2 Marks Questions

Question 12 Marks

This is the picture of an ice-cream cone.

The radius of the cone is 4 cm and the height is 15 cm.
An ice-cream seller keeps 1/4of it empty.
What is the volume (in cm³) of the empty part of the cone?

Answer

20π

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Question 22 Marks

Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find the

radius r' of the new sphere, and
ratio of S and S'.

Answer

Volume of 27 solid sphere, each of radius, $r = 27$$\times$$\frac{4}{3}$$\pi$$r^3 = 36$$\pi$$r^3 $
According to the question,
Volume of sphere of radius r' = Volume of 27 solid spheres
$\Rightarrow$$\frac{4}{3}$$\pi$$(r')^3 = 36$$\pi$$r^3$
$\Rightarrow$$(r')^3 = 27r^3 = (3r)^3$

$\Rightarrow$$r' = 3r$

We have,
S' = 4$\pi$r'² = 4$\pi$(3r)² = 36$\pi$r²
$\therefore$ $\frac{S}{S'}$ = $\frac{4 \pi r^2}{36 \pi r^2}$ = $\frac{1}{9}$
$\Rightarrow$$S : S' = 1 : 9.$

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Question 32 Marks

Find the volume of a sphere whose surface area is 154 cm².

Answer

Let the radius of the sphere be r cm.
Surface area = 154 cm²
$\Rightarrow\ 4\pi r^2$ = 154
$\Rightarrow\ 4\times{22\over7}\times r^2=154$
$\Rightarrow\ r^2={154\times7\over4\times22}\Rightarrow {r^2={49\over4} }$
$\Rightarrow r=\sqrt{49\over4}\Rightarrow r={7\over2}$ cm
$\therefore$Volume of the sphere =${4\over3}\pi r^3$
$={4\over3}\times{22\over7}\times({7\over2})^2={539\over3}\ cm^3$
$=179{2\over3}\ cm^3$

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Question 42 Marks

A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Answer

Inner radius of hemispherical tank (r) = 1 m = 100 cm
Thickness of sheet = 1 cm
$\therefore$ Outer radius of hemispherical tank (R) = 100 + 1 = 101 cm
Volume of iron of hemisphere = $\frac{2}{3}\pi$ [R³ - r³] cm²
= $\frac{2}{3}\times \frac{22}{7}\times \left[ {{\left( 101 \right)}^{3}}-{{\left( 100 \right)}^{3}} \right]$ cm²
= $\frac{44}{21}$ [1030301 - 1000000] cm²
= 0.06348 m²

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Question 52 Marks

The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the metal weighs 8.9 g per $cm^3$?

Answer

Diameter of metallic ball = 4.2 cm
$\therefore$ Radius of metallic ball $\left( r \right)$=$\frac{4.2}{2}$ = 2.1 cm
Volume of metallic ball =$\frac{4}{3}\pi {{r}^{3}}$=$\frac{4}{3}\times \frac{22}{7}\times 2.1\times 2.1\times 2.1$
=$\frac{4}{3}\times \frac{22}{7}\times \frac{21}{10}\times \frac{21}{10}\times \frac{21}{10} =38.808 c{{m}^{3}}$
Density of metal = 8.9 g per $c{{m}^{3}}$
DensityxVolume=Mass
$\because$ Mass of 1 $c{{m}^{3}}$ = 8.9 g
$\therefore$ Mass of 38.808 $c{{m}^{3}}4 =\text{ }8.9\times 38.808$ = 345.3912 g = 345.39 g (approx.)

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Question 62 Marks

A right triangle ABC with slides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.

Answer

The solid obtained will be a right circular cone whose radius of the base is 5 cm. and height is 12 cm
$\therefore$ r = 5 cm, h = 12 cm
$\therefore$ Volume =${1\over3}\pi r^2h$
${1\over3}\times\pi \times (5)^2\times12\ cm^3$
= 100$\pi$ cm³
The volume of the solid so obtained is 100$\pi$ cm³

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Question 72 Marks

A conical pit of top diameter 3.5 cm is 12 m deep. What is its capacity in kilolitres?

Answer

For conical pit : Diameter = 3.5 cm.
∴ Radius (r) = ${3.5}\over2$m = 1.75 m
Depth (h) = 12 m
∴ Capacity of the conical pit =${1\over3}\pi r^2h$
${1\over3}\times{22\over7}\times(1.75)^2\times12\ m^3$
= 38.5 m³ = 38.5 × 1000 l
= 38.5 kl.

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Question 82 Marks

If the volume of a right circular cone of height 9 cm is 48$\pi$ cm³, find the diameter of its base.

Answer

Let the radius of the base of the right circular cone be r cm.
h = 9 cm, volume = 48$\pi$ cm³
⇒^{${1\over3}\pi r^2h$}= 48$\pi$
⇒${1\over3}r^2h=48$
⇒ ${1\over3}\times r^2\times9=48$
⇒ r² = ${48\times3}\over9$
⇒ r² = 16 ⇒ r = $\sqrt{16}$ = 4 cm
⇒ 2r = 2(4) = 8 cm.
∴ the diameter of the base of the right circular cone is 8 cm.

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Question 92 Marks

The height of a cone is 15 cm. If its volume is 1570 cm³. Find the radius of the base.

Answer

Let the radius of the base of the cone be r cm.
h = 15 cm, Volume = 1570 cm³
⇒${1\over3}\pi r^2h$= 1570
⇒$1\over3$ × 3.14 × r² × 15 = 1570
⇒ $r^2={{1570\times3}\over3.14\times15}$ ⇒ $r^2=100$
⇒ $r=\sqrt{100}$ ⇒ r = 10 cm.
∴ the radius of the base of the cone is 10 cm.

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Question 102 Marks

Find the capacity in litres of a conical vessel with height 12 cm, slant height 13 cm.

Answer

12 cm, l = 13 cm.
r² + h² = l²
⇒ r² + (12)² = (13)²
⇒ r² + 144 = 169
⇒ r² = 169– 144
⇒ r² = 25
⇒ r = $\sqrt{25}$
⇒ r = 5 cm
∴ Capacity $={1\over3}\pi r^2h$
$={1\over3}\times{22\over7}\times(5)^2\times12$
$={2200\over7}\ cm^3={2200\over7000}l$
$={11\over35}l$

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Question 112 Marks

Find the capacity in litres of a conical vessel with radius 7 cm, slant height 25 cm.

Answer

r = 7 cm, l = 25 cm.
r² + h² = l²
⇒ (7)² + h² = (25)²
⇒ h² = (25)²– (7)²
⇒ h² = 625 – 49
⇒ h² = 576
⇒ h = $\sqrt{576}$
⇒ h = 24 cm
∴ Capacity $={1\over3}\pi r^2h$
$={1\over3}\times{22\over7}\times(7)^2\times24$
= 1232 cm³ = 1.232 l

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Question 122 Marks

A right circular cylinder just encloses a sphere of radius r. Find

surface area of the sphere.
curved surface of the cylinder.
ratio of the area obtained in (i) and (ii).

Answer

Surface area of the sphere =$4\pi r^2$
For cylinder
Radius of the base = r
Height = 2r
$\therefore$ Curved surface area of the cylinder = $2\pi rh=2\pi (r)(2r)=4\pi r^2$
Ratio of the areas obtained in (i) and (ii)
$={{surface\ area\ of\ the\ sphere}\over{curved\ surface\ area\ of\ the\ cylinder}}$
$={{4\pi r^2}\over{4\pi r^2}}={1\over1}=1:1$

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Question 132 Marks

Find the radius of sphere whose surface area is 154 cm².

Answer

Let the radius of the sphere be r cm.
Surface area = 154 cm²
⇒ $4\pi r^2$ = 154
⇒ $4\times{22\over7}\times r^2=154$
⇒ $r^2={{154\times7}\over4\times22}={49\over4}$
⇒ r = ${\sqrt{49\over4}}={7\over2}$
⇒ r = 3.5 cm
∴ the radius of the sphere is 3.5 cm.

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Question 142 Marks

A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹16 per 100cm².

Answer

Inner diameter of bowl = 10.5 cm
$\therefore $ Inner radius of bowl $\left( r \right)=\frac{10.5}{2}$ = 5.25 cm
Now, Inner surface area of bowl = $2\pi {{r}^{2}}$
$=2\times \frac{22}{7}\times 5.25\times 5.25$
$=2\times \frac{22}{7}\times \frac{21}{4}\times \frac{21}{4}$
=$\frac{693}{4}c{{m}^{2}}$
$\because $ Cost of tin-plating per$ 100\text{ }c{{m}^{2}}$ = ₹ 16
$\therefore $ Cost of tin-plating per$ 1\text{ }c{{m}^{2}}=\frac{16}{100}$
$\therefore $ Cost of tin-plating per $\frac{693}{4}c{{m}^{2}}=\frac{16}{100}\times \frac{693}{4}$ = ₹27.72

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Question 152 Marks

The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Answer

Case I : r = 7 cm
Surface area = ^{$4\pi r^2$}
= 4 × $22\over7$ × (7)² = 616 cm²
Case II : r = 14 cm
Surface area = ^{$4\pi r^2$}
= 4 ×$22\over7$ × (14)² = 2464 cm²
^∴Ratio of surface area of the balloon = 616 : 2464

=$1:4$

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Question 162 Marks

The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white washing its curved surface at the rate of ₹ 210 per 100 m².

Answer

Slant height (l) = 25 m
Base diameter (d) = 14 m
∴ Base radius (r) = $14\over2$m = 7 m
∴ Curved surface area of the tomb = $\pi rl$
= $22\over7$× 7 × 25 = 550 m²
∴ Cost of white-washing the curved surface of the tomb at the rate of Rs. 210 per 100 m²
= $210\over100$ × 550 = Rs. 1155.

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Question 172 Marks

Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Answer

Slant height (l) = 21 m
Diameter of base = 24 m
∴ Radius of base (r) = $24\over2$m = 12 m
∴ Total curved surface area of the cone =$\pi r(l+r)$
= $22\over7$× 12 × (21 + 12)
=${22\over7}\times12\times33={8712\over7}$
=$1244{4\over7}\ m^2$

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Question 182 Marks

The hollow sphere, in which the circus motorcyclist performs his stunts, has a diameter of 7 m. Find the area available to the motorcyclist for riding.

Answer

We are given,
The diameter of the sphere = 7 m
Therefore, the radius is 3.5 m
So, the riding space available for the motorcyclist is the surface area of the ‘sphere’ which is given by
$4\pi r^2 = 4 \times \frac {22}7 \times 3.5 \times 3.5 m^2$
= 154 m²

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Question 192 Marks

Hameed has built a cubical water tank with lid for his house, with each other edge 1.5 m long. He gets the outer surface of the tank excluding the base, covered with square tiles of side 25 cm. Find how much he would spend for the tiles if the cost of tiles is ₹ 360 per dozen.

Answer

Since Hameed is getting the five outer faces of the tank covered with tiles, he would need to know the surface area of the tank, to decide on the number of tiles required.
Edge of the cubic tank, a = 1.5m = 150 cm
So, surface area of the tank = 5 $\times$ 150 $\times$ 150 cm².
Area of each square title = $\frac{surface \ area \ of \ the \ tank}{area \ of \ each \ title}$ = $\frac{5 \times 150 \times 150}{25 \times 25}$ = 180
Cost of 1 dozen tiles, i.e., cost of 12 tiles = Rs. 360
Therefore, cost of one tile = Rs $\frac{360}{12}$ = Rs. 30
So the cost of 180 tiles = 180 $\times$ Rs.30 = Rs. 5400

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Question 202 Marks

The height of a cone is 16 cm and its base radius is 12 cm. Find the curved surface area and the total surface area of the cone (Use $\pi$ = 3.14).

Answer

We are given that,

Here, h = 16 cm and r = 12 cm.
So, from l² = h² + r²,we have
l = $\sqrt {16^2 + 12^2} cm = 20 cm$
So, curved surface area = $\pi rl$
= 3.14 $\times$ 12 $\times$ 20 cm² = 753.6 cm²
Further, total surface area = $\pi rl + \pi r^2 $
= (753.6 + 3.14 $\times$ 12 $\times$ 12) cm²
= (753.6 + 452.16) cm²
= 1205.76 cm²

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Question 212 Marks

A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per cm³, find the mass of the shot-put.

Answer

Since the shot-putt is a solid sphere made of metal and its mass is equal to the product of its volume and density, we need to find the volume of the sphere.
Now, volume of the sphere is given by= $\frac 43 \pi r^3$
= $\frac 43 \times \frac {22}7 \times 4.9 \times 4.9 \times 4.9 cm^3$
= 493 cm³ (nearly)
Further, mass of 1 cm³ of metal is 7.8 g.
Therefore, mass of the shot-putt will be = 7.8 $\times$ 493 g
= 3845.44 g = 3.85 kg (nearly)

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Question 222 Marks

This is the picture of a gas balloon illed with helium gas.

This balloon has 18 faces that are square in shape and 8 equilateral faces that are triangular.
questions:
1. How many square faces does the gas balloon have?
2. The side length of the square is 20 cm. What is the total surface area of the balloon?

Answer

1. The gas balloon has 18 square faces.
2. $ \begin{aligned} & (7200+820 \sqrt{3}) \\ & (7200+800 \sqrt{3}){cm^2} \\ & 8585.6 \mathrm{~cm}^2\end{aligned} $

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Question 232 Marks

Read the Source/ Text given below and answer the questions:

In Agra in a grinding mill, there were installed 5 types of mills. These mills used steel balls of radius 5mm, 7mm,10mm, 14mm and 16mm respectively. All the balls were in the spherical shape. For repairing purpose mills need 10 balls of 7mm radius and 20 balls of 3.5mm radius. The workshop was having 20000mm³ steel. This 20000mm³ steel was melted and 10 balls of 7mm radius and 20 balls of 3.5mm radius were made and the remaining steel was stored for future use.
questions:
1. How much steel was kept for future use?
2. What was the surface area of one ball of 7mm radius?

Answer

1. $2033.3 mm^3$
2. $600 mm^2$

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Question 242 Marks

Read the Source/ Text given below and answer the questions:

In Agra in a grinding mill, there were installed 5 types of mills. These mills used steel balls of radius 5mm, 7mm,10mm, 14mm and 16mm respectively. All the balls were in the spherical shape. For repairing purpose mills need 10 balls of 7mm radius and 20 balls of 3.5mm radius. The workshop was having 20000mm³ steel. This 20000mm³ steel was melted and 10 balls of 7mm radius and 20 balls of 3.5mm radius were made and the remaining steel was stored for future use.
questions:
1. What was the volume of 10 balls of radius 7mm?
2. What was the volume of 20 balls of radius 3.5mm?

Answer

1. $14373.3 mm^3$
2. $1796.6mm^3$

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Question 252 Marks

Read the Source/ Text given below and answer the questions:
Veena planned to make a jewellery box to gift her friend Reeta on her marriage. She made the jewellery box of wood in the shape of a cuboid. The jewellery box has the dimensions as shown in the figure below. The rate of painting the exterior of the box is Rs. 2 per cm². After making the box she took help from his friends to decorate the box. The blue colour was painted by Deepak, Black by Suresh, green by Harsh and the yellow was painted by Naresh.

1. How much area did Deepak paint?
2. What amount did Harsh charge?

Answer

1. $1200 cm^2$
2. Rs. 1600

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Question 262 Marks

1. What is the volume of the box?
2. How much area did Suresh paint?

Answer

1. $24000 cm^3$
2. $1200 cm^2$

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Question 272 Marks

Read the passage given below and answer the questions:
Once four friends Rahul, Arun, Ajay and Vijay went for a picnic at a hill station. Due to peak season, they did not get a proper hotel in the city. The weather was fine so they decided to make a conical tent at a park. They were carrying 300m² cloth with them. As shown in the figure they made the tent with height 10m and diameter 14m. The remaining cloth was used for the floor.

1. What was the area of the floor?
2. What was the total surface area of the tent?

Answer

1. $154 m^2$
2. $422.4 m^2$

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Question 282 Marks

1. How much Cloth was used for the floor?
2. What was the volume of the tent?

Answer

1. $31.6 m^2$
2. $513.3 m^2$

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Question 292 Marks

Read the passage given below and answer the questions:
Dev was doing an experiment to find the radius r of a sphere. For this he took a cylindrical container with radius R = 7cm and height 10cm. He filled the container almost half by water as shown in the left figure. Now he dropped the yellow sphere in the container. Now he observed as shown in the right figure the water level in the container raised from A to B equal to 3.40cm.

1. What is the volume of the sphere?
2. How many litres water can be filled in the full container? ( Take 1 litre = 1000cm³)

Answer

1. $523.8 cm^3$
2. 1.54

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Question 302 Marks

1. What is the approximate radius of the sphere?
2. What is the volume of the cylinder?

Answer

1. 5cm
2. $1540 cm^3$

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Question 312 Marks

Read the passage given below and answer the questions:
Sohan's house has one bedroom hall with kitchen. His son needed a separate room for study. Thus Sohan planned to construct a new room with length 4m, width 2m and the height 3m as shown in the following figure. The room was separate at the roof of the house. The dimensions of the bricks used are: 25cm × 10cm × 5cm.

1. How many bricks will be used on both walls along the width (width = 3m)?
2. What is the volume of the room?

Answer

1. 9600
2. $24 m^3$

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Question 322 Marks

1. Total how many bricks will be required? (1m³ =1000000cm³):
2. How many bricks will be used on both walls along the length (length = 4m)?

Answer

1. 28800
2. 19200

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Question 332 Marks

Raju designs a hut for homeless people. The hut is a combination of a cuboid and a right cone. The top of the hut is a cone with radius 4 m and height 1 m. It is made of economical material. The loor of the tent is covered with rugs.
The total height of the tent is 4.5 m. The cuboidal part of the tent is 6 m long and 5 m wide.
questions:
1. What is the outer surface area (in m²) of the hut?
2. The length and width of a rug used for the loor are 2.6 m and 2 m respectively.
What is the minimum number of rugs required to cover the loor of the tent house?

Answer

1. $ 77+4 \pi \sqrt{17} $
2. 6 rugs

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Question 342 Marks

Raghav bought this planter.

The radius of the rim is 14 cm. The curved surface area of the planter is 1848 cm².
1. What is the height of the planter?
2. What is the volume of the planter?

Answer

1. 21 cm
2. 4116 $ \pi cm ^3$

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Question 352 Marks

A company manufactures wooden boxes. Given below is the picture of an open wooden box.

questions:
1. A shopkeeper store cubes in it.
The side length of one cube is 9 cm.
What would be the maximum number of cubes the shopkeeper can store in a box? (All cubes should be inside the box.)
2. Rajan packs a football into a cubical cardboard box. The radius of the football is 11 cm. Rajan keeps a margin of 1 cm from all the sides of the box while packing.
What is the side length of the cardboard box?

Answer

1. 41 (30x40x25)/(9x9x9)= 41.15 . Exact answer = 41 as all cubes should it in it)
2. D. 24 cm

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