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

Exercise 12.3

Step 1: Write the formula for the volume of a cube.
Volume of a cube = (edge)³.

Step 2: Find the volume of each cube.

• Cube with edge 3 cm: Volume = (3^3 = 27 , ext{cm}^3)
• Cube with edge 4 cm: Volume = (4^3 = 64 , ext{cm}^3)
• Cube with edge 5 cm: Volume = (5^3 = 125 , ext{cm}^3)

Step 3: Add all the volumes to get the total volume of metal.
Total volume = (27 + 64 + 125 = 216 , ext{cm}^3).

Step 4: This total volume is used to make one new cube.
Let the edge of the new cube be (a).

Step 5: Volume of new cube = (a^3).
So, (a^3 = 216).

Step 6: Find cube root of 216.
(a = sqrt[3]{216} = 6 , ext{cm}).

Final Answer: The edge of the new cube is 6 cm.

Step 1: Find the volume of the cuboid (the big lead block).

Formula: \(\text{Volume of cuboid} = \text{length} \times \text{breadth} \times \text{height}\)

Here: \(9\,\text{cm} \times 11\,\text{cm} \times 12\,\text{cm} = 1188\,\text{cm}^3\).

Step 2: Find the radius of one spherical shot.

Diameter = 3 cm, so Radius = \(\dfrac{3}{2} = 1.5\,\text{cm}\).

Step 3: Find the volume of one spherical shot.

Formula: \(\text{Volume of sphere} = \dfrac{4}{3} \pi r^3\).

Substitute: \(\dfrac{4}{3} \pi (1.5)^3 = \dfrac{4}{3} \pi (3.375) = 4.5 \pi\,\text{cm}^3\).

Step 4: Find the number of spherical shots.

Formula: \(\text{Number of shots} = \dfrac{\text{Volume of cuboid}}{\text{Volume of one sphere}}\).

Substitute: \(\dfrac{1188}{4.5\pi} = \dfrac{1188}{\dfrac{99}{7}} = \dfrac{1188 \times 7}{99} = 84\).

Final Answer: 84 shots can be made.

Step 1: Write down the known values.

• Volume of water = \(28.490\,\text{litres}\).
• Top radius (\(R\)) = \(28\,\text{cm}\).
• Bottom radius (\(r\)) = \(21\,\text{cm}\).
• Height (\(h\)) = ? (to be found)

Step 2: Convert volume into cubic centimetres (SI unit for volume).

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

So, \(28.490\,\text{litres} = 28.490 \times 1000 = 28490\,\text{cm}^3\).

Step 3: Recall the formula for the volume of a frustum of a cone.

\[ V = \dfrac{1}{3}\pi h \big(R^2 + r^2 + Rr\big) \]

Step 4: Substitute the values of \(R\) and \(r\).

\(R^2 = 28^2 = 784\)

\(r^2 = 21^2 = 441\)

\(Rr = 28 \times 21 = 588\)

So, \(R^2 + r^2 + Rr = 784 + 441 + 588 = 1813\).

Step 5: Put values in the volume formula.

\(28490 = \dfrac{1}{3}\pi h (1813)\)

Step 6: Simplify the equation.

Multiply denominator first: \(\dfrac{1}{3} \times 1813 = 604.33\).

So, equation becomes: \(28490 = 604.33 \pi h\).

Step 7: Use \(\pi \approx 3.1416\).

\(604.33 \pi \approx 604.33 \times 3.1416 = 1899.56\).

So, \(28490 = 1899.56 h\).

Step 8: Solve for \(h\).

\(h = \dfrac{28490}{1899.56} \approx 15.0\,\text{cm}\).

Final Answer: The height of the bucket is 15 cm.

Step 1: Understand the cone dimensions.

The original cone has:

• Radius \(R = 8\,\text{cm}\)
• Height \(H = 12\,\text{cm}\)

Step 2: Where the plane cuts the cone.

The plane passes through the mid-point of the axis, so it divides the height into two equal parts:

• Upper small cone height = \(6\,\text{cm}\)
• Remaining lower part is a frustum.

Step 3: Find radius of the small cone at the top.

Because the plane is parallel to the base, the small cone formed at the top is similar to the whole cone.

Ratio of heights = \(\dfrac{6}{12} = \dfrac{1}{2}\).

So, its radius = \(\dfrac{1}{2} \times 8 = 4\,\text{cm}\).

Step 4: Volume of the full cone.

Formula: \(V = \tfrac{1}{3}\pi r^2 h\).

\(V_{\text{big}} = \tfrac{1}{3} \pi (8^2)(12) = \tfrac{1}{3} \pi (64)(12) = 256\pi\,\text{cm}^3\).

Step 5: Volume of the small top cone.

\(V_{\text{small}} = \tfrac{1}{3} \pi (4^2)(6) = \tfrac{1}{3} \pi (16)(6) = 32\pi\,\text{cm}^3\).

Step 6: Volume of the frustum (bottom part).

Frustum volume = (whole cone volume) − (small cone volume).

\(V_{\text{frustum}} = 256\pi - 32\pi = 224\pi\,\text{cm}^3\).

Step 7: Ratio of volumes.

\(V_{\text{small}} : V_{\text{frustum}} = 32\pi : 224\pi = 1 : 7\).

Final Answer: The ratio of the volumes of the two parts is 1 : 7.

Step 1: Find the edge of one cube.

The volume of a cube is given by: \( V = a^3 \), where \(a\) is the edge length.

Here, \( V = 64\,\text{cm}^3 \).

So, \( a^3 = 64 \).

Taking cube root: \( a = \sqrt[3]{64} = 4\,\text{cm} \).

Step 2: Dimensions of the new cuboid.

When two such cubes are joined end to end, one dimension doubles while the other two remain the same.

So the new cuboid has dimensions: \( 8\,\text{cm} \times 4\,\text{cm} \times 4\,\text{cm} \).

Step 3: Formula for surface area of a cuboid.

Surface area \( A = 2(lb + bh + hl) \).

Here: \( l = 8, b = 4, h = 4 \).

Step 4: Substitute values.

\( A = 2(8\times4 + 4\times4 + 8\times4) \).

\( A = 2(32 + 16 + 32) \).

\( A = 2(80) = 160\,\text{cm}^2 \).

Final Answer: The surface area of the cuboid is \(160\,\text{cm}^2\).

Step 1: Write down what is given.

• Cube side length = 7 cm
• Cone height (h) = 7 cm
• Cone radius (r) = 3 cm

Step 2: Find the volume of the cube.

Formula: \( V_{\text{cube}} = (\text{side})^3 \)

\( V_{\text{cube}} = 7^3 = 343\,\text{cm}^3 \)

Step 3: Find the volume of the cone (the hollowed part).

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

\( V_{\text{cone}} = \tfrac{1}{3}\pi (3^2)(7) \)

\( V_{\text{cone}} = \tfrac{1}{3}\pi (9)(7) = 21\pi\,\text{cm}^3 \)

Step 4: Subtract the cone volume from the cube volume.

Remaining volume = \( 343 - 21\pi \;\text{cm}^3 \)

Step 5: Approximate using \(\pi ≈ 3.1416\).

\( 21\pi ≈ 21 × 3.1416 ≈ 65.97 \)

Remaining volume = \( 343 - 65.97 ≈ 277.0\,\text{cm}^3 \)

Final Answer: The volume of the remaining solid is about 277 cm³.

Step 1: Write down the known values.

• Radius of base, \(r = 8\;\text{cm}\)
• Height of cone, \(h = 15\;\text{cm}\)

Step 2: Find the slant height \(l\) of the cone using Pythagoras theorem:

\[ l = \sqrt{r^2 + h^2} = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\;\text{cm} \]

Step 3: Since the two cones are joined along their bases, the base areas will not be visible. So, the surface area of the new solid = sum of curved surface areas of the two cones.

Step 4: Formula for curved surface area (CSA) of one cone:

\[ \text{CSA} = \pi r l \]

Substitute values for one cone:

\[ \pi r l = \pi \times 8 \times 17 = 136\pi\;\text{cm}^2 \]

Step 5: For two cones, total surface area:

\[ 2 \times 136\pi = 272\pi\;\text{cm}^2 \]

Final Answer: The surface area of the solid is \(272\pi\;\text{cm}^2\).

Step 1: The cylindrical tube has an inner diameter of \(6\,\text{cm}\). Therefore, the radius of the cylinder (and also of both cones) is: \(r = \dfrac{6}{2} = 3\,\text{cm}\).

Step 2: The length (or height) of the tube is given as \(21\,\text{cm}\). The two cones are placed one above the other inside the tube. So, the sum of their heights must be: \(h_A + h_B = 21\,\text{cm}\).

Step 3: The capacity (volume) of a cone is proportional to its height if the radius is the same. Formula: \(V = \tfrac{1}{3}\pi r^2 h\). Here, both cones have the same base radius (\(3\,\text{cm}\)), so the ratio of their volumes is the same as the ratio of their heights. Given: volume ratio \(= 2:1\). So, \(h_A : h_B = 2:1\).

Step 4: Let \(h_A = 2x\) and \(h_B = x\). Then, \(2x + x = 21 \Rightarrow 3x = 21 \Rightarrow x = 7\). Therefore: \(h_A = 14\,\text{cm}, h_B = 7\,\text{cm}\).

Step 5: Find the volumes of the cones using the formula \(V = \tfrac{1}{3}\pi r^2 h\).

• For cone A: \(V_A = \tfrac{1}{3} \pi (3^2)(14) = \tfrac{1}{3}\pi \times 9 \times 14 = 42\pi\,\text{cm}^3\).
• For cone B: \(V_B = \tfrac{1}{3} \pi (3^2)(7) = \tfrac{1}{3}\pi \times 9 \times 7 = 21\pi\,\text{cm}^3\).

Step 6: Find the volume of the entire cylinder. Formula: \(V = \pi r^2 h\). \(V_{cylinder} = \pi (3^2)(21) = \pi \times 9 \times 21 = 189\pi\,\text{cm}^3\).

Step 7: The remaining portion of the cylinder is the cylinder volume minus the volumes of the two cones. \(V_{remaining} = 189\pi - (42\pi + 21\pi) = 189\pi - 63\pi = 126\pi\,\text{cm}^3\).

Final Answer: Heights of the cones: \(14\,\text{cm}\) and \(7\,\text{cm}\). Capacities: \(42\pi\,\text{cm}^3\) and \(21\pi\,\text{cm}^3\). Remaining volume of cylinder: \(126\pi\,\text{cm}^3\).

Step 1: Identify the shape of the ice cream.

• The ice cream has two parts:
  • A hemisphere on the top.
  • A cone at the bottom.
• Radius of both = \(r = 5\,\text{cm}\).
• Height of cone = \(h = 10\,\text{cm}\).

Step 2: Volume of the hemisphere.

Formula for volume of hemisphere = \(\dfrac{2}{3}\pi r^3\).

Substitute values: \(\dfrac{2}{3}\pi (5^3) = \dfrac{2}{3}\pi (125) = \dfrac{250}{3}\pi\,\text{cm}^3\).

Step 3: Volume of the full cone.

Formula for volume of cone = \(\dfrac{1}{3}\pi r^2 h\).

Substitute values: \(\dfrac{1}{3}\pi (5^2)(10) = \dfrac{1}{3}\pi (25)(10) = \dfrac{250}{3}\pi\,\text{cm}^3\).

Step 4: Only \(\tfrac{5}{6}\) of the cone is filled.

So, filled cone volume = \(\dfrac{5}{6} \times \dfrac{250}{3}\pi = \dfrac{625}{9}\pi\,\text{cm}^3\).

Step 5: Total volume of ice cream.

Add hemisphere volume and filled cone volume:

\(\dfrac{250}{3}\pi + \dfrac{625}{9}\pi = \dfrac{750}{9}\pi + \dfrac{625}{9}\pi = \dfrac{1375}{9}\pi\,\text{cm}^3\).

Step 6: Approximate numerical value.

\(\dfrac{1375}{9}\pi \approx 480\,\text{cm}^3\).

Final Answer: The volume of the ice cream is about \(480\,\text{cm}^3\).

Step 1: Volume of water displaced (rise in cylinder).

The water level rises by 5.6 cm in the cylindrical beaker of diameter 7 cm.

Radius of beaker = \( \frac{7}{2} = 3.5 \) cm.

Volume of water displaced = volume of cylinder part filled = \( \pi r^2 h \).

\( V_{cylinder} = \pi (3.5)^2 (5.6) \).

\( V_{cylinder} = \pi \times 12.25 \times 5.6 = 68.6\pi \; cm^3 \).

Step 2: Volume of one marble.

Diameter of marble = 1.4 cm, so radius = \( 0.7 \) cm.

Volume of a sphere = \( \frac{4}{3}\pi r^3 \).

\( V_{marble} = \frac{4}{3}\pi (0.7)^3 \).

\( V_{marble} = \frac{4}{3}\pi (0.343) = 0.457\pi \; cm^3 \).

(In fraction form, \( V_{marble} = \frac{1372}{3000}\pi \; cm^3 \)).

Step 3: Number of marbles needed.

Total displaced volume must equal total volume of marbles.

So, Number of marbles = \( \frac{V_{cylinder}}{V_{marble}} \).

\( N = \frac{68.6\pi}{0.457\pi} \).

\( N = 150 \).

Final Answer: 150 marbles are required.

Step 1: Find the volume of the rectangular lead block.

Length = 66 cm, Breadth = 42 cm, Height = 21 cm.

Volume of block = \(66 \times 42 \times 21 = 58,212\,\text{cm}^3\).

Step 2: Find the radius of one spherical shot.

Diameter of one shot = 4.2 cm.

Radius = Diameter ÷ 2 = \(4.2 \div 2 = 2.1\,\text{cm}\).

Step 3: Find the volume of one spherical shot.

Formula: Volume of sphere = \(\dfrac{4}{3} \pi r^3\).

Here, \(r = 2.1\,\text{cm}\).

So, Volume = \(\dfrac{4}{3} \pi (2.1)^3\).

First calculate \((2.1)^3 = 2.1 \times 2.1 \times 2.1 = 9.261\).

Now, Volume = \(\dfrac{4}{3} \pi \times 9.261 = 38.808 \pi / 3 = 12.936 \pi\,\text{cm}^3\).

In fraction form: \(\dfrac{6174}{500} \pi\,\text{cm}^3\).

Step 4: Find how many such spheres can be made from the block.

Number of shots = (Volume of block) ÷ (Volume of one shot).

\(= \dfrac{58,212}{(6174/500)\pi}\).

Simplify: \(= \dfrac{58,212 \times 500}{6174 \pi}\).

\(= \dfrac{29,106,000}{6174 \pi}\).

\(= \dfrac{4715}{\pi}\).

Using \(\pi = 3.1416\), we get approximately 1500.

Final Answer: 1500 spherical lead shots can be obtained.

Step 1: Find the volume of the cube.

The formula for the volume of a cube is:

\( V = a^3 \), where \(a\) is the length of the edge.

Here, \( a = 44\,\text{cm} \).

So, volume of cube = \( 44^3 = 85184\,\text{cm}^3 \).

Step 2: Find the volume of one spherical lead shot.

We are told the diameter of each spherical shot = 4 cm.

So, radius \( r = \dfrac{4}{2} = 2\,\text{cm} \).

Formula for volume of a sphere is:

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

Substitute \( r = 2 \):

\( V = \dfrac{4}{3} \pi (2)^3 = \dfrac{4}{3} \pi \times 8 = \dfrac{32}{3} \pi \).

So, the volume of one spherical shot = \( \dfrac{32}{3}\pi\,\text{cm}^3 \).

Step 3: Find the number of shots.

Total number of shots = (Volume of cube) ÷ (Volume of one sphere)

= \( \dfrac{85184}{(32/3)\pi} \).

= \( \dfrac{85184 \times 3}{32 \pi} \).

= \( \dfrac{255552}{32 \pi} \).

= \( \dfrac{7986}{\pi} \).

Now, using \( \pi \approx 3.1416 \):

\( \dfrac{7986}{3.1416} \approx 2541 \).

Final Answer:

The cube of lead can be melted to make 2541 spherical shots.

Step 1: Find the volume of the wall.

The formula for volume is: length × thickness × height.

Wall volume = \(24 \times 0.4 \times 6 = 57.6\,\text{m}^3\).

Step 2: Account for mortar.

We are told that mortar occupies \(\dfrac{1}{10}\) of the total wall volume.

So, volume actually filled with bricks = \(\dfrac{9}{10} \times 57.6 = 51.84\,\text{m}^3\).

Step 3: Convert brick dimensions to metres (SI unit).

1 cm = 0.01 m.

Brick length = \(25\,\text{cm} = 0.25\,\text{m}\).

Brick width = \(16\,\text{cm} = 0.16\,\text{m}\).

Brick height = \(10\,\text{cm} = 0.10\,\text{m}\).

Step 4: Find the volume of one brick.

Brick volume = \(0.25 \times 0.16 \times 0.10 = 0.004\,\text{m}^3\).

Step 5: Find number of bricks.

Number of bricks = \(\dfrac{\text{Volume of bricks in wall}}{\text{Volume of one brick}}\).

\(= \dfrac{51.84}{0.004} = 12960\).

Final Answer: 12,960 bricks are needed.

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

The volume of a cylinder is given by:

\( V = \pi r^2 h \)

where \( r \) = radius of the base, and \( h \) = height of the cylinder.

Step 2: Find the volume of one small disc.

Given diameter of the disc = 1.5 cm, so radius \( r = \frac{1.5}{2} = 0.75\,cm \).

Height of disc \( h = 0.2\,cm \).

Now, volume of one disc:

\( V_{disc} = \pi (0.75)^2 (0.2) = \pi (0.5625)(0.2) = 0.1125\pi \; cm^3 \).

Step 3: Find the volume of the bigger cylinder.

Given diameter of big cylinder = 4.5 cm, so radius \( r = \frac{4.5}{2} = 2.25\,cm \).

Height of big cylinder \( h = 10\,cm \).

Now, volume of big cylinder:

\( V_{big} = \pi (2.25)^2 (10) = \pi (5.0625)(10) = 50.625\pi \; cm^3 \).

Step 4: Calculate how many discs are needed.

Since all the discs are melted and formed into the big cylinder, total number of discs =

\( \dfrac{\text{Volume of big cylinder}}{\text{Volume of one disc}} = \dfrac{50.625\pi}{0.1125\pi} \).

\( \pi \) cancels out.

So, \( \dfrac{50.625}{0.1125} = 450 \).

Final Answer: 450 discs are required.