NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 12: Surface Areas & Volumes

Exercise 12.1

Step 1: Think of a new pencil before sharpening. Its body is in the shape of a cylinder (like a long pipe). A cylinder has two flat circular faces at the ends and a curved surface all around.

Step 2: When we sharpen one end, the wood and lead are cut to form a pointed tip. This pointed tip is shaped like a cone. A cone has a flat circular base and a curved surface that comes to a sharp point (called the vertex).

Step 3: So now the pencil is made of two parts joined together: - The cylindrical body (main length of the pencil). - The conical tip (sharpened part).

Final Answer: The pencil is the combination of a cone and a cylinder.

Step 1: A surahi is a traditional water pot or bottle used in India.

Step 2: Look carefully at its shape. The bottom part is round and looks like a sphere.

Step 3: The upper part is long and straight, like the neck of a bottle. This looks like a cylinder.

Step 4: So, the surahi is made up of two solid shapes – a sphere (for the body) and a cylinder (for the neck).

Final Answer: Option A – a sphere and a cylinder.

Step 1: Look carefully at the figure of the plumbline (sahul).

Step 2: The lower portion of the plumbline is pointed, like an ice-cream cone. This shape is called a cone.

Step 3: The upper portion is not flat. Instead, it is smoothly rounded, like half of a ball. This shape is called a hemisphere.

Step 4: So, the plumbline is made up of two solid shapes joined together — a cone at the bottom and a hemisphere at the top.

Final Answer: A plumbline is a combination of a hemisphere and a cone.

Step 1: Look carefully at the glass (tumbler). It is wide at the top and becomes narrow as we go down.

Step 2: A cone also has this property — wide at one end and pointed at the other. But in a glass, the bottom is flat, not pointed.

Step 3: If we cut off the sharp tip of a cone and make the bottom flat, the shape we get is called a frustum of a cone.

Step 4: A cylinder has the same width from top to bottom, so it cannot be the shape of a glass.

Step 5: A sphere is completely round like a ball, which also does not match the shape of a tumbler.

Conclusion: Therefore, the glass (tumbler) is in the shape of a frustum of a cone.

Step 1: Look at the middle part of the gilli. It is like a straight pipe with the same circular cross-section everywhere. This is the shape of a cylinder.

Step 2: Now look at both ends of the gilli. Each end becomes narrower and comes to a pointy tip. This shape is like a cone.

Step 3: Since there are two ends, we have two cones.

Step 4: Together, the gilli is made of one cylinder in the centre and two cones (one at each end).

Final Answer: The shape is a combination of two cones and a cylinder.

Step 1: Look at the structure of a shuttlecock. It has two main parts:

• The head, which is the part we hit with the racket.
• The feathers (or plastic net), which spread outwards to give the shuttle its shape.

Step 2: The head of the shuttlecock is like half of a solid ball. In geometry, half of a sphere is called a hemisphere.

Step 3: The feathers form a cone-like shape, but the pointed tip is cut off (flat at the top). This shape is called the frustum of a cone.

Step 4: So the shuttlecock is made by combining:

• One hemisphere (for the head)
• One frustum of a cone (for the feather part)

Final Answer: The shuttlecock is a combination of a frustum of a cone and a hemisphere.

Step 1: Imagine a cone — it has a circular base and a pointed top (called the vertex).

Step 2: Now, take a plane (a flat surface) and cut the cone parallel to its base. "Parallel to the base" means the cutting plane is flat and at the same angle as the base, just at some height above it.

Step 3: After cutting, we remove the small cone (the top portion near the vertex).

Step 4: What remains is a shape with two circular surfaces: • one is the original big base (at the bottom), • the other is the new smaller circular face (at the top, where we cut). The sides are still slanting.

Step 5: This new solid (with two parallel circular ends and slanting sides) is called a frustum of a cone.

Therefore, the correct answer is: A frustum of a cone.

Step 1: Find the volume of the cube.

The internal edge of the cube = 22 cm.

Volume of cube = \(a^3 = 22^3 = 10648\,\text{cm}^3\).

Step 2: Account for the empty space.

Only \(\tfrac{7}{8}\) of the cube is filled with marbles.

Filled volume = \(\tfrac{7}{8} \times 10648 = 9317\,\text{cm}^3\).

Step 3: Find the volume of one marble.

Diameter of one marble = 0.5 cm, so radius = 0.25 cm.

Volume of one sphere = \(\tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\pi (0.25)^3\).

= \(\tfrac{4}{3}\pi (0.015625) = 0.020833\pi \approx 0.06545\,\text{cm}^3\).

Step 4: Find the number of marbles.

Total filled volume ÷ Volume of one marble = \(\tfrac{9317}{0.06545}\).

≈ 142396 marbles.

Final Answer: Option B (142396)

Step 1: Write the given data.

• Internal diameter of shell = 4 cm ⇒ radius = 2 cm.
• External diameter of shell = 8 cm ⇒ radius = 4 cm.
• Base diameter of cone = 8 cm ⇒ radius of cone base = 4 cm.
• Height of cone = ? (to be found).

Step 2: Find the volume of the spherical shell.

Volume of shell = Volume of outer sphere − Volume of inner sphere.

Outer radius = 4 cm, Inner radius = 2 cm.

So,

\[ V_{shell} = \dfrac{4}{3}\pi \big( R^3 - r^3 \big) = \dfrac{4}{3}\pi (4^3 - 2^3). \]

\(4^3 = 64, \; 2^3 = 8 \).

Therefore, \( V_{shell} = \dfrac{4}{3}\pi (64 - 8) = \dfrac{4}{3}\pi (56) = \dfrac{224}{3}\pi \; \text{cm}^3. \)

Step 3: Volume of the cone formed.

Formula: \( V_{cone} = \dfrac{1}{3} \pi r^2 h \)

Here, \(r = 4\,\text{cm}, h = ?\)

So, \( V_{cone} = \dfrac{1}{3} \pi (4^2) h = \dfrac{1}{3} \pi (16) h = \dfrac{16}{3}\pi h \).

Step 4: Equating the volumes.

Since the shell is melted to form the cone:

\( V_{shell} = V_{cone} \)

\( \dfrac{224}{3}\pi = \dfrac{16}{3}\pi h \)

Step 5: Simplify and solve for \(h\).

Cancel \( \dfrac{\pi}{3} \) on both sides:

\(224 = 16h \)

\( h = \dfrac{224}{16} = 14 \)

Final Answer: Height of the cone = 14 cm.

Step 1: Write the formula for the volume of a cuboid.

\(V_{\text{cuboid}} = \text{length} \times \text{breadth} \times \text{height}\)

Step 2: Put the given values (in cm).

\(V_{\text{cuboid}} = 49 \times 33 \times 24 = 38808 \, \text{cm}^3\)

Step 3: When the cuboid is melted and moulded into a sphere, the volume remains the same.

So, \(V_{\text{cuboid}} = V_{\text{sphere}}\)

Step 4: Formula for the volume of a sphere:

\(V_{\text{sphere}} = \dfrac{4}{3} \pi r^3\)

Step 5: Equating the volumes:

\(38808 = \dfrac{4}{3} \pi r^3\)

Step 6: Simplify.

Multiply both sides by 3:

\(3 \times 38808 = 4 \pi r^3\)

\(116424 = 4 \pi r^3\)

Step 7: Divide both sides by \(4\pi\).

\(r^3 = \dfrac{116424}{4\pi}\)

Taking \(\pi = \dfrac{22}{7}\):

\(r^3 = \dfrac{116424 \times 7}{4 \times 22} = \dfrac{815,000}{88} = 9261\)

Step 8: Find cube root.

\(r^3 = 9261 \implies r = \sqrt[3]{9261} = 21\)

Final Answer: The radius of the sphere is 21 cm.

Step 1: Find the volume of one brick.

Each brick has dimensions: \(22.5\,\text{cm} \times 11.25\,\text{cm} \times 8.75\,\text{cm}\).

So, volume of one brick = \(22.5 \times 11.25 \times 8.75 = 2214.84375\,\text{cm}^3\).

Step 2: Find the total volume of the wall (without considering mortar).

Wall dimensions = \(270\,\text{cm} \times 300\,\text{cm} \times 350\,\text{cm}\).

Total wall volume = \(270 \times 300 \times 350 = 28,350,000\,\text{cm}^3\).

Step 3: Adjust for mortar.

We are told \(\tfrac{1}{8}\) of the volume is filled with mortar, not bricks.

So, fraction available for bricks = \(1 - \tfrac{1}{8} = \tfrac{7}{8}\).

Usable volume = \(\tfrac{7}{8} \times 28,350,000 = 24,806,250\,\text{cm}^3\).

Step 4: Find how many bricks can fit in this usable volume.

Number of bricks = \(\dfrac{24,806,250}{2214.84375} \approx 11200\).

Final Answer: The number of bricks used is 11200.

Step 1: Write the given values.

• Base diameter of cylinder = 2 cm
• So, radius of cylinder \(r = \tfrac{2}{2} = 1\,\text{cm}\)
• Height of cylinder \(h = 16\,\text{cm}\)
• Number of spheres formed = 12

Step 2: Find the volume of the cylinder.

Formula: \(V_{\text{cylinder}} = \pi r^2 h\)

\(= \pi (1)^2 (16) = 16\pi\,\text{cm}^3\)

Step 3: Let the radius of each sphere be \(R\).

Volume of one sphere: \(V_{\text{sphere}} = \tfrac{4}{3}\pi R^3\)

Volume of 12 spheres: \(12 \times \tfrac{4}{3}\pi R^3\)

Step 4: Since the metal is melted, total volume remains the same.

So, \(16\pi = 12 \times \tfrac{4}{3}\pi R^3\)

Step 5: Cancel \(\pi\) from both sides.

\(16 = 16R^3\)

Step 6: Solve for \(R^3\).

\(R^3 = 1 \;\Rightarrow R = 1\,\text{cm}\)

Step 7: Find the diameter of each sphere.

Diameter = \(2R = 2 \times 1 = 2\,\text{cm}\)

Final Answer: Diameter of each sphere is 2 cm.

Step 1: Recall the formula for the curved surface area (CSA) of a frustum of a cone:

\[ CSA = \pi (R + r) l \]

where:

• \(R\) = radius of the bigger circular end (top)
• \(r\) = radius of the smaller circular end (bottom)
• \(l\) = slant height of the frustum

Step 2: Substitute the given values:

\(R = 28\,\text{cm},\; r = 7\,\text{cm},\; l = 45\,\text{cm}\)

Step 3: Add the radii:

\(R + r = 28 + 7 = 35\)

Step 4: Multiply with slant height:

\((R + r) \times l = 35 \times 45 = 1575\)

Step 5: Multiply with \(\pi\) (take \(\pi = 3.1416\)):

\(CSA = 1575 \times 3.1416 \approx 4949.98\,\text{cm}^2\)

Step 6: Round off to the nearest whole number:

\(CSA \approx 4950\,\text{cm}^2\)

Final Answer: The curved surface area is 4950 cm².

Step 1: The capsule is made of a cylinder with two hemispherical ends.

Step 2: Diameter of capsule = 0.5 cm.
Radius \(r = \frac{0.5}{2} = 0.25\,\text{cm}\).

Step 3: Total length of capsule = 2 cm.
The two hemispherical ends together make one full sphere of radius 0.25 cm.
So, cylindrical length \(h = 2 - 2r = 2 - 0.5 = 1.5\,\text{cm}\).

Step 4: Volume of cylinder = \(\pi r^2 h\).
\(= 3.1416 \times (0.25)^2 \times 1.5\).
\(= 3.1416 \times 0.0625 \times 1.5 = 0.2945\,\text{cm}^3\).

Step 5: Volume of sphere = \(\frac{4}{3}\pi r^3\).
\(= \frac{4}{3} \times 3.1416 \times (0.25)^3\).
\(= \frac{4}{3} \times 3.1416 \times 0.015625\).
\(= 0.0654\,\text{cm}^3\).

Step 6: Total volume = volume of cylinder + volume of sphere.
\(= 0.2945 + 0.0654 = 0.3599 \approx 0.36\,\text{cm}^3\).

Final Answer: The capacity of the capsule is 0.36 cm³ (Option A).

Step 1: Recall that a hemisphere is half of a sphere.

Step 2: Each hemisphere has a curved surface area (CSA) given by:

\( \text{CSA of one hemisphere} = 2\pi r^2 \).

Step 3: When two hemispheres of radius \(r\) are joined together along their flat bases, the flat parts disappear, and they form a complete sphere.

Step 4: The surface area of a sphere is entirely curved surface (no flat faces).

Formula: \( \text{CSA of a sphere} = 4\pi r^2 \).

Step 5: Therefore, the new solid (a sphere) has curved surface area:

\( 4\pi r^2 \).

Final Answer: Option A (\(4\pi r^2\)).

Step 1: A sphere is a perfectly round ball. If it is placed inside a cylinder, it will touch the cylinder from all sides.

Step 2: The widest part of the sphere is its diameter. This diameter must be the same as the inner width of the cylinder for the sphere to fit exactly.

Step 3: The inner width of the cylinder is measured across its circular base. That width is equal to the diameter of the base circle.

Step 4: The diameter of the base circle of the cylinder is \(2r\) (since radius is \(r\), diameter = \(2 \times r\)).

Step 5: Therefore, the diameter of the sphere = diameter of the cylinder’s base = \(2r\).

Final Answer: The diameter of the sphere is \(2r\) cm. (Option B)

Step 1: A solid is made of a certain amount of material. Example: If you have 100 cm³ of clay, that is the total material available.

Step 2: When we change the shape of the solid (say, from a cube to a sphere), we are only changing the form, not adding or removing any material.

Step 3: The volume means the amount of space the material occupies. Since the material is the same, the total space occupied will also remain the same.

Step 4: Therefore, during conversion from one shape to another, the volume of the solid remains unaltered.

Step 1: Identify the shape

The bucket is in the shape of a frustum of a cone (a cone with its top portion cut off).

Step 2: Write the formula for the volume of a frustum

\[ V = \dfrac{1}{3} \pi h (R^2 + Rr + r^2) \] where - \(R\) = radius of the bigger circular end,
- \(r\) = radius of the smaller circular end,
- \(h\) = height of the frustum.

Step 3: Find the radii

Given diameters: 44 cm and 24 cm.

\(R = \dfrac{44}{2} = 22\,\text{cm}\),
\(r = \dfrac{24}{2} = 12\,\text{cm}\).

Step 4: Height

Height of the bucket: \(h = 35\,\text{cm}\).

Step 5: Substitute values in the formula

\[ V = \dfrac{1}{3} \pi (35) (22^2 + 22 \times 12 + 12^2) \]

First calculate inside the brackets:

\(22^2 = 484\),
\(22 \times 12 = 264\),
\(12^2 = 144\).

So, \(484 + 264 + 144 = 892\).

Step 6: Multiply

\[ V = \dfrac{1}{3} \pi (35)(892) \]

\(35 \times 892 = 31,220\).

\[ V = \dfrac{1}{3} \pi (31,220) \]

\(31,220 \div 3 = 10,406.67\).

\[ V \approx 3.1416 \times 10,406.67 = 32,689.1\,\text{cm}^3 \]

Step 7: Convert to litres

We know \(1000\,\text{cm}^3 = 1\,\text{L}\).

\[ V = \dfrac{32,689.1}{1000} = 32.7\,\text{L} \]

Final Answer: The capacity of the bucket is 32.7 litres (Option A).

Step 1: A right circular cone has a circular base and a pointed top (vertex).

Step 2: The base of the cone is a circle. Now imagine a flat surface (a plane) cutting the cone, but this plane is kept parallel to the base.

Step 3: When the cutting plane is parallel to the base, the shape of the cut (cross-section) will be the same shape as the base.

Step 4: Since the base is a circle, the cross-section is also a circle.

Final Answer: The cross-section is a circle (Option A).

Step 1: Recall formulas in SI units

• Volume of a sphere: \(V = \tfrac{4}{3} \pi r^3\) (in cubic metres, \(m^3\))
• Surface area of a sphere: \(A = 4 \pi r^2\) (in square metres, \(m^2\))

Step 2: Use the volume ratio

Given ratio of volumes: \(64 : 27\).

Since volume is proportional to \(r^3\),

\(\dfrac{r_1}{r_2} = \sqrt[3]{\dfrac{64}{27}} = \dfrac{4}{3}.\)

So the radii are in the ratio \(4 : 3\).

Step 3: Find the ratio of surface areas

Surface area is proportional to \(r^2\).

So, \(\dfrac{A_1}{A_2} = \left(\dfrac{r_1}{r_2}\right)^2 = \left(\dfrac{4}{3}\right)^2 = \dfrac{16}{9}.\)

Final Answer: The ratio of surface areas is \(16:9\) (Option D).