$ and height $= 4.5\ cm$
AnswerHere, $l = 12\ cm, b = 8\ cm, h = 4.5\ cm$
Volume of the cuboid $= l \times b \times h$
$= (12 \times 8 \times 4.5)\ cm^3$
$= 432\ cm^3$
Total surface area $= 2(lb \times lh \times bh)$
$= 2(12 \times 8 + 12 \times 4.5 + 8 \times 4.5)\ cm^2$
$= 2(96 + 54 + 36)cm^2$
$= 2 \times 186\ cm^2$
$= 372\ cm^2$
Lateral surface area $= 2(l + b) \times h$
$= [2(12 + 8) \times 4.5]\ cm^2$
$= [2(20) \times 4.5]\ cm^2$
$= 40 \times 4.5\ cm^2$
$= 180\ cm^2$
View full question & answer→Question 324 Marks
A conical tent is $10m$ high and the radius of its base is $24m$. Find the slant height of the tent. If the cost of $1m^2$ canvas is $₹ \ 70,$ find the cost of canvas required to make the tent. $\Big($Use $\pi=\frac{22}{7}\Big).$
AnswerRadius of a conical tent$, r = 24m$
Height of a conical tent$, h = 10m$
$\therefore$ Slant height of a conical tent,
$\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$\text{l}=\sqrt{24^2+10^2}$
$\text{l}=\sqrt{576+100}$
$\text{l}=\sqrt{676}$
$\text{l}=26\ \text{cm}$
Curved surface area of a conical tent $=\pi\text{rl}$
$=\Big(\frac{22}{7}\times24\times26\Big)\text{m}^2$
$=\frac{13728}{7}\text{m}^2$
Cost of $1m^2$ canvas $= ₹ \ 70$
$\Rightarrow\text{Cost}\ \text{of}\ \frac{13728}{7}\text{m}^2\ \text{canvas}$
$=₹\ \Big(70\times\frac{13728}{7}\Big)=₹\ 137280$
View full question & answer→Question 334 Marks
A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is $10.4\ cm$ and its length is $25\ cm$. The thickness of the metal is $8mm$ everywhere. Calculate the volume of the metal.
AnswerInternal diameter of the tube $= 10.4\ cm$
Internal radius $=\Big(\frac{10.4}{2}\Big)\text{cm}=5.2\ \text{cm}$
and length $= 25\ cm$
and external radius $= (5.2 + 0.8)\ \text{cm}= 6\ \text{cm}$
Required volume $=\big[\pi\times(6)^2\times25-\pi\times(5.2)^2\times25\big]\text{cm}^2$
$=\pi\times25\big[(6)^2-(5.2)^2\big]\text{cm}^3$
$=\frac{22}{7}\times25\big[36-27.04\big]\text{cm}^3$
$=\Big(\frac{22}{7}\times25\times8.96\Big)\text{cm}^3$
$=704\ \text{cm}^3$
$\therefore$ the volume of the metal $= 704\ cm^3.$
View full question & answer→Question 344 Marks
A cylindrical bucket, 28cm in diameter and 72cm and high, is full of water. The water is emptied into a rectangular tank, 66cm long and 28cm wide. Find the height of the water level in the tank.
AnswerHere, cylindrical bucket has diameter = 28cm.
$\therefore$ radius $=\Big(\frac{28}{2}\Big)\text{cm}=14\text{cm}$ and height = 72cm.
Length of the tank = 66cm
Breadth of the tank = 28cm
$\therefore$ Volume of the tank = Volume of cylindrical bucket
$\Rightarrow\text{l}\times\text{b}\times\text{h}=\pi\text{r}^2\text{h}$
$\Rightarrow66\times28\times\text{h}=\frac{22}{7}\times14\times14\times72$
$\Rightarrow\text{h}=\Big(\frac{22\times2\times14\times72}{66\times28}\Big)\text{cm}$
$\Rightarrow\text{h}=24\text{cm}.$
$\therefore$ The height of the water level in the tank = 24cm.
View full question & answer→Question 354 Marks
The curved surface area of a cylinder is $4400\ cm^2$ and the circumference of its base is $110\ cm.$ Find the height and the volume of the cylinder.
AnswerLet base radius be $r$ and height be $h$
Then, $2\pi\text{rh}=4400\text{ cm}^2$
And, $2\pi\text{r}=110\text{ cm}$
$\Rightarrow\frac{2\pi\text{rh}}{2\pi\text{r}}=\frac{4400}{110}$
$\Rightarrow\text{h}=40\text{ cm}$
$\therefore2\times\frac{22}{7}\times\text{r}\times\text{h}\times40=4400\text{ cm}.$
$\Rightarrow\text{r}=\Big(\frac{4400\times7}{44\times40}\Big)\text{ cm}=\frac{35}{2}\text{ cm}.$
$\therefore$ Volume of the cylinder $=\pi\text{r}^2\text{h}$
$=\Big(\frac{22}{7}\times\frac{35}{2}\times\frac{35}{2}\times40\Big)\text{ cm}^3$
$=38500\text{ cm}^3.$
View full question & answer→Question 364 Marks
Find the weight of a solid cylinder of radius $10.5\ cm$ and height $60\ cm$ if the material of the cylinder weighs $5g$ per $ \ cm^3.$
AnswerHere, radius $(r) = 10.5\ cm$ and height $= 60\ cm.$
$\therefore$ Volume of the cylinder $=(\pi\text{r}^2\text{h})$
$=\Big(\frac{22}{7}\times10.5\times10.5\times60\Big)\text{ cm}^3$
$=20790\text{ cm}^3$
$\therefore$ Weight of the solid cylinder if the material of the cylinder.
Weighs $5g$ per $\ cm^3 = (20790 \times 5) = 103950g$
$=\frac{103950}{1000}$ $\big[\therefore1000\text{g}=1\text{ kg}\big]$
$=103.95\text{ kg}$
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