CBSE Class 10 Mathematics Surface Area and Volume Assignment Set 11

Question. A toy is in the form of a cone mounted on a hemisphere of radius 3.5cm. If the total height of the toy is 15.5cm. Find its T. S. A.
Answer: Radius of the hemisphere (\(r\)) = \(3.5\text{ cm}\)
Radius of the cone (\(r\)) = \(3.5\text{ cm}\)
Total height of the toy = \(15.5\text{ cm}\)
Height of the cone (\(h\)) = \(15.5 - 3.5 = 12\text{ cm}\)
Slant height of the cone (\(l\)) = \(\sqrt{h^2 + r^2} = \sqrt{12^2 + 3.5^2} = \sqrt{144 + 12.25} = \sqrt{156.25} = 12.5\text{ cm}\)
Total Surface Area (T.S.A.) of the toy = Curved Surface Area of cone + Curved Surface Area of hemisphere
\(\text{T.S.A.} = \pi r l + 2\pi r^2 = \pi r(l + 2r)\)
\(\text{T.S.A.} = \frac{22}{7} \times 3.5 \times (12.5 + 2 \times 3.5) = 11 \times 19.5 = 214.5\text{ cm}^2\)

Question. The internal and external diameters of a hollow hemispherical vessel are 24cm and 25cm respectively. If the cost of painting \(1\text{ cm}^2\) of the surface area is Rs 0.05, find the total cost of painting the vessel all over. [Hint:- Total s. A. of vessel = \(2\pi r^2 + 2\pi R^2 + \pi(R^2 - r^2)\)]
Answer: Internal radius (\(r\)) = \(\frac{24}{2} = 12\text{ cm}\)
External radius (\(R\)) = \(\frac{25}{2} = 12.5\text{ cm}\)
Total Surface Area of the vessel = \(2\pi r^2 + 2\pi R^2 + \pi(R^2 - r^2) = \pi(3R^2 + r^2)\)
Total Surface Area = \(\frac{22}{7} \times [3(12.5)^2 + (12)^2] = \frac{22}{7} \times [3(156.25) + 144] = \frac{22}{7} \times [468.75 + 144] = \frac{22}{7} \times 612.75 \approx 1925.79\text{ cm}^2\)
Cost of painting \(1\text{ cm}^2\) = \(\text{Rs } 0.05\)
Total cost of painting = \(1925.79 \times 0.05 \approx \text{Rs } 96.29\)

Question. A toy is in the form of a right circular cylinder with a hemisphere on one end and a cone on the other. The height and radius of base of the cylindrical part are 13cm and 5cm respectively. The radius of hemisphere and base of the conical part are same as that of cylinder. Calculate the S.A. of the toy, if the height of the cone is 12cm.
Answer: Radius of the hemisphere, cylinder, and cone (\(r\)) = \(5\text{ cm}\)
Height of the cylinder (\(h_{\text{cyl}}\)) = \(13\text{ cm}\)
Height of the cone (\(h_{\text{cone}}\)) = \(12\text{ cm}\)
Slant height of the cone (\(l\)) = \(\sqrt{r^2 + h_{\text{cone}}^2} = \sqrt{5^2 + 12^2} = 13\text{ cm}\)
Total Surface Area of the toy = Curved Surface Area of hemisphere + Curved Surface Area of cylinder + Curved Surface Area of cone
\(\text{S.A.} = 2\pi r^2 + 2\pi r h_{\text{cyl}} + \pi r l = \pi r(2r + 2h_{\text{cyl}} + l)\)
\(\text{S.A.} = \frac{22}{7} \times 5 \times (2(5) + 2(13) + 13) = \frac{110}{7} \times (10 + 26 + 13) = \frac{110}{7} \times 49 = 770\text{ cm}^2\)

Question. A right circular cone of height 4cm has a C.S.A \(47.1\text{ cm}^2\). Find its volume [Use \(\pi\) = 3.14]
Answer: Height of the cone (\(h\)) = \(4\text{ cm}\)
Curved Surface Area (\(\text{C.S.A.}\)) = \(\pi r l = 47.1\text{ cm}^2\)
We know slant height \(l = \sqrt{r^2 + h^2} = \sqrt{r^2 + 16}\)
Substituting \(\pi = 3.14\):
\(3.14 \times r \sqrt{r^2 + 16} = 47.1\)
\(r \sqrt{r^2 + 16} = 15\)
Squaring both sides:
\(r^2(r^2 + 16) = 225\)
\(r^4 + 16r^2 - 225 = 0\)
Let \(x = r^2\):
\(x^2 + 16x - 225 = 0 \implies (x + 25)(x - 9) = 0\)
Since radius cannot be negative, \(x = r^2 = 9 \implies r = 3\text{ cm}\)
Volume of the cone (\(V\)) = \(\frac{1}{3}\pi r^2 h = \frac{1}{3} \times 3.14 \times 9 \times 4 = 37.68\text{ cm}^3\)

Question. A semi circular thin sheet of metal of diameter 28cm is bent and an open conical cup is made. Find the capacity of the cup.
Answer: Diameter of the semicircular sheet = \(28\text{ cm} \implies\) Radius (\(R\)) = \(14\text{ cm}\)
The arc length of the semicircular sheet = \(\pi R = 14\pi\text{ cm}\)
When bent into a conical cup:
1. Slant height of the cone (\(l\)) = Radius of the sheet (\(R\)) = \(14\text{ cm}\)
2. Circumference of the cone's base = Arc length of the sheet \(\implies 2\pi r = 14\pi \implies r = 7\text{ cm}\)
3. Height of the cone (\(h\)) = \(\sqrt{l^2 - r^2} = \sqrt{14^2 - 7^2} = \sqrt{147} \approx 12.12\text{ cm}\)
Capacity of the cup (\(V\)) = \(\frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 12.12 \approx 622.16\text{ cm}^3\)

Question. A sector of a circle of radius 15cm has the angle 120°. It is rolled up so that two bounding radii are joined together to form a cone as shown in the diagram. Find the volume of the cone.
Answer: Radius of the sector (\(R\)) = \(15\text{ cm}\), Sector angle (\(\theta\)) = \(120^\circ\)
Arc length of the sector = \(\frac{\theta}{360^\circ} \times 2\pi R = \frac{120^\circ}{360^\circ} \times 2\pi \times 15 = 10\pi\text{ cm}\)
When rolled to form a cone:
1. Slant height of the cone (\(l\)) = Radius of the sector (\(R\)) = \(15\text{ cm}\)
2. Circumference of the cone's base = Arc length of the sector \(\implies 2\pi r = 10\pi \implies r = 5\text{ cm}\)
3. Height of the cone (\(h\)) = \(\sqrt{l^2 - r^2} = \sqrt{15^2 - 5^2} = \sqrt{200} = 10\sqrt{2}\text{ cm} \approx 14.14\text{ cm}\) (using \(\sqrt{2} \approx 1.414\))
Volume of the cone (\(V\)) = \(\frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 5^2 \times 14.14 \approx 370.33\text{ cm}^3\)

Question. If the \(\Delta ABC\) with sides 5cm, 12cm, 13cm is revolved about the side 5cm, then find the volume of the solid so obtained. Find also the volume of the solid obtained by revolving the \(\Delta ABC\) about the side 12cm. Find also the ratio of the volumes of the two solids obtained.
Answer: Let the right-angled \(\Delta ABC\) have sides \(5\text{ cm}\), \(12\text{ cm}\), and hypotenuse \(13\text{ cm}\).
1. When revolved about the side \(5\text{ cm}\):
Height (\(h_1\)) = \(5\text{ cm}\), Radius (\(r_1\)) = \(12\text{ cm}\)
Volume of the solid (\(V_1\)) = \(\frac{1}{3}\pi r_1^2 h_1 = \frac{1}{3}\pi (12)^2(5) = 240\pi\text{ cm}^3\)
2. When revolved about the side \(12\text{ cm}\):
Height (\(h_2\)) = \(12\text{ cm}\), Radius (\(r_2\)) = \(5\text{ cm}\)
Volume of the solid (\(V_2\)) = \(\frac{1}{3}\pi r_2^2 h_2 = \frac{1}{3}\pi (5)^2(12) = 100\pi\text{ cm}^3\)
3. Ratio of the volumes:
\(\frac{V_1}{V_2} = \frac{240\pi}{100\pi} = \frac{12}{5}\) or \(12:5\)

Question. The height of right circular cone is 20cm and the radius of its base 4.5cm. It is cut off through the mid point of its height parallel to the base. Find the ratio of the volume of the upper part to that of the lower part.
Answer: Let the height of the original cone be \(H = 20\text{ cm}\) and the radius of its base be \(R = 4.5\text{ cm}\).
The cone is cut at the midpoint of its height, so the height of the smaller upper cone is \(h = \frac{H}{2} = 10\text{ cm}\).
By similar triangles, the radius of the upper cone is \(r = \frac{R}{2} = 2.25\text{ cm}\).
Volume of the upper cone (\(V_{\text{upper}}\)) = \(\frac{1}{3}\pi r^2 h\)
Volume of the original cone (\(V_{\text{whole}}\)) = \(\frac{1}{3}\pi R^2 H = \frac{1}{3}\pi (2r)^2(2h) = 8 \times \left(\frac{1}{3}\pi r^2 h\right) = 8 V_{\text{upper}}\)
Volume of the lower part (\(V_{\text{lower}}\)) = \(V_{\text{whole}} - V_{\text{upper}} = 8 V_{\text{upper}} - V_{\text{upper}} = 7 V_{\text{upper}}\)
Ratio of the volume of the upper part to the lower part:
\(\frac{V_{\text{upper}}}{V_{\text{lower}}} = \frac{1}{7}\) or \(1:7\)

Question. A cylindrical pencil is sharpened to produce a perfect cone at one end with no overall loss of length. If the diameter of the pencil is 1cm and the length of the conical portion is 2cm. Calculate the volume of the shavings. Use \(\pi = \frac{355}{113}\).
Answer: Diameter of the pencil = \(1\text{ cm} \implies\) Radius (\(r\)) = \(0.5\text{ cm}\)
Length of the conical portion (\(h\)) = \(2\text{ cm}\)
The volume of the shavings is the difference between the volume of the cylinder (before sharpening) and the volume of the cone (after sharpening) for the same portion:
\(\text{Volume of shavings} = \pi r^2 h - \frac{1}{3}\pi r^2 h = \frac{2}{3}\pi r^2 h\)
Using \(\pi = \frac{355}{113}\):
\(\text{Volume of shavings} = \frac{2}{3} \times \frac{355}{113} \times (0.5)^2 \times 2 = \frac{2}{3} \times \frac{355}{113} \times 0.25 \times 2 = \frac{355}{339} \approx 1.05\text{ cm}^3\)

Question. Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 14cm. [Hint: The largest cone has \(r = \frac{14}{2}\text{ cm}\), \(h = 14\text{ cm}\)]
Answer: Edge of the cube = \(14\text{ cm}\)
For the largest cone cut from the cube:
Radius (\(r\)) = \(\frac{14}{2} = 7\text{ cm}\)
Height (\(h\)) = \(14\text{ cm}\)
Volume of the cone (\(V\)) = \(\frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 14 = \frac{2156}{3} \approx 718.67\text{ cm}^3\)

Question. If h, c, V respectively are the height, the curved surface and volume of cone Prove that \(3\pi V h^3 - c^2 h^2 + 9 V^2 = 0\)
Answer: Let \(r\) be the base radius and \(l\) be the slant height of the cone.
We know:
1. Volume (\(V\)) = \(\frac{1}{3}\pi r^2 h \implies r^2 = \frac{3V}{\pi h}\)
2. Curved Surface Area (\(c\)) = \(\pi r l \implies c^2 = \pi^2 r^2 l^2 = \pi^2 r^2 (r^2 + h^2)\)
Substituting \(r^2 = \frac{3V}{\pi h}\) into the expression for \(c^2\):
\(c^2 = \pi^2 \left(\frac{3V}{\pi h}\right) \left(\frac{3V}{\pi h} + h^2\right)\)
\(c^2 = \frac{3\pi V}{h} \left(\frac{3V + \pi h^3}{\pi h}\right)\)
\(c^2 = \frac{3V(3V + \pi h^3)}{h^2}\)
\(c^2 h^2 = 9V^2 + 3\pi V h^3\)
Rearranging the terms, we get:
\(3\pi V h^3 - c^2 h^2 + 9V^2 = 0\)
Hence proved.