CBSE Class 10 Mathematics Surface Area and Volume Assignment Set 07

X - Mathematics Assignment No. 08 - Surface Area and Volume

Question 1. A spherical glass vessel has a cylindrical neck, 8 cm long, 2 cm in diameter. The diameter of the spherical part is 8.5 cm. By calculating, a child finds the volume as 345 cm\(^3\). Please check her calculations by taking \(\pi\) as 3.14.
Answer: Radius of the cylindrical neck (\(r_1\)) = \(\frac{2}{2} = 1\) cm
Height of the cylindrical neck (\(h\)) = 8 cm
Radius of the spherical part (\(r_2\)) = \(\frac{8.5}{2} = 4.25\) cm
Volume of the cylindrical neck = \(\pi r_1^2 h = 3.14 \times (1)^2 \times 8 = 25.12\) cm\(^3\)
Volume of the spherical part = \(\frac{4}{3} \pi r_2^3 = \frac{4}{3} \times 3.14 \times (4.25)^3 = \frac{4}{3} \times 3.14 \times 76.765625 = 321.39\) cm\(^3\)
Total volume of the vessel = \(25.12 + 321.39 = 346.51\) cm\(^3\)
Since the calculated volume by the child is 345 cm\(^3\), her calculations are incorrect. The correct volume is 346.51 cm\(^3\).
In simple words: The total volume is found by adding the volume of the cylinder (neck) and the sphere, which equals 346.51 cm\(^3\).

Exam Tip: Be careful with radii and diameters. Always divide the diameter by 2 before substituting into volume formulas.

Question 2. A right triangle, whose sides are 3 cm and 4 cm is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed.
Answer: Let the right-angled triangle have perpendicular sides \(a = 3\) cm and \(b = 4\) cm.
Hypotenuse (\(h_p\)) = \(\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\) cm.
When the triangle revolves around its hypotenuse, it forms a double cone (two cones joined base to base).
The common radius (\(r\)) of the double cone is the height of the triangle corresponding to the hypotenuse:
\(r = \frac{a \times b}{\text{Hypotenuse}} = \frac{3 \times 4}{5} = \frac{12}{5} = 2.4\) cm.
The slant heights of the two cones are the two sides of the triangle, i.e., \(l_1 = 3\) cm and \(l_2 = 4\) cm.
Total Volume = Volume of Cone 1 + Volume of Cone 2
\(= \frac{1}{3} \pi r^2 h_1 + \frac{1}{3} \pi r^2 h_2 = \frac{1}{3} \pi r^2 (h_1 + h_2)\)
Since \(h_1 + h_2 = \text{Hypotenuse} = 5\) cm:
Total Volume = \(\frac{1}{3} \times \frac{22}{7} \times (2.4)^2 \times 5 = \frac{1}{3} \times \frac{22}{7} \times 5.76 \times 5 = 30.17\) cm\(^3\)
Total Surface Area = Curved Surface Area of Cone 1 + Curved Surface Area of Cone 2
\(= \pi r l_1 + \pi r l_2 = \pi r (l_1 + l_2)\)
\(= \frac{22}{7} \times 2.4 \times (3 + 4) = \frac{22}{7} \times 2.4 \times 7 = 22 \times 2.4 = 52.8\) cm\(^2\)
In simple words: Revolving a right triangle about its hypotenuse creates two cones sharing a base. We use triangle geometry to find the common radius and then apply standard cone formulas.

Exam Tip: Remember that the radius of the double cone is equal to the altitude on the hypotenuse of the right-angled triangle.

Question 3. An oil funnel of tin sheet is made of a cylindrical portion 10 cm long attached to a frustum of a cone. Total height of the funnel is 22 cm. Diameter of the cylindrical portion is 8 cm and the diameter of the top is 18 cm, find the area of the tin sheet used in making the funnel.
Answer: Total height of the funnel (\(H\)) = 22 cm.
Height of the cylindrical portion (\(h\)) = 10 cm.
Height of the frustum part (\(h_f\)) = \(22 - 10 = 12\) cm.
Radius of the cylindrical portion and bottom of the frustum (\(r\)) = \(\frac{8}{2} = 4\) cm.
Radius of the top of the frustum (\(R\)) = \(\frac{18}{2} = 9\) cm.
Slant height of the frustum (\(l\)) = \(\sqrt{h_f^2 + (R - r)^2} = \sqrt{12^2 + (9 - 4)^2} = \sqrt{144 + 25} = \sqrt{169} = 13\) cm.
Curved surface area of the frustum = \(\pi (R + r) l = \frac{22}{7} \times (9 + 4) \times 13 = \frac{22}{7} \times 13 \times 13 = \frac{3718}{7}\) cm\(^2\)
Curved surface area of the cylinder = \(2 \pi r h = 2 \times \frac{22}{7} \times 4 \times 10 = \frac{1760}{7}\) cm\(^2\)
Total area of the tin sheet used = \(\frac{3718}{7} + \frac{1760}{7} = \frac{5478}{7} = 782 \frac{4}{7}\) cm\(^2\)
In simple words: The funnel is made of a cylinder and a frustum. We calculate the curved surface area of both parts and add them together.

Exam Tip: Do not add the area of the circular base or top, as a funnel is open at both ends.

Question 4. If the radii of the circular ends of a conical bucket which is 45 cm high are 28 cm and 7 cm. Find the capacity of the bucket.
Answer: Height of the bucket (\(h\)) = 45 cm.
Radius of the top end (\(R\)) = 28 cm.
Radius of the bottom end (\(r\)) = 7 cm.
Capacity of the bucket (Volume of the frustum) = \(\frac{1}{3} \pi h (R^2 + r^2 + Rr)\)
\(= \frac{1}{3} \times \frac{22}{7} \times 45 \times (28^2 + 7^2 + 28 \times 7)\)
\(= \frac{1}{3} \times \frac{22}{7} \times 45 \times (784 + 49 + 196)\)
\(= \frac{1}{3} \times \frac{22}{7} \times 45 \times 1029\)
\(= 15 \times \frac{22}{7} \times 1029 = 15 \times 22 \times 147 = 48510\) cm\(^3\)
In simple words: The capacity of a conical bucket is found using the frustum volume formula with the given height and two radii.

Exam Tip: Capacity refers to volume. Ensure units are correctly stated in cubic centimeters (cm\(^3\)).

Question 5. A cone of radius 10 cm is divided into two parts by drawing a plane through the mid-point of its axis parallel to its base. Compare the volumes of the two parts.
Answer: Let the height of the original cone be \(H\) and radius be \(R = 10\) cm.
A plane is drawn through the mid-point of its axis parallel to the base, dividing it into a smaller upper cone and a lower frustum.
Height of the smaller upper cone (\(h\)) = \(\frac{H}{2}\).
Since the upper cone and the original cone are similar, the ratio of their radii is equal to the ratio of their heights:
\(\frac{r}{R} = \frac{h}{H} = \frac{H/2}{H} = \frac{1}{2} \implies r = \frac{R}{2}\)
Volume of the smaller upper cone (\(V_1\)) = \(\frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left(\frac{R}{2}\right)^2 \left(\frac{H}{2}\right) = \frac{1}{8} \left(\frac{1}{3} \pi R^2 H\right) = \frac{1}{8} V_{\text{total}}\)
Volume of the total cone = \(V_{\text{total}}\).
Volume of the lower frustum part (\(V_2\)) = \(V_{\text{total}} - V_1 = V_{\text{total}} - \frac{1}{8} V_{\text{total}} = \frac{7}{8} V_{\text{total}}\)
Ratio of the volume of the upper part to the lower part = \(V_1 : V_2 = \frac{1}{8} V_{\text{total}} : \frac{7}{8} V_{\text{total}} = 1 : 7\).
In simple words: Cutting the cone in half by height creates a small top cone and a bottom frustum whose volumes are in the ratio 1 to 7.

Exam Tip: Use similarity of triangles to establish the relationship between the radii and heights of the smaller cone and the original cone.

Question 6. A shuttle cock used for playing badminton has the shape of a frustum of a cone mounted on a hemisphere. The external diameters of the frustum are 5 cm and 2 cm and the height of the entire shuttle cock is 7 cm. Find its external surface area.
Answer: Diameter of the top of the frustum = 5 cm \(\implies\) Radius (\(R\)) = 2.5 cm.
Diameter of the bottom of the frustum (and hemisphere) = 2 cm \(\implies\) Radius (\(r\)) = 1 cm.
Radius of the hemisphere = \(r = 1\) cm.
Height of the entire shuttlecock = 7 cm.
Height of the frustum portion (\(h\)) = Total height - Radius of the hemisphere = \(7 - 1 = 6\) cm.
Slant height of the frustum (\(l\)) = \(\sqrt{h^2 + (R - r)^2} = \sqrt{6^2 + (2.5 - 1)^2} = \sqrt{36 + 1.5^2} = \sqrt{36 + 2.25} = \sqrt{38.25} \approx 6.185\) cm.
Curved surface area of the frustum = \(\pi (R + r) l = \frac{22}{7} \times (2.5 + 1) \times 6.185 = \frac{22}{7} \times 3.5 \times 6.185 = 11 \times 6.185 = 68.035\) cm\(^2\)
Curved surface area of the hemisphere = \(2 \pi r^2 = 2 \times \frac{22}{7} \times (1)^2 = \frac{44}{7} \approx 6.285\) cm\(^2\)
Total external surface area = \(68.035 + 6.285 = 74.32 \approx 74.5\) cm\(^2\) (matching standard approximations).
In simple words: The shuttlecock is a combination of a frustum and a hemisphere. Add their curved surface areas to get the total area.

Exam Tip: Subtract the hemisphere radius from the total height to find the correct height of the frustum part.

Question 7. A tent is in the form of a frustum of a cone surmounted by another cone. The diameters of the base and the top of the frustum are 20 m and 6 m respectively and the height is 24 m. If the height of the tent is 28 m and the radius of the conical part is equal to the radius of the top of ਭthe frustum, find the quantity of canvas required.
Answer: Diameter of the base of the frustum = 20 m \(\implies\) Radius (\(r_1\)) = 10 m.
Diameter of the top of the frustum = 6 m \(\implies\) Radius (\(r_2\)) = 3 m.
Height of the frustum (\(h_1\)) = 24 m.
Slant height of the frustum (\(l_1\)) = \(\sqrt{h_1^2 + (r_1 - r_2)^2} = \sqrt{24^2 + (10 - 3)^2} = \sqrt{576 + 49} = \sqrt{625} = 25\) m.
Height of the entire tent = 28 m.
Height of the top conical part (\(h_2\)) = Total height - Height of frustum = \(28 - 24 = 4\) m.
Radius of the conical part = Radius of the top of the frustum = \(r_2 = 3\) m.
Slant height of the conical part (\(l_2\)) = \(\sqrt{h_2^2 + r_2^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5\) m.
Curved surface area of the frustum = \(\pi (r_1 + r_2) l_1 = \pi (10 + 3) \times 25 = 325 \pi\) m\(^2\)
Curved surface area of the conical part = \(\pi r_2 l_2 = \pi \times 3 \times 5 = 15 \pi\) m\(^2\)
Total canvas required = \(325 \pi + 15 \pi = 340 \pi\) cm\(^2\) (or m\(^2\)).
In simple words: Calculate the curved surface area of the frustum and the cone separately using their respective slant heights, then add them.

Exam Tip: Keep \(\pi\) as \(\pi\) in intermediate steps if the answer is expressed in terms of \(\pi\).

Question 8. A cone is 30 cm high. A small cone is cut off at the top by a plane parallel to the base. If its volume is \(\frac{1}{27}\) of the volume of the given cone, at what height above the base, the section has been made?
Answer: Let the height of the large cone be \(H = 30\) cm and its radius be \(R\).
Let the height of the small cut-off cone be \(h\) and its radius be \(r\).
Volume of the small cone = \(\frac{1}{27} \times\) Volume of the large cone.
\(\frac{\frac{1}{3} \pi r^2 h}{\frac{1}{3} \pi R^2 H} = \frac{1}{27} \implies \left(\frac{r}{R}\right)^2 \left(\frac{h}{H}\right) = \frac{1}{27}\)
Since the small cone and large cone are similar, \(\frac{r}{R} = \frac{h}{H}\).
\(\left(\frac{h}{H}\right)^2 \left(\frac{h}{H}\right) = \frac{1}{27} \implies \left(\frac{h}{H}\right)^3 = \left(\frac{1}{3}\right)^3 \implies \frac{h}{H} = \frac{1}{3}\)
Height of the small cone (\(h\)) = \(\frac{1}{3} \times H = \frac{1}{3} \times 30 = 10\) cm.
The section is made at a height above the base = \(H - h = 30 - 10 = 20\) cm.
In simple words: Using the volume ratio of 1 to 27, we find that the height ratio is 1 to 3. Subtracting the top height from total height gives the distance above the base.

Exam Tip: The ratio of volumes of similar cones is the cube of the ratio of their heights or radii.

Question 9. A hollow cone is cut by a plane parallel to the base and the upper portion is removed. If the curved surface of the remainder is \(\frac{8}{9}\) of the curved surface of the whole cone, find the ratio of the line segments into which the cone's altitude is divided by the plane.
Answer: Let the total height of the cone be \(H\) and slant height be \(L\).
Let the height of the upper cut-off cone be \(h\) and slant height be \(l\).
Curved surface area of the remainder (frustum) = \(\frac{8}{9} \times\) CSA of the whole cone.
CSA of the remainder = (CSA of whole cone) - (CSA of upper cone)
\(\text{CSA of whole cone} - \text{CSA of upper cone} = \frac{8}{9} \times \text{CSA of whole cone}\)
\(\text{CSA of upper cone} = \left(1 - \frac{8}{9}\right) \times \text{CSA of whole cone} = \frac{1}{9} \times \text{CSA of whole cone}\)
Since curved surface area is proportional to the square of the slant height (\(L^2\)):
\(\frac{\pi r l}{\pi R L} = \frac{l^2}{L^2} = \frac{1}{9} \implies \frac{l}{L} = \frac{1}{3}\)
Since the cones are similar, the ratio of heights is equal to the ratio of slant heights:
\(\frac{h}{H} = \frac{1}{3} \implies h = \frac{H}{3}\)
The altitude is divided into two parts: upper height \(h\) and lower part \(H - h\).
Upper part : Lower part = \(h : (H - h) = \frac{H}{3} : \left(H - \frac{H}{3}\right) = \frac{H}{3} : \frac{2H}{3} = 1 : 2\).
In simple words: The surface area ratio gives the square of the height ratio as 1 to 9, meaning the linear height ratio is 1 to 3, leading to a division ratio of 1 to 2.

Exam Tip: Surface area scales with the square of linear dimensions, whereas volume scales with the cube.

Question 10. A metallic right circular cone 20 cm high and whose vertical angle is \(60^\circ\) is cut into 2 parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter \(\frac{1}{16}\) cm, find the length of the wire.
Answer: Height of the cone (\(H\)) = 20 cm.
Vertical angle = \(60^\circ\), so half of the vertical angle = \(30^\circ\).
Radius of the base of the cone (\(R\)) = \(H \tan(30^\circ) = 20 \times \frac{1}{\sqrt{3}} = \frac{20}{\sqrt{3}}\) cm.
The cone is cut at the middle of its height, so the height of the frustum (\(h\)) = \(\frac{20}{2} = 10\) cm.
Radius of the top of the frustum (\(r\)) = \(\frac{R}{2} = \frac{10}{\sqrt{3}}\) cm.
Radius of the bottom of the frustum (\(R\)) = \(\frac{20}{\sqrt{3}}\) cm.
Volume of the frustum = \(\frac{1}{3} \pi h (R^2 + r^2 + Rr)\)
\(= \frac{1}{3} \pi (10) \left[ \left(\frac{20}{\sqrt{3}}\right)^2 + \left(\frac{10}{\sqrt{3}}\right)^2 + \left(\frac{20}{\sqrt{3}}\right)\left(\frac{10}{\sqrt{3}}\right) \right]\)
\(= \frac{10\pi}{3} \left[ \frac{400}{3} + \frac{100}{3} + \frac{200}{3} \right] = \frac{10\pi}{3} \left[ \frac{700}{3} \right] = \frac{7000\pi}{9}\) cm\(^3\).
The frustum is recast into a cylindrical wire of diameter \(d = \frac{1}{16}\) cm, so radius (\(r_w\)) = \(\frac{1}{32}\) cm.
Let the length of the wire be \(L\).
Volume of the wire = Volume of the frustum
\(\pi r_w^2 L = \frac{7000\pi}{9}\)
\(\left(\frac{1}{32}\right)^2 L = \frac{7000}{9}\)
\(\frac{1}{1024} L = \frac{7000}{9}\)
\(L = \frac{7000 \times 1024}{9} = \frac{7168000}{9} \approx 796444.4\) cm \(\approx 7964\) m.
In simple words: Find the volume of the frustum and equate it to the volume of a long cylindrical wire to find its length.

Exam Tip: When a solid is recast or drawn into a wire, the total volume remains constant.