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Maths 4 Marks Questions

Surface Areas and Volumes · CBSE STD 9 (CBSE - English Medium). 19 questions.

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Question 14 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 1cm3 of wood has a mass of 0.6g

$\Big[$Hint: Assume $\pi=\frac{22}{7},$ unless stated otherwise$\Big]$

Answer
Inner radius (r1) of cylindrical pipe $=\Big(\frac{24}{2}\Big)\text{cm}=12\text{cm}$
Outer radius (r2) of cylindrical pipe $=\Big(\frac{28}{2}\Big)\text{cm}=14\text{cm}$
Height (h) of pipe = Length of pipe = 35cm
Volume of pipe $=\pi\big(\text{r}^2_2-\text{r}^2_1\big)\text{h}$
$=\Big[\frac{22}{7}\times\big(14^2-12^2\big)\times35\Big]\text{cm}^3$
$=110\times52\text{cm}^3$
$=5720\text{cm}^3$
Mass of 1cm3 wood = 0.6g
Mass of 5720cm3 wood = (5720 × 0.6)g
= 3432g
= 3.432kg
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Question 24 Marks
A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹ 4989.60. If the cost of white-washing is ₹ 20 per square metre, find the:

  1. Inside surface area of the dome.
  2. Volume of the air inside the dome.

$\Big[$Hint: Assume $\pi=\frac{22}{7},$ unless stated otherwise$\Big]$

Answer
$\text{Inside surface area of the dome}=\frac{\text{Total cost of white washing}}{\text{Rate of white washing}}$
$\Big(\frac{498.96}{2.00}\Big)\text{m}^2=249.48\text{m}^2$
Let r be the radius of the dome.
Surface area $=2\pi\text{r}^2$
$\Rightarrow2\times\frac{22}{7}\times\text{r}^2=249.48$
$\Rightarrow\text{r}^2=\frac{(249.48\times7)}{(2\times22)}=39.69$
$\Rightarrow\text{r}^2=39.69$
$\Rightarrow\text{r}^2=6.3\text{m}$
Volume of the air inside the dome = Volume of the dome
$=\frac{2}{3}\pi\text{r}^3$
$=\Big(\frac{2}{3}\times\frac{22}{7}\times6.3\times6.3\times6.3\Big)\text{m}^3$
$=523.9\text{m}^3\text{(approx)}$
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Question 34 Marks
The floor of a rectangular hall has a perimeter 250m. If the cost of painting the four walls at the rate of ₹ 10 per m2 is ₹ 15000, find the height of the hall.
[Hint: Area of the four walls = Lateral surface area]
Answer
Let length, breadth, and height of the rectangular hall be l m, b m, and h m respectively.
Area of four walls = 2lh + 2bh
= 2(l + b)h
Perimeter of the floor of hall = 2(l + b)
= 250m
$\therefore$ Area of four walls = 2(l + b) h = 250h m2
Cost of painting per m2 area = Rs. 10
Cost of painting 250h m2 area = Rs. (250h × 10) = Rs. 2500h
However, it is given that the cost of paining the walls is Rs. 15000
$\therefore$ 15000 = 2500h
h = 6
Therefore, the height of the hall is 6m
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Question 44 Marks
If the triangle ABC in the Question 7 above is revolved about the side 5cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

$\Big[$Hint: Assume $\pi=\frac{22}{7},$ unless stated otherwise$\Big]$

Answer
When right-angled $\triangle\text{ABC}$ is revolved about its side 5cm, a cone will be formed having radius (r) as 12cm, height (h) as 5cm, and slant height (l) as 13cm.
Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\Big[\frac{1}{3}\times\pi\times(12)^2\times5\Big]\text{cm}^3$
$=240\pi\text{cm}^3$
Therefore, the volume of the cone so formed is $240\pi\text{cm}^3$
$=\frac{100\pi}{240\pi}$
Ratio of the volumes of the two solids obtained in Questions 7 and 8-
$=\frac{5}{12}=5:12$
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Question 54 Marks
A heap of wheat is in the form of a cone whose diameter is 10.5m and height is 3m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
$\Big[$Hint: Assume $\pi=\frac{22}{7},$ unless stated otherwise$\Big]$
Answer
Radius (r) of heap $=\Big(\frac{10.5}{2}\Big)\text{m}=5.25\text{m}$
Height (h) of heap = 3m
Volume of heap $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\Big(\frac{1}{3}\times\frac{22}{7}\times(5.25)^2\times3\Big)\text{m}^3$
$=86.625\text{m}^3$
Therefore, the volume of the heap of wheat is 86.625m3
Area of canvas required = CSA of cone
$=\pi\text{rl}=\pi\text{r}\sqrt{\text{r}^2+\text{h}^2}$
$=\Big[\frac{22}{7}\times5.25\times\sqrt{(5.25)^2+(3)^2}\Big]\text{m}^2$
$=\Big(\frac{22}{7}\times5.25\times6.05\Big)\text{m}^2$
$=99.825\text{m}^2$
Therefore, 99.825m2 canvas will be required to protect the heap from rain.
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Question 64 Marks
A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40cm and height 1m. If the outer side of each of the cones is to be painted and the cost of painting is ₹ 12 per m2, what will be the cost of painting all these cones?
$\big(\text{Use}\pi=3.14\text{ and take }\sqrt{1.04}=1.02\big)$
Answer
Radius (r) of cone $=\frac{40}{2}=20\text{cm}=0.2\text{m}$
Height (h) of cone = 1m
Slant height (l) of cone $=\sqrt{\text{h}^2+\text{r}^2}$
$=\Big[\sqrt{(1)^2+(0.2)^2}\Big]\text{m}=\big(\sqrt{1.04}\big)\text{m}=1.02\text{m}$
CSA of each cone $=\pi\text{rl}$
= (3.14 × 0.2 × 1.02)m2 = 0.64056m2
CSA of 50 such cones = (50 × 0.64056)m2
= 32.028m2
Cost of painting 1m2 area = Rs. 12
Cost of painting 32.028m2 area = Rs. (32.028 × 12)
= Rs. 384.336
= Rs. 384.34 (approximately)
Therefore, it will cost Rs. 384.34 in painting 50 such hollow cones.
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Question 74 Marks
A cubical box has each edge 10cm and another cuboidal box is 12.5cm long, 10cm wide and 8cm high:
Which box has the greater lateral surface area and by how much?
Answer
Edge of cube = 10cm
Length (l) of box = 12.5cm
Breadth (b) of box = 10cm
Height (h) of box = 8cm
Lateral surface area of cubical box = 4(edge)2
= 4(10cm)2
= 400cm2
Lateral surface area of cuboidal box = 2[lh + bh]
= [2(12.5 × 8 + 10 × 8)]cm2
= (2 × 180)cm2
= 360cm2
Clearly, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box.
Lateral surface area of cubical box - Lateral surface area of cuboidal box = 400cm2 - 360cm2 = 40cm2
Therefore, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box by 40cm2
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Question 84 Marks
A solid cube of side 12cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.
Answer
Side (a) of cube = 12cm
Volume of cube = (a)3 = (12cm)3 = 1728cm3
Let the side of the smaller cube be a1
Volume of 1 smaller cube $=\Big(\frac{1728}{8}\Big)\text{cm}^3=216\text{cm}^3$
$(\text{a}_1)^3=216\text{cm}^3$
$\Rightarrow\text{a}_1=6\text{cm}$
Therefore, the side of the smaller cubes will be 6cm
Ratio between surface areas of cubes $=\frac{\text{surface area of bigger cube}}{\text{surface area of smaller cube}}$
$=\frac{6\text{a}^2}{6\text{a}^2_1}=\frac{(12)^2}{(6)^2}=\frac{4}{1}$
Therefore, the ratio between the surface areas of these cubes is 4 : 1
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Question 94 Marks
What length of tarpaulin 3m wide will be required to make conical tent of height 8m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20cm $\big(\text{Use }\pi=3.14\big).$
Answer
Height (h) of conical tent = 8m
Radius (r) of base of tent = 6m
Slant height (l) of tent $=\sqrt{\text{r}^2+\text{h}^2}$
$=\Big(\sqrt{6^2+8^2}\Big)\text{m}=\big(\sqrt{100}\big)\text{m}=10\text{m}$
CSA of conical tent $=\pi\text{rl}$
= (3.14 × 6 × 10)m2
= 188.4m2
Let the length of tarpaulin sheet required be l.
As 20cm will be wasted, therefore, the effective length will be (l - 0.2m).
Breadth of tarpaulin = 3m
Area of sheet = CSA of tent
[(l − 0.2m) × 3] m = 188.4m2
l − 0.2m = 62.8m
l = 63m
Therefore, the length of the required tarpaulin sheet will be 63m.
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Question 104 Marks
The capacity of a closed cylindrical vessel of height 1m is 15.4 litres. How many square metres of metal sheet would be needed to make it?

$\Big[$Hint: Assume $\pi=\frac{22}{7},$ unless stated otherwise$\Big]$

Answer
Let the radius of the circular end be r.
Height (h) of cylindrical vessel = 1m
Volume of cylindrical vessel = 15.4 litres = 0.0154m3
$\pi\text{r}^2\text{h}=0.0154\text{m}^3$
$\Big(\frac{22}{7}\times\text{r}^2\times1\Big)\text{m}=0.0154\text{m}^3$
$\text{r}=0.07\text{m}$
Total surface area of vessel $=2\pi\text{r}(\text{r}+\text{h})$
$=\Big[2\times\frac{22}{7}\times0.07(0.07+1)\Big]\text{m}^2$
$=0.44\times1.07\text{m}^2$
$=0.4708\text{m}^2$
Therefore, 0.4708m2 of the metal sheet would be required to make the cylindrical vessel.
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Question 114 Marks
The paint in a certain container is sufficient to paint an area equal to 9.375m2. How many bricks of dimensions 22.5cm × 10cm × 7.5cm can be painted out of this container?
Answer
Total surface area of one brick = 2(lb + bh + lh)
= [2(22.5 ×10 + 10 × 7.5 + 22.5 × 7.5)]cm2
= 2(225 + 75 + 168.75)cm2
= (2 × 468.75)cm2
= 937.5cm2
Let n bricks can be painted out by the paint of the container.
Area of n bricks = (n × 937.5)cm2 = 937.5n cm2
Area that can be painted by the paint of the container = 9.375m2 = 93750cm2
$\therefore$ 93750 = 937.5n
n = 100
Therefore, 100 bricks can be painted out by the paint of the container.
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Question 124 Marks
The circumference of the base of a cylindrical vessel is 132cm and its height is 25cm. How many litres of water can it hold? (1000cm3 = 1l)

$\Big[$Hint: Assume $\pi=\frac{22}{7},$ unless stated otherwise$\Big]$

Answer
Let the radius of the cylindrical vessel be r.
Height (h) of vessel = 25cm
Circumference of vessel = 132cm
$2\pi\text{r}=132\text{cm}$
$\text{r}=\Big(\frac{132\times7}{2\times22}\Big)\text{cm}=21\text{cm}$
Volume of cylindrical vessel $=\pi\text{r}^2\text{h}$
$=\Big[\frac{22}{7}\times(21)^2\times25\Big]\text{cm}^2$
$=34650\text{cm}^2$
$=\Big(\frac{34650}{1000}\Big)\text{ litres}$ $\big[\because\text{litre}=1000\text{cm}^3\Big]$
$=34.65\text{ litres}$
Therefore, such vessel can hold 34.65 litres of water.
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Question 134 Marks
A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7cm. If the bowl is filled with soup to a height of 4cm, how much soup the hospital has to prepare daily to serve 250 patients?
$\Big[$Hint: Assume $\pi=\frac{22}{7},$ unless stated otherwise$\Big]$
Answer


Radius (r) of cylindrical bowl $=\Big(\frac{7}{2}\Big)\text{cm}=3.5\text{cm}$
Height (h) of bowl, up to which bowl is filled with soup = 4cm
Volume of soup in 1 bowl $=\pi\text{r}^2\text{h}$
$=\Big(\frac{22}{7}\times(3.5)^2\times4\Big)\text{cm}^3$
= (11 × 3.5 × 4)cm3
= 154cm3
Volume of soup given to 250 patients = (250 × 154)cm3
= 38500cm3
= 38.5 litres.
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Question 144 Marks
A cuboidal vessel is 10m long and 8m wide. How high must it be made to hold 380 cubic metres of a liquid?
Answer
Let the height of the cuboidal vessel be h.
Length (l) of vessel = 10m
Width (b) of vessel = 8m
Volume of vessel = 380m3
$\therefore$ l × b × h = 380
[(10)(8)h]m2= 380m3
h = 4.75m
Therefore, the height of the vessel should be 4.75m.
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Question 154 Marks
The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3cm and height 10.5cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?
$\Big[$Hint: Assume $\pi=\frac{22}{7},$ unless stated otherwise$\Big]$
Answer
Radius (r) of the circular end of cylindrical penholder = 3cm
Height (h) of penholder = 10.5cm
Surface area of 1 penholder = CSA of penholder + Area of base of penholder.
$=2\pi\text{rh}+\pi\text{r}^2$
$=\Big[2\times\frac{22}{7}\times3\times10.5+\frac{22}{7}\times(3)^2\Big]\text{cm}^2$
$=\Big(132\times1.5+\frac{198}{7}\Big)\text{cm}^2$
$=\Big(198+\frac{198}{7}\Big)\text{cm}^2$
$=\frac{1584}{7}\text{cm}^2$
Area of cardboard sheet used by 1 competitor $=\frac{1584}{7}\text{cm}^2$
Area of cardboard sheet used by 35 competitors,
$=\Big(\frac{1584}{7}\times35\Big)\text{cm}^2$
$=7920\text{cm}^2$
Therefore, 7920cm2 cardboard sheet will be bought.
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Question 164 Marks
A godown measures 40m × 25m × 15m. Find the maximum number of wooden crates each measuring 1.5m × 1.25m × 0.5m that can be stored in the godown.
Answer
The godown has its length (l1) as 60m, breadth (b1) as 25m, height (h1) as 10m, while the wooden crate has its length (l2) as 1.5m, breadth (b2) as 1.25m, and height (h2) as 0.5m.
Therefore, volume of godown = l1 × b1 × h1
= (60 × 25 × 10)m3
= 15000m3
Volume of 1 wooden crate = l2 × b2 × h2
= (1.5 × 1.25 × 0.5)m3
= 0.9375m3
Let n wooden crates can be stored in the godown.
Therefore, volume of n wooden crates = Volume of godown
0.9375 × n = 15000
$\text{n} = \frac{15000}{9375}=16000$
Therefore, 16,000 wooden crates can be stored in the godown.
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Question 174 Marks
Find the capacity in litres of a conical vessel with:
Radius 7cm, slant height 25cm
$\Big[$Hint: Assume $\pi=\frac{22}{7},$ unless stated otherwise$\Big]$
Answer
Radius (r) of cone = 7cm
Slant height (l) of cone = 25cm
Height (h) of cone $=\sqrt{\text{l}^2-\text{r}^2}$
$=\Big(\sqrt{25^2-7^2}\Big)\text{cm}$
$=24\text{cm}$
Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\Big(\frac{1}{3}\times\frac{22}{7}\times(7)^2\times24\Big)\text{cm}^3$
$=\big(154\times8\big)\text{cm}^3$
$=1232\text{cm}^3$
Therefore, capacity of the conical vessel,
$=\Big(\frac{1232}{1000}\Big)\text{litres}$ (1 litre = 1000cm3)
$=1.232\text{ litres}$
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Question 184 Marks
Find the capacity in litres of a conical vessel with:
Height 12cm, slant height 13cm
$\Big[$Hint: Assume $\pi=\frac{22}{7},$ unless stated otherwise$\Big]$
Answer
Height (h) of cone = 12cm
Slant height (l) of cone = 13cm
Radius (r) of cone $=\sqrt{\text{l}^2-\text{h}^2}$
$=\Big(\sqrt{13^2-12^2}\Big)\text{cm}$
$=5\text{cm}$
Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\Big[\frac{1}{3}\times\frac{22}{7}\times(5)^2\times12\Big]\text{cm}^3$
$=\Big(4\times\frac{22}{7}\times25\Big)\text{cm}^3$
$=\Big(\frac{2200}{7}\Big)\text{cm}^3$
Therefore, capacity of the conical vessel,
$=\Big(\frac{2200}{7000}\Big)\text{litres}$ (1 litre = 1000cm3)
$=\frac{11}{35}\text{ litres}$
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Question 194 Marks
It costs ₹ 2200 to paint the inner curved surface of a cylindrical vessel 10m deep. If the cost of painting is at the rate of ₹ 20 per m2, find:
  1. Inner curved surface area of the vessel.
  2. Radius of the base.
  3. Capacity of the vessel.
$\Big[$Hint: Assume $\pi=\frac{22}{7},$ unless stated otherwise$\Big]$
Answer
Rs. 20 is the cost of painting 1m2 area.
Rs. 2200 is the cost of painting $=\Big(\frac{1}{20}\times2200\Big)\text{m}^2\text{ area}$
$=110\text{m}^2\text{ area}$
Therefore, the inner surface area of the vessel is 110m2
Let the radius of the base of the vessel be r.
Height (h) of vessel = 10m
Surface area $=2\pi\text{rh}=110\text{m}^2$
$\Big(2\times\frac{22}{7}\times\text{r}\times10\Big)\text{m}=110\text{m}^2$
$\text{r}=\Big(\frac{7}{4}\Big)\text{m}=1.75\text{m}$
Volume of vessel $=\pi\text{r}^2\text{h}$
$=\Big[\frac{22}{7}\times(1.75)^2\times10\Big]\text{m}^3$
$=96.25\text{m}^3$
Therefore, the capacity of the vessel is 96.25m3 or 96250 litres.
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