Cylinder, Cone, Sphere ICSE Class 10

Our detailed page provides the most accurate ICSE Class 10 Cylinder Cone Sphere Solutions, helping you master this important chapter from the Selina Concise Mathematics textbook. This chapter, officially titled Cylinder, Cone and Sphere (Surface Area and Volume), is a crucial part of Mensuration. Here, you will move beyond simple shapes and learn to calculate the curved surface area, total surface area, and volume of cylinders, cones, and spheres. You will also tackle more complex problems involving combinations of these solids, such as finding the capacity of a tent shaped like a cone surmounted on a cylinder, or the amount of metal needed to make a hollow sphere. Mastering these formulas and their applications is key to scoring well in your board exams.

If you are stuck on a particular question about surface area or volume, you have come to the right place. This chapter contains a large number of problems—a total of 130 questions spread across Exercise 20(A), 20(B), 20(C), 20(D), 20(E), 20(F), and the Test Yourself section. It’s easy to get confused with the formulas or the multi-step calculations. Our solutions are designed to guide you through each problem using the exact method expected by the ICSE board. Here you will find clear, step-by-step solutions for every single question.

Exercise 20(A)

Question 1(a)

An open hollow cylindrical container, made of very thin metal sheet, has radius r cm and height h cm, its total surface area (in cm^2) is :

(a) πr^2h + 2πrh
(b) 4πrh + πr^2
(c) 4πrh + 2πr^2
(d) 2πrh + 2πr^2

Answer: (b) 4πrh + πr^2

We have an open hollow cylindrical container. This means it is open on one end and closed on the other. The cylinder is constructed from a thin metal sheet, implying that the external and internal radii are both r cm.

The height of this cylinder is given as h cm.

To find the total surface area of this hollow cylinder, we need to consider the external curved surface area, the internal curved surface area, and the area of the closed end.

Thus, the total surface area is calculated as:

Total surface area = External curved surface area + Internal curved surface area + Area of the closed end

= 2πrh + 2πrh + πr^2

= 4πrh + πr^2.

Hence, Option 2 is the correct option.

Question 1(b)

The curved area of a solid cylinder is S cm^2 and the circumference of its base is C cm; the height of the cylinder is :

(a) C × S
(b) C + S
(c) S ÷ C
(d) S – C

Answer: (c) S ÷ C

Assume the height of the cylinder is $h$ cm and the radius is $r$ cm.

We know that the curved surface area of the cylinder is $S$ cm².

Recall the formula for the circumference of the base:

\text{Circumference of base} = 2\pi r

The formula for the curved surface area is:

\text{Curved surface area} = 2\pi rh

Given $S = 2\pi rh$ , we can substitute the circumference $C = 2\pi r$ into this equation:

S = C \times h

Solving for $h$ , we find:

h = \dfrac{S}{C}

Thus, the height of the cylinder is $S \div C$ .

Hence, Option 3 is the correct option.

Question 1(c)

A cylindrical container has sufficient water to submerge a solid cubical object of its each edge a cm. If the radius of the container is r cm, the rise in level (h cm) of water in the container is :

$\sqrt{\dfrac{a^3}{πr}}$
$\dfrac{a}{\sqrt{πr}}$
$\dfrac{a^3}{πr^2}$
$\dfrac{πr^2}{a}$

Answer:

Assume the water level rises by $h$ cm in the cylindrical container.

The additional volume in the cylinder due to this rise is given by $\pi r^2 h$ .

We’re told that the cube, with each side measuring $a$ cm, is fully submerged.

The volume of this submerged cube is $a^3$ .

Since the cube’s volume is equal to the increase in the cylinder’s volume:

a^3 = \pi r^2 h

Solving for $h$ , we get:

h = \dfrac{a^3}{\pi r^2}

Thus, the rise in water level is $\dfrac{a^3}{\pi r^2}$ .

Hence, Option 3 is the correct option.

Question 1(d)

A solid cylinder of radius R cm and height H cm is melted and recast into smaller identical solid cylinders of height h cm and radius r cm. The number of smaller solid cylinders obtained is :

(a) $\dfrac{RH}{rh}$
(b) $\dfrac{R^2H}{r^2h}$
(c) R^2H ÷ r^2h^2
(d) $\dfrac{RH^2}{rh}$

Answer: (b)

\dfrac{R^2H}{r^2h}

We start by considering a solid cylinder with a radius of $R$ cm and a height of $H$ cm, which is melted down and transformed into smaller cylinders, each having a radius of $r$ cm and a height of $h$ cm. Let the number of these smaller cylinders be $n$ .

The volume of the larger cylinder is given by $\pi R^2 H$ . Since the entire volume of the larger cylinder is used to form the smaller ones, the total volume of the smaller cylinders is $n \times \pi r^2 h$ .

Setting the volume of the larger cylinder equal to the total volume of the smaller cylinders, we have:

\pi R^2 H = n \times \pi r^2 h

Cancelling $\pi$ from both sides, we get:

R^2 H = n \times r^2 h

Solving for $n$ , we find:

n = \dfrac{R^2 H}{r^2 h}

Thus, the number of smaller cylinders formed is $\dfrac{R^2 H}{r^2 h}$ . Hence, Option 2 is the correct option.

Question 1(e)

The total surface area of an open pipe of length H cm, external radius R cm and internal radius r cm, is :

(a) πR^2 – πr^2 + 2π(R – r)H
(b) 2πR^2 – 2πr^2 + 2πRH
(c) 2πRH – 2πrh + πR^2 – πr^2
(d) 2(πR^2 – πr^2) + 2πRH + 2πrH

Answer: (d) 2(πR^2 – πr^2) + 2πRH + 2πrH

Consider an open pipe with an external radius of $R$ cm, an internal radius of $r$ cm, and a height of $H$ cm. To find the total surface area of this hollow cylinder, we need to account for three components: the external curved surface area, the internal curved surface area, and the area of the two circular ends (cross sections).

The external curved surface area is calculated as $2\pi RH$ .

The internal curved surface area is $2\pi rH$ .

The area of the cross sections, which are the differences between the areas of two circles, is $2(\pi R^2 - \pi r^2)$ .

Thus, the total surface area is:

2\pi RH + 2\pi rH + 2(\pi R^2 - \pi r^2).

Hence, Option 4 is the correct option.

Question 2

The inner radius of a pipe is 2.1 cm. How much water can 12 m of this pipe hold ?

Answer:

To find out how much water the pipe can hold, we use the formula for the volume of a cylinder:

Volume of a cylinder = πr²h

Here, the pipe resembles a cylindrical shape. We have the following measurements: the radius (r) is 2.1 cm, and the height (h) is given as 12 m, which converts to 1200 cm.

Thus, the volume of water the pipe can contain is calculated as:

\text{Volume} = \dfrac{22}{7} \times (2.1)^2 \times 1200

Calculating step-by-step:

= 22 \times 0.3 \times 2.1 \times 1200

= 16632 \text{ cm}^3

Therefore, the pipe can hold 16632 cm³ of water.

Question 3

How many cubic meters of earth must be dug out to make a well 28 m deep and 2.8 m in diameter ? Also, find the cost of plastering its inner surface at ₹ 4.50 per sq meter.

Answer:

To solve this, we start by noting the dimensions of the well. The diameter is 2.8 m, which gives us a radius of $\dfrac{2.8}{2} = 1.4$ m. The depth of the well is 28 m.

The volume of earth that needs to be removed is the same as the volume of the cylindrical well. Using the formula for the volume of a cylinder, $\pi r^2 h$ , we substitute the values:

\text{Volume} = \dfrac{22}{7} \times 1.4 \times 1.4 \times 28

Calculating this gives us a volume of 172.48 cubic meters.

Next, we determine the area of the curved surface of the well, which is given by the formula $2\pi rh$ :

\text{Curved Surface Area} = 2 \times \dfrac{22}{7} \times 1.4 \times 28 = 2 \times 22 \times 0.2 \times 28

Simplifying this, we find the area to be 246.4 square meters.

Now, considering the cost of plastering, which is ₹ 4.50 per square meter, the total cost for plastering the inner surface becomes:

\text{Cost} = 246.40 \times 4.50 = ₹ 1108.80

Thus, the volume of earth to be dug out is 172.48 m³, and the cost of plastering the inner surface is ₹ 1108.80.

Question 4

What length of solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of external diameter 20 cm, 0.25 cm thick and 15 cm long?

Answer:

We have the external diameter of the hollow cylinder as 20 cm, which gives us an external radius (R) of $\dfrac{20}{2} = 10$ cm. The thickness of the cylinder is 0.25 cm, so the internal radius (r) is $10 - 0.25 = 9.75$ cm. The height of the hollow cylinder is 15 cm.

The volume of the material used in the hollow cylinder can be calculated using the formula:

\text{Volume of material} = \pi h(R^2 - r^2)

Substituting the values, we get:

\begin{aligned}= \pi \times 15 \times [10^2 - (9.75)^2] \\= \pi \times 15 \times [100 - 95.0625] \\= \pi \times 15 \times 4.9375.\end{aligned}

For the solid cylinder, the diameter is 2 cm, giving a radius (a) of $\dfrac{2}{2} = 1$ cm. Let the length of this solid cylinder be h. Thus, the volume is:

\text{Volume} = \pi a^2 h = \pi (1)^2 h = \pi h \text{ cm}^3.

Since the volume of the solid cylinder must match the volume of the material in the hollow cylinder for recasting:

\pi h = 15 \pi \times 4.9375

Solving for h, we find:

h = 15 \times 4.9375

h = 74.0625 \text{ cm}.

Hence, the length of solid cylinder = 74.06 cm.

Question 5

A cylinder has a diameter of 20 cm. The area of curved surface is 100 cm^2. Find :

(i) the height of the cylinder correct to one decimal place.

(ii) the volume of the cylinder correct to one decimal place.

Answer:

We have a cylinder with a diameter of 20 cm.

This means the radius $r$ is $\frac{20}{2} = 10$ cm.

The curved surface area given is 100 $\text{cm}^2$ .

Let the height be $h$ cm.

(i) Using the formula for the curved surface area of a cylinder:

\text{Curved surface area} = 2\pi rh

Thus, $2\pi rh = 100$ .

Substituting the values:

2 \times \frac{22}{7} \times 10 \times h = 100

Solving for $h$ :

h = \frac{100 \times 7}{22 \times 10 \times 2}

h = \frac{700}{440} = 1.6 \text{ cm}

Therefore, the height of the cylinder is 1.6 cm.

(ii) To find the volume of the cylinder, we use:

\text{Volume} = \pi r^2 h

Substituting the known values:

\text{Volume} = \frac{22}{7} \times 10 \times 10 \times 1.6

Calculating gives:

\text{Volume} = 502.9 \text{ cm}^3

Hence, the volume of the cylinder is 502.9 $\text{cm}^3$ .

Question 6

A metal pipe has a bore (inner diameter) of 5 cm. The pipe is 5 mm thick all round. Find the weight, in kilogram, of 2 metres of the pipe if 1 cm^3 of the metal weights 7.7 g.

Answer:

We have the inner diameter of the metal pipe as 5 cm, and the thickness of the pipe is 5 mm, which is equivalent to 0.5 cm.

This gives the inner radius, $r$ , as $\dfrac{5}{2}$ cm = 2.5 cm.

To find the external radius, $R$ , we add the thickness to the inner radius:

$R = 2.5 \text{ cm} + 0.5 \text{ cm} = 3 \text{ cm}$ .

The length of the pipe is given as 2 metres, which converts to 200 cm.

The volume of the metal used in the pipe can be calculated as the difference between the external and internal volumes:

Volume of the pipe = $\pi R^2 h - \pi r^2 h$

This simplifies to:

= \pi (R^2 - r^2) h

Further simplifying:

= \pi (R + r)(R - r) h

Substituting the known values:

= \dfrac{22}{7}(3 + 2.5)(3 - 2.5) \times 200

$= \dfrac{22}{7} \times 5.5 \times 0.5 \times 200 = 1728.6 \text{ cm}^3$ .

Given that each cubic centimeter of the metal weighs 7.7 grams, we can find the total weight of the pipe:

Weight of the pipe = $1728.6 \times 7.7$ grams

Convert this to kilograms:

$= \dfrac{(1728.6 \times 7.7)}{1000}$ kg = 13.31 kg.

Thus, the weight of the pipe is 13.31 kg.

Question 7

A cylindrical container with diameter of base 42 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 22 cm x 14 cm x 10.5 cm. Find the rise in the level of the water when the solid is submerged.

Answer:

We are given that the diameter of the cylindrical container’s base is 42 cm, which means the radius is $\dfrac{42}{2} = 21$ cm.

Let the rise in water level be denoted by $h$ cm.

When the rectangular solid is submerged, the water level increases due to the displacement of water. The volume of water displaced is equal to the volume of the rectangular solid.

Therefore, we have the equation:

\pi r^2 h = 22 \times 14 \times 10.5

Substituting the values, we get:

\dfrac{22}{7} \times 21^2 \times h = 3234

Solving for $h$ :

h = \dfrac{3234 \times 7}{22 \times 21^2} = \dfrac{22638}{9702} = 2\dfrac{1}{3} \text{ cm}

Hence, the water level will be raised to a level of $2\dfrac{1}{3}$ cm when the solid is submerged.

Question 8

A cylindrical container with internal radius of its base 10 cm, contains water up to a height of 7 cm. Find the area of wet surface of the cylinder.

Answer:

We have a cylindrical container where the internal radius, $r$ , is 10 cm, and the water reaches a height, $h$ , of 7 cm.

To find the wet surface area of the cylinder, we need to consider both the curved surface area that is in contact with the water and the base area. The formula for the wet surface area is:

\text{Wet Surface Area} = 2\pi rh + \pi r^2 = \pi r(2h + r)

Substituting the given values, we have:

= \dfrac{22}{7} \times 10 \times (2 \times 7 + 10)

This simplifies to:

= \dfrac{220}{7} \times 24

Calculating this gives us:

= 754.29 \text{ cm}^2.

Therefore, the area of the wet surface of the cylinder is 754.29 cm².

Question 9

Find the total surface area of an open pipe of length 50 cm, external diameter 20 cm and internal diameter 6 cm.

Answer:

We have an open pipe with a length of 50 cm.

The external diameter of the pipe is 20 cm, which means the external radius $R$ is $\dfrac{20}{2} = 10$ cm.

Similarly, the internal diameter is 6 cm, giving an internal radius $r$ of $\dfrac{6}{2} = 3$ cm.

The surface area of the open pipe, which is open from both ends, is calculated using the formula:

Surface area = $2\pi Rh + 2\pi rh = 2\pi h(R + r)$

Substituting the values, we get:

= $2 \times \dfrac{22}{7} \times 50 \times (10 + 3)$

= 4085.71 cm $^2$

Next, we calculate the area of the upper and lower parts:

Area = $2\pi R^2 - 2\pi r^2$

= $2 \times \dfrac{22}{7} \times (10^2 - 3^2)$

= $2 \times \dfrac{22}{7} \times 91$

= 572 cm $^2$ .

Adding these areas gives us the total surface area:

Total surface area = 4085.71 + 572 = 4657.71 cm $^2$ .

Hence, total surface area of open pipe = 4657.71 cm $^2$ .

Question 10

The height and the radius of the base of a cylinder are in the ratio 3 : 1. If its volume is 1029 π cm^3; find its total surface area.

Answer:

We know that the height and radius of a cylinder are in the ratio 3:1.

The volume given is 1029π cm³. Let’s denote the radius of the base by $r$ . Consequently, the height becomes $3r$ .

Using the formula for the volume of a cylinder, $\text{Volume} = \pi r^2 h$ , we have:

\pi r^2 \times 3r = 1029\pi

Canceling $\pi$ from both sides gives:

r^3 = \dfrac{1029}{3}

r^3 = 343

Taking the cube root, $r = \sqrt[3]{343}$

r = 7 \text{ cm}

Thus, the radius $r$ is 7 cm, and the height $h$ is $3r = 3 \times 7 = 21 \text{ cm}$ .

The formula for total surface area of a cylinder is:

\text{Total Surface Area} = 2\pi r(h + r)

Substituting the known values:

= 2 \times \dfrac{22}{7} \times 7 \times (21 + 7)

= 2 \times 22 \times 28

= 1232 \text{ cm}^2

Hence, the total surface area of the cylinder is 1232 cm².

Question 11

The radius of a solid right circular cylinder decreases by 20% and its height increases by 10%. Find the percentage change in its:

(i) volume (ii) curved surface area

Answer:

Consider the initial measurements of the solid right circular cylinder: radius = r cm and height = h cm.

The formula for volume is given by:

Volume = πr²h.

The formula for the curved surface area (CSA) is:

CSA = 2πrh.

After the modifications, the new dimensions become:

New radius (r’) = r – $\dfrac{20}{100} \times r$ = r – 0.2r = 0.8r.

New height (h’) = h + $\dfrac{10}{100} \times h$ = h + 0.1h = 1.1h.

Therefore, the updated volume is calculated as:

New volume = π(r’)²h’ = π(0.8r)²(1.1h) = 0.704 πr²h.

For the curved surface area, we have:

New CSA = 2πr’h’ = 2π(0.8r)(1.1h) = 1.76πrh.

Let’s determine the changes:

(i) Change in volume:

Decrease in volume = Original volume – New volume = πr²h – 0.704 πr²h = 0.296 πr²h.

The percentage change in volume is:

\dfrac{\text{Decrease in volume}}{\text{Original volume}} \times 100\% = \dfrac{0.296πr²h}{πr²h} \times 100\% = 0.296 \times 100\% = 29.6\%.

Hence, the volume decreases by 29.6%.

(ii) Change in curved surface area:

Decrease in CSA = Original CSA – New CSA = 2πrh – 1.76πrh = 0.24πrh.

The percentage change in CSA is:

\dfrac{\text{Decreased CSA}}{\text{Original CSA}} \times 100\% = \dfrac{0.24πrh}{2πrh} \times 100\% = 0.12 \times 100\% = 12\%.

Thus, the curved surface area decreases by 12%.

Question 12

Find the minimum length in cm and correct to nearest whole number of the thin metal sheet required to make a hollow and closed cylindrical box of diameter 20 cm and height 35 cm. Given that the width of the metal sheet is 1 m. Also, find the cost of the sheet at the rate of Rs. 56 per m.

Find the area of metal sheet required, if 10% of it is wasted in cutting, overlapping, etc.

Answer:

We are provided with the following details:

Height of the cylindrical box (h) = 35 cm
Diameter of the base = 20 cm, thus the radius (r) = $\dfrac{20}{2}$ = 10 cm
Width of the metal sheet = 1 m = 100 cm

To find the area of the metal sheet needed, we calculate the total surface area of the cylindrical box, which is given by the formula:

∴ Length $\times$ width = $2\pi r(r + h)$

Substituting the values, we have:

⇒ Length $\times$ 100 = $2 \times \dfrac{22}{7} \times 10(10 + 35)$

⇒ Length $\times$ 100 = $2 \times \dfrac{22}{7} \times 10 \times 45$

∴ Length = $\dfrac{2 \times 22 \times 10 \times 45}{7 \times 100}$ = 28.28 cm, which rounds to 28 cm when corrected to the nearest whole number.

Thus, the area of the metal sheet required is:

Area = Length $\times$ width = 28 $\times$ 100 = 2800 cm $^2$

Converting this to square meters, we have:

2800 cm $^2$ = $\dfrac{2800}{100 \times 100}$ = 0.28 m $^2$ .

The cost of the metal sheet at ₹ 56 per m $^2$ is calculated as:

Cost = ₹ (56 $\times$ 0.28) = ₹ 15.68

Now, to account for 10% wastage in the metal sheet due to cutting and overlapping, let the total sheet required be x cm $^2$ .

Then, $x - 10\%$ of $x$ = 2800 cm $^2$

⇒ $x - \dfrac{10}{100} \times x$ = 2800

⇒ $\dfrac{100x - 10x}{100}$ = 2800

⇒ $\dfrac{90x}{100}$ = 2800

⇒ $\dfrac{9x}{10}$ = 2800

Solving for x, we get:

⇒ $x = \dfrac{2800 \times 10}{9}$

⇒ x = 3111.11 cm $^2$ , which rounds to 3111 cm $^2$ when corrected to the nearest whole number.

Hence, length = 28 cm, cost of sheet = ₹15.68 and area of sheet required = 3111 cm $^2$ .

Question 13

3080 cm^3 of water is required to fill a cylindrical vessel completely and 2310 cm^3 of water is required to fill it upto 5 cm below the top. Find :

(i) radius of the vessel.

(ii) height of the vessel.

(iii) wetted surface area of the vessel when it is half-filled with water.

Answer:

Let’s denote the radius of the cylindrical vessel as $r$ cm and its height as $h$ cm.

We know that the volume of the entire cylinder is given as 3080 cm³. Thus, we have:

\pi r^2 h = 3080 \quad \text{...(1)}

It’s also given that the volume of water needed to fill the cylinder up to 5 cm below the top is 2310 cm³. Therefore, we can write:

\pi r^2 (h - 5) = 2310 \quad \text{...(2)}

Now, by dividing equation (1) by equation (2), we obtain:

\begin{aligned}\Rightarrow \dfrac{\pi r^2 h}{\pi r^2 (h - 5)} = \dfrac{3080}{2310} \\\Rightarrow \dfrac{h}{h - 5} = \dfrac{308}{231} \\\Rightarrow 231h = 308(h - 5) \\\Rightarrow 231h = 308h - 1540 \\\Rightarrow 308h - 231h = 1540 \\\Rightarrow 77h = 1540 \\\Rightarrow h = \dfrac{1540}{77} \\\Rightarrow h = 20 \text{ cm}\end{aligned}

(i) Substituting the value of $h$ into equation (1), we find:

\begin{aligned}\Rightarrow \pi r^2 \times 20 = 3080 \\\Rightarrow \dfrac{22}{7} \times r^2 \times 20 = 3080 \\\Rightarrow r^2 = \dfrac{3080 \times 7}{20 \times 22} \\\Rightarrow r^2 = 49 \\\Rightarrow r = 7 \text{ cm}.\end{aligned}

Hence, radius of vessel = 7 cm.

(ii) From our calculations above, we have:

h = 20 \text{ cm.}

Hence, height of vessel = 20 cm.

(iii) When the vessel is half-filled, the water reaches a height of $\dfrac{20}{2} = 10$ cm.

The wetted surface area in this case is given by:

\begin{aligned}\text{Wetted surface area} = 2\pi rh + \pi r^2 \\= \pi r(2h + r) \\= \dfrac{22}{7} \times 7 \times (2 \times 10 + 7) \\= 22 \times 27 \\= 594 \text{ cm}^2.\end{aligned}

Hence, wetted surface area = 594 cm².

Question 14

Find the volume of the largest cylinder formed when a rectangular piece of paper 44 cm by 33 cm is rolled along its :

(i) shorter side

(ii) longer side.

Answer:

(i) Consider rolling the paper along its shorter side. Here, the height of the cylinder becomes 44 cm, and the circumference of the base is 33 cm.

∴ Circumference = 33 cm

⇒ $2πr = 33$

⇒ $2 \times \dfrac{22}{7} \times r = 33$

⇒ $r = \dfrac{33 \times 7}{2 \times 22} = 5.25$ cm.

Now, calculate the volume using the formula $\pi r^2 h$ :

Volume = $\dfrac{22}{7} \times (5.25)^2 \times 44$

= $\dfrac{22}{7} \times 27.56 \times 44$

= $\dfrac{26680.5}{7} = 3811.5$ cm $^3$ .

Thus, the volume of the cylinder is 3811.5 cm $^3$ .

(ii) Now, roll the paper along its longer side. This makes the height of the cylinder 33 cm, with the circumference of the base being 44 cm.

∴ Circumference = 44 cm

⇒ $2πr = 44$

⇒ $2 \times \dfrac{22}{7} \times r = 44$

⇒ $r = \dfrac{44 \times 7}{2 \times 22} = 7$ cm.

Using the formula for volume $\pi r^2 h$ :

Volume = $\dfrac{22}{7} \times (7)^2 \times 33$

= $\dfrac{22}{7} \times 49 \times 33$

= 5082 cm $^3$ .

Thus, the volume of the cylinder is 5082 cm $^3$ .

Question 15

A metal cube of side 11 cm is completely submerged in water contained in a cylindrical vessel with diameter 28 cm. Find the rise in the level of water.

Answer:

We have a cylindrical vessel with a diameter of 28 cm, which gives us a radius of 14 cm since the radius is half of the diameter.

The side of the cube is 11 cm.

Let the water level rise by $h$ cm.

The volume of water displaced by the cube is equal to the volume of the cube itself.

∴ Volume of water displaced = Volume of the cube

This gives us the equation:

\pi r^2 h = a^3

Substituting the values, we have:

\dfrac{22}{7} \times (14)^2 \times h = (11)^3

Solving for $h$ :

h = \dfrac{11 \times 11 \times 11 \times 7}{22 \times 14 \times 14}

Simplifying further:

h = \dfrac{9317}{4312}

Thus, $h = 2.16$ cm.

Therefore, the rise in the water level is 2.16 cm.

Question 16

A circular tank of diameter 2 m is dug and the earth removed is spread uniformly all around the tank to form an embarkment 2 m in width and 1.6 m in height. Find the depth of the circular tank.

Answer:

Let’s start by noting the given measurements:

The diameter of the circular tank is 2 m, which gives us a radius (r) of $\dfrac{2}{2} = 1$ m.
The embarkment has a width of 2 m, leading to an external radius (R) of $r + w = 1 + 2 = 3$ m.
The height of the embarkment is 1.6 m.

Assume the depth of the tank is $h'$ meters.

The volume of the earth removed from the tank must equal the volume of the embarkment formed. Therefore, we have:

\pi r^2 h' = \pi (R^2 - r^2) h

Substitute the known values:

1^2 h' = (3^2 - 1^2) \times 1.6

Simplify the expression:

h' = (9 - 1) \times 1.6

h' = 12.8 \text{ m}

Thus, the depth of the circular tank is 12.8 meters.

Question 17

The sum of the height and the radius of a solid cylinder is 35 cm and its total surface area is 3080 cm^2; find the volume of the cylinder.

Answer:

We have the equation for the sum of the height and radius of the cylinder:

h + r = 35

The total surface area of the cylinder is given as 3080 cm². The formula for the total surface area is:

2\pi r(h + r) = 3080

Substituting the known value of $h + r$ from the first equation, we get:

2\pi r \times 35 = 3080

Simplifying, we have:

70 \times \dfrac{22}{7} \times r = 3080

This simplifies to:

220r = 3080

Solving for $r$ :

r = \dfrac{3080}{220}

r = 14 \text{ cm}

Substitute $r = 14$ back into the equation $h + r = 35$ :

h + 14 = 35

Solving for $h$ :

h = 35 - 14

h = 21 \text{ cm}

Now, to find the volume of the cylinder, use the formula:

\text{Volume} = \pi r^2 h

Substitute the values of $r$ and $h$ :

\dfrac{22}{7} \times (14)^2 \times 21

Calculating further, we have:

22 \times 2 \times 14 \times 21

= 12936 \text{ cm}^3

Therefore, the volume of the cylinder is 12936 cm³.

Question 18

The total surface area of a solid cylinder is 616 cm^2. If the ratio between its curved surface area and total surface area is 1 : 2; find the volume of the cylinder.

Answer:

We know that the total surface area of the cylinder is 616 cm².

∴ $2\pi r(h + r) = 616$

⇒ $\pi r(h + r) = \dfrac{616}{2}$

⇒ $\pi r(h + r) = 308$ ……….(1)

The given ratio of the curved surface area to the total surface area is 1 : 2.

This implies:

\Rightarrow \dfrac{\text{Curved surface area}}{\text{Total surface area}} = \dfrac{1}{2}

\Rightarrow \dfrac{2\pi rh}{2\pi r(h + r)} = \dfrac{1}{2}

\Rightarrow \dfrac{h}{h + r} = \dfrac{1}{2}

\Rightarrow 2h = h + r

\Rightarrow 2h - h = r

\Rightarrow h = r.

Now, substituting $h = r$ in equation (1), we have:

⇒ $\pi r(r + r) = 308$

⇒ $\pi r \cdot 2r = 308$

⇒ $2\pi r^2 = 308$

⇒ $\pi r^2 = 154$

Using $\pi = \dfrac{22}{7}$ , we find:

⇒ $\dfrac{22}{7} \times r^2 = 154$

⇒ $r^2 = \dfrac{154 \times 7}{22}$

⇒ $r^2 = 49$

⇒ $r = \sqrt{49}$

⇒ $r = 7$ cm

Thus, $h = 7$ cm as well.

Now, let’s calculate the volume of the cylinder:

Volume = $\pi r^2 h$

= $\dfrac{22}{7} \times (7)^2 \times 7$

= $22 \times 49$

= 1078 cm³.

Therefore, the volume of the cylinder is 1078 cm³.

Question 19

A cylindrical vessel of height 24 cm and diameter 40 cm is full of water. Find the exact number of small cylindrical bottles, each of height 10 cm and diameter 8 cm, which can be filled with this water.

Answer:

Let’s start by noting the dimensions of the larger cylindrical vessel. Its height is 24 cm, and its diameter is 40 cm. This means the radius, $r_1$ , is $\frac{40}{2} = 20$ cm.

For the smaller cylindrical bottles, each has a height of 10 cm and a diameter of 8 cm. Thus, the radius, $r_2$ , is $\frac{8}{2} = 4$ cm.

Assuming the number of smaller bottles that can be filled is $n$ , we equate the volume of the large vessel to the total volume of the $n$ small bottles:

Volume of the large vessel = $n \times$ Volume of one small bottle.

This gives us the equation:

\pi r_1^2 h_1 = n \times \pi r_2^2 h_2

Simplifying, we find:

\therefore n = \frac{r_1^2 h_1}{r_2^2 h_2}

Substituting the given values:

\Rightarrow n = \frac{20^2 \times 24}{4^2 \times 10}

\Rightarrow n = \frac{400 \times 24}{16 \times 10}

\Rightarrow n = 60.

Therefore, the number of small cylindrical bottles that can be filled is 60.

Question 20

Two solid cylinders, one with diameter 60 cm and height 30 cm and the other with radius 30 cm and height 60 cm, are melted and recasted into a third solid cylinder of height 10 cm. Find the diameter of the cylinder formed.

Answer:

To determine the diameter of the newly formed cylinder, let’s denote its radius as $r$ and its volume as $V$ . We know the height $h$ of this new cylinder is 10 cm.

For the first cylinder being melted down:
– Diameter is 60 cm, giving a radius $r_1 = 30$ cm.
– Height $h_1 = 30$ cm.
– Let its volume be $V_1$ .

For the second cylinder being melted down:
– Radius $r_2 = 30$ cm.
– Height $h_2 = 60$ cm.
– Let its volume be $V_2$ .

The volume of the newly formed cylinder is the sum of the volumes of these two original cylinders:

V = V_1 + V_2

Using the formula for the volume of a cylinder, we have:

\Rightarrow \pi r^2 h = \pi r_1^2 h_1 + \pi r_2^2 h_2

This simplifies to:

\Rightarrow r^2 \times 10 = (30)^2 \times 30 + (30)^2 \times 60

Calculating further:

\Rightarrow r^2 \times 10 = 27000 + 54000

\Rightarrow r^2 = \frac{81000}{10}

\Rightarrow r^2 = 8100

Taking the square root gives:

\Rightarrow r = \sqrt{8100} = 90 \text{ cm}

Thus, the diameter of the new cylinder is:

\text{Diameter} = 2r = 2 \times 90 = 180 \text{ cm}

Hence, the diameter of the new cylinder is 180 cm.

Question 21

The total surface area of a hollow cylinder, which is open from both the sides, is 3575 cm^2; area of its base ring is 357.5 cm^2 and its height is 14 cm. Find the thickness of the cylinder.

Answer:

Consider the external radius of the cylinder as $R$ cm and the internal radius as $r$ cm.

From the problem, we know:

The area of the base ring is 357.5 cm².

∴ $\pi(R^2 - r^2) = 357.5$

This leads to:

\begin{aligned}\Rightarrow \dfrac{22}{7} \times (R^2 - r^2) = 357.5 \\\Rightarrow (R^2 - r^2) = \dfrac{357.5 \times 7}{22} \\\Rightarrow (R^2 - r^2) = 113.75 \space ..........(1)\end{aligned}

Additionally, the total surface area of the hollow cylinder is 3575 cm².

Thus, we have:

\begin{aligned}\Rightarrow 2\pi Rh + 2\pi rh + 2\pi(R^2 - r^2) = 3575 \\\Rightarrow 2\pi h(R + r) + 2\pi \times 113.75 = 3575 \space ..........\text{[From (1)]} \\\Rightarrow 2 \times \dfrac{22}{7} \times 14 (R + r) + 2 \times \dfrac{22}{7} \times 113.75 = 3575 \\\Rightarrow 88(R + r) + 715 = 3575 \\\Rightarrow 88(R + r) = 3575 - 715 \\\Rightarrow 88(R + r) = 2860 \\\Rightarrow (R + r) = \dfrac{2860}{88} \\\Rightarrow (R + r) = 32.5 \space ..........(2)\end{aligned}

Now, dividing equation (1) by equation (2), we find:

\begin{aligned}\Rightarrow \dfrac{R^2 - r^2}{R + r} = \dfrac{113.75}{32.5} \\\Rightarrow \dfrac{(R - r)(R + r)}{(R + r)} = 3.5 \\\Rightarrow (R - r) = 3.5 \text{ cm}.\end{aligned}

Therefore, the thickness of the hollow cylinder is 3.5 cm.

Question 22

Two right circular solid cylinders have radii in the ratio 3 : 5 and heights in the ratio 2 : 3. Find the ratio between their :

(i) curved surface areas.

(ii) volumes.

Answer:

(i) We are given the ratios of the radii and heights of two cylinders as follows:

$r_1 : r_2 = 3 : 5$ and $h_1 : h_2 = 2 : 3$ .

Assume $r_1 = 3x$ and $r_2 = 5x$ . Similarly, let $h_1 = 2y$ and $h_2 = 3y$ .

The formula for the curved surface area (CSA) of a cylinder is $2πrh$ . So, the ratio of the curved surface areas of the two cylinders is:

\dfrac{\text{CSA of 1st cylinder}}{\text{CSA of 2nd cylinder}} = \dfrac{2πr_1h_1}{2πr_2h_2} = \dfrac{r_1h_1}{r_2h_2} = \dfrac{3x \times 2y}{5x \times 3y} = \dfrac{6xy}{15xy} = \dfrac{2}{5} = 2 : 5.

∴ The ratio of the curved surface areas is 2 : 5.

(ii) Now, let’s find the ratio of their volumes. The volume of a cylinder is given by $πr^2h$ .

Thus, the ratio of the volumes of the two cylinders is:

\dfrac{\text{Vol. of 1st cylinder}}{\text{Vol. of 2nd cylinder}} = \dfrac{πr_1^2h_1}{πr_2^2h_2} = \dfrac{r_1^2h_1}{r_2^2h_2} = \dfrac{(3x)^2 \times 2y}{(5x)^2 \times 3y} = \dfrac{18x^2y}{75x^2y} = \dfrac{6}{25} = 6 : 25.

∴ The ratio of the volumes is 6 : 25.

Question 23

A closed cylindrical tank, made of thin iron-sheet, has diameter = 8.4 m and height 5.4 m. How much metal sheet, to the nearest m^2, is used in making this tank, if $\dfrac{1}{15}$ of the sheet actually used was wasted in making the tank ?

Answer:

The radius of the cylindrical tank can be calculated as half of the diameter: $\dfrac{8.4}{2} = 4.2$ m.

The formula for the total surface area (TSA) of a closed cylinder is given by $2\pi r(h + r)$ . Substituting the known values:

2 \times \dfrac{22}{7} \times 4.2 \times (5.4 + 4.2)

Calculating further:

2 \times 22 \times 0.6 \times 9.6 = 253.44 \, \text{m}^2

We know that $\dfrac{1}{15}$ of the sheet was wasted. Therefore, the fraction of the sheet actually used to make the tank is:

1 - \dfrac{1}{15} = \dfrac{15 - 1}{15} = \dfrac{14}{15}

Let $x$ be the total area of the metal sheet used. Then:

\dfrac{14}{15} \times x = 253.44

Solving for $x$ :

x = \dfrac{253.44}{\dfrac{14}{15}} = 253.44 \times \dfrac{15}{14} \approx 271.54 \, \text{m}^2

Rounding to the nearest square meter, the area of the metal sheet used is 272 m $^2$ .

Hence, total metal (iron) sheet used in nearest m $^2$ is 272 m $^2$ .

Exercise 20(B)

Question 1(a)

The radius and height of a solid metallic cone are r cm each. The volume of the cone is :

(a) $\dfrac{1}{3}πr^2$
(b) $\dfrac{4}{3}πr^2$
(c) $\dfrac{1}{3}πr^3$
(d) $\dfrac{4}{3}πr^3$

Answer: (c)

\dfrac{1}{3}πr^3

The formula for finding the volume of a cone is given by:

Volume of cone = $\dfrac{1}{3}π \times (\text{radius})^2 \times \text{height}$ .

In this scenario, both the radius and the height of the cone are $r$ cm. Therefore, we substitute these values into the formula:

Volume of cone = $\dfrac{1}{3}π \times r^2 \times r = \dfrac{1}{3}πr^3$ .

Hence, Option 3 is the correct option.

Question 1(b)

The radius of the base of a solid cone is r cm and its height is h cm; its curved surface area is :

(a) π × r × $\sqrt{h^2 - r^2}$
(b) π × r × $\sqrt{h^2 + r^2}$
(c) π × r × (h + r)
(d) πr^2(h + r)

Answer: (b) π × r ×

\sqrt{h^2 + r^2}

To find the curved surface area of a cone, we use the formula $\pi r l$ , where $l$ is the slant height. The slant height $l$ can be determined using Pythagoras’ theorem in the right-angled triangle formed by the radius, height, and slant height of the cone. Thus, $l = \sqrt{h^2 + r^2}$ . Substituting this into the formula, we have:

Curved surface area = $\pi r \sqrt{h^2 + r^2}$ .

Hence, Option 2 is the correct option.

Question 1(c)

A conical toy tent-house, 28 cm in radius and 21 cm in height, is made from a rectangular sheet of paper 22 cm wide. The smallest length of the paper sheet required is :

(a) 70 cm
(b) 105 cm
(c) 140 cm
(d) 280 cm

Answer: (c) 140 cm

The conical toy is crafted from a rectangular sheet, so we equate the surface area of the cone to the area of the rectangular sheet.

Thus, $\pi rl = \text{length} \times \text{breadth}$ .

Given:
– Radius $r = 28$ cm
– Height $h = 21$ cm
– Breadth of the sheet = 22 cm

The slant height $l$ of the cone can be calculated using the Pythagorean theorem:

l = \sqrt{r^2 + h^2} = \sqrt{28^2 + 21^2}

Substituting the values, we have:

\dfrac{22}{7} \times 28 \times \sqrt{28^2 + 21^2} = a \times 22

Calculating further:

22 \times 4 \times \sqrt{784 + 441} = a \times 22

88 \times \sqrt{1225} = a \times 22

88 \times 35 = a \times 22

Solving for $a$ :

a = \dfrac{88 \times 35}{22}

a = 4 \times 35 = 140 \text{ cm}

Hence, Option 3 is the correct option.

Question 1(d)

For a cone, the ratio between the volume and area of its base is 11 : 6. The height of the cone is :

(a) $1\dfrac{5}{6}$ units
(b) 5.5 units
(c) 11 units
(d) 5 units

Answer: (b) 5.5 units

Assume the radius of the cone is $r$ cm and the height is $h$ cm.

We know that the ratio of the cone’s volume to the area of its base is given as 11 : 6.

∴ $\dfrac{\text{Volume of cone}}{\text{Area of base}} = \dfrac{11}{6}$

The volume of the cone is $\dfrac{1}{3} \pi r^2 h$ and the area of the base is $\pi r^2$ .

⇒ $\dfrac{\dfrac{1}{3} \pi r^2 h}{\pi r^2} = \dfrac{11}{6}$

This simplifies to:

⇒ $\dfrac{h}{3} = \dfrac{11}{6}$

Solving for $h$ , we have:

⇒ $h = \dfrac{11}{6} \times 3$

⇒ $h = \dfrac{11}{2} = 5.5 \text{ units}$ .

Hence, Option 2 is the correct option.

Question 1(e)

The radii of two solid cones are equal and their slant heights are in the ratio 7 : 4. The ratio between their curved surface areas is :

(a) 4 : 7
(b) 7 : 4
(c) 16 : 49
(d) 49 : 16

Answer: (b) 7 : 4

Assume the radii of both cones are $r$ units.

Let the slant heights of the cones be $l_1$ and $l_2$ .

The curved surface area (CSA) of the first cone is given by $\pi r l_1$ .

The curved surface area of the second cone is $\pi r l_2$ .

Thus, the ratio of the curved surface areas of the first cone to the second cone is:

\dfrac{\text{CSA of 1st cone}}{\text{CSA of 2nd cone}} = \dfrac{\pi r l_1}{\pi r l_2} = \dfrac{l_1}{l_2} = \dfrac{7}{4} = 7 : 4.

Hence, Option 2 is the correct option.

Question 2

Find the volume of a cone whose slant height is 17 cm and radius of base is 8 cm.

Answer:

We are given the slant height $l = 17$ cm and the radius $r = 8$ cm of the cone. We need to find the height $h$ of the cone.

Using the Pythagorean theorem for the cone, we have:

l^2 = h^2 + r^2

Substituting the known values:

17^2 = h^2 + 8^2

This simplifies to:

289 = h^2 + 64

Solving for $h^2$ :

h^2 = 225

Thus, $h = \sqrt{225} = 15$ cm.

Now, let’s calculate the volume of the cone using the formula:

\text{Volume of cone} = \dfrac{1}{3} \pi r^2 h

Substitute the values:

\dfrac{1}{3} \times \dfrac{22}{7} \times 8^2 \times 15

This becomes:

= \dfrac{22 \times 64 \times 5}{7}

Calculating further:

= \dfrac{7040}{7}

Which gives:

= 1005.71 \text{ cm}^3

Hence, volume of cone = 1005.71 cm^3.

Question 3

The curved surface area of a cone is 12320 cm^2. If the radius of its base is 56 cm, find its height.

Answer:

To find the height of the cone, we start with the formula for the curved surface area of a cone, which is given by:

Curved surface area = $\pi rl$

Given:

\therefore 12320 = \dfrac{22}{7} \times 56 \times l

Solving for $l$ :

\Rightarrow l = \dfrac{12320 \times 7}{22 \times 56}

\Rightarrow l = \dfrac{86240}{1232} = 70.

Now, using the Pythagorean theorem in the context of a cone, we know:

\Rightarrow l^2 = h^2 + r^2

Substituting the known values:

\Rightarrow 70^2 = h^2 + 56^2

\Rightarrow 4900 = h^2 + 3136

\Rightarrow h^2 = 4900 - 3136

\Rightarrow h^2 = 1764

Taking the square root on both sides gives:

$\Rightarrow h = \sqrt{1764} = 42$ cm.

Hence, height of cone = 42 cm.

Question 4

The circumference of the base of a 12 m high conical tent is 66 m. Find the volume of the air contained in it.

Answer:

We have the circumference of the conical tent’s base as 66 m.

∴ $2πr = 66$

Substituting the value of π as $\dfrac{22}{7}$ , we get:

2 \times \dfrac{22}{7} \times r = 66

⇒ $r = \dfrac{66 \times 7}{2 \times 22} = \dfrac{21}{2}$

Thus, $r = 10.5$ m.

Now, using the formula for the volume of a cone:

Volume of cone = $\dfrac{1}{3}πr^2h$

Substitute the values:

= \dfrac{1}{3} \times \dfrac{22}{7} \times (10.5)^2 \times 12

= \dfrac{1}{3} \times \dfrac{22}{7} \times 110.25 \times 12

= \dfrac{22 \times 110.25 \times 4}{7}

= \dfrac{9702}{7}

= 1386 \text{ m}^3.

Hence, volume of air contained in cone = 1386 m^3.

Question 5

The radius and the height of a right circular cone are in the ratio 5 : 12 and its volume is 2512 cubic cm. Find the radius and slant height of the cone. (Take π = 3.14)

Answer:

We are given that the ratio of the radius to the height of the cone is 5:12.

Assume the radius $r = 5x$ and the height $h = 12x$ .

The volume of a cone is calculated using the formula:

Volume of cone = $\dfrac{1}{3} \pi r^2 h$

Given that the volume is 2512 cubic cm, we have:

\Rightarrow 2512 = \dfrac{1}{3} \times 3.14 \times (5x)^2 \times (12x)

\Rightarrow 2512 = \dfrac{1}{3} \times 3.14 \times 300x^3

\Rightarrow x^3 = \dfrac{2512 \times 3}{3.14 \times 300}

\Rightarrow x^3 = \dfrac{7536}{942}

\Rightarrow x^3 = 8

\Rightarrow x^3 = 2^3

$\Rightarrow x = 2 \text{ cm}$ .

Thus, the radius $r = 5x = 5(2) = 10$ cm and the height $h = 12x = 12(2) = 24$ cm.

Now, to find the slant height $l$ , use the Pythagorean theorem:

l^2 = h^2 + r^2

\Rightarrow l^2 = (24)^2 + (10)^2

\Rightarrow l^2 = 576 + 100

\Rightarrow l^2 = 676

$\Rightarrow l = \sqrt{676} = 26$ cm.

Hence, radius = 10 cm and slant height = 26 cm.

Question 6

The diameters of two cones are equal. If their slant heights are in the ratio 5 : 4, find the ratio of their curved surface areas.

Answer:

We know the ratio of the slant heights is 5 : 4.

Assume the slant height of the first cone is 5x cm and the second cone is 4x cm.

For the first cone:

⇒ Diameter = d~1

⇒ Radius = r~1

⇒ Slant height (l~1) = 5x

For the second cone:

⇒ Diameter = d~2

⇒ Radius = r~2

⇒ Slant height (l~2) = 4x

Since the diameters are equal, we have:

⇒ d~1 = d~2

∴ r~1 = r~2.

The curved surface area (CSA) of a cone is given by the formula $πrl$ . Therefore, the ratio of the CSAs of the two cones is:

\Rightarrow \dfrac{\text{CSA of 1st cone}}{\text{CSA of 2nd cone}} = \dfrac{πr_1l_1}{πr_2l_2} = \dfrac{l_1}{l_2} = \dfrac{5x}{4x} = \dfrac{5}{4} = 5 : 4.

Hence, ratio of curved surface areas = 5 : 4.

Question 7

There are two cones. The curved surface area of one is twice that of the other. The slant height of the latter is twice that of the former. Find the ratio of their radii.

Answer:

Consider the curved surface area of the first cone to be double that of the second cone.

For the first cone:

∴ Slant height ( (l_1) = l )

∴ Radius $= r_1$

∴ Curved surface area ( (C_1) = 2C ) ………(1)

Now, for the second cone:

∴ Slant height ( (l_2) = 2l )

∴ Radius $= r_2$

∴ Curved surface area ( (C_2) = C ) ………(2)

By dividing equation (1) by equation (2), we obtain:

\begin{aligned}\Rightarrow \dfrac{\text{CSA of 1st cone}}{\text{CSA of 2nd cone}} = \dfrac{2C}{C} \\\Rightarrow \dfrac{\pi r_1 l_1}{\pi r_2 l_2} = \dfrac{2C}{C} \\\Rightarrow \dfrac{\pi r_1.l}{\pi r_2.2l} = 2 \\\Rightarrow \dfrac{r_1}{2r_2} = 2 \\\Rightarrow \dfrac{r_1}{r_2} = \dfrac{4}{1}.\end{aligned}

∴ The ratio of their radii is $r_1 : r_2 = 4 : 1$ .

Hence, ratio of radii = 4 : 1.

Question 8

A heap of wheat is in the form of a cone of diameter 16.8 m and height 3.5 m. Find its volume. How much cloth is required to just cover the heap?

Answer:

Let’s start by noting the dimensions of the cone-shaped heap of wheat. The diameter is given as 16.8 m, which means the radius (r) is half of that: $\dfrac{16.8}{2} = 8.4$ m. The height (h) of the cone is 3.5 m.

To find the volume of the cone, we use the formula:

\text{Volume of cone} = \dfrac{1}{3} \pi r^2 h

Substituting the given values:

= \dfrac{1}{3} \times \dfrac{22}{7} \times (8.4)^2 \times 3.5

Calculating further:

= \dfrac{22}{21} \times 246.96

= \dfrac{5433.12}{21}

= 258.72 \text{ m}^3

Next, to determine how much cloth is needed to cover the heap, we need the slant height (l) of the cone. The slant height is found using the Pythagorean theorem:

l^2 = r^2 + h^2

Substitute the values:

l^2 = (8.4)^2 + (3.5)^2

l^2 = 70.56 + 12.25

l^2 = 82.81

l = \sqrt{82.81} = 9.1 \text{ m}

The cloth required is equal to the curved surface area of the cone, calculated as:

\text{Curved surface area} = \pi r l

Substituting the values:

= \dfrac{22}{7} \times 8.4 \times 9.1

= 240.24 \text{ m}^2

Hence, volume = 258.72 m $^3$ and cloth required to cover the heap = 240.24 m $^2$ .

Question 9

If you are given a rectangular canvas of 1.5 m in width, what length of this canvas would you require to make a conical tent that is 48 m in diameter and 7 m in height? Note that 10% of the canvas is used (wasted) in folds and stitching.

Also, find the cost of the canvas at the rate of ₹ 24 per meter.

Answer:

We start by noting the diameter of the conical tent is 48 m. This gives us a radius $r$ of $\dfrac{48}{2} = 24$ m. The height $h$ is given as 7 m.

To find the slant height $l$ of the cone, we use the Pythagorean theorem:

$l = \sqrt{r^2 + h^2}$
$\Rightarrow l = \sqrt{(24)^2 + 7^2}$
$\Rightarrow l = \sqrt{576 + 49}$
$\Rightarrow l = \sqrt{625}$
$\Rightarrow l = 25 \text{ m}$

The surface area $S$ of the conical tent is calculated using the formula:

$S = \pi r l$
$\Rightarrow S = \dfrac{22}{7} \times 24 \times 25$
$\Rightarrow S = \dfrac{13200}{7} \text{ m}^2$

Since 10% of the canvas is wasted, only 90% is used for the tent. Therefore, we have:

$\dfrac{90}{100} \times \text{ Total canvas area} = \text{Surface area of conical tent}$
$\Rightarrow \dfrac{90}{100} \times \text{ Total canvas area} = \dfrac{13200}{7}$
$\Rightarrow \text{ Total canvas area} = \dfrac{13200}{7} \times \dfrac{100}{90}$
$\Rightarrow \text{ Total canvas area} = \dfrac{132000}{63} \text{ m}^2$

The total canvas area is given by the product of the length and width of the canvas:

$\Rightarrow \dfrac{132000}{63} = \text{Length of canvas} \times 1.5$
$\Rightarrow \text{Length of canvas} = \dfrac{132000}{63 \times 1.5} = \dfrac{132000}{94.5} = 1396.83 \text{ m}$

Finally, to find the cost of the canvas:

$\text{Cost of canvas} = \text{Rate per m} \times \text{Length}$
$= 24 \times 1396.83$
$= \text{₹ 33,523.92}$

Hence, length of canvas required = 1396.83 m and cost of canvas = ₹ 33,523.92

Question 10

A solid cone of height 8 cm and base radius 6 cm is melted and recast into identical cones, each of height 2 cm and diameter 1 cm. Find the number of cones formed.

Answer:

Consider the dimensions of the larger cone: its height is 8 cm and its base radius is 6 cm. Now, for the smaller cones, each has a height of 2 cm and a radius of 0.5 cm (since the diameter is 1 cm, the radius is $\frac{1}{2}$ cm).

Let the number of smaller cones formed be $n$ . The volume of the larger cone is equal to the total volume of all the smaller cones combined.

The formula for the volume of a cone is $\frac{1}{3} \pi r^2 h$ . Therefore, we equate the volume of the larger cone to $n$ times the volume of a smaller cone:

\Rightarrow \dfrac{1}{3}\pi r_1^2h_1 = n \times \dfrac{1}{3}\pi r_2^2h_2

Simplifying, we get:

\Rightarrow r_1^2h_1 = n \times r_2^2h_2

Solving for $n$ , we have:

\Rightarrow n = \dfrac{r_1^2h_1}{r_2^2h_2}

Substitute the given values:

\Rightarrow n = \dfrac{6^2 \times 8}{(0.5)^2 \times 2}

Calculate the expression:

\Rightarrow n = \dfrac{288}{0.5}

Finally, we find:

\Rightarrow n = 576.

Hence, the number of cones formed = 576.

Question 11

The total surface area of a right circular cone of slant height 13 cm is 90π cm^2. Calculate :

(i) its radius in cm

(ii) its volume in cm^3.

[Take π = 3.14]

Answer:

(i) We know the total surface area of a right circular cone is given by:

\pi rl + \pi r^2 = 90\pi

This simplifies to:

\pi r(l + r) = 90\pi

Dividing both sides by $\pi$ , we get:

r(l + r) = 90

Substituting the slant height $l = 13$ cm:

r(13 + r) = 90

Expanding, we have:

r^2 + 13r - 90 = 0

We can factor this quadratic equation as:

r^2 + 18r - 5r - 90 = 0

Grouping terms, we get:

r(r + 18) - 5(r + 18) = 0

Thus, it factors to:

(r - 5)(r + 18) = 0

Setting each factor to zero gives:

r - 5 = 0 \text{ or } r + 18 = 0

This results in:

r = 5 \text{ or } r = -18

The radius cannot be negative, so:

∴ radius = 5 cm.

(ii) To find the height $h$ , use the Pythagorean theorem for the cone:

l^2 = r^2 + h^2

Substituting $l = 13$ cm and $r = 5$ cm:

13^2 = 5^2 + h^2

This simplifies to:

h^2 = 169 - 25

h^2 = 144

Taking the square root:

h = \sqrt{144}

h = 12 \text{ cm.}

Now, calculate the volume of the cone using:

\text{Volume} = \dfrac{1}{3}\pi r^2 h

Substitute $r = 5$ cm, $h = 12$ cm, and $\pi = 3.14$ :

= \dfrac{1}{3} \times 3.14 \times (5)^2 \times 12

This simplifies to:

= \dfrac{942}{3}

= 314 \text{ cm}^3.

Hence, volume of circular cone = 314 cm³.

Question 12

The area of the base of a conical solid is 38.5 cm^2 and its volume is 154 cm^3. Find the curved surface area of the solid.

Answer:

We start with the area of the base of the conical solid, which is 38.5 cm².

∴ πr² = 38.5 …………(1)

Substituting the value of π as $\dfrac{22}{7}$ , we have:

⇒ $\dfrac{22}{7}r^2 = 38.5$

Solving for $r^2$ , we get:

⇒ $r^2 = \dfrac{38.5 \times 7}{22} = 12.25$

Taking the square root gives:

⇒ $r = \sqrt{12.25} = 3.5 \text{ cm}$

Next, consider the volume of the cone given as 154 cm³:

\Rightarrow \text{Volume} = 154

Using the formula for the volume of a cone:

\Rightarrow \dfrac{1}{3}πr^2h = 154

Substituting from equation (1):

\Rightarrow \dfrac{1}{3} \times 38.5 \times h = 154

Solving for $h$ , we find:

\Rightarrow h = \dfrac{154 \times 3}{38.5}

\Rightarrow h = 12 \text{ cm}

To find the slant height $l$ , use the Pythagorean theorem:

⇒ $l^2 = r^2 + h^2$

Substitute the known values:

⇒ ( l^2 = (3.5)^2 + (12)^2 )

⇒ $l^2 = 12.25 + 144$

⇒ $l^2 = 156.25$

Taking the square root:

⇒ $l = \sqrt{156.25} = 12.5 \text{ cm}$

Now, calculate the curved surface area of the cone:

Curved surface area = πrl

Substitute the values:

= $\dfrac{22}{7} \times 3.5 \times 12.5$

= 137.5 cm².

Hence, curved surface area = 137.5 cm².

Question 13

A vessel, in the form of an inverted cone, is filled with water to the brim. Its height is 32 cm and diameter of the base is 25.2 cm. Six equal solid cones are dropped in it, so that they are fully submerged. As a result one-fourth of water in the original cone overflows. What is the volume of each of the solid cones submerged?

Answer:

The radius of the vessel, which is in the shape of an inverted cone, is calculated as $R = \dfrac{25.2}{2} = 12.6$ cm.

The total volume of water that fills the vessel to the brim is given by the formula for the volume of a cone: $\dfrac{1}{3} \pi R^2 H$ , where $H$ is the height of the cone, which is 32 cm.

When six identical solid cones are submerged in the vessel, they cause one-fourth of the water to overflow.

Let us denote the radius of these smaller cones as $r$ and their height as $h$ . The total volume of these six cones is $6 \times \text{Vol. of each cone} = \dfrac{1}{4} \times \dfrac{1}{3} \pi R^2 H$ .

This implies that the volume of each cone is:

\text{Vol. of each cone} = \dfrac{1}{72} \pi R^2 H

Substituting the values, we get:

\text{Vol. of each cone} = \dfrac{1}{72} \times \dfrac{22}{7} \times (12.6)^2 \times 32

After calculating, the volume of each cone comes out to be $221.76$ cm $^3$ .

Hence, volume of each cone = 221.76 cm $^3$ .

Question 14

The volume of a conical tent is 1232 m^3 and the area of the base floor is 154 m^2. Calculate the :

(i) radius of the floor,

(ii) height of the tent,

(iii) length of the canvas required to cover this conical tent if its width is 2 m.

Answer:

(i) We know the area of the base floor is 154 m².

∴ $πr^2 = 154$

⇒ $\dfrac{22}{7} \times r^2 = 154$

⇒ $r^2 = \dfrac{154 \times 7}{22}$

⇒ $r^2 = 7 \times 7$

⇒ $r = 7$ m.

Thus, the radius of the floor is 7 m.

(ii) The volume of the tent is given as 1232 m³.

∴ $\dfrac{1}{3}πr^2h = 1232$

⇒ $\dfrac{1}{3} \times \dfrac{22}{7} \times (7)^2 \times h = 1232$

⇒ $\dfrac{154}{3} \times h = 1232$

⇒ $h = \dfrac{1232 \times 3}{154}$

⇒ $h = 24$ m.

Thus, the height of the tent is 24 m.

(iii) For the slant height $l$ , use the Pythagorean theorem:

⇒ $l^2 = r^2 + h^2$

⇒ $l^2 = (7)^2 + (24)^2$

⇒ $l^2 = 49 + 576$

⇒ $l^2 = 625$

⇒ $l = \sqrt{625}$

⇒ $l = 25$ m.

The curved surface area of the cone is:

⇒ $πrl = \dfrac{22}{7} \times 7 \times 25$

⇒ $= 550$ m².

To find the length of the canvas, set the area of the canvas equal to the curved surface area:

⇒ $l \times b = 550$

⇒ $l \times 2 = 550$

⇒ $l = \dfrac{550}{2}$

⇒ $l = 275$ m.

Thus, the length of canvas required is 275 m.

Exercise 20(C)

Question 1(a)

A solid metallic sphere of radius 16 cm is melted to form small identical spheres each of diameter 4 cm. The number of spheres formed is :

(a) 64
(b) 252
(c) 512
(d) 1024

Answer: (c) 512

We have a large metallic sphere with a radius of 16 cm that is melted down to create smaller spheres, each with a diameter of 4 cm.

For the larger sphere, the radius (R) is 16 cm. The smaller spheres have a radius (r) of $\dfrac{4}{2} = 2$ cm.

If we let the number of smaller spheres formed be n, then:

∴ The volume of the large sphere is equal to the total volume of the smaller spheres combined.

⇒ $\dfrac{4}{3}πR^3 = n \times \dfrac{4}{3}πr^3$

This simplifies to:

\Rightarrow n = \dfrac{\dfrac{4}{3}πR^3}{\dfrac{4}{3}πr^3}

Canceling out common terms, we have:

\Rightarrow n = \dfrac{R^3}{r^3}

Substitute the values of R and r:

\Rightarrow n = \dfrac{16^3}{2^3}

Calculate the values:

\Rightarrow n = \dfrac{4096}{8} = 512.

Thus, the number of smaller spheres formed is 512.

Hence, Option 3 is the correct option.

Question 1(b)

A hemi-spherical bowl (as shown) has external radius R and internal radius r, the outer surface area of the bowl is :

2πR^2 + πr^2
3πR^2 – πr^2
2πR^2 + 2πr^2
2πR^2 – πr^2

Answer:

To find the outer surface area of the hemi-spherical bowl, consider the following components:

The surface area of the outer hemisphere is given by $2\pi R^2$ .
The area of the outer circular cross-section is $\pi R^2$ .
The area of the inner circular cross-section, which needs to be subtracted, is $\pi r^2$ .

Thus, the total outer surface area of the bowl is:

2\pi R^2 + \pi R^2 - \pi r^2

Simplifying this expression, we have:

3\pi R^2 - \pi r^2

Hence, Option 2 is the correct option.

Question 1(c)

The ratio between the volumes of two spherical solids is 27 : 8. The ratio between their curved surface areas is :

(a) 27 : 8
(b) 8 : 27
(c) 3 : 2
(d) 9 : 4

Answer: (d) 9 : 4

Consider two spherical solids with radii $R$ and $r$ .

We know the ratio of their volumes is given as 27 : 8.

\begin{aligned}\Rightarrow \dfrac{\text{Volume of 1st solid}}{\text{Volume of 2nd solid}} = \dfrac{27}{8} \\\Rightarrow \dfrac{\dfrac{4}{3}πR^3}{\dfrac{4}{3}πr^3} = \dfrac{27}{8} \\\Rightarrow \dfrac{R^3}{r^3} = \dfrac{27}{8} \\\Rightarrow \Big(\dfrac{R}{r}\Big)^3 = \Big(\dfrac{3}{2}\Big)^3 \\\Rightarrow \dfrac{R}{r} = \dfrac{3}{2} \text{ .........(1)}\end{aligned}

Next, let’s find the ratio of their curved surface areas (CSA):

\begin{aligned}\Rightarrow \dfrac{\text{CSA of 1st solid}}{\text{CSA of 2nd solid}} = \dfrac{4πR^2}{4πr^2} \\= \dfrac{R^2}{r^2} \\= \Big(\dfrac{R}{r}\Big)^2 \\= \Big(\dfrac{3}{2}\Big)^2 \\= \dfrac{9}{4} \\= 9 : 4.\end{aligned}

Thus, the ratio of their curved surface areas is 9 : 4.

Hence, Option 4 is the correct option.

Question 1(d)

A solid metallic sphere of radius 8 cm is melted and recast into 64 identical solid spheres. The diameter of each smaller sphere formed is :

(a) 4 cm
(b) 2 cm
(c) 8 cm
(d) 1 cm

Answer: (a) 4 cm

We start with the radius of the larger metallic sphere, which is 8 cm. Let’s assume the radius of each smaller sphere is $r$ cm. It’s given that the larger sphere is melted and recast into 64 identical smaller spheres.

∴ The volume of the larger sphere must equal the total volume of the 64 smaller spheres.

This gives us:

\begin{align} \dfrac{4}{3}πR^3 &= 64 \times \dfrac{4}{3}πr^3 \ \Rightarrow 64 &= \dfrac{\dfrac{4}{3}πR^3}{\dfrac{4}{3}πr^3} \ \Rightarrow 64 &= \dfrac{R^3}{r^3} \ \Rightarrow 64 &= \dfrac{8^3}{r^3} \ \Rightarrow r^3 &= \dfrac{8^3}{64} \ \Rightarrow r^3 &= \dfrac{512}{64} \ \Rightarrow r^3 &= 8 \ \Rightarrow r &= \sqrt[3]{8} = 2 \text{ cm}\end{align}

The diameter of each smaller sphere is therefore $2 \times 2 = 4$ cm.

Hence, Option 1 is the correct option.

Question 1(e)

r~1, r~2 and r~3 are the radii of three metallic spheres. If these spheres are melted to form a single solid sphere, the radius of sphere formed is :

(a) $\sqrt{r_1^3 + r_2^3 + r_3^3}$
(b) r~1 + r~2 + r~3
(c) $r_1^3 + r_2^3 + r_3^3$
(d) $\sqrt[3]{r_1^3 + r_2^3 + r_3^3}$

Answer: (d)

\sqrt[3]{r_1^3 + r_2^3 + r_3^3}

We have three metallic spheres with radii $r_1$ , $r_2$ , and $r_3$ . These spheres are melted and reshaped into a single larger sphere. Let the radius of this new sphere be $R$ .

The principle of conservation of volume tells us that the volume of the new sphere is equal to the total volume of the three original spheres.

∴ Volume of the new sphere = Sum of the volumes of the three original spheres.

This gives us the equation:

\dfrac{4}{3}πR^3 = \dfrac{4}{3}πr_1^3 + \dfrac{4}{3}πr_2^3 + \dfrac{4}{3}πr_3^3

Simplifying, we find:

\dfrac{4}{3}πR^3 = \dfrac{4}{3}π(r_1^3 + r_2^3 + r_3^3)

Cancelling $\dfrac{4}{3}π$ from both sides, we have:

R^3 = r_1^3 + r_2^3 + r_3^3

Taking the cube root on both sides, we obtain:

R = \sqrt[3]{r_1^3 + r_2^3 + r_3^3}.

Hence, Option 4 is the correct option.

Question 2

The volume of a sphere is 38808 cm^3; find its diameter and the surface area.

Answer:

We have the volume of the sphere given as 38808 cm³.

Assume the radius of the sphere is $r$ .

Using the formula for the volume of a sphere:

\text{Volume of sphere} = \dfrac{4}{3}πr^3

Thus, $\dfrac{4}{3} πr^3 = 38808$

Substitute $π$ with $\dfrac{22}{7}$ :

\dfrac{4}{3} \times \dfrac{22}{7} \times r^3 = 38808

To find $r^3$ , rearrange the equation:

r^3 = \dfrac{38808 \times 7 \times 3}{4 \times 22}

Simplifying further:

r^3 = \dfrac{814968}{88}

r^3 = 9261

Taking the cube root:

r = \sqrt[3]{9261}

r = 21 \text{ cm}

Therefore, the diameter is:

\text{Diameter} = 2r = 21 \times 2 = 42 \text{ cm}

Now, calculate the surface area using the formula:

\text{Surface area} = 4πr^2

Substitute the values:

4 \times \dfrac{22}{7} \times 21 \times 21

This gives:

\text{Surface area} = 5544 \text{ cm}^2

Hence, diameter of ball = 42 cm and surface area = 5544 cm².

Question 3

A spherical ball of lead has been melted and made into identical smaller balls with radius equal to half the radius of the original one. How many such balls can be made?

Answer:

Assume the radius of the original spherical ball is $r$ cm. The volume of this sphere is given by $\dfrac{4}{3} \pi r^3$ .

For the smaller balls, the radius is $\dfrac{r}{2}$ cm.

The problem states that the volume of the large spherical ball is equal to the total volume of all the smaller balls formed. Let the number of smaller balls be $n$ .

Thus, we have:

\dfrac{4}{3}\pi r^3 = n \times \dfrac{4}{3}\pi\Big(\dfrac{r}{2}\Big)^3

This simplifies to:

n = \dfrac{\dfrac{4}{3}\pi r^3}{\dfrac{4}{3}\pi\Big(\dfrac{r}{2}\Big)^3}

Further simplifying, we find:

n = \dfrac{\dfrac{4}{3}\pi r^3 \times 2^3}{\dfrac{4}{3}\pi r^3}

Finally, this results in:

n = 2^3 = 8.

Hence, 8 balls can be made.

Question 4

How many balls each of radius 1 cm can be made by melting a bigger ball whose diameter is 8 cm.

Answer:

Let’s start by identifying the given dimensions. The diameter of the larger ball is 8 cm, which means its radius (R) is calculated as $\dfrac{8}{2} = 4$ cm.

Now, we calculate the volume of the larger ball using the formula for the volume of a sphere, $\dfrac{4}{3} \pi R^3$ :

\text{Volume of larger ball} = \dfrac{4}{3} \times \pi \times (4)^3 = \dfrac{4}{3} \times \pi \times 64 = \dfrac{256\pi}{3} \text{ cm}^3.

Next, consider the smaller balls, each with a radius (r) of 1 cm. The volume of one smaller ball is given by $\dfrac{4}{3} \pi r^3$ :

\text{Volume of each smaller ball} = \dfrac{4}{3} \times \pi \times (1)^3 = \dfrac{4}{3} \pi \text{ cm}^3.

Assume that we can create $n$ smaller balls from the larger one. Therefore, the volume of the larger ball is equal to the total volume of $n$ smaller balls:

\dfrac{256\pi}{3} = n \times \dfrac{4}{3} \pi

Solving for $n$ , we have:

n = \dfrac{\dfrac{256\pi}{3}}{\dfrac{4}{3} \pi} = \dfrac{256 \times \pi \times 3}{4 \times 3 \times \pi} = 64.

Hence, 64 balls can be made.

Question 5

The volume of one sphere is 27 times that of another sphere. Calculate the ratio of their:

(i) radii

(ii) surface areas

Answer:

Given that the volume of the first sphere is 27 times the volume of the second sphere, let’s denote the radius of the first sphere as $r_1$ and the radius of the second sphere as $r_2$ .

(i) From the problem statement, we can write:

\Rightarrow \dfrac{4}{3} \pi r_1^3 = 27 \times \dfrac{4}{3} \pi r_2^3

By simplifying, we get:

\Rightarrow r_1^3 = 27 \times r_2^3

Thus, the ratio of the cubes of their radii is:

\Rightarrow \dfrac{r_1^3}{r_2^3} = \dfrac{27}{1}

Taking the cube root on both sides, we obtain:

\Rightarrow \Big(\dfrac{r_1}{r_2}\Big)^3 = \Big(\dfrac{3}{1}\Big)^3

Therefore, the ratio of their radii is:

\Rightarrow \dfrac{r_1}{r_2} = \dfrac{3}{1}.

Hence, $r_1 : r_2 = 3 : 1$ .

(ii) The surface area of the first sphere is given by $4\pi(r_1)^2$ , and for the second sphere, it is $4\pi(r_2)^2$ .

The ratio of their surface areas becomes:

\text{Ratio of surface areas} = \dfrac{4\pi r_1^2}{4\pi r_2^2}

Simplifying this, we find:

= \dfrac{r_1^2}{r_2^2} = \Big(\dfrac{3}{1}\Big)^2

Which gives us:

= \dfrac{9}{1}.

Hence, the ratio of surface areas = 9 : 1.

Question 6

If the number of square centimeters on the surface of a sphere is equal to the number of cubic centimeters in its volume, what is the diameter of the sphere ?

Answer:

Consider the radius of the sphere as $r$ .

According to the problem, the surface area of the sphere equals its volume.

Thus, we have:

4\pi r^2 = \dfrac{4}{3}\pi r^3

\Rightarrow r^2 = \dfrac{r^3}{3}

\Rightarrow \dfrac{r^3}{r^2} = 3

$\Rightarrow r = 3\text{ cm}$ .

Therefore, the diameter of the sphere is calculated as $2r = 2 \times 3 = 6\text{ cm}$ .

Hence, diameter of sphere = 6 cm.

Question 7

A solid metal sphere is cut through its center into 2 equal parts. If the diameter of the sphere is $3\dfrac{1}{2}$ cm, find the total surface area of each part correct to two decimal places.

Answer:

The diameter of the sphere is given as $3\dfrac{1}{2}$ cm, which can be expressed as $\dfrac{7}{2}$ cm.

To find the radius $r$ of the sphere, divide the diameter by 2:

Radius, $r = \dfrac{\dfrac{7}{2}}{2} = \dfrac{7}{4}$ cm.

When the sphere is cut into two equal hemispheres, each hemisphere’s total surface area includes its curved surface area plus the area of its circular base.

The formula for the total surface area of a hemisphere is:

$\text{Total surface area of hemisphere} = \dfrac{1}{2}$ (Curved surface area of sphere) + Area of circular base.

Calculating this, we have:

= \dfrac{1}{2} \times 4\pi r^2 + \pi r^2

= 2\pi r^2 + \pi r^2

= 3\pi r^2

Substitute $r = \dfrac{7}{4}$ cm into the formula:

= 3 \times \dfrac{22}{7} \times \dfrac{7}{4} \times \dfrac{7}{4}

= 3 \times \dfrac{22 \times 7}{16}

= 3 \times \dfrac{77}{8}

= 28.88 \text{ cm}^2.

Therefore, the total surface area of each hemisphere is 28.88 cm $^2$ .

Question 8

The internal and external diameters of a hollow hemispherical vessel are 21 cm and 28 cm respectively. Find :

(i) internal curved surface area,

(ii) external curved surface area,

(iii) total surface area,

(iv) volume of material of the vessel.

Answer:

(i) We know that the internal diameter of the vessel is 21 cm.

This gives us an internal radius $r$ as $\dfrac{21}{2}$ cm.

To find the internal curved surface area, we apply the formula:

\text{Internal curved surface area} = 2\pi r^2

Substituting the values, we have:

= 2 \times \dfrac{22}{7} \times \dfrac{21}{2} \times \dfrac{21}{2} = \dfrac{19404}{28} = 693 \text{ cm}^2

Thus, the internal curved surface area is 693 cm².

(ii) The external diameter is 28 cm, leading to an external radius $R$ of 14 cm.

Using the formula for the external curved surface area:

\text{External curved surface area} = 2\pi R^2

We substitute the values:

= 2 \times \dfrac{22}{7} \times 14 \times 14 = 1232 \text{ cm}^2

Thus, the external curved surface area is 1232 cm².

(iii) The total surface area of the hemisphere is given by:

2\pi r^2 + 2\pi R^2 + \pi(R^2 - r^2)

Plugging in the values:

$= 693 + 1232 + \dfrac{22}{7} \times \left$ (14)^2 – \left(\dfrac{21}{2}\right)^2\right$$$$

$= 1925 + \dfrac{22}{7} \times \left$ 196 – \dfrac{441}{4}\right$$$$

= 1925 + \dfrac{22}{7} \times \dfrac{784 - 441}{4}

= 1925 + \dfrac{22}{7} \times \dfrac{343}{4}

= 1925 + \dfrac{11 \times 49}{2}

= 1925 + \dfrac{539}{2}

= 1925 + 269.5 = 2194.5 \text{ cm}^2

Thus, the total surface area of the hemisphere is 2194.5 cm².

(iv) To find the volume of the material of the vessel, we use:

\text{Volume} = \dfrac{2}{3}\pi(R^3 - r^3)

Substituting the given values:

$= \dfrac{2}{3} \times \dfrac{22}{7} \times$ (14)^3 – \left(\dfrac{21}{2}\right)^3$$$$

= \dfrac{2}{3} \times \dfrac{22}{7} \times [2744 - 1157.625]

= \dfrac{2}{3} \times \dfrac{22}{7} \times 1586.375

= \dfrac{44}{21} \times 1586.375 = 3323.83 \text{ cm}^3

Thus, the volume of the vessel is 3323.83 cm³.

Question 9

A solid sphere and a solid hemi-sphere have the same total surface area. Find the ratio between their volumes.

Answer:

Assume the radius of the sphere is $R$ cm and that of the hemi-sphere is $r$ cm.

We know that the total surface area of the solid sphere is equal to that of the solid hemi-sphere.

∴ $4πR^2 = 3πr^2$ \
⇒ $\dfrac{R^2}{r^2} = \dfrac{3π}{4π}$ \
⇒ $\dfrac{R^2}{r^2} = \dfrac{3}{4}$ \
⇒ $\dfrac{R}{r} = \sqrt{\dfrac{3}{4}}$ \
⇒ $\dfrac{R}{r} = \dfrac{\sqrt{3}}{2}$ .

Next, let’s determine the ratio of their volumes.

$\dfrac{\text{Vol. of sphere}}{\text{Vol. of hemi-sphere}} = \dfrac{\dfrac{4}{3}πR^3}{\dfrac{2}{3}πr^3}$ \
$= \dfrac{4πR^3 \times 3}{2πr^3 \times 3}$ \
$= 2 \times \dfrac{R^3}{r^3}$ \
$= 2 \times \Big(\dfrac{\sqrt{3}}{2}\Big)^3$ \
$= 2 \times \dfrac{3\sqrt{3}}{8}$ \
$= \dfrac{3\sqrt{3}}{4}$ .

Hence, ratio between volumes = $3\sqrt{3} : 4$ .

Question 10

Metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted and recasted into a single solid sphere. Taking π = 3.1, find the surface area of the solid sphere formed.

Answer:

Consider the radii of the three original spheres as $r_1 = 6$ cm, $r_2 = 8$ cm, and $r_3 = 10$ cm.

Let $r$ be the radius of the new sphere formed after melting and recasting.

The volume of the newly formed sphere is equal to the combined volume of the three original spheres.

\Rightarrow \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi (r_1)^3 + \dfrac{4}{3}\pi (r_2)^3 + \dfrac{4}{3}\pi (r_3)^3

Simplifying, we have:

\Rightarrow \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi (r_1^3 + r_2^3 + r_3^3)

Cancel $\dfrac{4}{3}\pi$ from both sides:

\Rightarrow r^3 = r_1^3 + r_2^3 + r_3^3

Substituting the values, we get:

\Rightarrow r^3 = (6)^3 + (8)^3 + (10)^3

Computing the cubes:

\Rightarrow r^3 = 216 + 512 + 1000

Adding these, we find:

\Rightarrow r^3 = 1728

Recognizing that ( 1728 = (12)^3 ), we have:

\Rightarrow r = 12 \text{ cm.}

Now, let’s calculate the surface area of the new sphere using the formula:

Surface area = $4\pi r^2$

Substitute $r = 12$ cm and $\pi = 3.1$ :

= 4 \times 3.1 \times 12^2

Calculating this, we find:

= 1785.6 \text{ cm}^2.

Exercise 20(D)

Question 1(a)

A cone and a sphere have equal volumes. Their radii are also equal each being 10 cm; the height of the cone is :

(a) 70 cm
(b) 40 cm
(c) 50 cm
(d) 20 cm

Answer: (b) 40 cm

Consider the radius of both the cone and the sphere as $r$ cm, and let the height of the cone be $h$ cm.

Given that the volumes of the cone and sphere are equal and their radii are the same, we have:

∴ Volume of the cone = Volume of the sphere

\Rightarrow \dfrac{1}{3} \pi r^2 h = \dfrac{4}{3} \pi r^3

By simplifying, we find:

\Rightarrow h = \dfrac{4 \times 3\pi r^3}{3\pi r^2}

This reduces to:

\Rightarrow h = 4r

Substituting the given radius $r = 10$ cm:

\Rightarrow h = 4 \times 10 = 40 \text{ cm}.

Hence, Option 2 is the correct option.

Question 1(b)

A sphere and a cone have equal volumes and equal radii. The ratio between the radius and height of the cone is :

(a) 3 : 4
(b) 4 : 3
(c) 4 : 1
(d) 1 : 4

Answer: (d) 1 : 4

Consider the radius of both the cone and the sphere to be $r$ cm, and let the height of the cone be $h$ cm.

We know that the volumes of the cone and sphere are equal, and they share the same radius.

∴ Volume of the cone = Volume of the sphere

\Rightarrow \dfrac{1}{3}πr^2h = \dfrac{4}{3}πr^3

Simplifying this equation, we get:

\Rightarrow h = \dfrac{4 \times 3πr^3}{3πr^2}

\Rightarrow h = 4r

Thus, the ratio of the radius to the height of the cone is:

\Rightarrow \dfrac{r}{h} = \dfrac{1}{4}

$\Rightarrow r : h = 1 : 4$ .

Hence, Option 4 is the correct option.

Question 1(c)

The radius and the height of a cone are in the ratio 2 : 1. The ratio between the volumes of a sphere and this cone (both having equal radii) is :

(a) 1 : 4
(b) 4 : 1
(c) 8 : 1
(d) 1 : 8

Answer: (c) 8 : 1

We know the radius and height of the cone are in the ratio 2:1.

Let’s assume the radius of the cone is 2x and the height is x.

The radius of the sphere is given to be the same as the radius of the cone.

∴ The radius of the sphere, r = 2x.

Now, let’s find the ratio of the volumes of the sphere and the cone:

\dfrac{\text{Volume of sphere}}{\text{Volume of cone}} = \dfrac{\dfrac{4}{3} \pi r^3}{\dfrac{1}{3} \pi r^2 h}

Simplifying, we have:

\Rightarrow \dfrac{\text{Volume of sphere}}{\text{Volume of cone}} = \dfrac{4 \times 3 \times \pi r^3}{3 \times \pi r^2 h}

This reduces to:

\Rightarrow \dfrac{\text{Volume of sphere}}{\text{Volume of cone}} = \dfrac{4r}{h}

Substituting the values of r and h:

\Rightarrow \dfrac{\text{Volume of sphere}}{\text{Volume of cone}} = \dfrac{4 \times 2x}{x}

Further simplifying:

\Rightarrow \dfrac{\text{Volume of sphere}}{\text{Volume of cone}} = \dfrac{8x}{x}

Finally, we get:

\Rightarrow \dfrac{\text{Volume of sphere}}{\text{Volume of cone}} = \dfrac{8}{1} = 8 : 1.

Hence, the ratio of the volumes is 8:1.

Hence, Option 3 is the correct option.

Question 1(d)

A solid metallic cylinder is melted and formed into identical solid cones each having same radius and same height as that of the cylinder. The number of cones formed is :

(a) 3
(b) 6
(c) 9
(d) 2

Answer: (a) 3

We have a solid metallic cylinder that is melted down to create identical cones. Each cone shares the same radius and height as the original cylinder.

Let the number of cones formed be denoted by $n$ .

Assume the radius of both the cylinder and each cone is $r$ units, and their height is $h$ units.

∴ The volume of the cylinder is equal to the total volume of all the cones formed.

Thus, the equation becomes:

\pi r^2 h = n \times \dfrac{1}{3} \pi r^2 h

By solving this equation, we get:

n = \dfrac{3\pi r^2 h}{\pi r^2 h} = 3

Hence, option 1 is the correct option.

Question 1(e)

A solid metallic cone of radius 10 cm and height 12 cm is melted to form a wire of uniform cross-section 1 cm^2. The length of wire formed is :

(a) 40 × π cm
(b) 4 × π m
(c) 400 × π m
(d) 2 × π m

Answer: (b) 4 × π m

We start with a solid metallic cone having a radius of 10 cm and a height of 12 cm. This cone is melted to create a wire with a uniform cross-section of 1 cm $^2$ . Let’s determine the length of this wire, denoted as $l$ cm.

Since the volume of the cone is equal to the volume of the wire, we have:

\text{Volume of cone} = \text{Volume of wire}

The formula for the volume of a cone is $\dfrac{1}{3} \pi r^2 h$ . Therefore:

\dfrac{1}{3} \pi \times 10^2 \times 12 = 1 \times l

Solving for $l$ , we get:

l = 4 \times \pi \times 10^2

l = 400 \pi \text{ cm}

To convert the length from centimeters to meters, we divide by 100:

l = \dfrac{400\pi}{100} \text{ m}

Thus, the length of the wire is:

l = 4 \times \pi \text{ m}

Hence, option 2 is the correct option.

Question 2

A solid sphere of radius 15 cm is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate the number of cones recast.

Answer:

We have a sphere with a radius of 15 cm. The volume of this sphere can be calculated using the formula $\dfrac{4}{3}πR^3$ . Substituting the given radius, we find:

\dfrac{4}{3} \times π \times 15 \times 15 \times 15 = 4500π \text{ cm}^3.

Next, we consider the cones that are formed. Each cone has a radius of 2.5 cm and a height of 8 cm. The formula for the volume of a cone is $\dfrac{1}{3}πr^2h$ . Therefore, the volume of one cone is:

\dfrac{1}{3} \times π \times 2.5 \times 2.5 \times 8 = \dfrac{1}{3} \times 50π = \dfrac{50}{3}π.

Let the number of cones formed be denoted by $n$ . Since the total volume of the sphere is equal to the combined volume of all the cones, we have:

4500π = n \times \dfrac{50}{3}π.

Solving for $n$ , we find:

n = \dfrac{4500π \times 3}{50π}.

This simplifies to:

n = 270.

Thus, the number of cones formed is 270.

Question 3

A hollow sphere of internal and external diameters 4 cm and 8 cm, respectively is melted into a cone of base diameter 8 cm. Find the height of the cone.

Answer:

We are given that the external diameter of the hollow sphere is 8 cm, which means the external radius $R$ is $\dfrac{8}{2} = 4$ cm. Similarly, the internal diameter is 4 cm, so the internal radius $r$ is $\dfrac{4}{2} = 2$ cm.

For the cone, the base diameter is 8 cm, giving us a radius $r_1$ of 4 cm. Let’s denote the height of the cone as $h$ cm.

Since the hollow sphere is melted and reshaped into a cone, the volumes must be equal.

\text{Volume of the sphere} = \text{Volume of the cone}

This implies:

\dfrac{4}{3} \pi (R^3 - r^3) = \dfrac{1}{3} \pi (r_1)^2 h

Removing $\pi$ and simplifying, we get:

4(R^3 - r^3) = r_1^2 h

Substituting the values:

4 \times [(4)^3 - (2)^3] = (4)^2 \times h

Calculating further:

4 \times [64 - 8] = 16h

4 \times 56 = 16h

Solving for $h$ :

h = \dfrac{4 \times 56}{16}

h = 14 \text{ cm}

Hence, the height of the cone = 14 cm.

Question 4

The radii of the internal and external surfaces of a metallic spherical shell are 3 cm and 5 cm respectively. It is melted and recast into a solid right circular cone of height 32 cm. Find the diameter of the base of the cone.

Answer:

We have a solid right circular cone with a height of 32 cm. The internal radius of the metallic spherical shell is 3 cm, while the external radius is 5 cm. Let the radius of the cone be denoted as $r_1$ cm.

Since the spherical shell is melted and reshaped into the cone, the volumes of both must be equal.

∴ Volume of the spherical shell = Volume of the cone.

This gives us the equation:

\Rightarrow \dfrac{4}{3}π(R^3 - r^3) = \dfrac{1}{3}πr_1^2h

Simplifying further:

\Rightarrow 4(R^3 - r^3) = r_1^2h

Substituting the known values:

\Rightarrow r_1^2 = \dfrac{4(R^3 - r^3)}{h}

\Rightarrow r_1^2 = \dfrac{4 \times (5^3 - 3^3)}{32}

Calculating the cube of the radii:

\Rightarrow r_1^2 = \dfrac{4 \times (125 - 27)}{32}

\Rightarrow r_1^2 = \dfrac{4 \times 98}{32}

\Rightarrow r_1^2 = \dfrac{49}{4}

To find $r_1$ , take the square root:

\Rightarrow r_1 = \sqrt{\dfrac{49}{4}}

\Rightarrow r_1 = \dfrac{7}{2}

Thus, the diameter of the cone’s base is:

Diameter = $2r = 2 \times \dfrac{7}{2} = 7$ cm.

Hence, diameter = 7 cm.

Question 5

Total volume of three identical cones is the same as that of a bigger cone whose height is 9 cm and diameter 40 cm. Find the radius of the base of each smaller cone, if height of each is 108 cm.

Answer:

Assume the radius of each smaller cone is $r$ cm.

We know:

Height of each smaller cone ( $h$ ) = 108 cm
Diameter of the larger cone = 40 cm, thus its radius ( $R$ ) = $\dfrac{40}{2}$ = 20 cm
Height of the larger cone ( $H$ ) = 9 cm

The problem states that the volume of the larger cone equals the total volume of the three smaller cones.

Therefore,

\dfrac{1}{3}πR^2H = 3 \times \dfrac{1}{3}πr^2h

Simplifying this, we have:

R^2H = 3 \times r^2h

Substitute the known values:

(20)^2 \times 9 = 3 \times r^2 \times 108

Solving for $r^2$ :

r^2 = \dfrac{20^2 \times 9}{3 \times 108}

r^2 = \dfrac{3600}{324}

r^2 = \dfrac{100}{9}

To find $r$ , take the square root:

r = \sqrt{\dfrac{100}{9}}

r = \dfrac{10}{3} = 3\dfrac{1}{3}\text{ cm}.

Thus, the radius of the base of each smaller cone is $3\dfrac{1}{3}$ cm.

Question 6

A solid rectangular block of metal 49 cm by 44 cm by 18 cm is melted and formed into a solid sphere. Calculate the radius of the sphere.

Answer:

Consider the radius of the sphere to be $r$ cm.

The metal block is reshaped into a sphere, so their volumes must be equal.

∴ Volume of the rectangular block = Volume of the sphere

\begin{aligned}\Rightarrow 49 \times 44 \times 18 = \dfrac{4}{3}πr^3 \\\Rightarrow 38808 = \dfrac{4}{3} \times \dfrac{22}{7} \times r^3 \\\Rightarrow r^3 = \dfrac{38808 \times 21}{4 \times 22} \\\Rightarrow r^3 = 9261 \\\Rightarrow r^3 = (21)^3 \\\Rightarrow r = 21 \text{ cm}.\end{aligned}

Hence, the radius of sphere = 21 cm.

Question 7

A hemispherical bowl of internal radius 9 cm is full of liquid. This liquid is to be filled into conical shaped small containers each of diameter 3 cm and height 4 cm. How many containers are necessary to empty the bowl?

Answer:

We have a hemispherical bowl with an internal radius of 9 cm.

The conical containers each have a diameter of 3 cm, which means their radius (r) is $\dfrac{3}{2}$ cm or 1.5 cm. The height (h) of each cone is 4 cm.

Suppose the number of conical containers required is n.

∴ The volume of the hemispherical bowl must equal the total volume of all the conical containers:

\Rightarrow \dfrac{2}{3}πR^3 = n \times \dfrac{1}{3}πr^2h

Rearranging for n gives:

\Rightarrow n = \dfrac{\dfrac{2}{3}πR^3}{\dfrac{1}{3}πr^2h}

Substitute the given values into the equation:

\Rightarrow n = \dfrac{2 \times (9)^3 \times 3}{1 \times (1.5)^2 \times 4 \times 3}

Simplifying further:

\Rightarrow n = \dfrac{2 \times 729 \times 3}{2.25 \times 4 \times 3}

\Rightarrow n = 162.

Therefore, 162 containers are necessary to empty the bowl.

Question 8

The total area of a solid metallic sphere is 1256 cm^3. It is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate :

(i) the radius of the solid sphere,

(ii) the number of cones recast.

[Take π = 3.14]

Answer:

(i) We have the total surface area of the solid metallic sphere as 1256 cm².

Let the radius of the sphere be denoted by $R$ .

Using the formula for the surface area of a sphere:

Total surface area = $4\pi R^2$

∴ $4\pi R^2 = 1256$

⇒ $4 \times 3.14 \times R^2 = 1256$

⇒ $R^2 = \dfrac{1256}{4 \times 3.14}$

⇒ $R^2 = \dfrac{1256}{12.56}$

⇒ $R^2 = 100$

⇒ $R = \sqrt{100} = 10 \text{ cm}$ .

Therefore, the radius of the sphere is 10 cm.

(ii) Now, consider the cone with a radius ( $r$ ) of 2.5 cm and a height ( $h$ ) of 8 cm.

Let $n$ represent the number of cones formed.

Since the sphere is melted and recast into these cones, the volume of the sphere equals the total volume of the cones.

∴ Volume of the sphere = $n \times$ Volume of one cone

⇒ $\dfrac{4}{3}\pi R^3 = n \times \dfrac{1}{3}\pi r^2h$

⇒ $4R^3 = n \times r^2h$

⇒ $n = \dfrac{4R^3}{r^2h}$

Substitute the known values:

⇒ $n = \dfrac{4 \times 10^3}{(2.5)^2 \times 8}$

⇒ $n = \dfrac{4 \times 1000}{6.25 \times 8}$

⇒ $n = \dfrac{4000}{50}$

⇒ $n = 80$ .

Thus, the number of cones formed is 80.

Question 9

The surface area of a solid metallic sphere is 2464 cm^2. It is melted and recast into solid right circular cones of radius 3.5 cm and height 7 cm. Calculate :

(i) the radius of the sphere.

(ii) the number of cones recast.

Answer:

(i) We know the surface area of the metallic sphere is 2464 cm².

Let’s denote the radius of the sphere as $R$ cm.

According to the formula for the surface area of a sphere:

Surface area = $4\pi R^2$

Thus, $4\pi R^2 = 2464$ .

Substituting the value of $\pi$ as $\dfrac{22}{7}$ , we have:

4 \times \dfrac{22}{7} \times R^2 = 2464

⇒ $R^2 = \dfrac{2464 \times 7}{22 \times 4}$

⇒ $R^2 = \dfrac{17248}{88}$

⇒ $R^2 = 196$

Taking the square root on both sides, we find:

R = \sqrt{196}

⇒ $R = 14$ cm.

Therefore, the radius of the sphere is 14 cm.

(ii) Now, consider the cones being formed.

The radius of each cone ( $r$ ) is 3.5 cm, and the height ( $h$ ) is 7 cm.

Let $n$ be the number of cones formed.

Since the sphere is melted and recast into cones, the volume of the sphere equals the total volume of the cones.

Thus, Volume of sphere = $n \times$ Volume of cone

Using the volume formulas:

\Rightarrow \dfrac{4}{3}\pi R^3 = n \times \dfrac{1}{3}\pi r^2 h

\Rightarrow 4R^3 = n \times r^2 h

Solving for $n$ :

\Rightarrow n = \dfrac{4R^3}{r^2 h}

Substituting the known values:

\Rightarrow n = \dfrac{4 \times (14)^3}{(3.5)^2 \times 7}

\Rightarrow n = \dfrac{4 \times 2744}{12.25 \times 7}

\Rightarrow n = \dfrac{10976}{85.75}

\Rightarrow n = 128.

Hence, the number of cones formed is 128.

Exercise 20(E)

Question 1(a)

In the given figure, the radius of the cone is same as its height equal to 2 cm each. The sum of the volumes of the whole body is :

8π cm^3
16 cm^3
32π cm^3
32 cm^3

Answer:

Let’s consider the given data:

The cone has a radius ((r)) of 2 cm and a height ((h)) of 2 cm. The radius of the hemisphere is also 2 cm, as it matches the cone’s radius.

∴ The total volume of the shape is the sum of the cone’s volume and the hemisphere’s volume.

⇒ Total Volume = Volume of cone + Volume of hemisphere

⇒ Total Volume = $\dfrac{1}{3}πr^2h + \dfrac{2}{3}πr^3$

Notice that we can factor out $\dfrac{πr^2}{3}$ :

= \dfrac{πr^2}{3}(h + 2r)

Substitute the given values:

= \dfrac{π \times 2^2}{3} \times (2 + 2 \times 2)

= \dfrac{4π}{3} \times (2 + 4)

= \dfrac{4π}{3} \times 6

= \dfrac{24π}{3}

= 8π \text{ cm}^3.

Hence, Option 1 is the correct option.

Question 1(b)

In the given figure, height of the conical part is 3 cm. The radius and the height of the cylindrical part are 1 cm each. The total volume of the body is :

9π cm^3
18π cm^3
2π cm^3
4π cm^3

Answer:

We have a cylindrical part and a conical part to consider. The radius of both the cylindrical and conical sections is 1 cm. The height of the cylindrical section is also 1 cm, while the conical section has a height of 3 cm.

The total volume of the entire structure can be calculated by adding the volume of the cone to the volume of the cylinder:

Volume of the cone: $\dfrac{1}{3} \pi r^2 H$

Volume of the cylinder: $\pi r^2 h$

Substituting the given values, we have:

\dfrac{1}{3} \pi \times 1^2 \times 3 + \pi \times 1^2 \times 1

This simplifies to:

\pi + \pi = 2\pi \text{ cm}^3.

Question 1(c)

The given figure shows a solid cylinder of height h cm and radius r cm. If the slant height of the conical cavity is l cm, the total surface area of the remaining solid is :

πrl + 2πrh + 2πr^2
πr^2 + πrl + 2πrh
2πr^2 – πrl + 2πrh
(πrl + πr^2) + 2πr^2 + 2πrh

Answer:

Consider the figure provided. To find the total surface area of the remaining solid, we need to account for the following components:

The curved surface area of the cylinder, which is given by the formula $2\pi rh$ .
The area of the upper surface of the cylinder, which is a circle with area $\pi r^2$ .
The curved surface area of the cone, calculated using $\pi rl$ .

Thus, the total surface area of the remaining solid becomes:

2\pi rh + \pi r^2 + \pi rl.

Hence, Option 2 is the correct option.

Question 1(d)

In the given figure, a solid cone is kept inverted in a closed cylindrical container such that the height of cone = height of cylinder = h, radius of cone = radius of cylinder = r and slant height of cone = l. If the remaining of cylinder is completely filled with water, the wetted surface area of the whole body is :

2πrh + πr^2 + πrl
2πrh + πrl
2πrh + 2πr^2 + πrl
2πrh – πr^2 + πrl

Answer:

Notice that the wetted surface area of the entire structure consists of three parts: the curved surface area of the cylinder, the area of the top surface of the cylinder, and the curved surface area of the cone.

The curved surface area of the cylinder is given by $2\pi rh$ . The top surface of the cylinder, which is a circle, has an area of $\pi r^2$ . The curved surface area of the cone is $\pi rl$ .

Therefore, the total wetted surface area of the body is:

2\pi rh + \pi r^2 + \pi rl.

Question 1(e)

The given figure shows a solid sphere and a closed cylindrical container, both having the same heights and same radii. The volume of air left in the cylinder is :

6πr^3
$\dfrac{3}{2}πr^3$
$\dfrac{2}{3}πr^3$
4πr^3

Answer:

Consider the diagram provided. We observe that both the solid sphere and the cylindrical container share the same radius, denoted as $r$ , and the same height. For the cylinder, the height $h$ is given by $2r$ .

To determine the volume of air remaining inside the cylinder, we subtract the volume of the sphere from the volume of the cylinder:

Volume of the cylinder is $πr^2h$ , and the volume of the sphere is $\dfrac{4}{3}πr^3$ .

Thus, the volume of air left is:

πr^2h - \dfrac{4}{3}πr^3

Substituting the height of the cylinder, $h = 2r$ , we have:

πr^2.2r - \dfrac{4}{3}πr^3

This simplifies to:

2πr^3 - \dfrac{4}{3}πr^3

Further simplifying, we get:

\dfrac{6πr^3 - 4πr^3}{3}

Finally, simplifying this expression gives:

\dfrac{2πr^3}{3} = \dfrac{2}{3}πr^3.

Hence, Option 3 is the correct option.

Question 2

A cone of height 15 cm and diameter 7 cm is mounted on a hemisphere of same diameter. Determine the volume of the solid thus formed.

Answer:

We have a cone with a height of 15 cm and a diameter of 7 cm. This cone is placed on a hemisphere that shares the same diameter.

To find the volume of this composite solid, we need the radius of both the cone and the hemisphere. Since the diameter is 7 cm, the radius (r) for both shapes is $\dfrac{7}{2}$ cm, which is 3.5 cm.

For the volume of the entire solid, we combine the volumes of the cone and the hemisphere:

Volume of the solid (V) = Volume of the cone + Volume of the hemisphere

V = \dfrac{1}{3}πr^2h + \dfrac{2}{3}πR^3

Substituting the values:

V = \dfrac{1}{3} \times \dfrac{22}{7} \times (3.5)^2 \times 15 + \dfrac{2}{3} \times \dfrac{22}{7} \times (3.5)^3

Calculating further:

= \dfrac{1}{3} \times \dfrac{22}{7} \times 12.25 \times 15 + \dfrac{2}{3} \times \dfrac{22}{7} \times 42.875

Breaking it down:

= 22 \times 1.75 \times 5 + \dfrac{1886.5}{21}

This simplifies to:

= 192.5 + 89.833

Thus, the total volume of the solid formed is $282.33\text{ cm}^3$ .**

Question 3

A buoy is made in the form of a hemisphere surmounted by a right cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 metres and its volume is two-third of the hemisphere. Calculate the height of the cone and the surface area of the buoy, correct to two places of decimal.

Answer:

Consider the buoy, which consists of a hemisphere and a cone sharing the same base radius. Here, both the radius of the hemisphere and the base of the cone are 3.5 m.

The volume of the hemisphere is calculated using the formula $\dfrac{2}{3}πr^3$ :

\text{Volume of hemisphere} = \dfrac{2}{3} \times \dfrac{22}{7} \times (3.5)^3 = \dfrac{2}{3} \times 22 \times 0.5 \times 3.5 \times 3.5 = \dfrac{269.5}{3} \text{ m}^3.

Let the height of the cone be denoted by $h$ . The formula for the volume of a cone is $\dfrac{1}{3}πR^2h$ . Given that the cone’s volume is two-thirds of the hemisphere’s volume, we have:

\text{Volume of cone} = \dfrac{2}{3} \times \dfrac{269.5}{3} = \dfrac{539}{9} \text{ m}^3.

Equating the two expressions for the volume of the cone:

\dfrac{1}{3}πR^2h = \dfrac{539}{9}

Substituting the values:

\dfrac{1}{3} \times \dfrac{22}{7} \times (3.5)^2 \times h = \dfrac{539}{9}

Solving for $h$ :

h = \dfrac{539 \times 7 \times 3}{9 \times (3.5)^2 \times 22}

h = \dfrac{11319}{2425.5}

h = 4.67 \text{ m}.

Next, to find the slant height $l$ of the cone, we use the Pythagorean theorem:

l^2 = R^2 + h^2

l^2 = (3.5)^2 + (4.67)^2

l^2 = 12.25 + 21.81

l^2 = 34.06

l = \sqrt{34.06} = 5.83 \text{ m}.

The total surface area of the buoy is the sum of the surface areas of the hemisphere and the cone:

\text{Surface area of buoy} = 2πr^2 + πRl

= 2 \times \dfrac{22}{7} \times (3.5)^2 + \dfrac{22}{7} \times 3.5 \times 5.83

= 2 \times 22 \times 0.5 \times 3.5 + 22 \times 0.5 \times 5.83

= 77 + 64.13

= 141.13 \text{ m}^2.

Thus, the height of the cone is 4.67 m and the surface area of the buoy is 141.13 m².

Question 4

From a rectangular solid of metal 42 cm by 30 cm by 20 cm, a conical cavity of diameter 14 cm and depth 24 cm is drilled out. Find:

(i) the surface area of the remaining solid

(ii) the volume of remaining solid

(iii) the weight of the material drilled out if it weighs 7 gm per cm^3.

Answer:

For the given problem, we start with the dimensions of the rectangular solid, which are 42 cm, 30 cm, and 20 cm. The conical cavity drilled into this solid has a diameter of 14 cm, giving it a radius of 7 cm, and a depth of 24 cm.

(i) Surface Area of the Remaining Solid

First, calculate the total surface area of the cuboid:

\text{Surface Area} = 2(lb + bh + lh) = 2 (42 \times 30 + 30 \times 20 + 20 \times 42)

= 2 (1260 + 600 + 840) = 2 (2700) = 5400 \text{ cm}^2

Next, find the area of the circular base of the conical cavity:

\text{Area of base} = \pi r^2 = \dfrac{22}{7} \times 7 \times 7 = 154 \text{ cm}^2

To find the slant height $l$ of the cone, use the Pythagorean theorem:

l^2 = r^2 + h^2 = (7)^2 + (24)^2 = 49 + 576 = 625

l = \sqrt{625} = 25 \text{ cm}

Now, calculate the curved surface area of the cone:

\text{Curved Surface Area} = \pi r l = \dfrac{22}{7} \times 7 \times 25 = 22 \times 25 = 550 \text{ cm}^2

The surface area of the remaining solid is:

\text{Remaining Surface Area} = 5400 + 550 - 154 = 5796 \text{ cm}^2

Thus, the surface area of the remaining solid is 5796 cm².

(ii) Volume of the Remaining Solid

Calculate the volume of the rectangular solid:

\text{Volume} = l \times b \times h = 42 \times 30 \times 20 = 25200 \text{ cm}^3

Next, find the volume of the conical cavity:

\text{Volume of cone} = \dfrac{1}{3} \pi r^2 h = \dfrac{1}{3} \times \dfrac{22}{7} \times 7 \times 7 \times 24 = 22 \times 7 \times 8 = 1232 \text{ cm}^3

Subtract the volume of the cone from the volume of the solid:

\text{Volume of remaining solid} = 25200 - 1232 = 23968 \text{ cm}^3

Therefore, the volume of the remaining solid is 23968 cm³.

(iii) Weight of the Material Drilled Out

The volume of the material drilled out is the volume of the cone, which is 1232 cm³. Given that the material weighs 7 gm per cm³:

\text{Weight} = 1232 \times 7 = 8624 \text{ g} = 8.624 \text{ kg}

Thus, the weight of the material drilled out is 8.624 kg.

Question 5

The cubical block of side 7 cm is surmounted by a hemisphere of the largest size. Find the surface area of the resulting solid.

Answer:

Consider a cube with a side length of 7 cm. The largest hemisphere that can sit on its face will have a diameter equal to the side of the cube, which is 7 cm.

Thus, the radius of the hemisphere is $r = \frac{7}{2} = 3.5 \text{ cm}$ .

The curved surface area of the hemisphere is calculated as:

2\pi r^2 = 2 \times \frac{22}{7} \times (3.5)^2 = 77 \text{ cm}^2.

Next, find the area of the base of the hemisphere:

\pi r^2 = \frac{22}{7} \times (3.5)^2 = 38.5 \text{ cm}^2.

To determine the surface area of the entire solid, we need to consider the surface area of the cube minus the base area of the hemisphere, and then add the curved surface area of the hemisphere. This gives us:

6 \times (7)^2 - 38.5 + 77 = 294 + 38.5 = 332.5 \text{ cm}^2.

Therefore, the surface area of the resulting solid is 332.5 $\text{cm}^2$ .

Question 6

A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of the top which is open is 5 cm. It is filled with water up to the rim. When lead shots, each of which is a sphere of radius 0.5 cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Answer:

We’re given the dimensions of the cone and the spheres. The cone has a height (H) of 8 cm and a radius (R) of 5 cm. Each lead shot is a sphere with a radius (r) of 0.5 cm.

To find the number of lead shots, let’s denote it as $n$ .

Since one-fourth of the water overflows when the lead shots are added, we have:

\dfrac{1}{4} \times \text{Volume of cone} = n \times \text{Volume of each sphere}

The formula for the volume of a cone is $\dfrac{1}{3} \pi R^2 H$ , and for a sphere, it’s $\dfrac{4}{3} \pi r^3$ .

Substituting these into the equation:

\dfrac{1}{4} \times \dfrac{1}{3} \pi R^2 H = n \times \dfrac{4}{3} \pi r^3

Simplifying, we get:

\dfrac{R^2 H}{4} = n \times 4r^3

Rearranging to solve for $n$ :

n = \dfrac{R^2 H}{16r^3}

Now, substitute the given values:

n = \dfrac{5^2 \times 8}{16 \times (0.5)^3}

n = \dfrac{25 \times 8}{16 \times 0.125}

n = \dfrac{200}{2}

n = 100

Therefore, the number of lead shots dropped is 100.

Question 7

A hemi-spherical bowl has negligible thickness and the length of its circumference is 198 cm. Find the capacity of the bowl.

Answer:

We know that the circumference of the hemi-spherical bowl is 198 cm.

The relationship between the circumference and the radius $r$ of a circle is given by:

2\pi r = 198

To find the radius, solve for $r$ :

r = \dfrac{198}{2\pi} = \dfrac{198}{2 \times \dfrac{22}{7}} = \dfrac{198 \times 7}{44} = \dfrac{1386}{44} = 31.5 \text{ cm}

Now, the capacity of the bowl, which is a hemisphere, can be calculated using the formula for the volume of a hemisphere:

\text{Capacity of bowl} = \dfrac{2}{3}\pi r^3

Substitute the value of $r$ :

\text{Capacity of bowl} = \dfrac{2}{3} \times \dfrac{22}{7} \times (31.5)^3

Calculate further:

= \dfrac{2 \times 22 \times 31255.875}{21} = \dfrac{1375258.5}{21} = 65488.5 \text{ cm}^3

Thus, the capacity of the bowl is 65488.5 cm³.

Question 8

Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r cm.

Answer:

To determine the largest volume of a cone that can fit within a hemisphere of radius $r$ cm, note that both the height and the radius of the cone must each be $r$ cm.

The formula for the volume of a cone is given by:

\text{Volume of cone} = \dfrac{1}{3} \pi r^2 h

Substituting the maximum values for the radius and height of the cone, we have:

= \dfrac{1}{3} \times \pi \times r^2 \times r

This simplifies to:

= \dfrac{1}{3} \pi r^3 \text{ cm}^3.

Thus, the maximum volume of the cone that can be carved from the solid hemisphere with radius $r$ cm is $\dfrac{1}{3} \pi r^3$ cm $^3$ .

Question 9

The radii of the bases of two right circular cones of same height are r~1 and r~2 respectively. The cones are melted and recast into a solid sphere of radius R. Find the height of each cone in terms of r~1, r~2 and R.

Answer:

Assume the height of each cone is $h$ cm.

Since the cones are melted and reshaped into a sphere, their combined volumes equal the volume of the sphere.

∴ Volume of the first cone $+$ Volume of the second cone $=$ Volume of the sphere

⇒ $\dfrac{1}{3}πr_1^2h + \dfrac{1}{3}πr_2^2h = \dfrac{4}{3}πR^3$

Simplifying, we have:

⇒ $\dfrac{πh}{3}(r_1^2 + r_2^2) = \dfrac{π}{3} \times 4R^3$

⇒ $h(r_1^2 + r_2^2) = 4R^3$

⇒ $h = \dfrac{4R^3}{r_1^2 + r_2^2}$ .

Hence, $h = \dfrac{4R^3}{r_1^2 + r_2^2}$ .

Question 10

A cone and a hemisphere have the same base and the same height. Find the ratio of their volumes.

Answer:

Assume the radius and height of both the cone and the hemisphere are each ‘a’ cm.

The volume of the cone can be calculated using the formula:

\text{Volume of cone} = \dfrac{1}{3} \pi (\text{radius})^2 \times \text{height} = \dfrac{1}{3} \pi a^3.

For the hemisphere, the volume is given by:

\text{Volume of hemisphere} = \dfrac{2}{3} \pi (\text{radius})^3 = \dfrac{4}{3} \pi a^3.

To find the ratio of their volumes, we set up the following expression:

\text{Ratio} = \dfrac{\text{Volume of cone}}{\text{Volume of hemisphere}} = \dfrac{\dfrac{1}{3} \pi a^3}{\dfrac{2}{3} \pi a^3}.

Notice that the $\pi a^3$ terms cancel out, simplifying the ratio to:

\dfrac{1}{2} = 1 : 2.

Hence, ratio between volumes = 1 : 2.

Exercise 20(F)

Question 1(a)

The volume of a solid cylinder and a solid cone are same. If their radii are also same, the ratio between the heights of cylinder and cone is :

(a) 3 : 1
(b) 2 : 3
(c) 1 : 3
(d) 3 : 2

Answer: (c) 1 : 3

Let’s assume the radius of both the cylinder and the cone is $r$ . Denote the height of the cylinder as $h_1$ and the height of the cone as $h_2$ .

Given that the volumes of the cylinder and the cone are identical, we can express this as:

\pi r^2 h_1 = \dfrac{1}{3} \pi r^2 h_2

Notice that the $\pi r^2$ term appears on both sides, allowing us to simplify the equation:

h_1 = \dfrac{1}{3} h_2

This leads to the ratio of their heights:

\dfrac{h_1}{h_2} = \dfrac{1}{3}

Thus, the ratio of the height of the cylinder to the height of the cone is:

h_1 : h_2 = 1 : 3

Hence, Option 3 is the correct option.

Question 1(b)

The volume of the given solid is :

$\dfrac{4}{3}πr^3$
$\dfrac{4}{3}πr^2$
2πr^3
4πr^3

Answer:

Examining the diagram, we observe:

The cone’s height (h) = r
The cylinder’s height (H) = r
The radius for the hemispherical, cylindrical, and conical sections is uniformly r.

The total volume of the solid is the sum of the volumes of the cone, cylinder, and hemisphere:

Volume of the cone = $\dfrac{1}{3}πr^2h$

Volume of the cylinder = $πr^2H$

Volume of the hemisphere = $\dfrac{2}{3}πr^3$

Thus, the combined volume is:

= \dfrac{1}{3}πr^2.r + πr^2.r + \dfrac{2}{3}πr^3

= \dfrac{1}{3}πr^3 + πr^3 + \dfrac{2}{3}πr^3

= \dfrac{πr^3 + 3πr^3 + 2πr^3}{3}

= \dfrac{6πr^3}{3}

= 2πr^3.

Hence, Option 3 is the correct option.

Question 1(c)

Eight identical metallic spheres, each of radius 1 cm are melted and recast into a solid sphere. The radius of the solid sphere formed is :

(a) 3 cm
(b) 2 cm
(c) 1.5 cm
(d) 2.5 cm

Answer: (b) 2 cm

We start with the radius of each small metallic sphere, which is 1 cm.

When these eight identical spheres are melted and reshaped into a single solid sphere, we need to determine the radius of this new sphere, denoted as $R$ .

The key idea here is that the total volume of the new solid sphere equals the combined volume of the eight smaller spheres. Therefore, we have:

\text{Volume of large sphere} = 8 \times \text{Volume of one small sphere}

This translates mathematically to:

\Rightarrow \dfrac{4}{3} \pi R^3 = 8 \times \dfrac{4}{3} \pi r^3

The $\dfrac{4}{3} \pi$ terms cancel out, leading to:

\Rightarrow R^3 = 8r^3

Substituting $r = 1$ cm, we get:

\Rightarrow R^3 = 8 \times 1^3

Simplifying further:

\Rightarrow R^3 = 8

Recognizing that $8$ is $2^3$ , we find:

\Rightarrow R^3 = 2^3

Thus, the radius $R$ of the large sphere is:

\Rightarrow R = 2 \text{ cm}

Hence, Option 2 is the correct option.

Question 1(d)

The ratio between the heights of two solid cones is 2 : 3 and ratio between their radii is 9 : 8. The ratio between their volumes is :

(a) 27 : 32
(b) 32 : 27
(c) 3 : 2
(d) 2 : 3

Answer: (a) 27 : 32

We know the ratio of the heights of two solid cones is 2:3. Let’s denote the height of the first cone as $h_1 = 2a$ and the second cone as $h_2 = 3a$ .

The ratio of their radii is given as 9:8, so let the radius of the first cone be $r_1 = 9x$ and the second cone be $r_2 = 8x$ .

To find the ratio of their volumes, use the formula for the volume of a cone: $V = \frac{1}{3} \pi r^2 h$ . Thus, the ratio of the volumes is:

\dfrac{\text{Vol. of 1st cone}}{\text{Vol. of 2nd cone}} = \dfrac{\frac{1}{3}\pi r_1^2 h_1}{\frac{1}{3}\pi r_2^2 h_2} = \dfrac{r_1^2 h_1}{r_2^2 h_2}

Substitute the values for $r_1$ , $r_2$ , $h_1$ , and $h_2$ :

= \dfrac{(9x)^2 \times 2a}{(8x)^2 \times 3a}

This simplifies to:

= \dfrac{81x^2 \times 2a}{64x^2 \times 3a}

Further simplifying, we have:

= \dfrac{162x^2a}{192x^2a}

Notice that $x^2a$ cancels out, leaving:

= \dfrac{27}{32}

Thus, the ratio of their volumes is 27:32.

Hence, Option 1 is the correct option.

Question 1(e)

The radius of a solid cylinder is doubled keeping the height same. The percentage increase in its volume is :

(a) 200%
(b) 100%
(c) 400%
(d) 300%

Answer: (d) 300%

Initially, consider the cylinder with a radius of $r$ and a height of $h$ . The volume of this cylinder is given by the formula $\pi r^2 h$ .

Now, if the radius is doubled to $2r$ while the height remains unchanged, the new volume becomes ( \pi (2r)^2 h = 4\pi r^2 h ).

To find the change in volume, subtract the original volume from the new volume:
$\text{Difference in volume} = 4\pi r^2 h - \pi r^2 h = 3\pi r^2 h.$

To calculate the percentage increase in volume, use the formula:
$\text{Percentage increase} = \dfrac{\text{Difference in volume}}{\text{Original volume}} \times 100.$
Substituting the values, we get:
$= \dfrac{3\pi r^2 h}{\pi r^2 h} \times 100 = 3 \times 100 = 300\%.$

Hence, Option 4 is the correct option.

Question 2

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 same base are removed. Find the volume of the remaining solid.

Answer:

We are provided with the dimensions of a right circular cylinder and a right circular cone, both having identical heights and base radii.

Height of the cylinder, $H = 10$ cm
Radius of the base of the cylinder, $R = 6$ cm
Height of the cone, $h = 10$ cm
Radius of the base of the cone, $r = 6$ cm

To determine the volume of the solid remaining after removing the cone from the cylinder, we subtract the volume of the cone from the volume of the cylinder.

The formula for the volume of the remaining solid is:

V = \text{Volume of cylinder} - \text{Volume of cone}

Substituting the formulas for the volumes, we have:

V = \pi R^2 H - \dfrac{1}{3} \pi r^2 h

Plugging in the given values:

V = \dfrac{22}{7} \times 6^2 \times 10 - \dfrac{1}{3} \times \dfrac{22}{7} \times 6^2 \times 10

Simplifying further:

V = \dfrac{22}{7} \times 6^2 \times 10 \times \left(1 - \dfrac{1}{3}\right)

V = \dfrac{22}{7} \times 36 \times 10 \times \dfrac{2}{3}

V = \dfrac{22 \times 12 \times 10 \times 2}{7}

V = \dfrac{5280}{7}

V = 754\dfrac{2}{7} \text{ cm}^3

∴ The volume of the remaining solid is $754\dfrac{2}{7}$ cm $^3$ .

Question 3

From a solid cylinder whose height is 16 cm and radius is 12 cm, a conical cavity of height 8 cm and of base radius 6 cm is hollowed out. Find the volume and total surface area of the remaining solid.

Answer:

Let’s consider the problem with the given dimensions. We have a cylinder with a height of 16 cm and a radius of 12 cm. Inside this cylinder, a conical cavity is carved out, where the cone has a height of 8 cm and a base radius of 6 cm.

(i) Calculating the Volume of the Remaining Solid

To find the volume of the remaining solid, we subtract the volume of the cone from the volume of the cylinder.

The formula for the volume of a cylinder is $\pi R^2 H$ , and for a cone, it is $\frac{1}{3} \pi r^2 h$ .

Thus, the volume $V$ of the remaining solid is:

V = \pi R^2 H - \dfrac{1}{3} \pi r^2 h

Substituting the given values:

V = \dfrac{22}{7} \times 12^2 \times 16 - \dfrac{1}{3} \times \dfrac{22}{7} \times 6^2 \times 8

Simplifying further:

V = \dfrac{22}{7} \times \left(12^2 \times 16 - \dfrac{1}{3} \times 6^2 \times 8\right)

= \dfrac{22}{7} \times \left(2304 - \dfrac{288}{3}\right)

= \dfrac{22}{7} \times \dfrac{6912 - 288}{3}

= \dfrac{22}{7} \times \dfrac{6624}{3}

= \dfrac{22}{7} \times 2208

= \dfrac{48576}{7}

= 6939.43 \text{ cm}^3.

Thus, the volume of the remaining solid is 6939.43 cm³.

(ii) Determining the Total Surface Area of the Remaining Solid

First, we need to find the slant height $l$ of the cone using the Pythagorean theorem:

l^2 = r^2 + h^2

Substituting the values:

l^2 = 6^2 + 8^2

l^2 = 36 + 64

l^2 = 100

l = 10 \text{ cm}.

Now, to find the total surface area $T$ of the remaining solid, we sum the curved surface area of the cylinder, the curved surface area of the cone, the base area of the cylinder, and the area of the circular ring on the upper side of the cylinder.

T = \text{Curved surface area of cylinder} + \text{Curved surface area of cone} + \text{Base area of cylinder} + \text{Area of circular ring}

This can be expressed as:

T = 2\pi RH + \pi rl + \pi R^2 + \pi(R^2 - r^2)

Substituting the given values:

T = \pi(2RH + rl + R^2 + R^2 - r^2)

= \pi(2RH + rl + 2R^2 - r^2)

= \dfrac{22}{7} \times (2 \times 12 \times 16 + 6 \times 10 + 2 \times 12^2 - 6^2)

= \dfrac{22}{7} \times (384 + 60 + 288 - 36)

= \dfrac{22}{7} \times 696

= \dfrac{15312}{7}

= 2187.43 \text{ cm}^2.

Therefore, the total surface area of the remaining solid is 2187.43 cm².

Question 4

A circus tent is cylindrical to a height of 4 m and conical above it. If its diameter is 105 m and its slant height is 80 m, calculate the total area of canvas required. Also, find the total cost of canvas used at ₹ 15 per meter if the width is 1.5 m.

Answer:

Let’s determine the dimensions of the circus tent. The radius of both the cylindrical and conical sections is $\dfrac{105}{2}$ m, since the diameter is 105 m. The slant height of the cone is given as 80 m.

The total curved surface area of the tent is the sum of the curved surface areas of the cylinder and the cone. This is calculated as:

\text{Total curved surface area} = 2\pi Rh + \pi rl

Substituting the known values:

\begin{aligned}= \left(2 \times \dfrac{22}{7} \times \dfrac{105}{2} \times 4\right) + \left(\dfrac{22}{7} \times \dfrac{105}{2} \times 80\right) \\= \dfrac{18480}{14} + \dfrac{184800}{14} \\= 1320 + 13200 \\= 14520 \text{ m}^2.\end{aligned}

Next, we calculate the length of the canvas used. Given the width of the canvas is 1.5 m, the length is:

\text{Length of canvas} = \dfrac{\text{Area of canvas}}{\text{Width of canvas}} = \dfrac{14520}{1.5} = 9680 \text{ m}.

The total cost of the canvas, at ₹ 15 per meter, is:

\text{Total cost} = 9680 \times 15 = ₹ 145200.

Hence, total area of canvas required = 14520 m^2 and cost = ₹ 145200.

Question 5

A circus tent is cylindrical to a height of 8 m surmounted by a conical part. If total height of the tent is 13 m and the diameter of its base is 24 m; calculate:

(i) total surface area of the tent

(ii) area of canvas, required to make this tent allowing 10% of the canvas used for folds and stitching.

Answer:

Let’s start by identifying the given dimensions:

The cylindrical section of the tent has a height of 8 m.
The conical section’s height is calculated as the total height minus the cylindrical height: 13 m – 8 m = 5 m.
The base diameter is 24 m, making the radius $r = \dfrac{24}{2} = 12$ m.

For the slant height $l$ of the cone, we use the Pythagorean theorem:

(l)^2 = r^2 + h^2

Substituting the values:

$l^2 = 12^2 + 5^2$
$l^2 = 144 + 25$
$l^2 = 169$
$l = \sqrt{169}$
$l = 13$ m.

(i) Calculating the total surface area of the tent:

The formula for the total surface area is:

\text{Total Surface Area} = 2\pi rH + \pi rl = \pi r(2H + l)

Substituting the values:

$= \dfrac{22}{7} \times 12 \times (2 \times 8 + 13)$
$= \dfrac{264}{7} (16 + 13)$
$= \dfrac{7656}{7} \text{ m}^2$
$= 1093.71 \text{ m}^2$

Thus, the total surface area of the tent is 1093.71 m².

(ii) Calculating the area of canvas required:

To account for folds and stitching, 10% extra canvas is needed.

Let the total area of canvas be $A$ . Then,

\text{Area for stitching} = \dfrac{10}{100} \times A

Thus,

A = \text{Total Surface Area} + \text{Area for stitching}

Substituting:

A = \dfrac{7656}{7} + \dfrac{1}{10} A

Rearranging gives:

$A - \dfrac{1}{10} A = \dfrac{7656}{7}$
$\dfrac{9}{10} A = \dfrac{7656}{7}$

Solving for $A$ :

$A = \dfrac{7656}{7} \times \dfrac{10}{9}$
$A = 1215.23 \text{ m}^2$

Therefore, the total area of canvas required is 1215.23 m².

Question 6

A cylindrical container with diameter of base 42 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 22 cm × 14 cm × 10.5 cm. Find the rise in level of the water when the solid is submerged.

Answer:

We start with the diameter of the cylindrical container’s base, which is 42 cm. Thus, the radius ( $r$ ) is calculated as:

r = \dfrac{42}{2} = 21 \text{ cm}

Let the increase in water level be denoted as $h$ cm.

The principle here is that the volume of the submerged solid is equal to the volume of the water displaced. Therefore:

22 \text{ cm} \times 14 \text{ cm} \times 10.5 \text{ cm} = \pi r^2 h

Substituting the known values, we have:

3234 = \dfrac{22}{7} \times 21^2 \times h

To find $h$ , rearrange the equation:

h = \dfrac{3234 \times 7}{22 \times 21^2}

Simplifying, we find:

h = \dfrac{22638}{9702} = 2\dfrac{1}{3} \text{ cm}

Thus, the water level rises by $2\dfrac{1}{3}$ cm in the container.

Question 7

Spherical marbles of diameter 1.4 cm are dropped into a beaker containing some water and are fully submerged. The diameter of the beaker is 7 cm. Find how many marbles have been dropped in it if the water rises by 5.6 cm?

Answer:

We have the following information:

Diameter of each spherical marble is 1.4 cm.
Therefore, the radius $r$ of a marble is $\frac{1.4}{2} = 0.7$ cm.
The diameter of the beaker is 7 cm.
Thus, the radius $R$ of the beaker is $\frac{7}{2} = 3.5$ cm.
The water level in the beaker rises by 5.6 cm.

Suppose $n$ marbles are dropped into the beaker. The increase in water volume is equal to the total volume of the marbles submerged.

∴ Volume of water displaced in the beaker = $n \times$ Volume of one marble.

This gives us the equation:

\pi R^2 h = n \times \dfrac{4}{3} \pi r^3

⇒ $R^2h = n \times \dfrac{4}{3}r^3$

Solving for $n$ , we have:

n = \dfrac{3R^2h}{4r^3}

Substituting the values:

n = \dfrac{3 \times (3.5)^2 \times 5.6}{4 \times (0.7)^3}

Calculating further:

n = \dfrac{3 \times 12.25 \times 5.6}{4 \times 0.343}

Finally, we find:

n = 150.

Hence, 150 marbles are dropped in the beaker.

Question 8

The given figure shows the cross-section of a water channel consisting of a rectangle and a semi-circle. Assuming that the channel is always full, find the volume of water discharged through it in one minute if water is flowing at the rate of 20 cm per second. Give your answer in cubic meters correct to one place of decimal.

Answer:

Consider the radius of the semi-circle as $r$ cm.

From the figure, we have:

∴ $2r = 21$ cm

⇒ $r = \dfrac{21}{2}$ cm.

The area of the cross-section of the water channel can be calculated by adding the area of the rectangle to the area of the semi-circle:

Area = $l \times b + \dfrac{1}{2}πr^2$

= $21 \times 7 + \dfrac{1}{2} \times \dfrac{22}{7} \times \dfrac{21}{2} \times \dfrac{21}{2}$

= $147 + \dfrac{693}{4}$

= $\dfrac{588 + 693}{4} = \dfrac{1281}{4}$

= $320.25$ cm $^2$ .

The length of the water column is determined by multiplying the water flow rate by the time:

Length = Water flow rate $\times$ Time

= $20$ cm/s $\times 60$ s

= $1200$ cm.

Now, the volume of water discharged is given by the product of the area of the cross-section and the length of the water column:

Volume = Area $\times$ Length

= $320.25 \times 1200$

= $384300$ cm $^3$

Convert the volume from cubic centimeters to cubic meters:

= ( 384300 \times \Big(\dfrac{1}{100}\Big)^3 ) m $^3$

= $0.3843$

≈ $0.4$ m $^3$ .

Thus, the volume of water discharged in one minute is $0.4$ m $^3$ .

Question 9

An open cylindrical vessel of internal diameter 7 cm and height 8 cm stands on a horizontal table. Inside this is placed a solid metallic right circular cone, the diameter of whose base is $3\dfrac{1}{2}$ cm and height 8 cm. Find the volume of water required to fill the vessel.

If this cone is replaced by another cone, whose height is $1\dfrac{3}{4}$ cm and the radius of whose base is 2 cm, find the drop in the water level.

Answer:

To solve this problem, we start with the cylindrical vessel.

The internal diameter of the cylinder is 7 cm, which means the radius (R) is $\dfrac{7}{2} = 3.5$ cm. The height (H) of the cylinder is given as 8 cm.

The volume of the cylindrical vessel is calculated using the formula:

Volume of cylinder = $\pi R^2 H = \dfrac{22}{7} \times (3.5)^2 \times 8$ .

Breaking it down:

$= 22 \times 0.5 \times 3.5 \times 8$ ,

which results in 308 cm³.

Next, we consider the solid cone inside the cylinder. The diameter of the cone’s base is $3\dfrac{1}{2} = \dfrac{7}{2}$ cm, giving us a radius (r) of $\dfrac{7}{4}$ cm. The height (h) of the cone is also 8 cm.

The volume of this cone is given by:

Volume of cone = $\dfrac{1}{3} \pi r^2 h = \dfrac{1}{3} \times \dfrac{22}{7} \times \Big(\dfrac{7}{4}\Big)^2 \times 8$ .

Simplifying:

$= \dfrac{1}{3} \times 22 \times \dfrac{1}{4} \times \dfrac{7}{4} \times 8$ ,

which gives $\dfrac{308}{12}$ cm³.

To find the volume of water needed to fill the vessel, we subtract the volume of the cone from the volume of the cylinder:

Volume of water = $308 - \dfrac{308}{12}$

= \dfrac{12 \times 308 - 308}{12}

= \dfrac{3388}{12}

$= 282.33$ cm³.

Now, replacing the original cone with a new one, we have a radius ( $r_1$ ) of 2 cm and a height ( $h_1$ ) of $1\dfrac{3}{4} = \dfrac{7}{4}$ cm.

The volume of the new cone is:

Volume of new cone = $\dfrac{1}{3} \pi r_1^2 h_1$

= \dfrac{1}{3} \times \dfrac{22}{7} \times 2^2 \times \dfrac{7}{4}

= \dfrac{1}{3} \times 22 \times 4 \times \dfrac{1}{4}

$= \dfrac{22}{3}$ cm³.

The change in water volume, which is the difference between the volumes of the original and new cones, is:

Volume change = $\dfrac{308}{12} - \dfrac{22}{3}$

= \dfrac{77}{3} - \dfrac{22}{3}

$= \dfrac{55}{3}$ cm³.

Let the drop in water level be $h_2$ cm. The drop in volume is $\dfrac{55}{3}$ cm³, so:

\pi R^2 h_2 = \dfrac{55}{3}

\Rightarrow \dfrac{22}{7} \times (3.5)^2 \times h_2 = \dfrac{55}{3}

\Rightarrow 22 \times 0.5 \times 3.5 \times h_2 = \dfrac{55}{3}

\Rightarrow 38.5 \times h_2 = \dfrac{55}{3}

\Rightarrow h_2 = \dfrac{55}{115.5}

$\Rightarrow h_2 = \dfrac{10}{21}$ cm.

Therefore, the volume of water required to fill the vessel is 282.33 cm³, and the drop in the water level is $\dfrac{10}{21}$ cm.

Question 10

A cylindrical can, whose base is horizontal and of radius 3.5 cm, contains sufficient water so that when a sphere is placed in the can, the water just covers the sphere. Given that the sphere just fits into the can, calculate :

(i) the total surface area of the can in contact with water when the sphere is in it;

(ii) the depth of water in the can before the sphere was put into the can.

Answer:

(i) We know that the radius of the base of the cylindrical can is 3.5 cm. Since the sphere fits perfectly inside the can, the height of the cylinder, which also equals the diameter of the sphere, is twice the radius of the base.

∴ Height ( (H) = 2R = 7 \text{ cm} ).

The total surface area of the can that comes into contact with the water consists of the lateral surface area plus the area of the base. The formula for this surface area is:

2\pi RH + \pi R^2 = \pi R(2H + R)

Substituting the known values:

\dfrac{22}{7} \times 3.5 \times (2 \times 7 + 3.5)

Simplifying, we get:

22 \times 0.5 \times (14 + 3.5) = 192.5 \text{ cm}^2.

Thus, the total surface area of the can in contact with water when the sphere is in it is 192.5 cm².

(ii) Let $h$ cm be the depth of the water in the can before the sphere was added. The volume of the water initially is equal to the volume of the cylinder minus the volume of the sphere:

\pi R^2h = \pi R^2H - \dfrac{4}{3}\pi R^3

This simplifies to:

\pi R^2h = \pi R^2\left(H - \dfrac{4}{3}R\right)

Thus, the depth $h$ is:

h = H - \dfrac{4}{3}R

Substituting the values:

h = 7 - \dfrac{4}{3} \times 3.5

Calculating further:

h = 7 - \dfrac{14}{3}

h = \dfrac{21 - 14}{3}

h = \dfrac{7}{3} = 2\dfrac{1}{3} \text{ cm}.

Therefore, the depth of the water in the can before the sphere was placed in it was $2\dfrac{1}{3} \text{ cm}$ .

Question 11

A hollow cylinder has solid hemisphere inward at one end and on the other end it is closed with a flat circular plate. The height of water is 10 cm when flat circular surface is downward. Find the level of water, when it is inverted upside down, common diameter is 7 cm and height of cylinder is 20 cm.

Answer:

We have the following measurements for the hollow cylinder and the hemisphere:

For the cylinder:
– Height (H) = 20 cm
– Radius (R) = 3.5 cm

For the hemisphere:
– Radius (r) = Radius of cylinder (R) = 3.5 cm

When the flat circular surface is at the bottom, the water reaches a height of 10 cm.

Now, consider what happens when the cylinder is inverted, making the circular surface the base. Let the new water height be denoted as $h_1$ .

The volume of water when the flat surface is downward is equal to the volume of water when the circular surface is downward plus the volume of the hemisphere.

\Rightarrow πR^2h_1 = πR^2h + \dfrac{2}{3}πr^3

Since $r = R$ , this becomes:

\Rightarrow πR^2h_1 = πR^2h + \dfrac{2}{3}πR^3

Simplifying further:

\Rightarrow πR^2h_1 = πR^2\left(h + \dfrac{2}{3}R\right)

Thus, we find $h_1$ :

\Rightarrow h_1 = h + \dfrac{2}{3}R

Substituting the values:

\Rightarrow h_1 = 10 + \dfrac{2}{3} \times 3.5

Calculating:

\Rightarrow h_1 = 10 + \dfrac{7}{3}

\Rightarrow h_1 = \dfrac{30 + 7}{3}

\Rightarrow h_1 = \dfrac{37}{3} = 12\dfrac{1}{3} = 12.33\text{ cm}

Therefore, the level of water when the cylinder is flipped upside down is 12.33 cm.

Test Yourself

Question 1(a)

The given figure shows a solid cylinder and a solid cone on it according to the given measurement, the total surface area of the solid in terms of π and r is :

(a) 4πr^2
(b) 5πr^2
(c) 6πr^2
(d) 7πr^2

Answer: (b) 5πr^2

Consider the total surface area of the solid, which consists of a cylinder and a cone. The total surface area is calculated by summing the curved surface area of the cylinder, the curved surface area of the cone, and the area of the circular base.

The formula for the curved surface area of the cylinder is given by $2\pi rh$ . For the cone, the curved surface area is $\pi rl$ . Additionally, the base of the cylinder contributes an area of $\pi r^2$ .

Substitute the given dimensions: $h = r$ and $l = 2r$ . This gives us:

2\pi r \times r + \pi r \times 2r + \pi r^2

Simplifying, we obtain:

2\pi r^2 + 2\pi r^2 + \pi r^2 = 5\pi r^2

Notice that the total surface area simplifies to $5\pi r^2$ . Hence, Option 2 is the correct option.

Question 1(b)

The radii of two solid spheres are 10 cm and 20 cm. The ratio between their volumes is :

3 : 8
1 : 4
1 : 8
8 : 3

Answer:

The radius of the first sphere is given as 10 cm, while the radius of the second sphere is 20 cm.

To determine the ratio of their volumes, use the formula for the volume of a sphere, which is $\dfrac{4}{3} \pi r^3$ .

The ratio of the volumes of the two spheres is:

\Rightarrow \dfrac{\text{Vol. of 1st sphere}}{\text{Vol. of 2nd sphere}} = \dfrac{\dfrac{4}{3}\pi r^3}{\dfrac{4}{3}\pi R^3}

Notice that $\dfrac{4}{3}\pi$ cancels out from both the numerator and the denominator, simplifying to:

= \dfrac{r^3}{R^3}

Substituting the given radii, we have:

= \dfrac{10^3}{20^3}

Calculating further, this becomes:

= \dfrac{1000}{8000}

Simplifying this fraction gives:

= \dfrac{1}{8}

Thus, the ratio is $1 : 8$ .

Hence, Option 3 is the correct option.

Question 1(c)

A cone and a sphere have equal volumes. If height of the cone = radius of the sphere = 10 cm, the radius of the cone is :

(a) 40 cm
(b) 10 cm
(c) 30 cm
(d) 20 cm

Answer: (d) 20 cm

We have a situation where the volumes of a cone and a sphere are identical, with the height of the cone and the radius of the sphere both being 10 cm.

Let’s denote the radius of the cone as $R$ cm.

The formula for the volume of a cone is $\dfrac{1}{3} \pi R^2 H$ , and for a sphere, it is $\dfrac{4}{3} \pi r^3$ .

Since their volumes are equal:

\therefore \dfrac{1}{3} \pi R^2 H = \dfrac{4}{3} \pi r^3

Cancelling $\pi$ and multiplying through by 3 gives:

\Rightarrow R^2 H = 4r^3

Substitute $H = 10$ and $r = 10$ :

\Rightarrow R^2 \times 10 = 4 \times 10^3

Simplifying, we find:

\Rightarrow R^2 = \dfrac{4 \times 10^3}{10}

\Rightarrow R^2 = 4 \times 10^2

\Rightarrow R^2 = 400

Taking the square root, we get:

\Rightarrow R = \sqrt{400} = 20 \text{ cm}.

Hence, option 4 is the correct option.

Question 1(e)

By applying force on an uniform metallic wire, the wire is extended four times along its length with same width all around. Then :

(i) The volume of metal in both the cases is same.

(ii) The surface area of solid wires in both the cases is same.

Which of the above statements is/are true?

(a) only (i)
(b) only (ii)
(c) both (i) and (ii)
(d) neither (i) nor (ii)

Answer: (a) only (i)

When you change the shape of a solid by applying force, such as stretching a wire, the volume of the material remains unchanged.

Consider the original wire with length $l$ and radius $r$ . After being stretched, the wire’s new length becomes $4l$ , and let’s denote the new radius as $R$ .

Since the volume before and after stretching is the same, we have:

∴ $\pi r^2 l = \pi R^2.4l$

This simplifies to:

⇒ $r^2 = 4R^2$

⇒ $r = \sqrt{4R^2}$

⇒ $r = 2R$

⇒ $R = \dfrac{r}{2}$

Now, let’s compare the surface areas:

Initially, the surface area is:

⇒ $2\pi rl$

After stretching, the surface area becomes:

⇒ $2\pi R \times 4l$

⇒ $2\pi \times \dfrac{r}{2} \times 4l$

⇒ $4\pi rl$

Thus, the surface areas before and after are not the same.

Notice that the volume remains constant, but the surface area changes. Hence, Option 1 is the correct option.

Question 1(d)

A solid metal cuboid with dimension 10 cm, 11 cm and 8 cm is melted and identical solid spheres, each of radius 1 cm, are formed. The number of sphere formed is :

(a) 105
(b) 210
(c) 315
(d) 420

Answer: (b) 210

We start with a solid metal cuboid that has dimensions 10 cm, 11 cm, and 8 cm. This cuboid is melted to create identical solid spheres, each having a radius of 1 cm. Let the number of spheres formed be denoted by $n$ .

∴ The volume of the metal cuboid equals the total volume of the spheres formed.

⇒ $l \times b \times h = n \times \text{Volume of one sphere}$

⇒ $10 \times 11 \times 8 = n \times \dfrac{4}{3} \pi r^3$

Substitute the value of $\pi$ as $\dfrac{22}{7}$ and $r = 1$ :

⇒ $10 \times 11 \times 8 = n \times \dfrac{4}{3} \times \dfrac{22}{7} \times 1^3$

⇒ $880 = n \times \dfrac{88}{21}$

To find $n$ , rearrange the equation:

⇒ $n = \dfrac{880 \times 21}{88}$

Simplifying further:

⇒ $n = 10 \times 21 = 210$ .

Hence, option 2 is the correct option.

Question 1(f)

The given figure shows a toy made of very thin metal sheet. The toy has the hemisphere, a cylinder and a cone all of the same radius.

Assertion(A): The minimum area of metal sheet required is:

Curved surface area of conical part + curved surface area of cylinder part – curved surface area of hemispherical part

Reason(R): Metal sheet required = curved surface area of (conical part + cylindrical part + hemispherical part)

(a) A is true, R is false.
(b) A is false, R is true.
(c) Both A and R are true and R is correct reason for A.
(d) Both A and R are true and R is incorrect reason for A.

Answer: (b) A is false, R is true.

The toy consists of a hemisphere, a cylinder, and a cone, all sharing the same radius.

To determine the minimum area of metal sheet needed, we calculate the curved surface areas of all these parts together:

Curved surface area of the conical part
Curved surface area of the cylindrical part
Curved surface area of the hemispherical part

This gives us the total curved surface area required:

Curved surface area of (conical part + cylindrical part + hemispherical part).

∴ The assertion is incorrect, whereas the reason is accurate.

Hence, option 2 is the correct option.

Question 1(g)

The base radius of two right circular cone of the same height are in ratio 3 : 5.

Assertion(A): The ratio between their volume is 9 : 25.

Reason(R): As

\begin{aligned}\dfrac{r_1}{r_2} = \dfrac{3}{5} \\ \dfrac{πr_1^2h}{πr_2^2h} = \dfrac{r_1^2}{r_2^2} = \Big(\dfrac{3}{5}\Big)^2 \\\end{aligned}

(a) A is true, R is false.
(b) A is false, R is true.
(c) Both A and R are true and R is correct reason for A.
(d) Both A and R are true and R is incorrect reason for A.

Answer: (c) Both A and R are true and R is correct reason for A.

Consider the situation where the base radii of two right circular cones with identical heights are in the ratio 3:5.

The formula for the volume of a cone is given by $\dfrac{1}{3}πr^2h$ .

To find the ratio of their volumes, calculate:

\dfrac{\text{Volume of 1st cone}}{\text{Volume of 2nd cone}} = \dfrac{\dfrac{1}{3}πr_1^2h}{\dfrac{1}{3}πr_2^2h}

This simplifies to:

= \dfrac{πr_1^2h}{πr_2^2h}

Notice that the $π$ and $h$ terms cancel out, leaving:

= \dfrac{r_1^2}{r_2^2}

Given $\dfrac{r_1}{r_2} = \dfrac{3}{5}$ , substitute to find:

= \dfrac{3^2}{5^2}

= \Big(\dfrac{3}{5}\Big)^2

= \dfrac{9}{25}.

∴ Both the assertion and the reason are true, and the reason correctly explains the assertion.

Hence, option 3 is the correct option.

Question 1(h)

A solid cone of height 3 cm and radius 3 cm is recast into solid cylinder each of height 1 cm and radius 1 cm.

Assertion(A): Number of cylinders formed = $\dfrac{1}{3}$ x 3 x 3 x 3

Reason(R): Number of cylinders formed = $\dfrac{\dfrac{1}{3}π(3)^2 \times 3}{π(1)^2 \times 1}$

(a) A is true, R is false.
(b) A is false, R is true.
(c) Both A and R are true and R is correct reason for A.
(d) Both A and R are true and R is incorrect reason for A.

Answer: (c) Both A and R are true and R is correct reason for A.

Let’s consider the dimensions of the cone: the radius is 3 cm and the height is 3 cm. For the cylinder, the radius is 1 cm and the height is 1 cm.

The formula for the volume of a cone is $\dfrac{1}{3}πR^2H$ , and for a cylinder, it is $πr^2h$ .

When the cone is melted and reshaped into cylinders, the total volume remains the same. Let the number of cylinders formed be $n$ .

∴ Volume of the cone = $n$ × Volume of one cylinder

\Rightarrow \dfrac{1}{3}πR^2H = n \times πr^2h

Substituting the given values:

\Rightarrow \dfrac{1}{3}π(3)^2 \times 3 = n \times π(1)^2 \times 1

\Rightarrow \dfrac{1}{3} (3)^2 \times 3 = n \times 1 \times 1

\Rightarrow n = \dfrac{1}{3} \times 3 \times 3 \times 3

$\Rightarrow n = 9$ .

Notice that the assertion and reason both correctly describe the process and calculation involved. ∴ Both A and R are true, and R provides the correct reasoning for A.

Hence, option 3 is the correct option.

Question 1(i)

A solid wooden cylinder is of height h cm and radius r cm. A conical cavity of same height and the same radius is drill out of the solid cylinder.

Statement (1): The volume of the remaining wood = volume of the solid cylinder – volume of the cone drilled.

Statement (2): The volume of the remaining wood = πr^2h – $\dfrac{1}{3}$ πr^2h = $\dfrac{2}{3}$ πr^2h

(a) Both the statement are true.
(b) Both the statement are false.
(c) Statement 1 is true, and statement 2 is false.
(d) Statement 1 is false, and statement 2 is true.

Answer: (a) Both the statement are true.

To determine the volume of the wood left, subtract the volume of the cone from the volume of the cylinder.

Given that both the cylinder and the cone have the same radius $r$ and height $h$ , we find:

Volume of the cylinder = $\pi r^2 h$
Volume of the cone = $\dfrac{1}{3} \pi r^2 h$

Thus, the volume of the remaining wood is:

Volume of the solid cylinder – Volume of the cone drilled = $\pi r^2 h - \dfrac{1}{3} \pi r^2 h$

= $\Big(1 - \dfrac{1}{3}\Big) \pi r^2 h$

= $\Big(\dfrac{3 - 1}{3}\Big) \pi r^2 h$

= $\dfrac{2}{3} \pi r^2 h$ .

Notice that both statements correctly describe the situation. Therefore, both statements are true.

Hence, option 1 is the correct option.

Question 1(j)

A solid sphere of radius 6 cm is melted and recast into solid spheres of diameter 2 cm each.

Statement (1): The number of solid sphere formed is 6 x 6 x 6 = 216.

Statement (2): If the smaller spheres formed are identical, the number of solid sphere formed = $\dfrac{\text{Volume of sphere melted}}{\text{Volume of each of the sphere formed}}$

(a) Both the statement are true.
(b) Both the statement are false.
(c) Statement 1 is true, and statement 2 is false.
(d) Statement 1 is false, and statement 2 is true.

Answer: (a) Both the statement are true.

We’re given a solid sphere with a radius of 6 cm. When this sphere is melted down, it’s reshaped into smaller spheres, each with a diameter of 2 cm.

First, let’s determine the radius of these smaller spheres. Since their diameter is 2 cm, the radius $r$ is $\dfrac{2}{2}$ cm, which equals 1 cm.

The formula for the volume of a sphere is $\dfrac{4}{3} \pi r^3$ .

Now, let’s consider the problem:

Statement 1: We need to find the number of smaller spheres formed. The volume of the original sphere is distributed into these smaller spheres. If we let the number of smaller spheres be $n$ , then:

n \times \text{Volume of each new sphere} = \text{Volume of the original sphere}

This implies:

n = \dfrac{\text{Volume of sphere melted}}{\text{Volume of each new sphere formed}}

Therefore, Statement 2 correctly describes how to find the number of smaller spheres.

Statement 2: Let’s compute the number of smaller spheres. Using the formula for the volume of a sphere, we have:

n = \dfrac{\dfrac{4}{3}\pi R^3}{\dfrac{4}{3}\pi r^3}

This simplifies to:

n = \dfrac{R^3}{r^3}

Substituting the given radii:

n = \dfrac{6^3}{1^3}

n = \dfrac{216}{1}

n = 216

Thus, both statements are indeed true.

Hence, option 1 is the correct option.

Question 1(k)

Water in a canal 6 m wide and 2 m deep, is flowing with the speed of 18 km/h.

Statement (1): The volume of water that flows through the canal in 20 minutes = 6 x 2 x $\Big(18 \times \dfrac{5}{18}\Big)$ x 20 x 60 m^3.

Statement (2): The volume of water that flows through the canal = 6 x 2 x 18 x 20 m^3.

(a) Both the statement are true.
(b) Both the statement are false.
(c) Statement 1 is true, and statement 2 is false.
(d) Statement 1 is false, and statement 2 is true.

Answer: (c) Statement 1 is true, and statement 2 is false.

The canal has a width of 6 m and a depth of 2 m, with water flowing at a speed of 18 km/h.

To find the volume of water flowing through the canal, we use the formula:

Volume = Area of cross-section × speed × time

The area of the cross-section is calculated as width × depth, which gives us (6 × 2) m².

The speed needs to be converted from km/h to m/s. Thus, 18 km/h becomes 18 × $\dfrac{1000}{3600}$ = $\Big(18 \times \dfrac{5}{18}\Big)$ m/s.

The time of flow is 20 minutes, which we convert to seconds: 20 × 60 seconds.

Therefore, the volume is calculated as:

Volume = 6 × 2 × 18 × $\dfrac{5}{18}$ × 20 × 60 m³

This confirms that Statement 1 is accurate. However, Statement 2 is incorrect as it does not account for the time in seconds or the conversion of speed.

Hence, Statement 1 is true, and Statement 2 is false. Therefore, option 3 is the correct option.

Question 2

What is the least number of solid metallic spheres, each of 6 cm diameter, that should be melted and recast to form a solid metal cone whose height is 45 cm and diameter 12 cm ?

Answer:

We start with the given dimensions:

Diameter of each metallic sphere is 6 cm, so the radius (r) is $\dfrac{6}{2} = 3$ cm.
The cone has a height (H) of 45 cm and a diameter of 12 cm, giving it a radius (R) of $\dfrac{12}{2} = 6$ cm.

We need to find the number of spheres, denoted as $n$ , that must be melted to form the cone. The key is to equate the total volume of the spheres to the volume of the cone:

n \times \text{Volume of each sphere} = \text{Volume of the cone}

The formula for the volume of a sphere is $\dfrac{4}{3} \pi r^3$ , and for a cone, it is $\dfrac{1}{3} \pi R^2 H$ . Substituting these into our equation gives:

\Rightarrow n \times \dfrac{4}{3} \pi r^3 = \dfrac{1}{3} \pi R^2 H

To solve for $n$ , rearrange the equation:

\Rightarrow n = \dfrac{\dfrac{1}{3} \pi R^2 H}{\dfrac{4}{3} \pi r^3}

Simplifying further:

\Rightarrow n = \dfrac{\pi R^2 H \times 3}{3 \times 4 \times \pi r^3}

Cancel out $\pi$ and simplify the expression:

\Rightarrow n = \dfrac{R^2 H}{4 r^3}

Substitute the values of $R$ , $H$ , and $r$ :

\Rightarrow n = \dfrac{6^2 \times 45}{4 \times 3^3}

Calculate step-by-step:

\Rightarrow n = \dfrac{36 \times 45}{4 \times 27}

\Rightarrow n = \dfrac{1620}{108}

\Rightarrow n = 15

Thus, 15 solid metallic spheres are required to be melted to form the cone.

Question 3

A largest sphere is to be carved out of a right circular cylinder of radius 7 cm and height 14 cm. Find the volume of the sphere.
(Answer correct to the nearest integer)

Answer:

The radius of the largest sphere that can be carved from a right circular cylinder is the same as the radius of the cylinder, which is 7 cm.

To find the volume of the sphere, we use the formula:

Volume of sphere = $\dfrac{4}{3}πr^3$

Substituting the values, we have:

\begin{aligned}= \dfrac{4}{3} \times \dfrac{22}{7} \times 7^3 \\= \dfrac{4}{3} \times 22 \times 7^2 \\= 1437 \text{ m}^3.\end{aligned}

Thus, the volume of the sphere is 1437 m³.

Question 4

A right circular cylinder having diameter 12 cm and height 15 cm is full of ice-cream. The ice-cream is to be filled in identical cones of height 12 cm and diameter 6 cm having a hemi-spherical shape on the top. Find the number of cones required.

Answer:

We have a right circular cylinder with a diameter of 12 cm. Therefore, the radius $r$ is $\dfrac{12}{2} = 6$ cm. The height $h$ of the cylinder is 15 cm.

For the cone, the diameter is 6 cm, giving us a radius $R$ of $\dfrac{6}{2} = 3$ cm. The height $H$ of the cone is 12 cm. The radius of the hemisphere on top matches the cone’s radius, so $R = 3$ cm.

Assume we need $n$ cones.

The volume of the cylinder is calculated as:
$\text{Volume of the cylinder} = \pi r^2 h = \pi \times (6)^2 \times 15 = 540\pi \text{ cm}^3.$

Each ice-cream cone, with its hemispherical top, has a total volume given by:
$\text{Volume of an ice-cream cone} = \text{Volume of cone} + \text{Volume of hemisphere}$
$\begin{aligned}= \dfrac{1}{3}\pi R^2 H + \dfrac{2}{3} \times \pi \times R^3 \\= \dfrac{1}{3} \times \pi \times 3^2 \times 12 + \dfrac{2}{3} \times \pi \times 3^3 \\= 36 \pi + 18 \pi \\= 54 \pi \text{ cm}^3.\end{aligned}$

To find the number of cones $n$ required, set the volume of the cylinder equal to $n$ times the volume of one cone:
$n = \dfrac{\text{Vol. of cylinder}}{\text{Vol. of ice-cream cone}} = \dfrac{540\pi}{54\pi} = 10.$

Hence, the number of cones required = 10.

Question 5

A solid is in the form of a cone standing on a hemisphere with both their radii being equal to 8 cm and the height of cone is equal to its radius. Find in terms of π, the volume of the solid.

Answer:

We are given that the radius of the cone and the hemisphere, denoted as $r$ , is 8 cm. The height of the cone, $h$ , is also 8 cm, which matches the radius.

The total volume of the solid is the sum of the volumes of the cone and the hemisphere.

The volume of the cone is calculated using the formula $\dfrac{1}{3} \pi r^2 h$ . For our cone:

$= \dfrac{1}{3} \times \pi \times 8^2 \times 8$ .

The volume of the hemisphere is given by $\dfrac{2}{3} \pi r^3$ . Thus, for our hemisphere:

$= \dfrac{2}{3} \times \pi \times 8^3$ .

Adding these two volumes together, we have:

$= \dfrac{1}{3}.\pi.8^3 + \dfrac{2}{3}.\pi.8^3$ .

Combine the terms:

$= \Big(\dfrac{1 + 2}{3}\Big) \times \pi \times 512$ .

This simplifies to:

$= 512\pi \text{ cm}^3$ .

Therefore, the volume of the solid is $512\pi$ cm $^3$ .

Question 6

The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm. Find the length of wire.

Answer:

We start with a sphere whose diameter measures 6 cm. This means the radius (R) of the sphere is $\frac{6}{2} = 3$ cm.

Next, consider the wire, which is cylindrical and has a diameter of 0.2 cm. Therefore, the radius (r) of the wire is $\frac{0.2}{2} = 0.1$ cm.

Let’s denote the length of the wire by $h$ . The sphere is melted and reshaped into this wire, which implies:

∴ Volume of the sphere = Volume of the wire

This gives us the equation:

\Rightarrow \dfrac{4}{3}πR^3 = πr^2h

By simplifying, we find:

\Rightarrow h = \dfrac{\dfrac{4}{3}πR^3}{πr^2}

Cancelling $π$ from both sides, we have:

\Rightarrow h = \dfrac{4R^3}{3r^2}

Substituting the values of $R$ and $r$ :

\Rightarrow h = \dfrac{4 \times 3^3}{3 \times (0.1)^2}

Calculating further:

\Rightarrow h = \dfrac{4 \times 3^2}{0.01}

Simplifying gives:

\Rightarrow h = 4 \times 3^2 \times 100

This results in:

\Rightarrow h = 3600 \text{ cm}

Converting to meters:

\Rightarrow h = 3600 \times \dfrac{1}{100} \text{ m}

Finally, we find:

$\Rightarrow h = 36 \text{ m}$ .

Hence, length of the wire = 36 m.

Question 7

Determine the ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube.

Answer:

Consider a cube with an edge length of $a$ units.

The volume of the cube is calculated as $a \times a \times a = a^3$ .

A sphere that fits perfectly inside this cube will have a diameter equal to the side length of the cube, which is $a$ units. Therefore, the radius $r$ of the sphere is $\dfrac{a}{2}$ units.

Now, to find the ratio of the volume of the cube to the volume of the sphere, we have:

\begin{aligned}\dfrac{\text{Volume of cube}}{\text{Volume of sphere}} = \dfrac{a^3}{\dfrac{4}{3}πr^3} \\= \dfrac{a^3}{\dfrac{4}{3}π \times \Big(\dfrac{a}{2}\Big)^3} \\= \dfrac{a^3 \times 3 \times 2^3}{4π \times a^3} \\= \dfrac{3 \times 8}{4 \times \dfrac{22}{7}} \\= \dfrac{3 \times 8 \times 7}{4 \times 22} \\= \dfrac{21}{11} \\= 21 : 11.\end{aligned}

Hence, the ratio of the volume of the cube to the volume of the sphere is 21 : 11.

Question 8

An iron pole consisting of a cylindrical portion 110 cm high and of base diameter 12 cm is surmounted by a cone 9 cm high. Find the mass of the pole, given that 1 cm^3 of iron has 8 gm of mass (approx). (Take π = $\dfrac{355}{113}$ )

Answer:

We are given the dimensions of a cylindrical section and a conical top of an iron pole. Let’s calculate the mass of this pole.

The diameter of the cylindrical section is 12 cm, hence the radius (r) is:

r = \dfrac{12}{2} = 6 \text{ cm}

The height of the cylindrical section (H) is 110 cm, and the height of the conical section (h) is 9 cm.

Since the cone sits atop the cylinder, both share the same radius of 6 cm.

The total volume of the pole is the sum of the volumes of the cylinder and the cone:

\text{Volume of the iron pole} = \text{Volume of the cylinder} + \text{Volume of the cone}

This can be expressed as:

\begin{aligned}= \pi r^2 H + \dfrac{1}{3} \pi r^2 h \\= \pi r^2 \Big(H + \dfrac{1}{3} h\Big) \\= \dfrac{355}{113} \times 6^2 \times \Big(110 + \dfrac{1}{3} \times 9\Big) \\= \dfrac{355}{113} \times 36 \times 113 \\= 12780 \text{ cm}^3.\end{aligned}

Knowing that each cubic centimeter of iron weighs 8 grams, the total mass of the iron pole is:

\text{Total weight} = 12780 \times 8 \text{ gm} = 102240 \text{ gm}

Converting grams to kilograms:

\dfrac{102240}{1000} \text{ kg} = 102.24 \text{ kg}

Therefore, the mass of the pole is 102.24 kg.

Question 9

The cross-section of a tunnel is a square of side 7 m surmounted by a semicircle as shown in the adjoining figure. The tunnel is 80 m long. Calculate:

(i) its volume,

(ii) the surface area of the tunnel (excluding the floor) and

(iii) its floor area.

Answer:

We are given that the side of the square in the tunnel’s cross-section is 7 m. The radius of the semicircle is half of this side, so:

⇒ 2r = 7

⇒ r = $\dfrac{7}{2}$ m.

The tunnel has a length of 80 m.

The cross-sectional area of the tunnel is the sum of the square’s area and the semicircle’s area:

= $a^2$ + $\dfrac{1}{2}πr^2$

= $7 \times 7$ + $\dfrac{1}{2} \times \dfrac{22}{7} \times \dfrac{7}{2} \times \dfrac{7}{2}$

= 49 + $\dfrac{77}{4}$

= $\dfrac{196 + 77}{4}$ = $\dfrac{273}{4}$ m $^2$ .

(i) To find the volume of the tunnel, we multiply the cross-sectional area by the tunnel’s length:

Volume = $\dfrac{273}{4}$ \times 80

= 5460 m $^3$ .

Thus, the volume of the tunnel is 5460 m $^3$ .

(ii) The surface area of the tunnel, excluding the floor, includes the lateral surfaces of the semicircle and the square:

= πrh + ah + ah

= $\dfrac{22}{7} \times \dfrac{7}{2} \times 80$ + $7 \times 80$ + $7 \times 80$

= 880 + 560 + 560

= 2000 m $^2$ .

Therefore, the surface area of the tunnel is 2000 m $^2$ .

(iii) The floor area of the tunnel is calculated by multiplying the breadth by the length:

Floor area = b \times h = 80 \times 7 = 560 m $^2$ .

Hence, the floor area is 560 m $^2$ .

Question 10

A cylindrical water tank of diameter 2.8 m and height 4.2 m is being fed by a pipe of diameter 7 cm through which water flows at the rate of 4 m s^-1. Calculate, in minutes, the time it takes to fill the tank.

Answer:

Let’s start by noting the measurements of the cylindrical tank. It has a diameter of 2.8 m, which means its radius (R) is $\frac{2.8}{2}$ = 1.4 m. The height (H) of the tank is 4.2 m.

The volume of the tank can be calculated using the formula for the volume of a cylinder, $\pi R^2 H$ :

= \dfrac{22}{7} \times 1.4 \times 1.4 \times 4.2

= 25.872 \text{ m}^3

Converting this to cubic centimeters, we have:

= 25.872 \times (100)^3 \text{ cm}^3

Now, consider the pipe. It has a diameter of 7 cm, giving it a radius (r) of $\dfrac{7}{2} = 3.5$ cm. The cross-sectional area of the pipe is $\pi r^2$ :

= \dfrac{22}{7} \times (3.5)^2

= 22 \times 0.5 \times 3.5

= 38.5 \text{ cm}^2

Water flows through the pipe at a speed of 4 m/s, so the volume of water discharged per second is:

= 38.5 \text{ cm}^2 \times 4 \text{ m/s}

= 38.5 \times 400 \text{ cm/s}

Let the time taken to fill the tank be $n$ seconds. Thus, we have:

n \times 38.5 \times 400 = 25.872 \times (100)^3

Solving for $n$ , we get:

n = \dfrac{25.872 \times 100^3}{38.5 \times 400}

n = \dfrac{25872000}{15400}

n = 1680 \text{ seconds}

Converting this time into minutes:

n = 1680 \times \dfrac{1}{60} \text{ minutes}

n = 28 \text{ minutes}

Hence, it takes 28 minutes to fill the tank.

Question 11

Water flows, at 9 km per hour, through a cylindrical pipe of cross-sectional area 25 cm^2. If this water is collected into a rectangular cistern of dimensions 7.5 m by 5 m by 4 m; calculate the rise in level in the cistern in 1 hour 15 minutes.

Answer:

First, let’s convert the flow rate of water. We have the flow rate as $9 \text{ km/hr}$ , which can be expressed in centimeters per hour as $9 \times 10^5$ cm/hr, since $1 \text{ km} = 10^5 \text{ cm}$ .

Next, consider the time duration given: $1$ hour and $15$ minutes. We convert this into hours:

1 + \frac{15}{60} = 1 + \frac{1}{4} = \frac{5}{4} \text{ hours}.

Now, calculate the volume of water flowing through the pipe in $\frac{5}{4}$ hours. This is given by:

\text{Volume} = \text{Area of cross-section} \times \text{Rate of flow} \times \text{Time}

Substitute the values:

= 25 \text{ cm}^2 \times 9 \times 10^5 \text{ cm/hr} \times \frac{5}{4} \text{ hr}

= \frac{1125 \times 10^5}{4} \text{ cm}^3.

Let the rise in water level in the cistern be $h$ cm. The dimensions of the cistern are given as $7.5 \text{ m}$ by $5 \text{ m}$ , which convert to centimeters as follows:

Length = $7.5 \text{ m} = 750 \text{ cm}$
Breadth = $5 \text{ m} = 500 \text{ cm}$

The volume of water increase in the cistern is:

750 \times 500 \times h \text{ cm}^3.

Since the volume of water flowing into the cistern equals the volume of water flowing through the pipe:

750 \times 500 \times h = \frac{1125 \times 10^5}{4}

Solve for $h$ :

h = \frac{1125 \times 10^5}{4 \times 750 \times 500}

h = 0.00075 \times 10^5

h = 75 \text{ cm}.

Thus, the water level in the cistern rises by $75 \text{ cm}$ .

Question 12

The given figure shows the cross-section of a cone, a cylinder and a hemisphere all with the same diameter 10 cm, and the other dimensions are as shown.

Calculate :

(i) the total surface area,

(ii) the total volume of the solid and

(iii) the density of the material if its total weight is 1.7 kg.

Answer:

(i) Let’s begin by identifying the given dimensions:

The diameter of the cone, cylinder, and hemisphere is 10 cm.

Thus, the radius $r$ is $\frac{10}{2} = 5$ cm.

From the figure, the height of the cone $h$ is 12 cm.

To find the slant height $l$ of the cone, we use the Pythagorean theorem:

l^2 = r^2 + h^2

l^2 = 5^2 + 12^2

l^2 = 25 + 144

l^2 = 169

l = \sqrt{169} = 13 \text{ cm}

Now, calculate the total surface area:

Total surface area = Surface area of cone + Surface area of cylinder + Surface area of hemisphere

= \pi rl + 2\pi rh + 2\pi r^2

= \pi r(l + 2h + 2r)

= \frac{22}{7} \times 5 \times (13 + 2 \times 12 + 2 \times 5)

= \frac{110}{7} \times (13 + 24 + 10)

= \frac{110 \times 47}{7}

= 738.57 \text{ cm}^2

∴ The surface area of the solid is 738.57 cm $^2$ .

(ii) Next, let’s find the total volume of the solid:

Volume of the solid = Volume of cone + Volume of cylinder + Volume of hemisphere

= \frac{1}{3}\pi r^2h + \pi r^2h + \frac{2}{3}\pi r^3

= \pi r^2\left(\frac{1}{3}h + h + \frac{2}{3}r\right)

= \frac{22}{7} \times 5^2 \times \left(\frac{1}{3} \times 12 + 12 + \frac{2}{3} \times 5\right)

= \frac{550}{7} \times \left(4 + 12 + \frac{10}{3}\right)

= \frac{550}{7} \times \frac{58}{3}

= \frac{31900}{21}

= 1519.05 \text{ cm}^3

∴ The volume of the solid is 1519.05 cm $^3$ .

(iii) Finally, to find the density of the material:

Density is given by the formula:

\text{Density} = \frac{\text{Mass}}{\text{Volume}}

The mass is given as 1.7 kg, which is 1700 grams.

Substituting the values, we have:

\text{Density} = \frac{1700}{1519.05}

= 1.12 \text{ g/cm}^3

∴ The density of the material is 1.12 g/cm $^3$ .

Question 13

A solid, consisting of a right circular cone standing on a hemisphere, is placed upright in a right circular cylinder, full of water, and touches the bottom. Find the volume of water left in the cylinder, having given that the radius of the cylinder is 3 cm and its height is 6 cm; the radius of the hemisphere is 2 cm and the height of cone is 4 cm. Give your answer to the nearest cubic centimeter.

Answer:

First, we need to determine the volume of water remaining in the cylinder, which can be calculated by subtracting the volumes of the cone and the hemisphere from the volume of the cylinder.

Let’s start by identifying the given measurements:

The height of the cylinder, $H$ , is 6 cm.
The radius of the cylinder, $R$ , is 3 cm.
The height of the cone, $h$ , is 4 cm.
The radius of both the cone and the hemisphere, $r$ , is 2 cm.

The formula for the volume of water left in the cylinder is:

ext{Volume of water left in cylinder} = ext{Volume of cylinder} - ext{Volume of cone} - ext{Volume of hemisphere}

Mathematically, this can be expressed as:

= \pi R^2H - \dfrac{1}{3}\pi r^2h - \dfrac{2}{3}\pi r^3

Substituting the given values, we have:

= \dfrac{22}{7} \times \left(3^2 \times 6 - \dfrac{1}{3} \times 2^2 \times 4 - \dfrac{2}{3} \times 2^3\right)

Simplifying inside the brackets:

= \dfrac{22}{7} \times \left(54 - \dfrac{16}{3} - \dfrac{16}{3}\right)

Combine the fractions:

= \dfrac{22}{7} \times \left(54 - \dfrac{32}{3}\right)

Convert 54 to a fraction with a denominator of 3:

= \dfrac{22}{7} \times \left(\dfrac{162 - 32}{3}\right)

This simplifies to:

= \dfrac{22}{7} \times \dfrac{130}{3}

Multiply the fractions:

= \dfrac{2860}{21}

Finally, when calculated, this equals approximately 136 cubic centimeters.

Therefore, the volume of water left in the cylinder is 136 cm³.

Question 14

A metal container in the form of a cylinder is surmounted by a hemisphere of the same radius. The internal height of the cylinder is 7 m and the internal radius is 3.5 m. Calculate :

(i) the total area of the internal surface, excluding the base;

(ii) the internal volume of the container in m^3.

Answer:

We have a container where the radius of both the cylindrical and hemispherical parts is 3.5 m. The height of the cylindrical section is given as 7 m.

(i) Calculating the total area of the internal surface, excluding the base:

The internal surface area is the sum of the lateral surface area of the cylinder and the curved surface area of the hemisphere. This can be expressed as:

Surface area of cylinder = $2\pi rh$

Surface area of hemisphere = $2\pi r^2$

Thus, the total internal surface area is:

2\pi r(h + r)

Substituting the given values, we have:

2 \times \dfrac{22}{7} \times 3.5 \times (7 + 3.5)

This simplifies to:

2 \times 22 \times 0.5 \times 10.5

Carrying out the multiplication, we find:

231 \text{ m}^2

∴ The total area of the internal surface is 231 m².

(ii) Calculating the internal volume of the container in cubic meters:

The volume of the container consists of the volume of the hemisphere plus the volume of the cylinder.

Volume of hemisphere = $\dfrac{2}{3}\pi r^3$

Volume of cylinder = $\pi r^2h$

The total volume is:

\pi r^2\left(\dfrac{2r}{3} + h\right)

Substituting the values, this becomes:

\dfrac{22}{7} \times (3.5)^2 \times \left(\dfrac{2 \times 3.5}{3} + 7\right)

Simplifying further gives:

22 \times 0.5 \times 3.5 \times \left(\dfrac{7}{3} + 7\right)

Which evaluates to:

38.5 \times \dfrac{28}{3}

Finally, calculating the above expression results in:

359.33 \text{ m}^3

∴ The internal volume of the container is 359.33 m³.

Question 15

An exhibition tent is in the form of a cylinder surmounted by a cone. The height of the tent above ground is 85 m and height of the cylindrical part is 50 m. If the diameter of the base is 168 m, find the quantity of canvas required to make the tent. Allow 20% extra for fold and for stitching. Give your answer to the nearest m^2.

Answer:

Let’s analyze the problem by first considering the dimensions given:

The diameter of the base of both the cylinder and the cone is 168 m.

To find the radius, we use:

r = \dfrac{168}{2} = 84 \text{ m}

The height of the cylindrical part is noted as 50 m, while the total height of the tent is 85 m. Thus, the height of the conical part is:

h = 85 - 50 = 35 \text{ m}

To find the slant height $l$ of the cone, apply the Pythagorean theorem:

$l^2 = r^2 + h^2$
$l^2 = 84^2 + 35^2$
$l^2 = 7056 + 1225$
$l^2 = 8281$
$l = \sqrt{8281} = 91 \text{ m}$

Now, calculate the surface area of the tent, which includes both the cylindrical and conical parts:

Surface area of the cylindrical part is:

2\pi r H = 2 \times \dfrac{22}{7} \times 84 \times 50

Surface area of the conical part is:

\pi r l = \dfrac{22}{7} \times 84 \times 91

Adding these gives:

$= 26400 + 24024$
$= 50424 \text{ m}^2$

To account for folds and stitching, we add 20% more to the calculated area:

\text{Extra area} = \dfrac{20}{100} \times 50424 = 10084.8 \text{ m}^2

Thus, the total canvas required is:

50424 + 10084.8 = 60508.8 \text{ m}^2

Rounding to the nearest square meter, we find:

Hence, the area of canvas required = 60509 m^2.

Question 16

A test tube consists of a hemisphere and a cylinder of the same radius. The volume of the water required to fill the whole tube is $\dfrac{5159}{6}$ cm^3, and $\dfrac{4235}{6}$ cm^3 of water is required to fill the tube to a level which is 4 cm below the top of the tube. Find the radius of the tube and the length of its cylindrical part.

Answer:

Let’s consider the test tube, which is made up of a hemisphere and a cylindrical section, both sharing the same radius. Denote the radius as $r$ cm and the height of the cylindrical portion as $h$ cm.

The complete volume of the test tube combines the volume of the hemisphere with that of the cylinder:

\Rightarrow \dfrac{5159}{6} = \dfrac{2}{3}\pi r^3 + \pi r^2 h

This can be rewritten as:

\Rightarrow \dfrac{5159}{6} = \pi r^2\left(\dfrac{2}{3}r + h\right)

Simplifying further, we have:

\Rightarrow \dfrac{5159}{6} = \pi r^2\left(\dfrac{2r + 3h}{3}\right)

Multiplying through by 3, we get:

\Rightarrow \pi r^2(2r + 3h) = \dfrac{5159}{6} \times 3

Thus,

\Rightarrow \pi r^2(2r + 3h) = \dfrac{5159}{2} \quad \text{...(1)}

Now, given that $\dfrac{4235}{6}$ cm $^3$ of water fills the tube up to 4 cm below the top, we have:

\therefore \dfrac{4235}{6} = \text{Volume of hemisphere + Volume of cylinder up to } (h - 4) \text{ cm}

This leads to:

\Rightarrow \dfrac{4235}{6} = \dfrac{2}{3}\pi r^3 + \pi r^2(h - 4)

Which simplifies to:

\Rightarrow \dfrac{4235}{6} = \pi r^2\left(\dfrac{2}{3}r + h - 4\right)

Or:

\Rightarrow \dfrac{4235}{6} = \pi r^2\left(\dfrac{2r + 3h - 12}{3}\right)

Multiplying by 3, we have:

\Rightarrow \dfrac{4235 \times 3}{6} = \pi r^2(2r + 3h) - 12\pi r^2

This simplifies to:

\Rightarrow \dfrac{4235}{2} = \pi r^2(2r + 3h) - 12\pi r^2

Substituting the value of $\pi r^2(2r + 3h)$ from equation (1), we get:

\Rightarrow \dfrac{4235}{2} = \dfrac{5159}{2} - 12\pi r^2

Thus:

\Rightarrow 12\pi r^2 = \dfrac{5159}{2} - \dfrac{4235}{2}

This leads to:

\Rightarrow 12\pi r^2 = \dfrac{924}{2}

Solving for $r^2$ , we have:

\Rightarrow r^2 = \dfrac{924}{2 \times 12\pi}

Simplifying further:

\Rightarrow r^2 = \dfrac{924}{24 \times \dfrac{22}{7}}

Which results in:

\Rightarrow r^2 = \dfrac{924 \times 7}{24 \times 22}

Hence:

\Rightarrow r^2 = \dfrac{6468}{528}

Simplifying gives:

\Rightarrow r^2 = 12.25

Taking the square root:

\Rightarrow r = \sqrt{12.25}

Thus, $r = 3.5$ cm.

Now, substituting $r = 3.5$ cm back into equation (1), we obtain:

\Rightarrow \pi r^2(2r + 3h) = \dfrac{5159}{2}

Substituting the known values:

\Rightarrow \dfrac{22}{7} \times (3.5)^2 \times (2 \times 3.5 + 3h) = \dfrac{5159}{2}

This simplifies to:

\Rightarrow 22 \times 0.5 \times 3.5 \times (7 + 3h) = \dfrac{5159}{2}

Further simplification gives:

\Rightarrow 38.5 \times (7 + 3h) = \dfrac{5159}{2}

Solving for $h$ :

\Rightarrow 7 + 3h = \dfrac{5159}{2 \times 38.5}

This results in:

\Rightarrow 7 + 3h = \dfrac{5159}{77}

Thus:

\Rightarrow 7 + 3h = 67

Solving for $h$ , we find:

\Rightarrow 3h = 67 - 7

Which gives:

\Rightarrow 3h = 60

Finally:

\Rightarrow h = \dfrac{60}{3}

Therefore, $h = 20$ cm.

Hence, the radius of the cylindrical part is 3.5 cm and the height is 20 cm.

Question 17

A solid is in the form of a right circular cone mounted on a hemisphere. The diameter of the base of the cone, which exactly coincides with hemisphere, is 7 cm and its height is 8 cm. The solid is placed in a cylindrical vessel of internal radius 7 cm and height 10 cm. How much water, in cm^3, will be required to fill the vessel completely?

Answer:

Let’s analyze the problem with the given dimensions. We have a cone sitting on a hemisphere where the diameter of both the cone and the hemisphere is 7 cm. This gives us a radius ( $r$ ) of $\frac{7}{2} = 3.5$ cm for both shapes. The height of the cone ( $h$ ) is 8 cm.

For the cylindrical vessel, the internal radius ( $R$ ) is 7 cm, and the height ( $H$ ) is 10 cm.

To find the volume of the solid, we need to add the volume of the cone and the hemisphere:

\text{Volume of solid} = \text{Volume of cone} + \text{Volume of hemisphere}

The formula for the volume of a cone is $\frac{1}{3} \pi r^2 h$ and for a hemisphere is $\frac{2}{3} \pi r^3$ . Thus, the combined volume becomes:

= \pi r^2 \left(\frac{1}{3}h + \frac{2}{3}r\right)

Substituting the known values:

= \frac{22}{7} \times (3.5)^2 \times \left(\frac{1}{3} \times 8 + \frac{2}{3} \times 3.5\right)

This simplifies to:

= 22 \times 0.5 \times 3.5 \times \left(\frac{8}{3} + \frac{7}{3}\right)

= 38.5 \times \frac{15}{3}

= 38.5 \times 5

= 192.5 \text{ cm}^3.

Now, calculate the volume of the cylindrical vessel using the formula $\pi R^2 H$ :

= \frac{22}{7} \times 7^2 \times 10

Simplifying further:

= 22 \times 7 \times 10

= 1540 \text{ cm}^3.

To find out how much water is needed to fill the vessel, subtract the volume of the solid from the volume of the cylindrical vessel:

\text{Volume of water required} = 1540 - 192.5

= 1347.5 \text{ cm}^3.

Hence, volume of water required to fill the vessel completely = 1347.5 cm^3.

Question 19

A certain number of metallic cones, each of radius 2 cm and height 3 cm, are melted and recast into a solid sphere of radius 6 cm. Find the number of cones used.

Answer:

We have a cone with a radius of 2 cm and a height of 3 cm. Additionally, we have a sphere with a radius of 6 cm. The task is to determine how many such cones are needed to form this sphere when melted.

The volume of a single cone is given by the formula $\dfrac{1}{3} \pi r^2 h$ . The volume of the sphere is $\dfrac{4}{3} \pi R^3$ .

If $n$ cones are melted to form the sphere, then:

n \times \dfrac{1}{3} \pi r^2 h = \dfrac{4}{3} \pi R^3

Solving for $n$ , we have:

n = \dfrac{\dfrac{4}{3} \pi R^3}{\dfrac{1}{3} \pi r^2 h}

Simplifying further:

n = \dfrac{4 \pi R^3 \times 3}{\pi r^2 h \times 3}

n = \dfrac{4R^3}{r^2h}

Substituting the given values:

n = \dfrac{4 \times 6^3}{2^2 \times 3}

n = \dfrac{864}{12}

n = 72

Therefore, 72 cones need to be melted to form a solid sphere.

Question 20

A conical tent is to accommodate 77 persons. Each person must have 16 m^3 of air to breathe. Given the radius of the tent as 7 m, find the height of the tent and also its curved surface area.

Answer:

Each individual needs 16 m³ of air for breathing.

∴ For 77 individuals, the total air volume required is 77 × 16 m³ = 1232 m³.

The radius of the tent (r) is 7 m.

Assume the height of the conical tent is h meters.

Since the tent must house 77 people, its volume must match the air volume needed for these individuals.

⇒ $\dfrac{1}{3}πr^2h = 1232$

\Rightarrow \dfrac{1}{3} \times \dfrac{22}{7} \times 7^2 \times h = 1232

\Rightarrow \dfrac{1}{3} \times 22 \times 7 \times h = 1232

\Rightarrow h = \dfrac{1232 \times 3}{22 \times 7}

\Rightarrow h = \dfrac{3696}{154}

\Rightarrow h = 24 \text{ m}.

Using the Pythagorean theorem for the slant height (l):

⇒ $l^2 = r^2 + h^2$

⇒ $l^2 = 7^2 + 24^2$

⇒ $l^2 = 49 + 576$

⇒ $l^2 = 625$

⇒ $l = \sqrt{625} = 25 \text{ m}.$

The curved surface area of the tent is calculated as:

Curved surface area = $πrl$

= $\dfrac{22}{7} \times 7 \times 25$

= 22 × 25

= 550 m².

Thus, the height of the tent is 24 m and its curved surface area is 550 m².

Frequently Asked Questions

Curved Surface Area (CSA) is the area of only the curved or lateral surface of a 3D object, excluding its flat bases. For example, it's the area of the label on a cylindrical can. Total Surface Area (TSA) is the sum of the CSA and the area of all its flat bases. For a cylinder, TSA = CSA + area of two circular bases.

To solve problems with combined solids, like a cone on a cylinder, you must break the problem down. For volume, you simply add the volumes of the individual shapes. For surface area, you must be careful to only add the areas of the exposed surfaces. For example, in a cone-cylinder tent, you would add the CSA of the cone and the CSA of the cylinder, but not the circular base where they join.

For a cylinder with radius 'r' and height 'h', CSA is 2πrh and Volume is πr²h. For a cone with radius 'r', height 'h', and slant height 'l', CSA is πrl and Volume is (1/3)πr²h. For a sphere with radius 'r', Surface Area is 4πr² and Volume is (4/3)πr³.

The chapter on Cylinder, Cone and Sphere in the Selina Concise Mathematics textbook for Class 10 is quite comprehensive. It contains a total of 130 questions. These are distributed across six main exercises, from Exercise 20(A) to Exercise 20(F), and a final 'Test Yourself' section to consolidate your learning.

ICSE Board

Exercise 20(A)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Question 23

Exercise 20(B)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Exercise 20(C)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Exercise 20(D)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Exercise 20(E)

Question 1(a)

Question 1(b)