CBSE Class 10 Mathematics Surface Area and Volume Assignment Set 10

X - Mathematics Assignment No - 05 - Surface Area and Volume

Q1. The radius and height of a right circular cone are in the ratio 3 : 4. If its volume is 301.44 cm³, find the radius and the slant height of the cone (use \(\pi = 3.14\))
Answer: Let the radius of the cone be \(r = 3x\) cm and the height of the cone be \(h = 4x\) cm.
The volume of a cone is given by the formula:
\(V = \frac{1}{3} \pi r^2 h\)
Substitute the given values into the formula:
\(301.44 = \frac{1}{3} \times 3.14 \times (3x)^2 \times (4x)\)
\(301.44 = \frac{1}{3} \times 3.14 \times 9x^2 \times 4x\)
\(301.44 = 3.14 \times 12 x^3\)
\(301.44 = 37.68 x^3\)
\(x^3 = \frac{301.44}{37.68} = 8\)
\(x = \sqrt[3]{8} = 2\)
Therefore, radius \(r = 3 \times 2 = 6\) cm and height \(h = 4 \times 2 = 8\) cm.
Slant height \(l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\) cm.
In simple words: Using the ratio variables, we found the actual radius and height by equating the volume formula to the given volume, then calculated the slant height using Pythagoras theorem.

Exam Tip: Whenever dimensions are given in a ratio, introduce a common variable like \(x\) to express the actual lengths before substituting into formulas.

Q2. A conical tent is to accommodate 11 persons. Each person must have 4 sq. meter of space on the ground and 20 m³ of air to breathe. Find the height of cone
Answer: Total base area for 11 persons = \(11 \times 4 = 44\) m².
Since the base of a conical tent is a circle, Base Area \(= \pi r^2 = 44\) m².
Total volume of air for 11 persons = \(11 \times 20 = 220\) m³.
The volume of a cone is given by:
\(V = \frac{1}{3} \times (\text{Base Area}) \times h\)
\(220 = \frac{1}{3} \times 44 \times h\)
\(h = \frac{220 \times 3}{44}\)
\(h = 5 \times 3 = 15\) m.
In simple words: The ground space gives us the base area of the cone, and the air volume gives the total volume. Using the cone volume formula, we can easily solve for the height.

Exam Tip: Remember that the floor space of a tent represents the circular base area (\(\pi r^2\)) and the air space represents the total volume (\(\frac{1}{3}\pi r^2 h\)).

Q3. The vertical height of a conical tent is 14 m and the diameter of its base is 18 m. How many persons can it accommodate, if each person is to be allowed 108 m³ of space
Answer: Height of the cone \(h = 14\) m.
Diameter \(= 18\) m, so radius \(r = \frac{18}{2} = 9\) m.
Volume of the conical tent \(V = \frac{1}{3} \pi r^2 h\)
\(V = \frac{1}{3} \times \frac{22}{7} \times 9^2 \times 14\)
\(V = \frac{1}{3} \times \frac{22}{7} \times 81 \times 14\)
\(V = 22 \times 27 \times 2 = 1188\) m³.
Number of persons that can be accommodated = \(\frac{\text{Total volume}}{\text{Space required per person}}\)
\(\text{Number of persons} = \frac{1188}{108} = 11\)
In simple words: First calculate the total volume of the tent, then divide it by the space required by each person to find the total capacity.

Exam Tip: Always check whether dimensions are given as radius or diameter before substituting into volume formulas.

Q4. An iron piller consists of a cylindrical portion 2.8 m high and 20 cm in diameter and a cone 42 cm high is surmounting it. Find the weight of the piller, if 1 cm³ of iron weighs 7.5 g.
Answer: For the cylindrical part:
Height \(h_1 = 2.8 \text{ m} = 280\) cm.
Diameter \(= 20\) cm, so radius \(r = \frac{20}{2} = 10\) cm.
Volume of cylinder \(V_1 = \pi r^2 h_1 = \frac{22}{7} \times (10)^2 \times 280 = \frac{22}{7} \times 100 \times 280 = 22 \times 100 \times 40 = 88000\) cm³.
For the conical part:
Height \(h_2 = 42\) cm.
Radius \(r = 10\) cm.
Volume of cone \(V_2 = \frac{1}{3} \pi r^2 h_2 = \frac{1}{3} \times \frac{22}{7} \times (10)^2 \times 42 = \frac{1}{3} \times \frac{22}{7} \times 100 \times 42 = 22 \times 100 \times 2 = 4400\) cm³.
Total volume of the pillar \(V = V_1 + V_2 = 88000 + 4400 = 92400\) cm³.
Weight of 1 cm³ of iron = 7.5 g.
Total weight \(= 92400 \times 7.5 = 693000\) g \(= 693\) kg.
In simple words: Find the volume of the cylinder and the cone separately, add them to get total volume, and multiply by the density to get the total weight in kg.

Exam Tip: Convert all dimensions to the same unit (centimetres) before calculating volumes to avoid unit mismatch errors.

Q5. A heap of wheat is in the form of a cone of diameter 9 m and height 3.5 m. Find its volume. How much canvas is required to just cover the heap ?
Answer: Diameter \(= 9\) m, so radius \(r = \frac{9}{2} = 4.5\) m.
Height \(h = 3.5\) m.
Volume of the conical heap \(V = \frac{1}{3} \pi r^2 h\)
\(V = \frac{1}{3} \times \frac{22}{7} \times (4.5)^2 \times 3.5\)
\(V = \frac{1}{3} \times \frac{22}{7} \times 20.25 \times 3.5 = 74.18\) m³.
Slant height \(l = \sqrt{r^2 + h^2} = \sqrt{(4.5)^2 + (3.5)^2} = \sqrt{20.25 + 12.25} = \sqrt{32.5} = 5.70\) m.
Canvas required to cover the heap = Curved Surface Area of the cone (\(\pi r l\))
\(\text{CSA} = \frac{22}{7} \times 4.5 \times 5.70 = 80.54\) m².
In simple words: Find the volume using the standard formula. To cover the heap, we only need the curved surface area, which requires finding the slant height first.

Exam Tip: To cover a solid heap with canvas, calculate the curved surface area (CSA), not the total surface area (TSA), since the base resting on the ground is not covered.

Q6. The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be \(\frac{1}{27}\) of the volume of the given cone, at what height above the base, the section has been made?
Answer: Let the height of the original cone be \(H = 30\) cm and radius be \(R\).
Let the height of the smaller cut-out cone be \(h\) and radius be \(r\).
By similar triangles, the ratio of radii is equal to the ratio of heights:
\(\frac{r}{R} = \frac{h}{H}\)
Given that the volume of the small cone is \(\frac{1}{27}\) of the volume of the large cone:
\(\frac{\frac{1}{3} \pi r^2 h}{\frac{1}{3} \pi R^2 H} = \frac{1}{27}\)
\(\left(\frac{r}{R}\right)^2 \times \left(\frac{h}{H}\right) = \frac{1}{27}\)
Substitute \(\frac{r}{R} = \frac{h}{H}\):
\(\left(\frac{h}{H}\right)^3 = \frac{1}{27}\)
\(\frac{h}{H} = \sqrt[3]{\frac{1}{27}} = \frac{1}{3}\)
Since \(H = 30\) cm, height of the small cone \(h = \frac{1}{3} \times 30 = 10\) cm.
The question asks for the height *above the base* where the section has been made:
Height above base = \(H - h = 30 - 10 = 20\) cm.
In simple words: Using similar triangles, the ratio of volumes is the cube of the ratio of heights. Once we find the height of the small cone from the vertex, subtracting it from the total height gives the height above the base.

Exam Tip: Read carefully whether the question asks for the height from the vertex or the height above the base.

Q7. A conical vessel whose internal radius is 10 cm and height 36 cm is full of water. The water is emptied into a cylindrical vessel with internal radius 20 cm. Find the height to which the water rises.
Answer: For the conical vessel:
Radius \(r = 10\) cm, Height \(h = 36\) cm.
Volume of water in cone \(V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \pi \times (10)^2 \times 36 = \frac{1}{3} \times \pi \times 100 \times 36 = 1200\pi\) cm³.
Let the height to which water rises in the cylindrical vessel be \(H\).
Radius of cylinder \(R = 20\) cm.
Volume of cylinder = Volume of water emptied from cone
\(\pi R^2 H = 1200\pi\)
\(\pi \times (20)^2 \times H = 1200\pi\)
\(400 \times H = 1200\)
\(H = \frac{1200}{400} = 3\) cm.
In simple words: Equate the volume of water in the cone to the volume of water in the cylinder to find the required height.

Exam Tip: When liquid is poured from one vessel into another, equate their volumes directly without calculating the numerical value of \(\pi\) until the final step.

Q8. A cylindrical bucket, 32 cm high and 18 cm of radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Answer: For the cylindrical bucket:
Radius \(r_1 = 18\) cm, Height \(h_1 = 32\) cm.
Volume of sand \(V = \pi r_1^2 h_1 = \pi \times (18)^2 \times 32 = \pi \times 324 \times 32 = 10368\pi\) cm³.
For the conical heap:
Height \(h_2 = 24\) cm.
Let the radius of the heap be \(r_2\).
Volume of conical heap = Volume of cylindrical bucket
\(\frac{1}{3} \pi r_2^2 h_2 = 10368\pi\)
\(\frac{1}{3} \times \pi \times r_2^2 \times 24 = 10368\pi\)
\(8 \pi r_2^2 = 10368\pi\)
\(r_2^2 = \frac{10368}{8} = 1296\)
\(r_2 = \sqrt{1296} = 36\) cm.
Slant height of the heap \(l = \sqrt{r_2^2 + h_2^2} = \sqrt{36^2 + 24^2} = \sqrt{1296 + 576} = \sqrt{1872} = 43.27\) cm.
In simple words: Equate the volume of the cylinder to the volume of the cone to find the radius of the heap, then use Pythagoras theorem to find the slant height.

Exam Tip: \(\pi\) gets cancelled out on both sides when equating volumes, making calculations much simpler.

Q9. Drive the formula for volume of frustum, in terms of \(r_1, r_2, h, r_1 > r_2\)
Answer: Consider a complete cone of radius \(r_1\) and height \(H\). A small cone of radius \(r_2\) and height \(H - h\) is cut off from the top to form a frustum of height \(h$.
By similar triangles, the ratio of the radii is equal to the ratio of the heights:
\(\frac{r_2}{r_1} = \frac{H - h}{H} = 1 - \frac{h}{H}\)
\(\frac{h}{H} = 1 - \frac{r_2}{r_1} = \frac{r_1 - r_2}{r_1}\)
From this, we get the total height \(H\):
\(H = \frac{h r_1}{r_1 - r_2}\)
And the height of the small cone \(H - h\):
\(H - h = \frac{h r_2}{r_1 - r_2}\)
Volume of the frustum \(V\) = Volume of large cone - Volume of small cone
\(V = \frac{1}{3} \pi r_1^2 H - \frac{1}{3} \pi r_2^2 (H - h)\)
\(V = \frac{1}{3} \pi \left[ r_1^2 \left(\frac{h r_1}{r_1 - r_2}\right) - r_2^2 \left(\frac{h r_2}{r_1 - r_2}\right) \right]\)
\(V = \frac{1}{3} \pi \left( \frac{h}{r_1 - r_2} \right) (r_1^3 - r_2^3)\)
Using the algebraic identity \(r_1^3 - r_2^3 = (r_1 - r_2)(r_1^2 + r_2^2 + r_1 r_2)\):
\(V = \frac{1}{3} \pi \left( \frac{h}{r_1 - r_2} \right) (r_1 - r_2)(r_1^2 + r_2^2 + r_1 r_2)\)
\(V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)\)
In simple words: The frustum volume is derived by subtracting the volume of the small cut-out cone from the large original cone using similar triangles and algebraic factorization.

Exam Tip: Remember the algebraic formula for the difference of cubes: \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\) as it is essential for simplifying frustum derivations.

Q10. Drive the formula for CSA and T.S.A of frustum in terms of \(r_1, r_2, h, l, r_1 > r_2\)
Answer: Let \(l\) be the slant height of the frustum, given by:
\(l = \sqrt{h^2 + (r_1 - r_2)^2}\)
Curved Surface Area (CSA) of the frustum = CSA of large cone - CSA of small cut-out cone.
Let \(l_1\) be the slant height of the large cone and \(l_2\) be the slant height of the small cone, so \(l_1 = l + l_2\).
\(\text{CSA} = \pi r_1 l_1 - \pi r_2 l_2 = \pi r_1 (l + l_2) - \pi r_2 l_2 = \pi r_1 l + \pi l_2 (r_1 - r_2)\)
Using similar triangles for slant heights: \(\frac{l_2}{r_2} = \frac{l}{r_1 - r_2}\), which gives \(\frac{l_2 (r_1 - r_2)}{r_2} = l\) or similar proportion relationships leading to:
\(\text{CSA} = \pi (r_1 + r_2) l\)
Total Surface Area (TSA) of the frustum = CSA + Area of top circular base + Area of bottom circular base
\(\text{TSA} = \pi (r_1 + r_2) l + \pi r_1^2 + \pi r_2^2\)
In simple words: The curved surface area is obtained by taking the sum of the circumferences of both circular ends multiplied by the slant height divided by 2, and the total surface area includes the top and bottom circular bases as well.

Exam Tip: In formulas for frustum areas, always remember that \((r_1 + r_2)\) is used in place of \(2r\) from the standard cone formulas.