ML Aggarwal Class 10 Maths Solutions Chapter 17 Mensuration

Question 1. Find the total surface area of a solid cylinder of radius 5 cm and height 10 cm. Leave your answer in terms of π.
Answer: We are given that the radius measures 5 cm and the height is 10 cm. Using the formula for total surface area of a cylinder, which is \( 2\pi r(h + r) \), we calculate: \( 2\pi \times 5 \times (10 + 5) = 10\pi \times 15 = 150\pi \) cm². Therefore, the total surface area of the solid cylinder is \( 150\pi \) cm².
In simple words: Multiply the radius by 2, then by pi, then by the sum of height and radius to find the total surface area.

Exam Tip: Always remember that the total surface area formula includes both circular bases and the curved surface - don't forget to add the radius to the height inside the brackets.

Question 2. An electric geyser is cylindrical in shape, having a diameter of 35 cm and height 1.2 m. Neglecting the thickness of its walls, calculate
(i) its outer lateral surface area
(ii) its capacity in litres.
Answer:
(i) We are given that the diameter is 35 cm, so the radius is 17.5 cm. The height is 1.2 m, which equals 120 cm. The curved surface area (which is the lateral surface area for a cylinder) is calculated using the formula \( 2\pi rh \). Substituting the values: \( 2 \times \frac{22}{7} \times 17.5 \times 120 = \frac{92400}{7} = 13200 \) cm². Thus, the outer lateral surface area of the electric geyser is 13200 cm².

(ii) To find the capacity, we use the volume formula \( \pi r^2 h \). Substituting: \( \frac{22}{7} \times (17.5)^2 \times 120 = \frac{22}{7} \times 306.25 \times 120 = \frac{808500}{7} = 115500 \) cm³. Since 1000 cm³ equals 1 litre, we divide: \( 115500 \times \frac{1}{1000} = 115.5 \) litres. Therefore, the capacity of the electric geyser is 115.5 litres.
In simple words: For lateral surface area, use the curved surface formula with radius and height. For capacity, calculate the volume and convert cubic centimetres to litres by dividing by 1000.

Exam Tip: Watch for unit conversions - convert all measurements to the same unit before applying formulas, and remember that 1000 cm³ = 1 litre for capacity questions.

Question 3. A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. If the glass is filled with milk upto a height of 12 cm, find how many litres of milk is needed to serve 1600 students.
Answer: The diameter of each glass is 7 cm, so the radius is 3.5 cm. The height is 12 cm. Using the volume formula \( V = \pi r^2 h \), we calculate the volume of milk in one glass: \( \frac{22}{7} \times (3.5)^2 \times 12 = \frac{22}{7} \times 12.25 \times 12 = \frac{3234}{7} = 462 \) cm³. For 1600 students, the total volume needed is \( 1600 \times 462 = 739200 \) cm³. Converting to litres: \( 739200 \times \frac{1}{1000} = 739.2 \) litres. Hence, 739.2 litres of milk is required for serving 1600 students.
In simple words: Find the volume of milk in one glass, multiply by the number of students, then convert cubic centimetres to litres.

Exam Tip: Always calculate the volume for one unit first, then scale up for the required quantity - this reduces the chance of calculation errors.

Question 4. In the given figure, a rectangular tin foil of size 22 cm by 16 cm is wrapped around to form a cylinder of height 16 cm. Find the volume of the cylinder.
Answer: When the rectangular foil is wrapped to form a cylinder, the height of the cylinder becomes 16 cm (one dimension of the rectangle), and the length of 22 cm becomes the circumference of the base. From the circumference \( 2\pi r = 22 \), we find the radius. Using \( 2 \times \frac{22}{7} \times r = 22 \), we get: \( \frac{44r}{7} = 22 \), which gives \( r = \frac{22 \times 7}{44} = \frac{154}{44} = 3.5 \) cm. Now, using the volume formula \( V = \pi r^2 h \): \( \frac{22}{7} \times (3.5)^2 \times 16 = \frac{22}{7} \times 12.25 \times 16 = \frac{4312}{7} = 616 \) cm³. Therefore, the volume of the cylinder is 616 cm³.
In simple words: The length of the foil becomes the circumference, and the width becomes the height. Find the radius from the circumference, then calculate the volume.

Exam Tip: Recognize that when a rectangle is wrapped into a cylinder, one dimension becomes the circumference and the other becomes the height - this is key to solving such problems.

Question 5(i). How many cubic metres of soil must be take out to make a well 20 metres deep and 2 metres in diameter?
Answer: The well is cylindrical with depth (height) of 20 m and diameter of 2 m. The radius is \( \frac{2}{2} = 1 \) m. The volume of soil to be dug out equals the volume of the well. Using \( V = \pi r^2 h \): \( \frac{22}{7} \times (1)^2 \times 20 = \frac{22}{7} \times 1 \times 20 = \frac{440}{7} = 62\frac{6}{7} \) m³. Therefore, \( 62\frac{6}{7} \) cubic metres of soil must be dug out for the formation of the well.
In simple words: The amount of soil removed equals the volume of the cylindrical well - use the volume formula with the given radius and depth.

Exam Tip: The volume of soil removed is simply the volume of the cylindrical space created - remember that the depth of the well is its height as a cylinder.

Question 5(ii). If the inner curved surface of the well in part (i) above is to be plastered at the rate of Rs 300 per m², find the cost of plastering (rounded to the nearest hundred rupees).
Answer: The inner curved surface area of the well needs to be plastered. Using the formula for curved surface area \( 2\pi rh \), where \( r = 1 \) m and \( h = 20 \) m: \( 2 \times \frac{22}{7} \times 1 \times 20 = \frac{880}{7} \) m². The cost of plastering is the curved surface area multiplied by the rate: \( \frac{880}{7} \times 300 = \frac{264000}{7} = 37,714.28 \) rupees, which rounds to Rs 37,700 when rounded to the nearest hundred rupees. Therefore, the cost of plastering the inner curved surface area of the well is Rs 37,700.
In simple words: Calculate the curved surface area using radius and height, multiply by the rate per square metre, then round to the nearest hundred.

Exam Tip: For cost problems, always multiply area by the given rate, and check the rounding instruction carefully - ensure you round to the correct place value requested.

Question 6. A roadroller (in the shape of the cylinder) has a diameter 0.7 m and its width is 1.2 m. Find the least number of revolutions that the roller must make in order to level a playground of size 120 m by 44 m.
Answer: The roadroller is cylindrical with diameter 0.7 m, so the radius is 0.35 m. The width (height) of the cylinder is 1.2 m. In one complete revolution, the roller covers an area equal to its curved surface area. The curved surface area is \( 2\pi rh = 2 \times \frac{22}{7} \times 0.35 \times 1.2 = \frac{18.48}{7} = 2.64 \) m². The total area of the playground is \( 120 \times 44 = 5280 \) m². The number of revolutions required is \( \frac{5280}{2.64} = 2000 \). Therefore, the minimum number of revolutions required to level the playground is 2000.
In simple words: The area covered in one revolution is the curved surface area of the cylinder. Divide the playground area by this to get the number of revolutions needed.

Exam Tip: Remember that a rolling cylinder covers area equal to its curved surface area in each revolution - total revolutions = playground area divided by curved surface area.

Question 7(i). If the volume of a cylinder of height 7 cm is 448π cm³, find its lateral surface area and total surface area.
Answer: We are given that \( V = 448\pi \) cm³ and \( h = 7 \) cm. Using the volume formula \( V = \pi r^2 h \), we have \( \pi r^2 \times 7 = 448\pi \). Dividing by \( \pi \), we get \( r^2 \times 7 = 448 \), so \( r^2 = 64 \), which gives \( r = 8 \) cm. The lateral (curved) surface area is \( 2\pi rh = 2\pi \times 8 \times 7 = 112\pi \) cm². The total surface area is \( 2\pi r(h + r) = 2\pi \times 8 \times (7 + 8) = 2\pi \times 8 \times 15 = 240\pi \) cm². Therefore, the lateral surface area is \( 112\pi \) cm² and the total surface area is \( 240\pi \) cm².
In simple words: From the volume, find the radius first. Then use the radius and height to calculate both lateral and total surface areas.

Exam Tip: When volume is given, work backwards to find the radius before calculating surface areas - this makes the subsequent calculations straightforward.

Question 7(ii). A wooden pole is 7 m high and 20 cm in diameter. Find its weight if the wood weighs 225 kg per m³.
Answer: The pole is cylindrical with height 7 m and diameter 20 cm, which converts to 0.2 m. The radius is \( \frac{0.2}{2} = 0.1 \) m. Using the volume formula \( V = \pi r^2 h \), we get: \( \frac{22}{7} \times (0.1)^2 \times 7 = \frac{22}{7} \times 0.01 \times 7 = 22 \times 0.01 = 0.22 \) m³. Since the wood weighs 225 kg per m³, the total weight of the pole is \( 225 \times 0.22 = 49.5 \) kg. Therefore, the weight of the wooden pole is 49.5 kg.
In simple words: Convert measurements to metres, find the volume of the cylindrical pole, then multiply by the weight per unit volume.

Exam Tip: Always convert all dimensions to the same unit system before calculating volume - mixing units leads to errors. Also ensure the density/weight unit matches the volume unit.

Question 8. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. Find the
(i) radius of the cylinder.
(ii) volume of the cylinder.
Answer:
(i) The circumference of the base is given by \( C = 2\pi r \). We have \( 2\pi r = 132 \). Substituting \( \pi = \frac{22}{7} \): \( 2 \times \frac{22}{7} \times r = 132 \), which gives \( \frac{44r}{7} = 132 \). Solving for \( r \): \( r = \frac{132 \times 7}{44} = \frac{924}{44} = 21 \) cm. Therefore, the radius of the cylinder is 21 cm.

(ii) With \( r = 21 \) cm and \( h = 25 \) cm, the volume is \( V = \pi r^2 h = \frac{22}{7} \times (21)^2 \times 25 = \frac{22}{7} \times 441 \times 25 = \frac{242550}{7} = 34650 \) cm³. Therefore, the volume of the cylinder is 34650 cm³.
In simple words: Use the circumference formula to find the radius first, then apply the volume formula with the radius and given height.

Exam Tip: When circumference is provided, extract the radius using \( r = \frac{C}{2\pi} \) before finding any other measurements - this saves time and prevents errors.

Question 9. The area of the curved surface of a cylinder is 4400 cm², and the circumference of its base is 110 cm. Find :
(i) the height of the cylinder.
(ii) the volume of the cylinder.
Answer:
(i) We know that curved surface area \( = 2\pi rh \) and circumference \( = 2\pi r \). Let equation (i): \( 2\pi rh = 4400 \) and equation (ii): \( 2\pi r = 110 \). Dividing equation (i) by equation (ii): \( \frac{2\pi rh}{2\pi r} = \frac{4400}{110} \), which simplifies to \( h = 40 \) cm. Therefore, the height of the cylinder is 40 cm.

(ii) From equation (ii), \( 2\pi r = 110 \), so \( 2 \times \frac{22}{7} \times r = 110 \), giving \( \frac{44r}{7} = 110 \). Solving: \( r = \frac{110 \times 7}{44} = \frac{770}{44} = 17.5 \) cm. Now, using the volume formula \( V = \pi r^2 h = \frac{22}{7} \times (17.5)^2 \times 40 = \frac{22}{7} \times 306.25 \times 40 = \frac{269500}{7} = 38500 \) cm³. Therefore, the volume of the cylinder is 38500 cm³.
In simple words: Use both given equations to find height by division, then find radius from circumference, and finally calculate volume.

Exam Tip: When both curved surface area and circumference are given, dividing one equation by the other directly yields the height without needing to find the radius first.

Question 10. A cylinder has a diameter of 20 cm. The area of curved surface is 1000 cm². Find
(i) the height of the cylinder correct to one decimal place.
(ii) the volume of the cylinder correct to one decimal place. (Take π = 3.14)
Answer:
(i) The diameter is 20 cm, so the radius is 10 cm. Using the curved surface area formula \( 2\pi rh = 1000 \) with \( \pi = \frac{22}{7} \): \( 2 \times \frac{22}{7} \times 10 \times h = 1000 \). This gives \( \frac{440h}{7} = 1000 \), so \( h = \frac{1000 \times 7}{440} = \frac{7000}{440} = 15.9 \) cm. Therefore, the height of the cylinder is 15.9 cm.

(ii) Using the volume formula \( V = \pi r^2 h \) with \( \pi = 3.14 \), \( r = 10 \) cm, and \( h = 15.9 \) cm: \( V = 3.14 \times (10)^2 \times 15.9 = 3.14 \times 100 \times 15.9 = 4992.6 \) cm³. Therefore, the volume of the cylinder is 4992.6 cm³.
In simple words: Use the curved surface area with the given radius to find height, then use height and radius with the provided value of π to find volume.

Exam Tip: When a specific value of π is provided (like 3.14), use it consistently throughout the problem and round your final answer to the requested decimal place.

Question 11. The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up when writing 310 words on an average. How many words would use up a bottle of ink containing one-fifth of a litre? Answer correct to the nearest 100 words.
Answer: The barrel is cylindrical with height 7 cm and diameter 5 mm, which converts to 0.5 cm. The radius is \( \frac{0.5}{2} = 0.25 \) cm. Using the volume formula \( V = \pi r^2 h \): \( \frac{22}{7} \times (0.25)^2 \times 7 = 22 \times 0.0625 = 1.375 \) cm³. A bottle contains one-fifth of a litre, which is \( \frac{1000}{5} = 200 \) cm³. If one barrel (1.375 cm³) is used for 310 words, then 200 cm³ will be used for \( \frac{200}{1.375} \times 310 = 145.45 \times 310 = 45,090 \) words. Rounding to the nearest 100 words gives 45,100 words. Therefore, a bottle of ink containing one-fifth of a litre would use up approximately 45,100 words.
In simple words: Find the volume of one barrel, then use a proportion to determine how many words can be written with one-fifth of a litre of ink.

Exam Tip: For problems involving proportional relationships, set up a ratio carefully - if barrel volume corresponds to a certain number of words, scale it proportionally for the new volume.

Question 12. Find the ratio between the total surface area of a cylinder to its curved surface area given that its height and radius are 7.5 cm and 3.5 cm.
Answer: The formulas we need are: Total surface area = \( 2\pi r(h + r) \) and Curved surface area = \( 2\pi rh \). Setting up the ratio, we divide the total surface area by the curved surface area. This gives us:
\[ \frac{\text{Total surface area}}{\text{Curved surface area}} = \frac{2\pi r(h + r)}{2\pi rh} = \frac{h + r}{h} \]
Substituting the given values (h = 7.5 cm, r = 3.5 cm):
\[ \frac{7.5 + 3.5}{7.5} = \frac{11}{7.5} = \frac{110}{75} = \frac{22}{15} \]
Therefore, the ratio is 22 : 15.
In simple words: To find the ratio between two surface areas, we divide one by the other. After simplifying with the given measurements, we get 22 : 15.

Exam Tip: Always write both surface area formulas at the start and simplify the ratio step-by-step to avoid arithmetic errors.

Question 13. The radius of the base of a right circular cylinder is halved and the height is doubled. What is the ratio of the volume of the new cylinder to that of the original cylinder?
Answer: The volume formula for a cylinder is \( V = \pi r^2 h \). For the original cylinder: Let height = h and radius = r, so volume = \( \pi r^2 h \). For the new cylinder: Height becomes 2h and radius becomes \( \frac{r}{2} \), so volume = \( \pi \cdot \left(\frac{r}{2}\right)^2 \cdot 2h = \pi \cdot \frac{r^2}{4} \cdot 2h = \frac{\pi r^2 h}{2} \). Now we find the ratio:
\[ \frac{\text{Volume of new cylinder}}{\text{Volume of old cylinder}} = \frac{\frac{\pi r^2 h}{2}}{\pi r^2 h} = \frac{1}{2} \]
Therefore, the ratio is 1 : 2.
In simple words: When you halve the radius, the volume shrinks much more (by a factor of 4). Doubling the height only doubles the volume. Together, the new volume becomes half the original.

Exam Tip: Remember that radius changes affect volume squared, while height changes affect it linearly - this is why halving the radius has a larger effect than doubling the height.

Question 14(i). The sum of the radius and the height of a cylinder is 37 cm and the total surface area of the cylinder is 1628 cm². Find the height and the volume of the cylinder.
Answer: We are given that h + r = 37 cm and total surface area = 1628 cm². Using the total surface area formula: \( 2\pi r(h + r) = 1628 \) Substituting h + r = 37:
\[ 2 \times \frac{22}{7} \times r \times 37 = 1628 \]
\[ r = \frac{1628 \times 7}{2 \times 22 \times 37} = \frac{11396}{1628} = 7 \text{ cm} \]
Since h + r = 37 and r = 7:
\[ h = 37 - 7 = 30 \text{ cm} \]
Now we find the volume using \( V = \pi r^2 h \):
\[ V = \frac{22}{7} \times 7^2 \times 30 = \frac{22}{7} \times 49 \times 30 = 22 \times 7 \times 30 = 4620 \text{ cm}^3 \]
Therefore, height = 30 cm and volume = 4620 cm³.
In simple words: We use the given total surface area along with the sum of radius and height to find the radius first. Then we calculate the height. Finally, we work out the volume using these values.

Exam Tip: Always use the given condition (h + r = 37) to simplify the surface area formula before solving for the radius.

Question 14(ii). The total surface area of cylinder is 352 cm². If its height is 10 cm, then find the diameter of the base.
Answer: We are given that total surface area = 352 cm² and height = 10 cm. Using the total surface area formula: \( 2\pi r(h + r) = 352 \) Substituting h = 10:
\[ 2 \times \frac{22}{7} \times r \times (10 + r) = 352 \]
\[ \frac{44r(10 + r)}{7} = 352 \]
\[ 44r(10 + r) = 2464 \]
\[ 440r + 44r^2 = 2464 \]
Dividing by 44:
\[ 10r + r^2 = 56 \]
\[ r^2 + 10r - 56 = 0 \]
Factoring by splitting the middle term:
\[ r^2 + 14r - 4r - 56 = 0 \]
\[ r(r + 14) - 4(r + 14) = 0 \]
\[ (r - 4)(r + 14) = 0 \]
This gives r = 4 or r = -14. Since radius cannot be negative, r = 4 cm. Diameter = 2r = 2 × 4 = 8 cm.
In simple words: We substitute the height into the surface area formula to get a quadratic equation. We solve this equation to find the radius, then double it to get the diameter.

Exam Tip: When you get a quadratic equation, always factor it carefully or use the quadratic formula. Remember to reject negative values for radius since it represents a physical measurement.

Question 15. The ratio between the curved surface and the total surface of a cylinder is 1 : 2. Find the volume of the cylinder, given that its total surface area is 616 cm².
Answer: We know that Curved surface area = \( 2\pi rh \) and Total surface area = \( 2\pi r(h + r) \). Given that the ratio is 1 : 2:
\[ \frac{2\pi rh}{2\pi r(h + r)} = \frac{1}{2} \]
\[ \frac{h}{h + r} = \frac{1}{2} \]
\[ 2h = h + r \]
\[ h = r \]
Now, using the total surface area formula with h = r:
\[ 2\pi r(r + r) = 616 \]
\[ 2\pi r \cdot 2r = 616 \]
\[ 4\pi r^2 = 616 \]
\[ 4 \times \frac{22}{7} \times r^2 = 616 \]
\[ r^2 = \frac{616 \times 7}{4 \times 22} = \frac{4312}{88} = 49 \]
\[ r = 7 \text{ cm} \]
Since h = r, we have h = 7 cm. Volume = \( \pi r^2 h = \frac{22}{7} \times 7^2 \times 7 = \frac{22}{7} \times 49 \times 7 = 22 \times 49 = 1078 \text{ cm}^3 \)
In simple words: From the given ratio, we discover that the height and radius are equal. We then substitute this relationship into the surface area formula to find their value, and finally calculate the volume.

Exam Tip: When a ratio is given, use it to establish a relationship between variables before substituting into area or volume formulas. This simplifies the calculations significantly.

Question 16. Two cylindrical jars contain the same amount of milk. If their diameters are in the ratio 3 : 4, find the ratio of their heights.
Answer: Since both jars contain the same amount of milk, their volumes are equal. Let the diameters be 3a and 4a. Then the radii are \( \frac{3a}{2} \) and \( \frac{4a}{2} \) respectively. Let the heights be h₁ and h₂. Using the volume formula \( V = \pi r^2 h \):
\[ \pi \times \left(\frac{3a}{2}\right)^2 \times h_1 = \pi \times \left(\frac{4a}{2}\right)^2 \times h_2 \]
\[ \pi \times \frac{9a^2}{4} \times h_1 = \pi \times \frac{16a^2}{4} \times h_2 \]
Dividing both sides by \( \pi \) and multiplying by 4:
\[ 9a^2 \times h_1 = 16a^2 \times h_2 \]
\[ \frac{h_1}{h_2} = \frac{16a^2}{9a^2} = \frac{16}{9} \]
Therefore, the ratio of heights is 16 : 9.
In simple words: When two cylinders hold the same volume but have different base sizes, the one with the smaller base must be taller. We set their volumes equal and solve for the height ratio.

Exam Tip: Always remember that equal volumes with different radii means the heights must be in an inverse relationship with the squares of the radii.

Question 17. A rectangular sheet of tin foil of size 30 cm × 18 cm can be rolled to form a cylinder in two ways along length and along breadth. Find the ratio of volumes of the two cylinders thus formed.
Answer: When rolled along the length (30 cm): the circumference of the base = 30 cm and height = 18 cm. For radius r₁:
\[ 2\pi r_1 = 30 \]
\[ 2 \times \frac{22}{7} \times r_1 = 30 \]
\[ r_1 = \frac{30 \times 7}{2 \times 22} = \frac{210}{44} = \frac{105}{22} \text{ cm} \]
When rolled along the breadth (18 cm): the circumference of the base = 18 cm and height = 30 cm. For radius r₂:
\[ 2\pi r_2 = 18 \]
\[ 2 \times \frac{22}{7} \times r_2 = 18 \]
\[ r_2 = \frac{18 \times 7}{2 \times 22} = \frac{126}{44} = \frac{63}{22} \text{ cm} \]
Now finding the ratio of volumes:
\[ \frac{V_1}{V_2} = \frac{\pi r_1^2 \times 18}{\pi r_2^2 \times 30} = \frac{\left(\frac{105}{22}\right)^2 \times 18}{\left(\frac{63}{22}\right)^2 \times 30} = \frac{105^2 \times 18}{63^2 \times 30} = \frac{11025 \times 18}{3969 \times 30} = \frac{198450}{119070} = \frac{5}{3} \]
Therefore, the ratio is 5 : 3.
In simple words: When a sheet is rolled in two different ways, the circumference becomes different in each case, which gives different radii. We calculate both volumes and compare them.

Exam Tip: Be careful to identify which dimension becomes the circumference and which becomes the height in each rolling scenario. Double-check your radius calculations before computing volumes.

Question 18. A cylindrical tube open at both ends is made of metal. The internal diameter of the tube is 11.2 cm and its length is 21 cm. The metal thickness is 0.4 cm. Calculate the volume of the metal.
Answer: Internal diameter = 11.2 cm, so internal radius = \( \frac{11.2}{2} = 5.6 \) cm. Since the thickness is 0.4 cm:
\[ \text{External radius} = 5.6 + 0.4 = 6.0 \text{ cm} \]
The length (height) = 21 cm. For a hollow cylinder, volume of metal = \( \pi(R^2 - r^2)h \), where R is the external radius and r is the internal radius. Substituting values:
\[ V = \frac{22}{7} \times (6^2 - 5.6^2) \times 21 \]
\[ V = \frac{22}{7} \times (36 - 31.36) \times 21 \]
\[ V = \frac{22}{7} \times 4.64 \times 21 \]
\[ V = 22 \times 4.64 \times 3 = 305.76 \approx 306.24 \text{ cm}^3 \]
Therefore, the volume of metal is approximately 306.24 cm³.
In simple words: To find how much metal is in the tube, we calculate the volume of the entire cylinder and subtract the volume of the hollow space inside.

Exam Tip: Always identify the internal and external radii clearly, especially when thickness is given. Use the hollow cylinder formula and not the solid cylinder formula.

Question 19. A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.
Answer: Radius of pencil = \( \frac{7}{2} = 3.5 \) mm. Radius of graphite = \( \frac{1}{2} = 0.5 \) mm. Height = 14 cm = 140 mm. Volume of wood = \( \pi(R^2 - r^2)h \), where R = radius of pencil and r = radius of graphite. Substituting values:
\[ V_{\text{wood}} = \frac{22}{7} \times (3.5^2 - 0.5^2) \times 140 \]
\[ V_{\text{wood}} = \frac{22}{7} \times (12.25 - 0.25) \times 140 \]
\[ V_{\text{wood}} = \frac{22}{7} \times 12 \times 140 = 22 \times 12 \times 20 = 5280 \text{ mm}^3 \]
\[ V_{\text{wood}} = 5280 \times \left(\frac{1}{10}\right)^3 = 5.28 \text{ cm}^3 \]
Volume of graphite = \( \pi r^2 h \):
\[ V_{\text{graphite}} = \frac{22}{7} \times 0.5^2 \times 140 \]
\[ V_{\text{graphite}} = \frac{22}{7} \times 0.25 \times 140 = \frac{770}{7} = 110 \text{ mm}^3 \]
\[ V_{\text{graphite}} = 110 \times \left(\frac{1}{10}\right)^3 = 0.11 \text{ cm}^3 \]
Therefore, volume of wood = 5.28 cm³ and volume of graphite = 0.11 cm³.
In simple words: The pencil is a hollow cylinder made of wood with graphite filling the center. We use the hollow cylinder formula for wood and the solid cylinder formula for graphite, then convert from mm³ to cm³.

Exam Tip: Remember to convert units properly when switching between mm and cm. Always use the correct formula for hollow vs. solid cylinders.

Question 20. A cylindrical roller made of iron is 2 m long. Its inner diameter is 35 cm and the thickness is 7 cm all round. Find the weight of the roller in kg, if 1 cm³ of iron weighs 8 g.
Answer: Internal radius = \( \frac{35}{2} = 17.5 \) cm. External radius = Internal radius + Thickness = 17.5 + 7 = 24.5 cm. Height = 2 m = 200 cm. Volume of hollow cylinder = \( \pi(R^2 - r^2)h \):
\[ V = \frac{22}{7} \times (24.5^2 - 17.5^2) \times 200 \]
\[ V = \frac{22}{7} \times (600.25 - 306.25) \times 200 \]
\[ V = \frac{22}{7} \times 294 \times 200 \]
\[ V = 22 \times 42 \times 200 = 184800 \text{ cm}^3 \]
Since 1 cm³ of iron weighs 8 g:
\[ \text{Weight} = 184800 \times 8 = 1478400 \text{ g} = 1478.4 \text{ kg} \]
Therefore, the weight of the roller is 1478.4 kg.
In simple words: We find the volume of the iron by using the hollow cylinder formula. Then we multiply this volume by the weight per unit volume to get the total weight. Finally, we convert grams to kilograms.

Exam Tip: Always convert all measurements to the same unit before calculating. Don't forget the final unit conversion when the answer is required in a different unit than your calculation.

Exercise 17.2

Question 1. Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.
Answer: The formula for curved surface area of a cone is \( \pi rl \), where r is the radius and l is the slant height. Substituting r = 7 cm and l = 10 cm: Curved surface area = \( \frac{22}{7} \times 7 \times 10 = 22 \times 10 = 220 \text{ cm}^2 \).
In simple words: The curved part of a cone has an area of 220 square centimetres.

Exam Tip: Remember the formula \( \pi rl \) for curved surface area; do not confuse it with total surface area which includes the base.

Question 2. Diameter of the base of a cone is 10.5 cm and slant height is 10 cm. Find its curved surface area.
Answer: First, calculate the radius from the diameter: \( \text{Radius} = \frac{\text{Diameter}}{2} = \frac{10.5}{2} = 5.25 \text{ cm} \). Given l = 10 cm, we apply the curved surface area formula: \( \pi rl = \frac{22}{7} \times 5.25 \times 10 = \frac{1155}{7} = 165 \text{ cm}^2 \).
In simple words: Convert the diameter to radius first, then use the formula to find the curved surface area as 165 square centimetres.

Exam Tip: Always find the radius when you are given the diameter; remember radius is half the diameter.

Question 3. Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find :
(i) radius of the base
(ii) total surface area of the cone.
Answer:
(i) We have the formula \( \pi rl = 308 \). Substituting the known values: \( \frac{22}{7} \times r \times 14 = 308 \). Simplifying: \( 22 \times r \times 2 = 308 \), which gives \( 44r = 308 \). Solving for r: \( r = \frac{308}{44} = 7 \text{ cm} \).
(ii) The total surface area includes both the curved surface and the circular base. Using the formula \( \pi r(l + r) = \frac{22}{7} \times 7 \times (14 + 7) = 22 \times 21 = 462 \text{ cm}^2 \).
In simple words: First work backwards from the curved surface area to find the radius as 7 cm. Then add the base area to find total surface area as 462 square centimetres.

Exam Tip: When finding radius from curved surface area, isolate r carefully; total surface area formula is \( \pi r(l + r) \), not just \( \pi rl \).

Question 4. Find the volume of the right circular cone with
(i) radius 6 cm and height 7 cm
(ii) radius 3.5 cm and height 12 cm.
Answer:
(i) The volume of a cone is given by \( V = \frac{1}{3}\pi r^2 h \). For r = 6 cm and h = 7 cm: \( V = \frac{1}{3} \times \frac{22}{7} \times 6^2 \times 7 = \frac{1}{3} \times \frac{22}{7} \times 36 \times 7 = \frac{22 \times 36 \times 7}{3 \times 7} = 22 \times 12 = 264 \text{ cm}^3 \).
(ii) For r = 3.5 cm and h = 12 cm: \( V = \frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times 12 = \frac{1}{3} \times \frac{22}{7} \times 12.25 \times 12 = \frac{22 \times 12.25 \times 12}{3 \times 7} = 22 \times 1.75 \times 4 = 154 \text{ cm}^3 \).
In simple words: For each cone, multiply one-third times pi times the radius squared times the height to get the volume.

Exam Tip: The volume formula \( \frac{1}{3}\pi r^2 h \) is different from the volume of a cylinder; the factor of \( \frac{1}{3} \) is crucial.

Question 5. Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm
(ii) height 12 cm, slant height 13 cm.
Answer:
(i) Given l = 25 cm and r = 7 cm. Using the Pythagorean relation \( l^2 = r^2 + h^2 \): \( 25^2 = 7^2 + h^2 \), so \( h^2 = 625 - 49 = 576 \), giving \( h = 24 \text{ cm} \). Volume = \( \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 24 = \frac{1}{3} \times \frac{22}{7} \times 49 \times 24 = 22 \times 7 \times 8 = 1232 \text{ cm}^3 \). Converting to litres: \( 1232 \text{ cm}^3 = 1232 \times \frac{1}{1000} = 1.232 \text{ litres} \).
(ii) Given l = 13 cm and h = 12 cm. Using \( l^2 = r^2 + h^2 \): \( 13^2 = r^2 + 12^2 \), so \( r^2 = 169 - 144 = 25 \), giving \( r = 5 \text{ cm} \). Volume = \( \frac{1}{3} \times \frac{22}{7} \times 5^2 \times 12 = \frac{22 \times 25 \times 12}{3 \times 7} = \frac{6600}{21} = \frac{66}{210/100} \) litres = \( \frac{11}{35} \) litres.
In simple words: Use the slant height to find the missing dimension, then calculate volume and convert to litres by dividing by 1000.

Exam Tip: Remember the relationship \( l^2 = r^2 + h^2 \); when slant height is given, use it to find the missing height or radius before calculating volume.

Question 6. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kiloliters?
Answer: The radius is half the diameter: \( \text{Radius} = \frac{3.5}{2} = 1.75 \text{ m} \). Height is 12 m. Volume = \( \frac{1}{3} \times \frac{22}{7} \times (1.75)^2 \times 12 = \frac{1}{3} \times \frac{22}{7} \times 3.0625 \times 12 = \frac{22 \times 3.0625 \times 12}{3 \times 7} = \frac{808.5}{21} = 38.5 \text{ m}^3 \). Since 1 m³ = 1 kilolitre, the capacity is 38.5 kilolitres.
In simple words: Find the radius from the diameter, use the volume formula, then remember that 1 cubic metre equals 1 kilolitre.

Exam Tip: When working with large measurements in metres, remember the unit conversion: 1 m³ = 1 kL directly, unlike the cm to litre conversion.

Question 7. If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.
Answer: Using the volume formula \( \frac{1}{3}\pi r^2 h = 48\pi \), we get \( \frac{1}{3}\pi r^2 \times 9 = 48\pi \). Simplifying: \( \pi r^2 \times 3 = 48\pi \), so \( r^2 = 16 \), giving \( r = 4 \text{ cm} \). Therefore, diameter = \( 2 \times 4 = 8 \text{ cm} \).
In simple words: Put the volume and height into the formula, solve for radius, then double it to get the diameter.

Exam Tip: When the volume is expressed in terms of π, the π cancels during calculation, simplifying the arithmetic.

Question 8. The height of the cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14)
Answer: Using \( \frac{1}{3}\pi r^2 h = 1570 \) with h = 15 cm and π = 3.14: \( \frac{1}{3} \times 3.14 \times r^2 \times 15 = 1570 \). Simplifying: \( r^2 = \frac{1570 \times 3}{15 \times 3.14} = \frac{4710}{47.1} = 100 \). Therefore, \( r = \sqrt{100} = 10 \text{ cm} \).
In simple words: Substitute the known values into the volume formula and solve for the radius step by step.

Exam Tip: When π = 3.14 is specified, use this value throughout; always check your arithmetic when dividing large numbers.

Question 9. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white washing its curved surface area at the rate of Rs 210 per 100 m².
Answer: Cost per square metre = \( \frac{210}{100} = 2.1 \text{ Rs/m}^2 \). From diameter = 14 m, radius = 7 m. Curved surface area = \( \pi rl = \frac{22}{7} \times 7 \times 25 = 22 \times 25 = 550 \text{ m}^2 \). Total cost = \( 2.1 \times 550 = 1155 \text{ Rs} \).
In simple words: Find the curved surface area first, then multiply by the cost per square metre.

Exam Tip: Convert the given rate per 100 m² to a rate per 1 m² before multiplying; breaking the rate down this way reduces calculation errors.

Question 10. A conical tent is 10 m high and the radius of its base is 24 m. Find :
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m² canvas is Rs 70.
Answer:
(i) Using \( l = \sqrt{r^2 + h^2} = \sqrt{24^2 + 10^2} = \sqrt{576 + 100} = \sqrt{676} = 26 \text{ m} \).
(ii) Curved surface area = \( \pi rl = \frac{22}{7} \times 24 \times 26 = \frac{13728}{7} \text{ m}^2 \). Cost = \( \frac{13728}{7} \times 70 = 13728 \times 10 = 137280 \text{ Rs} \).
In simple words: First find the slant height using the Pythagorean theorem. Then find curved surface area and multiply by the cost per square metre.

Exam Tip: For a tent, only the curved surface needs canvas (not the base); use the curved surface area formula, not total surface area.

Question 11. A Joker's cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the cloth required to make 10 such caps.
Answer: First, find slant height: \( l = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \text{ cm} \). Curved surface area of one cap = \( \pi rl = \frac{22}{7} \times 7 \times 25 = 22 \times 25 = 550 \text{ cm}^2 \). For 10 caps: \( 10 \times 550 = 5500 \text{ cm}^2 \).
In simple words: Find the slant height, then the curved surface area of one cap, then multiply by 10.

Exam Tip: Do not forget the multiplication by 10; always check whether the question asks for one item or multiple items.

Question 12(a). The ratio of the base radii of two right circular cones of the same height is 3 : 4. Find the ratio of their volumes.
Answer: Let the radii be 3a and 4a, with both cones having the same height h. The volume of a cone is \( V = \frac{1}{3}\pi r^2 h \). For cone 1: \( V_1 = \frac{1}{3}\pi (3a)^2 h = \frac{1}{3}\pi \times 9a^2 \times h \). For cone 2: \( V_2 = \frac{1}{3}\pi (4a)^2 h = \frac{1}{3}\pi \times 16a^2 \times h \). The ratio of volumes is \( \frac{V_1}{V_2} = \frac{9a^2}{16a^2} = \frac{9}{16} \). Hence, the ratio is 9 : 16.
In simple words: When two cones have the same height, their volumes are in the ratio of the squares of their radii.

Exam Tip: Remember that volume depends on the square of the radius; if radii are in ratio 3 : 4, volumes are in ratio 3² : 4² = 9 : 16.

Question 12(b). The ratio of the heights of two right circular cones is 5 : 2 and that of their base radii is 2 : 5. Find the ratio of their volumes.
Answer: Let heights be 5a and 2a, and radii be 2b and 5b respectively. Volume of cone 1: \( V_1 = \frac{1}{3}\pi (2b)^2 \times 5a = \frac{1}{3}\pi \times 4b^2 \times 5a \). Volume of cone 2: \( V_2 = \frac{1}{3}\pi (5b)^2 \times 2a = \frac{1}{3}\pi \times 25b^2 \times 2a \). The ratio is \( \frac{V_1}{V_2} = \frac{4b^2 \times 5a}{25b^2 \times 2a} = \frac{20a}{50a} = \frac{2}{5} \). Hence, the ratio is 2 : 5.
In simple words: When both radius and height ratios are different, multiply them together: radius ratio squared times height ratio gives the volume ratio.

Exam Tip: For volume ratio with two different proportions, compute \( \frac{r_1^2 h_1}{r_2^2 h_2} \) carefully, cancelling common factors as you go.

Question 12(c). The height and the radius of the base of a right circular cone is half the corresponding height and radius of another bigger cone. Find :
(i) the ratio of their volumes.
(ii) the ratio of their lateral surface areas.
Answer:
(i) Let the bigger cone have radius r and height h. The smaller cone has radius \( \frac{r}{2} \) and height \( \frac{h}{2} \). Volume of bigger cone: \( V_{\text{big}} = \frac{1}{3}\pi r^2 h \). Volume of smaller cone: \( V_{\text{small}} = \frac{1}{3}\pi \left(\frac{r}{2}\right)^2 \times \frac{h}{2} = \frac{1}{3}\pi \times \frac{r^2}{4} \times \frac{h}{2} = \frac{1}{3}\pi \times \frac{r^2 h}{8} \). The ratio is \( \frac{V_{\text{small}}}{V_{\text{big}}} = \frac{\frac{1}{8} \times \frac{1}{3}\pi r^2 h}{\frac{1}{3}\pi r^2 h} = \frac{1}{8} \). Hence, the ratio is 1 : 8.
(ii) For the bigger cone, slant height \( l = \sqrt{r^2 + h^2} \). For the smaller cone, slant height \( l_1 = \sqrt{\left(\frac{r}{2}\right)^2 + \left(\frac{h}{2}\right)^2} = \sqrt{\frac{r^2 + h^2}{4}} = \frac{1}{2}\sqrt{r^2 + h^2} = \frac{l}{2} \). Lateral surface area of bigger cone: \( A_{\text{big}} = \pi rl \). Lateral surface area of smaller cone: \( A_{\text{small}} = \pi \times \frac{r}{2} \times \frac{l}{2} = \frac{\pi rl}{4} \). The ratio is \( \frac{A_{\text{small}}}{A_{\text{big}}} = \frac{1}{4} \). Hence, the ratio is 1 : 4.
In simple words: When all dimensions are halved, volume reduces by a factor of 8 (because \( \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \)). Surface area reduces by a factor of 4 (because \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)).

Exam Tip: When a 3D shape is scaled down by a factor k, volumes scale by k³ and surface areas by k². Here k = \( \frac{1}{2} \), so volume scales by \( \left(\frac{1}{2}\right)^3 = \frac{1}{8} \) and area by \( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).

Question 13. Find what length of canvas 2 m in width is required to make a conical tent 20 m in diameter and 42 m in slant height allowing 10% for folds and the stitching. Also find the cost of the canvas at the rate of Rs 80 per meter.
Answer: The tent has a diameter of 20 m, so the radius is 10 m, and the slant height is 42 m. The curved surface area of a cone is found using \( \pi \times r \times l \), which gives us \( \frac{22}{7} \times 10 \times 42 = 1320 \, m^2 \). We must add 10% extra for folds and stitching: \( 10\% \text{ of } 1320 = 132 \, m^2 \). So the total canvas area needed is \( 1320 + 132 = 1452 \, m^2 \). Since the canvas width is 2 m, the length required is \( 1452 \div 2 = 726 \, m \). At Rs 80 per meter, the total cost is \( 726 \times 80 = \text{Rs } 58080 \).
In simple words: To find the canvas length, first work out the curved surface area of the tent. Add 10% more for folds, then divide by the width of 2 m to get 726 m. Multiply by the cost per meter to get Rs 58080.

Exam Tip: Always remember to add the extra percentage for folds and stitching before calculating the length - this is a common error.

Question 14. The perimeter of the base of a cone is 44 cm and the slant height is 25 cm. Find the volume and the curved surface of the cone.
Answer: The base perimeter is 44 cm, so \( 2\pi r = 44 \). Using \( \pi = \frac{22}{7} \), we get \( r = 7 \, cm \). With slant height \( l = 25 \, cm \), we find the height using \( l^2 = r^2 + h^2 \): \( 625 = 49 + h^2 \), giving \( h = 24 \, cm \). The volume is \( \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 49 \times 24 = 1232 \, cm^3 \). The curved surface area is \( \pi rl = \frac{22}{7} \times 7 \times 25 = 550 \, cm^2 \).
In simple words: Find the radius from the perimeter. Then use the Pythagoras formula to find the height. Use these to calculate both the volume and curved surface area.

Exam Tip: Remember the key relationship \( l^2 = r^2 + h^2 \) for cones - this appears in almost every cone question.

Question 15. The volume of a right circular cone is 9856 cm³ and the area of its base is 616 cm². Find (i) the slant height of the cone. (ii) total surface area of the cone.
Answer:
(i) From the base area, \( \pi r^2 = 616 \), we get \( r = 14 \, cm \). Using the volume formula \( \frac{1}{3}\pi r^2 h = 9856 \), we solve for height: \( h = 48 \, cm \). The slant height is found from \( l^2 = r^2 + h^2 = 196 + 2304 = 2500 \), so \( l = 50 \, cm \).
(ii) The total surface area is \( \pi r(l + r) = \frac{22}{7} \times 14 \times 64 = 2816 \, cm^2 \).
In simple words: First find the radius and height. Then use the Pythagoras formula to get the slant height. Finally, use the total surface area formula with these values.

Exam Tip: Total surface area of a cone includes the base - the formula is \( \pi r(l + r) \), not just the curved surface \( \pi rl \).

Question 16. A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. Find the volume and the curved surface of the cone so formed. (Take \( \pi = 3.14 \))
Answer: When the right triangle is spun about the 8 cm side, it creates a cone where the height is 8 cm, the radius is 6 cm (the other leg), and the slant height is 10 cm (the hypotenuse). The volume is \( \frac{1}{3}\pi r^2 h = \frac{1}{3} \times 3.14 \times 36 \times 8 = 301.44 \, cm^3 \). The curved surface area is \( \pi rl = 3.14 \times 6 \times 10 = 188.4 \, cm^2 \).
In simple words: When you spin a right triangle around one of its sides, that side becomes the cone's height, and the other leg becomes the radius. The hypotenuse is the slant height.

Exam Tip: In solid of revolution problems, clearly identify which side becomes the height, which becomes the radius, and which becomes the slant height.

Question 17. The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume be \( \frac{1}{27} \) of the volume of the given cone, at what height above the base is the section cut?
Answer: Let the original cone OAB have height 30 cm and base radius R cm. When cut by a plane parallel to the base, a smaller cone OCD is formed with height h cm and radius r cm. Since the triangles are similar, \( \frac{r}{R} = \frac{h}{30} \). Given that the volume of the small cone is \( \frac{1}{27} \) of the original: \( \frac{1}{3}\pi r^2 h = \frac{1}{27} \times \frac{1}{3}\pi R^2 \times 30 \). Simplifying and substituting the ratio, we get \( \left(\frac{h}{30}\right)^3 = \frac{1}{27} \), which gives \( h = 10 \, cm \). The section is cut at a height of \( 30 - 10 = 20 \, cm \) above the base.
In simple words: Use the fact that similar cones have volumes in the ratio of the cube of their heights. If the small cone is \( \frac{1}{27} \) the volume, its height is \( \frac{1}{3} \) the original height.

Exam Tip: When cones are cut by a plane parallel to the base, the resulting smaller cone is similar to the original - use this similarity property to set up your equations.

Question 18. A semi-circular lamina of radius 35 cm is folded so that the two bounding radii are joined together to form a cone. Find : (i) the radius of the cone. (ii) the (lateral) surface area of the cone.
Answer:
(i) The arc length of the semi-circular sheet becomes the circumference of the cone's base. The arc length is \( \frac{1}{2} \times 2\pi r = \pi \times 35 = \frac{22}{7} \times 35 = 110 \, cm \). When this arc forms the base of the cone, \( 2\pi r_{cone} = 110 \), giving \( r_{cone} = 17.5 \, cm \).
(ii) The curved surface area of the cone equals the area of the original semi-circular sheet: \( \frac{1}{2}\pi R^2 = \frac{1}{2} \times \frac{22}{7} \times 1225 = 1925 \, cm^2 \).
In simple words: The curved edge of the semi-circle becomes the circle around the base of the cone. The area of the flat semi-circle becomes the curved surface of the cone.

Exam Tip: In cone-folding problems, remember that the arc length of the flat sheet becomes the circumference of the cone's base, and the area is preserved.

Exercise 17.3

Question 1. Find the surface area and volume of a sphere of radius 14 cm.
Answer: The surface area of a sphere is \( 4\pi r^2 \). Substituting \( r = 14 \, cm \): \( 4 \times \frac{22}{7} \times 196 = \frac{17248}{7} = 2464 \, cm^2 \). The volume of a sphere is \( \frac{4}{3}\pi r^3 \). Substituting: \( \frac{4}{3} \times \frac{22}{7} \times 2744 = \frac{241472}{21} = 11498\frac{2}{3} \, cm^3 \).
In simple words: Use the sphere formulas: surface area is \( 4\pi r^2 \) and volume is \( \frac{4}{3}\pi r^3 \). Plug in the radius and calculate.

Exam Tip: Write formulas correctly - surface area has \( r^2 \) while volume has \( r^3 \). A common mistake is mixing these up.

Question 2. Find the surface area and volume of a sphere of diameter 21 cm.
Answer: With diameter 21 cm, the radius is 10.5 cm. The surface area is \( 4\pi r^2 = 4 \times \frac{22}{7} \times 110.25 = \frac{9702}{7} = 1386 \, cm^2 \). The volume is \( \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 1157.625 = \frac{101871}{21} = 4851 \, cm^3 \).
In simple words: Always find the radius first by dividing the diameter by 2. Then apply the surface area and volume formulas as usual.

Exam Tip: When given the diameter, never forget to divide by 2 first - this step is critical and often causes errors.

Question 3. A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per cm³, find the mass of the shot-put.
Answer: The volume of the sphere is \( \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (4.9)^3 = \frac{4 \times 22 \times 117.649}{21} = \frac{10353.112}{21} = 493 \, cm^3 \). Since mass equals volume times density: \( \text{Mass} = 493 \times 7.8 = 3845.4 \, g = 3.845 \, kg \approx 3.85 \, kg \).
In simple words: Find the volume using the sphere formula. Then multiply by the density to get the mass. Remember to convert grams to kilograms at the end.

Exam Tip: The relationship "Mass = Volume × Density" is fundamental - make sure you apply it correctly and convert units appropriately.

Question 4. Find the diameter of a sphere whose surface area is 154 cm².
Answer: The surface area of a sphere is \( 4\pi r^2 = 154 \). Solving for r: \( \frac{4 \times 22}{7} \times r^2 = 154 \), which gives \( r^2 = \frac{154 \times 7}{88} = \frac{1078}{88} = 12.25 \). Taking the square root, \( r = 3.5 \, cm \). The diameter is \( 2 \times 3.5 = 7 \, cm \).
In simple words: Start with the surface area formula. Rearrange it to find r squared. Take the square root to get r. Double it to get the diameter.

Exam Tip: When working backwards from surface area to find the radius, be careful with algebraic steps and always take the positive square root.

Question 5. Find (i) the curved surface area. (ii) the total surface area of a hemisphere of radius 21 cm.
Answer:
(i) The curved surface area of a hemisphere is \( 2\pi r^2 \). Substituting \( r = 21 \, cm \): \( 2 \times \frac{22}{7} \times 441 = \frac{2 \times 22 \times 441}{7} = \frac{19404}{7} = 2772 \, cm^2 \).
(ii) The total surface area includes both the curved part and the flat base: \( 3\pi r^2 = 3 \times \frac{22}{7} \times 441 = \frac{3 \times 22 \times 441}{7} = \frac{29106}{7} = 4158 \, cm^2 \).
In simple words: A hemisphere is half a sphere. Its curved surface is half a sphere's surface. Its total surface adds the flat circular base on top.

Exam Tip: Do not confuse curved surface area (\( 2\pi r^2 \)) with total surface area (\( 3\pi r^2 \)) for hemispheres - the difference is the flat base area (\( \pi r^2 \)).

Question 6. A hemispherical brass bowl has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs. 16 per 100 cm².
Answer: The internal diameter measures 10.5 cm, so the internal radius is 5.25 cm. The curved surface area of a hemispherical shell uses the formula \( 2\pi r^2 \). Substituting the values:
\( 2 \times \frac{22}{7} \times (5.25)^2 = \frac{2 \times 22 \times 27.5625}{7} = \frac{1212.75}{7} = 173.25 \text{ cm}^2 \)
The tin-plating cost is Rs. 16 per 100 cm², which equals Rs. 0.16 per cm². Therefore, for 173.25 cm²:
\( 173.25 \times 0.16 = \text{Rs. } 27.72 \)
In simple words: First find the inner curved surface by using \( 2\pi r^2 \). Then multiply that area by the cost per square centimeter.

Exam Tip: Remember that a hemispherical bowl has only a curved surface area (no flat base is plated), so use \( 2\pi r^2 \), not the total surface area formula.

Question 7. The radius of a spherical balloon increases from 7 cm to 14 cm as air is pumped into it. Find the ratio of the surface areas of the balloon in two cases.
Answer: The surface area of a sphere is given by \( 4\pi r^2 \). In the first case, the radius is 7 cm; in the second case, it is 14 cm. The ratio of surface areas equals:
\( \frac{4\pi(14)^2}{4\pi(7)^2} = \frac{14 \times 14}{7 \times 7} = \frac{196}{49} = \frac{4}{1} = 4:1 \)
In simple words: When you double the radius, the surface area becomes four times larger because the radius is squared in the formula.

Exam Tip: When comparing surface areas of spheres, the ratio simplifies to the square of the ratio of their radii - (r₁ : r₂)² = (7 : 14)² = 1 : 4.

Question 8. A sphere and a cube have the same surface. Show that the ratio of the volume of the sphere to that of the cube is \( \sqrt{6}:\sqrt{\pi} \).
Answer: Let the cube have side length a cm and the sphere have radius r cm. For a sphere, the surface area equals \( 4\pi r^2 \); for a cube, it equals \( 6a^2 \). Since both surfaces are equal:
\( 4\pi r^2 = 6a^2 \)
\( \implies \frac{r^2}{a^2} = \frac{6}{4\pi} \)
\( \implies \frac{r}{a} = \sqrt{\frac{6}{4\pi}} \)
The volume of a sphere is \( \frac{4}{3}\pi r^3 \) and the volume of a cube is \( a^3 \). The ratio of volumes becomes:
\( \frac{\text{Volume of sphere}}{\text{Volume of cube}} = \frac{\frac{4}{3}\pi r^3}{a^3} = \frac{4\pi}{3} \times \left(\frac{r}{a}\right)^3 = \frac{4\pi}{3} \times \left(\sqrt{\frac{6}{4\pi}}\right)^3 \)
\( = \frac{4\pi}{3} \times \frac{6\sqrt{6}}{8\pi\sqrt{\pi}} = \sqrt{6}:\sqrt{\pi} \)
In simple words: When a sphere and cube have equal surface areas, their volume ratio depends on the constants in their surface and volume formulas.

Exam Tip: This proof requires careful algebraic manipulation - set the surface areas equal, solve for the radius-to-side ratio, then substitute into the volume ratio formula.

Question 9(a). If the ratio of the radii of two spheres is 3 : 7, find: (i) the ratio of their volumes. (ii) the ratio of their surface areas.
Answer: (i) Let the radii be 3a and 7a. Using the volume formula \( \frac{4}{3}\pi r^3 \):
\( \frac{\text{Volume of Sphere 1}}{\text{Volume of Sphere 2}} = \frac{\frac{4}{3}\pi(3a)^3}{\frac{4}{3}\pi(7a)^3} = \frac{27a^3}{343a^3} = \frac{27}{343} \)
Therefore, the volume ratio is 27 : 343.
(ii) Using the surface area formula \( 4\pi r^2 \):
\( \frac{\text{Surface area of Sphere 1}}{\text{Surface area of Sphere 2}} = \frac{4\pi(3a)^2}{4\pi(7a)^2} = \frac{9a^2}{49a^2} = \frac{9}{49} \)
Therefore, the surface area ratio is 9 : 49.
In simple words: When two radii are in the ratio 3 : 7, their volumes are in the ratio 3³ : 7³ = 27 : 343, and their surface areas are in the ratio 3² : 7² = 9 : 49.

Exam Tip: Remember that volume ratio = (radius ratio)³ and surface area ratio = (radius ratio)² for similar spheres.

Question 9(b). If the ratio of the volumes of the two spheres is 125 : 64, find the ratio of their surface areas.
Answer: Given that the volume ratio is 125 : 64:
\( \frac{\text{Volume of Sphere 2}}{\text{Volume of Sphere 1}} = \frac{64}{125} \)
\( \implies \frac{(r_1)^3}{(r_2)^3} = \frac{5^3}{4^3} \)
\( \implies \frac{r_1}{r_2} = \frac{5}{4} \)
Now, for the surface area ratio:
\( \frac{\text{Surface area of Sphere 1}}{\text{Surface area of Sphere 2}} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{5}{4}\right)^2 = \frac{25}{16} \)
Therefore, the surface area ratio is 25 : 16.
In simple words: When volume ratio is known, first find the radius ratio by taking the cube root, then square it to get the surface area ratio.

Exam Tip: Volume ratio = (radius ratio)³, so radius ratio = ∛(volume ratio). Once you have the radius ratio, square it to find the surface area ratio.

Question 10. Find the volume of a sphere whose surface area is 154 cm².
Answer: The surface area formula for a sphere is \( 4\pi r^2 = 154 \). Solving for the radius:
\( 4 \times \frac{22}{7} \times r^2 = 154 \)
\( \implies r^2 = \frac{154 \times 7}{4 \times 22} = \frac{1078}{88} = 12.25 \)
\( \implies r = 3.5 \text{ cm} \)
Now, using the volume formula \( \frac{4}{3}\pi r^3 \):
\( V = \frac{4}{3} \times \frac{22}{7} \times (3.5)^3 = \frac{4 \times 22 \times 42.875}{21} = \frac{3773}{21} = \frac{539 \times 7}{3 \times 7} = \frac{539}{3} = 179\frac{2}{3} \text{ cm}^3 \)
In simple words: Use the surface area to find the radius, then plug that radius into the volume formula.

Exam Tip: Always solve for the radius first when given surface area, then use it to calculate volume - do not try to find volume directly from surface area without the radius.

Question 11. If the volume of a sphere is \( 179\frac{2}{3} \) cm³, find its radius and the surface area.
Answer: Given: volume = \( 179\frac{2}{3} = \frac{539}{3} \) cm³. Using the volume formula \( \frac{4}{3}\pi r^3 = \frac{539}{3} \):
\( \frac{4}{3} \times \frac{22}{7} \times r^3 = \frac{539}{3} \)
\( \implies \frac{88}{21} \times r^3 = \frac{539}{3} \)
\( \implies r^3 = \frac{539 \times 21}{3 \times 88} = \frac{539 \times 7}{88} = \frac{3773}{88} = 42.875 \)
\( \implies r = 3.5 \text{ cm} \)
For the surface area, using \( 4\pi r^2 \):
\( SA = 4 \times \frac{22}{7} \times (3.5)^2 = \frac{4 \times 22 \times 12.25}{7} = \frac{1078}{7} = 154 \text{ cm}^2 \)
In simple words: Take the cube root of the volume equation to find the radius, then use the radius to find the surface area.

Exam Tip: When volume is given as a mixed number, convert it to an improper fraction first to make calculations cleaner and less prone to rounding errors.

Question 12. A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?
Answer: The volume formula for a hemisphere is \( \frac{2}{3}\pi r^3 \). With radius 3.5 cm:
\( V = \frac{2}{3} \times \frac{22}{7} \times (3.5)^3 = \frac{2 \times 22 \times 42.875}{21} = \frac{1886.5}{21} = \frac{18865}{210} = \frac{539}{6} = 89\frac{5}{6} \text{ cm}^3 \)
In simple words: A hemisphere holds half the volume of a full sphere, so use the hemisphere formula with the given radius.

Exam Tip: The volume of a hemisphere is exactly half the volume of a sphere with the same radius - use \( \frac{2}{3}\pi r^3 \) not \( \frac{4}{3}\pi r^3 \).

Question 13. The surface area of a solid sphere is 1256 cm². It is cut into two hemispheres. Find the total surface area and the volume of a hemisphere. Take π = 3.14
Answer: From the sphere's surface area formula \( 4\pi r^2 = 1256 \):
\( 4 \times 3.14 \times r^2 = 1256 \)
\( \implies r^2 = \frac{1256}{12.56} = 100 \)
\( \implies r = 10 \text{ cm} \)
The total surface area of a hemisphere (curved surface plus flat circular base) is \( 3\pi r^2 \):
\( SA = 3 \times 3.14 \times (10)^2 = 3 \times 3.14 \times 100 = 942 \text{ cm}^2 \)
The volume of a hemisphere is \( \frac{2}{3}\pi r^3 \):
\( V = \frac{2}{3} \times 3.14 \times (10)^3 = \frac{2 \times 3.14 \times 1000}{3} = \frac{6280}{3} = 2093\frac{1}{3} \text{ cm}^3 \)
In simple words: Find the radius from the original sphere's surface area, then calculate the hemisphere's total surface area (which includes the flat circular base) and volume.

Exam Tip: When a sphere is cut into hemispheres, the total surface area of each hemisphere includes BOTH the curved surface and the newly exposed flat circular base - do not forget the base area.

Exercise 17.4

Question 1. The adjoining figure shows a cuboidal block of wood through which a circular cylindrical hole of the biggest size is drilled. Find the volume of the wood left in the block.
Answer: From the figure, the diameter of the largest possible hole is 30 cm, so the radius is 15 cm. The height of the cylinder is 70 cm. The volume of the cuboidal block is:
\( V_{\text{cuboid}} = l \times b \times h = 70 \times 30 \times 30 = 63000 \text{ cm}^3 \)
The volume of the cylindrical hole is:
\( V_{\text{cylinder}} = \pi r^2 h = \frac{22}{7} \times (15)^2 \times 70 = \frac{22 \times 225 \times 70}{7} = \frac{346500}{7} = 49500 \text{ cm}^3 \)
The remaining wood volume is:
\( V_{\text{remaining}} = 63000 - 49500 = 13500 \text{ cm}^3 \)
In simple words: Find the volume of the whole block, subtract the volume of the hole drilled out, and what is left is the wood remaining.

Exam Tip: The biggest cylindrical hole that fits has a diameter equal to the smallest dimension of the cuboid - here the sides are 30 cm, so the maximum hole diameter is 30 cm.

Question 2. The adjoining figure shows a solid trophy made of shining glass. If one cubic centimeter of glass costs Rs. 0.75, find the cost of the glass for making the trophy.
Answer: The trophy consists of a cube topped with a cylinder. From the figure, the cube has an edge of 28 cm and the cylinder has a radius of 14 cm (half the cube edge) and height 28 cm. The volume of the cube is:
\( V_{\text{cube}} = (28)^3 = 21952 \text{ cm}^3 \)
The volume of the cylinder is:
\( V_{\text{cylinder}} = \pi r^2 h = \frac{22}{7} \times (14)^2 \times 28 = \frac{22 \times 196 \times 28}{7} = \frac{120736}{7} = 17248 \text{ cm}^3 \)
Total volume is:
\( V_{\text{total}} = 21952 + 17248 = 39200 \text{ cm}^3 \)
The cost of glass at Rs. 0.75 per cm³:
\( \text{Cost} = 39200 \times 0.75 = \text{Rs. } 29400 \)
In simple words: Calculate the volume of each shape (cube and cylinder) separately, add them together, then multiply by the cost per cubic centimeter.

Exam Tip: When a solid is made of multiple shapes, always find the volume of each shape separately, add them, then apply any cost or other calculations to the total.

Question 3. From a cube of edge 14 cm, a cone of maximum size is carved out. Find the volume of the remaining material.
Answer: The cube has an edge length of 14 cm, giving it a volume of 2744 cm³. A cone with the largest possible dimensions is removed from it. This cone has a diameter of 14 cm (matching the cube's side), so its radius is 7 cm, and its height is also 14 cm. Using the formula for cone volume, we get \( \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 14 = \frac{2156}{3} \) cm³. The leftover material has a volume of \( 2744 - \frac{2156}{3} = \frac{6076}{3} = 2025\frac{1}{3} \) cm³.
In simple words: Subtract the cone's volume from the cube's volume to find what is left.

Exam Tip: Always match the cone's maximum diameter to the cube's side length, and ensure the height equals the cube's edge for maximum volume carved out.

Question 4. A cone of maximum volume is carved out of a block of wood of size 20 cm × 10 cm × 10 cm. Find the volume of the remaining wood.
Answer: The wooden block measures 20 cm × 10 cm × 10 cm, so its total volume is 2000 cm³. For maximum volume, the cone's diameter is set to 10 cm (the smaller dimension), giving a radius of 5 cm, and the height is 20 cm. The cone's volume equals \( \frac{1}{3} \times \frac{22}{7} \times 5^2 \times 20 = \frac{11000}{21} \) cm³. The wood remaining after the cone is removed has a volume of \( 2000 - \frac{11000}{21} = \frac{31000}{21} = 1476\frac{4}{21} \) cm³.
In simple words: Take the block's volume and subtract the cone's volume to get what stays.

Exam Tip: Identify the smallest cross-section dimension to determine the cone's maximum diameter; the height should use the longest block dimension.

Question 5. 16 glass spheres each of radius 2 cm are packed in a cuboidal box of internal dimensions 16 cm × 8 cm × 8 cm and then the box is filled with water. Find the volume of the water filled in the box.
Answer: The box has dimensions 16 cm × 8 cm × 8 cm, so its internal capacity is 1024 cm³. Each glass sphere has a radius of 2 cm, and its volume is \( \frac{4}{3} \times \frac{22}{7} \times 2^3 = \frac{704}{21} \) cm³. For 16 spheres, the combined volume is \( 16 \times \frac{704}{21} = \frac{11264}{21} \approx 536.38 \) cm³. When the box is filled with water after inserting the spheres, the water occupies the remaining space: \( 1024 - 536.38 = 487.62 \) cm³.
In simple words: Water fills the empty space left after all the spheres are placed in the box.

Exam Tip: Always subtract the total volume of solid objects from the container's capacity to find the liquid volume.

Question 6. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter that the hemisphere can have? Also, find the surface area of the solid.
Answer: The cube has a side of 7 cm. The largest hemisphere that can sit on top of it has a diameter equal to the cube's side, which is 7 cm, giving a radius of 3.5 cm. The curved surface area of this hemisphere is \( 2\pi r^2 = 2 \times \frac{22}{7} \times (3.5)^2 = 77 \) cm². The cube's surface area is \( 6 \times 7^2 = 294 \) cm². However, the base of the hemisphere (a circle of area \( \pi r^2 = \frac{22}{7} \times 12.25 = 38.5 \) cm²) overlaps with the top face of the cube, so this overlapped region must not be counted twice. The total surface area of the combined solid is \( 294 + 77 - 38.5 = 332.5 \) cm².
In simple words: Add the cube's outer surface and the hemisphere's curved surface, then subtract the overlapping circular base.

Exam Tip: Always subtract the area of contact between two solids to avoid double-counting when combining surface areas.

Question 7. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder (as shown in the adjoining figure). If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the total surface area of the article.
Answer: The cylinder has a height of 10 cm and a radius of 3.5 cm. When hemispheres are removed from both ends, the surface includes only the curved side of the cylinder and the two curved surfaces of the hemispheres. The formula for total surface area becomes \( T = 2\pi rh + 2 \times 2\pi r^2 = 2\pi r(h + 2r) \). Substituting values: \( T = 2 \times \frac{22}{7} \times 3.5 \times (10 + 7) = \frac{154}{7} \times 17 = 22 \times 17 = 374 \) cm².
In simple words: Find the curved surface of the cylinder plus the curved surfaces of both hemispheres scooped out from the ends.

Exam Tip: When material is removed from both ends symmetrically, the surface area formula simplifies; recognise this pattern to avoid calculation errors.

Question 8. From a solid wooden cylinder of height 28 cm and diameter 6 cm, two conical cavities are hollowed out. The diameters of the cone are also of 6 cm and height 10.5 cm. Taking π = 22/7, find the volume of remaining solid.
Answer: The solid cylinder has a diameter of 6 cm (radius 3 cm) and height 28 cm. Its volume is \( \pi r^2 h = \frac{22}{7} \times 3^2 \times 28 = \frac{22}{7} \times 9 \times 28 = 792 \) cm³. Each conical cavity has the same diameter (6 cm, radius 3 cm) and a height of 10.5 cm. The volume of one cone is \( \frac{1}{3} \times \frac{22}{7} \times 3^2 \times 10.5 = \frac{1}{3} \times \frac{22}{7} \times 9 \times 10.5 = 99 \) cm³. Since two identical cones are removed, their total volume is \( 2 \times 99 = 198 \) cm³. The remaining material has a volume of \( 792 - 198 = 594 \) cm³.
In simple words: Subtract the volume of both cones from the full cylinder's volume to get the remaining solid.

Exam Tip: When identical cavities are removed, calculate one cavity's volume and multiply by the number of cavities for efficiency.

Question 9. A hemispherical and conical hole are scooped out of a solid wooden cylinder. Find the volume of the remaining solid where the measurements are as follows: The height of the cylinder is 7 cm, radius of each hemisphere, cone and cylinder is 3 cm. Height of cone is 3 cm. Give your answer correct to nearest whole number. Take π = 22/7.
Answer: The cylinder has a radius of 3 cm and height 7 cm. A hemisphere and a cone, each with radius 3 cm, are removed. The cone has height 3 cm. Using the combined formula for remaining volume: \( V = \pi r^2\left(h - \frac{h_1}{3} - \frac{2r}{3}\right) = \frac{22}{7} \times 3^2 \times \left(7 - \frac{3}{3} - \frac{2 \times 3}{3}\right) = \frac{198}{7} \times (7 - 1 - 2) = \frac{198}{7} \times 4 = \frac{792}{7} = 113.14 \) cm³. Rounded to the nearest whole number, the remaining volume is 113 cm³.
In simple words: Subtract both the cone's and hemisphere's volumes from the cylinder to find what stays behind.

Exam Tip: Derive a single combined formula when multiple shapes are removed to reduce calculation steps and minimise rounding errors before the final answer.

Question 10. A toy is in form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. If the total height of the toy is 15.5 cm, find the total surface area and volume of the toy, giving your answer correct to one decimal place.
Answer: The toy combines a cone and hemisphere, both with radius 3.5 cm. Since the total height is 15.5 cm and the hemisphere's height equals its radius (3.5 cm), the cone's height is \( 15.5 - 3.5 = 12 \) cm. The slant height of the cone is \( l = \sqrt{r^2 + h^2} = \sqrt{3.5^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5 \) cm. The curved surface area of the cone is \( \pi rl = \frac{22}{7} \times 3.5 \times 12.5 = 137.5 \) cm². The curved surface area of the hemisphere is \( 2\pi r^2 = 2 \times \frac{22}{7} \times (3.5)^2 = 77 \) cm². Total surface area is \( 137.5 + 77 = 214.5 \) cm². The volume of the cone is \( \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times 12 = 154 \) cm³. The volume of the hemisphere is \( \frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times (3.5)^3 = 89.8 \) cm³. Total volume is \( 154 + 89.8 = 243.8 \) cm³.
In simple words: Add the curved surfaces of both shapes for total area, and add their volumes separately to get the combined volume.

Exam Tip: When a cone sits on a hemisphere, their shared base area is not counted in surface area - only curved surfaces contribute to the exterior.

Question 11. A circus tent is in the shape of a cylinder surmounted by a cone. The diameter of the cylindrical portion is 24 m and its height is 11 m. If the vertex of the cone is 16 m above the ground, find the area of the canvas used to make the tent.
Answer: The cylindrical part has a diameter of 24 m, so its radius is 12 m and its height is 11 m. Since the cone's apex is 16 m above ground and the cylinder is 11 m tall, the cone's height is 5 m. Using the formula for slant height, \( l = \sqrt{h^2 + r^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \) m. The canvas covers the curved surface of both shapes: \( \text{Area} = 2\pi rH + \pi rl = \pi r(2H + l) = \frac{22}{7} \times 12 \times (2 \times 11 + 13) = \frac{22}{7} \times 12 \times 35 = 22 \times 12 \times 5 = 1320 \) m\( ^2 \).
In simple words: The tent's canvas area equals the curved part of the cylinder plus the slanted surface of the cone. When you add these together using their formulas, you get 1320 m\( ^2 \).

Exam Tip: Always identify the cone height by subtracting the cylinder height from the total height given. The slant height formula \( l = \sqrt{h^2 + r^2} \) is essential - do not skip this step.

Question 12. An exhibition tent is in the form of a cylinder surmounted by a cone. The height of the tent above the ground is 85 m and the height of the cylindrical part is 50 m. If the diameter of the base is 168 m, find the quantity of canvas required to make the tent. Allow 20% extra for folds and stitching. Give your answers to the nearest m\( ^2 \).
Answer: The total height is 85 m, with the cylinder measuring 50 m. This means the cone's height is \( 85 - 50 = 35 \) m. The radius of the base is \( \frac{168}{2} = 84 \) m. The cone's slant height is \( l = \sqrt{35^2 + 84^2} = \sqrt{1225 + 7056} = \sqrt{8281} = 91 \) m. The surface area of the tent (without extra) is \( S = 2\pi rh_1 + \pi rl = \pi r(2h_1 + l) = \frac{22}{7} \times 84 \times (2 \times 50 + 91) = \frac{22}{7} \times 84 \times 191 = 22 \times 12 \times 191 = 50424 \) m\( ^2 \). Adding 20% for folds and stitching: \( 50424 + 0.2 \times 50424 = 50424 + 10084.8 = 60508.8 \approx 60509 \) m\( ^2 \).
In simple words: Calculate the curved surface area of both the cylinder and cone parts, then add an extra 20% to account for folds and stitches. The final answer is about 60509 m\( ^2 \).

Exam Tip: Remember to apply the percentage increase at the end - do not forget this step, as it is explicitly asked for in the question. Always round to the nearest whole number unless stated otherwise.

Question 13. From a solid cylinder of height 30 cm and radius 7 cm, a conical cavity of height 24 cm and of base radius 7 cm is drilled out. Find the volume and the total surface of the remaining solid.
Answer: Both the cylinder and cone share the same radius of 7 cm. The cone's slant height is \( l = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \) cm. For the remaining volume, subtract the cone's volume from the cylinder's volume: \( V = \pi r^2 H - \frac{1}{3}\pi r^2 h = \pi r^2(H - \frac{h}{3}) = \frac{22}{7} \times 7^2 \times (30 - 8) = 22 \times 7 \times 22 = 3388 \) cm\( ^3 \). The total surface area includes the curved surface of the cylinder, the base of the cylinder, and the curved surface of the cone: \( S = 2\pi rH + \pi r^2 + \pi rl = \pi r(2H + r + l) = \frac{22}{7} \times 7 \times (2 \times 30 + 7 + 25) = 22 \times (60 + 32) = 22 \times 92 = 2024 \) cm\( ^2 \).
In simple words: When you drill out a cone from inside the cylinder, the remaining solid has a volume that is the cylinder's volume minus the cone's volume. The surface area now includes all outer surfaces plus the cone-shaped cavity opening.

Exam Tip: Do not forget to include the base area of the cylinder in the surface area calculation. This is a common mistake. Also, verify that the cone's height is less than the cylinder's height so the cone fits entirely inside.

Question 14. A buoy is made in the form of a hemisphere surmounted by a right cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 metres and its volume is \( \frac{2}{3} \) of the hemisphere. Calculate the height of the cone and the surface area of the buoy correct to 2 places of decimal.
Answer: The hemisphere and cone both have radius 3.5 m. The volume of the hemisphere is \( V_h = \frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times (3.5)^3 = \frac{44}{21} \times 42.875 = 89.8 \) m\( ^3 \). Since the cone's volume equals \( \frac{2}{3} \) of this, the cone's volume is \( \frac{2}{3} \times 89.8 = 59.87 \) m\( ^3 \). Using \( \frac{1}{3}\pi r^2 h = 59.87 \), we get \( \frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times h = 59.87 \), which simplifies to \( \frac{22}{21} \times 12.25 \times h = 59.87 \). Solving: \( h = \frac{59.87 \times 21}{22 \times 12.25} = \frac{1257.27}{269.5} = 4.67 \) m. The slant height is \( l = \sqrt{h^2 + r^2} = \sqrt{(4.67)^2 + (3.5)^2} = \sqrt{21.81 + 12.25} = \sqrt{34.06} = 5.84 \) m. The surface area is \( S = \pi r(l + 2r) = \frac{22}{7} \times 3.5 \times (5.84 + 7) = \frac{22}{7} \times 3.5 \times 12.84 = 141.17 \) m\( ^2 \).
In simple words: The cone's height can be found by using the fact that its volume is two-thirds of the hemisphere's volume. Once you know the height, the slant height and surface area follow from standard formulas.

Exam Tip: Pay close attention to the fractional volume relationship given in the problem - this is the key constraint that allows you to find the cone's height. Round your final answers to exactly 2 decimal places as requested.

Question 15. A building is in the form of a cylinder surmounted by a hemisphere vaulted dome and contains \( 41\frac{19}{21} \) m\( ^3 \) of air. If the internal diameter of dome is equal to the total height of the building, find the height of the building.
Answer: Let the dome's radius be \( r \) m. Then the internal diameter is \( 2r \), which equals the building's total height, so \( h = 2r \). The hemisphere's height is \( r \), which means the cylindrical section has height \( h_1 = 2r - r = r \). The total volume is the sum of the cylindrical and hemispherical volumes: \( V = \pi r^2 h_1 + \frac{2}{3}\pi r^3 = \pi r^2 \cdot r + \frac{2}{3}\pi r^3 = \pi r^3 + \frac{2}{3}\pi r^3 = \frac{5}{3}\pi r^3 \). Given that \( V = 41\frac{19}{21} = \frac{880}{21} \) m\( ^3 \), we have \( \frac{5}{3} \times \frac{22}{7} \times r^3 = \frac{880}{21} \). Solving: \( r^3 = \frac{880 \times 3 \times 7}{5 \times 22 \times 21} = \frac{18480}{2310} = 8 \), so \( r = 2 \) m. Therefore, the building's height is \( h = 2r = 2(2) = 4 \) m.
In simple words: Since the diameter equals the total height, both the cylindrical and hemispherical parts have the same radius. Using the volume formula and the given total volume, you can solve for the radius, then double it to get the height.

Exam Tip: The key insight is recognizing that when the diameter equals the height, the cylindrical height and the hemisphere's radius are both equal to \( r \). This relationship greatly simplifies the volume calculation.

Question 16. A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and the height of the cylinder are 6 cm and 12 cm respectively. If the slant height of the conical portion is 5 cm, find the total surface area and the volume of the rocket. (Use \( \pi = 3.14 \)).
Answer: The cylinder has diameter 6 cm, so radius \( r = 3 \) cm and height \( H = 12 \) cm. The cone's slant height is \( l = 5 \) cm. The cone's perpendicular height is found from \( h^2 = l^2 - r^2 = 5^2 - 3^2 = 25 - 9 = 16 \), giving \( h = 4 \) cm. The total surface area consists of the curved cylinder surface, the cylinder's base, and the cone's curved surface: \( S = 2\pi rH + \pi r^2 + \pi rl = \pi r(2H + r + l) = 3.14 \times 3 \times (2 \times 12 + 3 + 5) = 9.42 \times 32 = 301.44 \) cm\( ^2 \). The total volume is the sum of the cone and cylinder volumes: \( V = \frac{1}{3}\pi r^2 h + \pi r^2 H = \pi r^2(\frac{h}{3} + H) = 3.14 \times 3^2 \times (\frac{4}{3} + 12) = 3.14 \times 9 \times \frac{40}{3} = 376.8 \) cm\( ^3 \).
In simple words: The rocket's total surface includes three parts: the curved side of the cylinder, the flat bottom of the cylinder, and the slanted cone surface. The volume is found by adding the volumes of the cone and cylinder together.

Exam Tip: Remember that the cylinder is closed at the lower end, so its base area must be included in the surface area calculation. Use the slant height formula to find the cone's perpendicular height before calculating volumes.

Question 17. The adjoining figure represents a solid consisting of a right circular cylinder with a hemisphere at one end and a cone at the other. Their common radius is 7 cm. The

Exam Tip: This question appears to be incomplete in the source document. The full question text and answer are not provided. Please verify the complete question statement and requirements to ensure an accurate solution can be given.

Question 17. A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. Their common radius is 7 cm, the height of the cone is 4 cm, and the height of the cylinder is also 4 cm. Find the volume of the solid.
Answer: Let us find the volumes of all three parts and add them together. The shared radius is 7 cm, the cone's height is 4 cm, and the cylinder's height is 4 cm. The volume formula for this composite solid is: Total Volume = \( \frac{1}{3}\pi r^2 h_{cone} + \pi r^2 h_{cylinder} + \frac{2}{3}\pi r^3 \). Factoring out \( \pi r^2 \), we get: \( V = \pi r^2 \left( \frac{h_{cone}}{3} + h_{cylinder} + \frac{2r}{3} \right) \)

Plugging in the values: \( V = \frac{22}{7} \times (7)^2 \times \left( \frac{4}{3} + 4 + \frac{2 \times 7}{3} \right) \)

\( = 22 \times 7 \times \left( \frac{4}{3} + 4 + \frac{14}{3} \right) \)

\( = 154 \times \left( \frac{4 + 12 + 14}{3} \right) \)

\( = 154 \times \frac{30}{3} \)

\( = 154 \times 10 = 1540 \text{ cm}^3 \)
In simple words: Add the three parts separately - the cone's volume, the cylinder's volume, and the hemisphere's volume. When you add them all, you get 1540 cubic centimeters.

Exam Tip: Always write the volume formula for the composite solid first by adding the individual formulas, then factor out common terms like \( \pi r^2 \) to simplify the calculation.

Question 18. A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. Their common diameter is 3.5 cm and the height of the cylindrical and conical portions are 10 cm and 6 cm respectively. Find the volume of the solid. (Take \( \pi = 3.14 \))
Answer: We start by identifying the given information. The shared diameter is 3.5 cm, so the shared radius is 1.75 cm. The cylinder has a height of 10 cm, and the cone has a height of 6 cm. The total volume is the sum of the cone's volume, the cylinder's volume, and the hemisphere's volume. Using the combined formula: \( V = \pi r^2 \left( \frac{h_{cone}}{3} + h_{cylinder} + \frac{2r}{3} \right) \)

Substituting the values: \( V = 3.14 \times (1.75)^2 \times \left( \frac{6}{3} + 10 + \frac{2 \times 1.75}{3} \right) \)

\( = 3.14 \times 3.0625 \times \left( 2 + 10 + 1.167 \right) \)

\( = 3.14 \times 3.0625 \times 13.167 \)

\( = 126.62 \text{ cm}^3 \)
In simple words: Find the volume by multiplying the radius squared by the height values of all three sections together with pi. The final answer is about 126.62 cubic centimeters.

Exam Tip: When the radius is given as a decimal (like 1.75), calculate \( r^2 \) carefully and keep sufficient decimal places throughout to ensure accuracy in the final result.

Question 19. A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The height and radius of the cylindrical part are 13 cm and 5 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Calculate the surface area of the toy if the height of the conical part is 12 cm.
Answer: Start by finding the slant height of the cone. Given that the radius is 5 cm and the height is 12 cm, using the Pythagorean theorem: \( l = \sqrt{r^2 + h^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ cm} \)

The total surface area consists of three parts. The curved surface area of the hemisphere is \( 2\pi r^2 \). The curved surface area of the cylinder is \( 2\pi r H \) where H is the cylinder's height. The curved surface area of the cone is \( \pi r l \). Combining: \( S = 2\pi r^2 + 2\pi r H + \pi r l = \pi r(2r + 2H + l) \)

Substituting values: \( S = \frac{22}{7} \times 5 \times (2 \times 5 + 2 \times 13 + 13) \)

\( = \frac{110}{7} \times (10 + 26 + 13) \)

\( = \frac{110}{7} \times 49 \)

\( = 110 \times 7 = 770 \text{ cm}^2 \)
In simple words: Find the slant height of the cone first using the Pythagorean theorem. Then add the curved surfaces of all three parts - the hemisphere, cylinder, and cone - to get the total surface area of 770 square centimeters.

Exam Tip: Calculate the slant height of the cone before computing the surface area - this is a common step that students sometimes skip, leading to incorrect answers.

Question 20. The adjoining figure shows a model of a solid consisting of a cylinder surmounted by a hemisphere at one end. If the model is drawn to a scale of 1:200, find (i) the total surface area of the solid in \( \pi \text{ m}^2 \). (ii) the volume of the solid in \( \pi \) litres.
Answer:
(i) From the given figure, the cylinder's height (H) is 8 cm and the radius of both the cylinder and hemisphere (r) is 3 cm. The scale factor is k = 200. The total surface area includes the curved surface area of the hemisphere (\( 2\pi r^2 \)), the curved surface area of the cylinder (\( 2\pi r H \)), and the base area of the cylinder (\( \pi r^2 \)). Therefore: \( S = 2\pi r^2 + 2\pi r H + \pi r^2 = 3\pi r^2 + 2\pi r H = \pi r(3r + 2H) \)

Substituting the measurements: \( S = 3\pi(3 \times 3 + 2 \times 8) = 3\pi(9 + 16) = 3\pi \times 25 = 75\pi \text{ cm}^2 \)

Since we're using a 1:200 scale, the actual surface area is: \( S_{actual} = 75\pi \times (200)^2 = 75\pi \times 40000 \text{ cm}^2 = 75\pi \times \frac{40000}{100 \times 100} \text{ m}^2 = 300\pi \text{ m}^2 \)

(ii) The volume of the solid includes the hemisphere (\( \frac{2}{3}\pi r^3 \)) and the cylinder (\( \pi r^2 H \)). Therefore: \( V = \frac{2}{3}\pi r^3 + \pi r^2 H = \pi r^2 \left( \frac{2r}{3} + H \right) \)

Substituting: \( V = \pi \times 3^2 \times \left( \frac{2 \times 3}{3} + 8 \right) = 9\pi \times (2 + 8) = 9\pi \times 10 = 90\pi \text{ cm}^3 \)

With the 1:200 scale: \( V_{actual} = 90\pi \times (200)^3 = 90\pi \times 8000000 \text{ cm}^3 = 720000000\pi \text{ cm}^3 = 720\pi \text{ m}^3 \)

Converting to litres (1 m³ = 1000 litres): \( V_{actual} = 720\pi \times 1000 = 720000\pi \text{ litres} \)
In simple words: To find the surface area and volume, first calculate them at the model's scale, then multiply by the scale factor squared for area (\( 200^2 \)) and cubed for volume (\( 200^3 \)) to get the actual measurements.

Exam Tip: When working with scale drawings, remember to use \( k^2 \) for areas and \( k^3 \) for volumes - forgetting this or mixing them up is a frequent error on scaled solid problems.

Question 21. A solid metallic cylinder is cut into two identical halves along its height. The diameter of the cylinder is 7 cm and the height is 10 cm. Find: (a) The total surface area (both the halves). (b) The total cost of painting the two halves at the rate of Rs 30 per cm².
Answer:
(a) Given that the cylinder's diameter is 7 cm, the radius is 3.5 cm, and the height is 10 cm. When the cylinder is cut into two equal halves along its height, each half exposes two rectangular faces. The total surface area of both halves includes the curved surface area of the original cylinder and the areas of the two rectangular cross-sections created by the cut. The formula is: Total surface area = \( 2\pi r(h + r) + 2(h \times d) \)

where the first part is the curved surface of the cylinder and the radius-based top/bottom, and the second part accounts for the two rectangular cuts (height × diameter, calculated twice). Substituting: \( S = \left[ 2 \times \frac{22}{7} \times 3.5 \times (3.5 + 10) \right] + \left[ 2 \times (10 \times 7) \right] \)

\( = (2 \times 22 \times 0.5 \times 13.5) + 140 \)

\( = 297 + 140 = 437 \text{ cm}^2 \)

(b) The total painting cost is calculated by multiplying the surface area by the rate per square centimeter: Cost = \( 437 \times 30 = \text{Rs } 13,110 \)
In simple words: When a cylinder is cut in half lengthwise, the new flat surfaces add to the total area. You add the original curved surface area to the area of the two rectangular faces created by the cut. Then multiply by the painting rate to find the cost.

Exam Tip: Don't forget to include the areas of the rectangular faces created when the cylinder is cut - many students only count the curved surface area and miss these new exposed faces.

Question 22. Oil is stored in a spherical vessel occupying \( \frac{3}{4} \) of its full capacity. Radius of this spherical vessel is 28 cm. This oil is then poured into a cylindrical vessel with a radius of 21 cm. Find the height of the oil in the cylindrical vessel (correct to the nearest cm). Take \( \pi = \frac{22}{7} \)
Answer: The volume of a complete sphere is \( V = \frac{4}{3}\pi r^3 \). With a radius of 28 cm: \( V = \frac{4}{3} \times \frac{22}{7} \times 28^3 \)

The oil occupies three-fourths of this capacity: \( V_{oil} = \frac{3}{4} \times \frac{4}{3} \times \frac{22}{7} \times 28^3 = \frac{22}{7} \times 28^3 \)

This oil is poured into a cylindrical vessel with radius 21 cm. If h is the height of oil in the cylinder: \( \pi r^2 h = V_{oil} \)

\( \frac{22}{7} \times 21^2 \times h = \frac{22}{7} \times 28^3 \)

Simplifying: \( 28^3 = 21^2 \times h \)

\( h = \frac{28^3}{21^2} = \frac{21952}{441} = 49.77 \approx 50 \text{ cm} \)
In simple words: Calculate how much oil fills three-fourths of the sphere. Then pour this amount into the cylinder and find how high the oil level reaches inside it.

Exam Tip: When working with fractions of a container's volume, simplify the calculation by cancelling common factors early - this makes the arithmetic easier and reduces rounding errors.

Exercise 17.5

Question 1. The diameter of a metallic sphere is 6 cm. The sphere is melted and drawn into a wire of uniform cross-section. If the length of the wire is 36 m, find its radius.
Answer: Let a represent the wire's radius. When the sphere is melted and reshaped into a wire, the volumes remain equal. The sphere has a radius of \( r = \frac{6}{2} = 3 \) cm. The sphere's volume is: \( V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \times 3^3 = 4\pi \times 9 = 36\pi \text{ cm}^3 \)

The wire is a cylinder with length (height) = 36 m = 3600 cm. The wire's volume is: \( V = \pi a^2 \times 3600 \)

Since the volumes are equal: \( \pi a^2 \times 3600 = 36\pi \)

\( a^2 \times 3600 = 36 \)

\( a^2 = \frac{36}{3600} = \frac{1}{100} \)

\( a = \frac{1}{10} \text{ cm} = 1 \text{ mm} \)
In simple words: The sphere and wire have the same amount of material, so their volumes are equal. Find the sphere's volume first, then use that to find the radius of the wire.

Exam Tip: Always convert units to the same system before equating volumes - here, converting 36 m to 3600 cm is crucial for getting the correct answer.

Question 2. A solid metallic sphere of radius 6 cm is melted and made into a solid cylinder of height 32 cm. Find the: (i) radius of the cylinder (ii) curved surface area of the cylinder. Take \( \pi = 3.1 \)
Answer:
(i) The sphere has a radius of 6 cm. Its volume is: \( V_{sphere} = \frac{4}{3}\pi \times 6^3 = \frac{4}{3}\pi \times 216 = 288\pi \text{ cm}^3 \)

This volume equals the cylinder's volume: \( \pi r^2 h = 288\pi \)

where h = 32 cm. Therefore: \( r^2 \times 32 = 288 \)

\( r^2 = \frac{288}{32} = 9 \)

\( r = 3 \text{ cm} \)

(ii) The curved surface area of the cylinder is: \( CSA = 2\pi r h = 2 \times 3.1 \times 3 \times 32 = 595.2 \text{ cm}^2 \)
In simple words: Start by finding the sphere's volume. Then set this equal to the cylinder's volume formula to find the radius. Use that radius to calculate the curved surface area.

Exam Tip: In two-part problems like this, always solve part (i) completely before attempting part (ii), as the answer to part (i) is usually needed for part (ii).

Question 3. A solid metallic hemisphere of radius 8 cm is melted and recasted into right circular cone of base radius 6 cm. Determine the height of the cone.
Answer: The hemisphere has a radius of 8 cm. When melted and poured into a cone with base radius 6 cm, the volumes stay the same.

Volume of hemisphere = \( \frac{2}{3}\pi r^3 = \frac{2}{3}\pi \times 8^3 = \frac{2}{3}\pi \times 512 = \frac{1024}{3}\pi \) cm³

Volume of cone = \( \frac{1}{3}\pi R^2 h \)

Setting them equal:

\( \frac{1}{3}\pi \times 6^2 \times h = \frac{1024}{3}\pi \)

\( \frac{1}{3}\pi \times 36 \times h = \frac{1024}{3}\pi \)

\( 36h = 1024 \)

\( h = \frac{1024}{36} = 28\frac{4}{9} \) cm

In simple words: When you melt a half-ball and pour it into a cone shape, the amount of material stays the same. You can work out the cone's height by saying the volumes are equal.

Exam Tip: Always set the volumes equal when material is melted and recast. Cancel \( \pi \) and simplify fractions carefully to avoid arithmetic errors.

Question 4. A rectangular water tank of base 11 m × 6 m contains water upto a height of 5 m. If the water in the tank is transferred to a cylindrical tank of radius 3.5 m, find the height of water level in the tank.
Answer: The rectangular tank holds water in a rectangular container. We need to find how high the water rises when it is moved to a cylindrical container.

Volume of water in rectangular tank = length × breadth × height = 11 × 6 × 5 = 330 m³

When this water moves to the cylinder, the volume stays the same.

Volume of cylindrical tank = \( \pi r^2 H \)

\( \pi r^2 H = 330 \)

\( \frac{22}{7} \times (3.5)^2 \times H = 330 \)

\( \frac{22}{7} \times 12.25 \times H = 330 \)

\( H = \frac{330 \times 7}{22 \times 12.25} = \frac{2310}{269.5} = 8\frac{4}{7} \) m

In simple words: Water moves from a rectangular box to a round cylinder. The amount of water is the same, so we use the volumes to find the new height.

Exam Tip: Remember to use the correct volume formula for each shape - rectangle uses \( l \times b \times h \) while a cylinder uses \( \pi r^2 h \). Keep \( \pi \) as a fraction unless told otherwise.

Question 5. The volume of a cone is the same as that of the cylinder whose height is 9 cm and diameter 40 cm. Find the radius of the base of the cone if its height is 108 cm.
Answer: First, find the volume of the cylinder. The diameter is 40 cm, so the radius is 20 cm and the height is 9 cm.

Volume of cylinder = \( \pi r^2 h = \pi \times 20 \times 20 \times 9 = 3600\pi \) cm³

Now use the cone volume. The cone has height 108 cm and unknown radius R.

Volume of cone = \( \frac{1}{3}\pi R^2 h = \frac{1}{3}\pi R^2 \times 108 = 36\pi R^2 \) cm³

Since volumes are equal:

\( 36\pi R^2 = 3600\pi \)

\( R^2 = 100 \)

\( R = 10 \) cm

In simple words: A cone and a cylinder have the same amount of space inside. Find the cylinder's volume first, then use it to find the cone's radius.

Exam Tip: Always identify which solid shapes you have and apply the right volume formula. When volumes are equal, the \( \pi \) cancels out, making the calculation easier.

Question 6. Solid spherical ball of radius 6 cm is melted and recast into 64 identical spherical marble. Find the radius of each marble.
Answer: A large sphere is broken into 64 small identical spheres. The total volume stays the same.

Volume of large sphere = \( \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \times 6^3 = \frac{4}{3}\pi \times 216 = 288\pi \) cm³

Total volume of 64 small spheres = 64 × (volume of one small sphere)

If each small sphere has radius r:

\( \frac{4}{3}\pi \times 6^3 = 64 \times \frac{4}{3}\pi r^3 \)

\( 6^3 = 64 \times r^3 \)

\( 216 = 64 \times r^3 \)

\( r^3 = \frac{216}{64} \)

\( r = \sqrt[3]{\frac{216}{64}} = \frac{6}{4} = 1.5 \) cm

In simple words: One big ball becomes 64 tiny balls. Divide the big ball's volume by 64 to get one small ball's volume, then find the radius.

Exam Tip: When a solid is melted into multiple identical pieces, divide the original volume by the number of pieces. Taking the cube root gives you the radius of each small sphere.

Question 7. A hemispherical bowl of diameter 7.2 cm is filled completely with chocolate sauce. This sauce is poured into an inverted cone of radius 4.8 cm. Find the height of the cone.
Answer: The sauce starts in a half-sphere bowl and is poured into a cone. The volume of sauce stays the same.

Diameter of bowl = 7.2 cm, so radius = 3.6 cm

Volume of hemisphere = \( \frac{2}{3}\pi r^3 = \frac{2}{3}\pi \times (3.6)^3 = \frac{2}{3}\pi \times 46.656 \) cm³

For the cone with radius 4.8 cm and unknown height h:

Volume of cone = \( \frac{1}{3}\pi R^2 h = \frac{1}{3}\pi \times (4.8)^2 \times h \)

Setting them equal:

\( \frac{1}{3}\pi \times (4.8)^2 \times h = \frac{2}{3}\pi \times (3.6)^3 \)

\( \frac{1}{3} \times 23.04 \times h = \frac{2}{3} \times 46.656 \)

\( h = \frac{2 \times 46.656}{23.04} = 4.05 \) cm

In simple words: Sauce in a curved bowl is poured into a cone shape. The amount of sauce stays the same, so their volumes match.

Exam Tip: For hemisphere problems, use \( \frac{2}{3}\pi r^3 \) not \( \frac{4}{3}\pi r^3 \). Always set volumes equal when liquid or material is transferred between containers.

Question 8. Two spheres of the same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the big sphere.
Answer: Two metal spheres are melted together. Since they are made of the same metal, their density is identical. The new sphere will have the same density as well.

For the small sphere: mass = 1 kg, radius = 3 cm

Volume of small sphere = \( \frac{4}{3}\pi \times 3^3 = 36\pi \) cm³

Density = \( \frac{\text{mass}}{\text{volume}} = \frac{1}{36\pi} \)

When both spheres are melted, total mass = 1 + 7 = 8 kg

Let R be the radius of the new sphere. Using the same density:

\( \frac{8}{\frac{4}{3}\pi R^3} = \frac{1}{36\pi} \)

\( \frac{8 \times 3}{4\pi R^3} = \frac{1}{36\pi} \)

\( \frac{24}{4\pi R^3} = \frac{1}{36\pi} \)

\( 24 \times 36\pi = 4\pi R^3 \)

\( 216 = R^3 \)

\( R = 6 \) cm

Diameter = 2 × 6 = 12 cm

In simple words: Two metal balls melt together. Since the metal is the same, we can use mass and volume to find how big the new ball is.

Exam Tip: In melting problems with same material, density stays constant. Use the ratio of masses and volumes to find the new radius without calculating density separately.

Question 9. A hollow copper pipe of inner diameter 6 cm and outer diameter 10 cm is melted and changed into a solid circular cylinder of the same height as that of the pipe. Find the diameter of the solid cylinder.
Answer: The hollow pipe has both an inner and outer radius. When melted, only the metal (the part between the two radii) is used to make the new solid cylinder.

Inner radius (r) = 3 cm, outer radius (R) = 5 cm

Volume of metal in hollow pipe = \( \pi(R^2 - r^2)h = \pi(5^2 - 3^2)h = \pi(25 - 9)h = 16\pi h \) cm³

This metal forms a solid cylinder with the same height h and new radius r₁:

Volume of solid cylinder = \( \pi r_1^2 h \)

Setting them equal:

\( \pi r_1^2 h = 16\pi h \)

\( r_1^2 = 16 \)

\( r_1 = 4 \) cm

Diameter = 2 × 4 = 8 cm

In simple words: A hollow tube is melted to make a solid rod of the same height. The metal inside the tube becomes the whole new rod.

Exam Tip: For hollow cylinders, subtract the inner volume from the outer volume. The height cancels when volumes are equal, making the calculation cleaner.

Question 10. A hollow sphere of internal and external diameters 4 cm and 8 cm respectively, is melted into a cone of base diameter 8 cm. Find the height of the cone.
Answer: A hollow sphere has metal between an inner and outer surface. This metal is melted and formed into a cone.

Internal radius (r) = 2 cm, external radius (R) = 4 cm

Volume of metal in hollow sphere = \( \frac{4}{3}\pi(R^3 - r^3) = \frac{4}{3}\pi(4^3 - 2^3) = \frac{4}{3}\pi(64 - 8) = \frac{4}{3}\pi \times 56 \) cm³

For the cone with base diameter 8 cm (radius = 4 cm) and unknown height h:

Volume of cone = \( \frac{1}{3}\pi r_1^2 h = \frac{1}{3}\pi \times 4^2 \times h = \frac{16\pi h}{3} \) cm³

Setting them equal:

\( \frac{16\pi h}{3} = \frac{4}{3}\pi \times 56 \)

\( 16h = 4 \times 56 \)

\( 16h = 224 \)

\( h = 14 \) cm

In simple words: A hollow ball has metal in its walls. When melted, that metal makes a solid cone. Find the volumes and make them equal.

Exam Tip: For hollow shapes, always subtract inner volume from outer volume to get the metal amount. Simplify by cancelling \( \pi \) early.

Question 11. A well with inner diameter 6 m is dug 22 m deep. Soil taken out of it has been spread evenly all round it to a width of 5 m to form an embankment. Find the height of the embankment.
Answer: When a well is dug, the soil is removed and spread as a ring around it. The volume of soil dug out equals the volume of the embankment ring.

Inner radius of well (r) = 3 m, depth (h) = 22 m

Volume of soil dug out = \( \pi r^2 h = \pi \times 3^2 \times 22 = 198\pi \) m³

The embankment is a ring with inner radius 3 m and outer radius 3 + 5 = 8 m, with height H:

Volume of embankment = \( \pi(R^2 - r^2)H = \pi(8^2 - 3^2)H = \pi(64 - 9)H = 55\pi H \) m³

Setting them equal:

\( 198\pi = 55\pi H \)

\( H = \frac{198}{55} = 3.6 \) m

In simple words: Dirt dug from a hole is piled around it as a ring. The amount of dirt stays the same, so find the ring's height.

Exam Tip: Draw a diagram to show the well and the embankment ring around it. The embankment is an annulus (ring shape), so use \( \pi(R^2 - r^2)H \) for its volume.

Question 12. A cylindrical can of internal diameter 21 cm contains water. A solid sphere whose diameter is 10.5 cm is lowered into the cylindrical can. The sphere is completely immersed in water. Calculate the rise in water level, assuming that no water overflows.
Answer: When a sphere sinks into water, the water level rises by an amount equal to the sphere's volume divided by the cylinder's cross-sectional area.

Radius of cylinder (R) = \( \frac{21}{2} \) cm

Radius of sphere (r) = \( \frac{10.5}{2} = \frac{21}{4} \) cm

Volume of sphere = \( \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \times \left(\frac{21}{4}\right)^3 \) cm³

The rise in water level h satisfies:

\( \pi R^2 h = \frac{4}{3}\pi r^3 \)

\( \left(\frac{21}{2}\right)^2 h = \frac{4}{3}\left(\frac{21}{4}\right)^3 \)

\( \frac{441}{4}h = \frac{4}{3} \times \frac{9261}{64} \)

\( h = \frac{4 \times 9261}{3 \times 64 \times 441} = \frac{36864}{84672} = 1.75 \) cm

In simple words: A ball sinks into water in a container. The water goes up because the ball takes up space. Calculate how much using the sphere's volume and the cylinder's area.

Exam Tip: When an object is immersed in a cylinder, volume of object = \( \pi R^2 \times \) height rise. Cancel \( \pi \) early and simplify fractions before multiplying.

Question 13. There is water to a height of 14 cm in a cylindrical glass jar of radius 8 cm. Inside the water there is a sphere of diameter 12 cm completely immersed. By what height will the water go down when the sphere is removed?
Answer: When the sphere is removed, the water level drops. The drop equals the sphere's volume divided by the cylinder's base area.

Radius of jar (R) = 8 cm

Radius of sphere (r) = 6 cm

Volume of sphere = \( \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \times 6^3 = \frac{4}{3}\pi \times 216 = 288\pi \) cm³

The water level drops by height h where:

\( \pi R^2 h = \frac{4}{3}\pi r^3 \)

\( \pi \times 8^2 \times h = 288\pi \)

\( 64h = 288 \)

\( h = \frac{288}{64} = 4.5 \) cm

In simple words: A ball is in water. When you take the ball out, the water goes down. The water drops by an amount based on the ball's size and the jar's width.

Exam Tip: When an object is removed from a cylinder of water, the height drop is the object's volume divided by the cylinder's base area \( (\pi R^2) \). This is the reverse of the immersion problem.

Question 14. A vessel in the form of an inverted cone is filled with water to the brim. Its height is 20 cm and diameter is 16.8 cm. Two equal solid cones are dropped in it so that they are fully submerged. As a result, one third of the water in the original cone overflows. What is the volume of each of the solid cone submerged?
Answer: The cone's height is 20 cm and its radius is 8.4 cm. Using the cone volume formula, the water volume is \( \frac{1}{3} \times \frac{22}{7} \times (8.4)^2 \times 20 = 1478.4 \text{ cm}^3 \). When the solid cones are placed inside, one third of the water spills out, which means the displaced water volume equals \( \frac{1}{3} \times 1478.4 = 492.8 \text{ cm}^3 \). Since two equal cones cause this displacement, each cone has a volume of \( \frac{492.8}{2} = 246.4 \text{ cm}^3 \).
In simple words: The water that flows out equals the space taken up by the two cones. Each cone takes up the same amount of space, so divide the overflowed water volume by 2.

Exam Tip: Remember that the volume of water displaced equals the volume of the submerged objects. Use the one-third fraction to find the overflow amount, then divide by the number of cones.

Question 15. A solid metallic circular cylinder of radius 14 cm and height 12 cm is melted and recast into small cubes of edge 2 cm. How many such cubes can be made from the solid cylinder?
Answer: The cylinder has radius 14 cm and height 12 cm. Its volume is \( \pi r^2 h = \frac{22}{7} \times (14)^2 \times 12 = \frac{22}{7} \times 196 \times 12 = 7392 \text{ cm}^3 \). Each small cube has edge 2 cm, so its volume is \( (2)^3 = 8 \text{ cm}^3 \). The number of cubes that can be made is \( \frac{7392}{8} = 924 \).
In simple words: Divide the cylinder's volume by the volume of one cube. This tells you how many cubes fit inside.

Exam Tip: Always match units and ensure you are dividing the total volume by the individual object's volume to find the count.

Question 16. How many shots each having diameter 3 cm can be made from a cuboidal lead solid of dimensions 9 cm × 11 cm × 12 cm?
Answer: Each shot is a sphere with diameter 3 cm, so its radius is 1.5 cm. The cuboidal lead solid has volume \( 9 \times 11 \times 12 = 1188 \text{ cm}^3 \). The volume of one spherical shot is \( \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (1.5)^3 = \frac{4}{3} \times \frac{22}{7} \times 3.375 = \frac{297}{21} \text{ cm}^3 \). Therefore, the number of shots is \( \frac{1188}{\frac{297}{21}} = 1188 \times \frac{21}{297} = 84 \).
In simple words: Find how much space one sphere uses, then divide the cuboid's total volume by that amount.

Exam Tip: Use the exact sphere volume formula and simplify fractions carefully to avoid calculation errors.

Question 17. A solid metal cylinder of radius 14 cm and height 21 cm is melted down and recast into spheres of radius 3.5 cm. Calculate the number of spheres that can be made.
Answer: The metal cylinder has volume \( \pi r^2 h = \frac{22}{7} \times (14)^2 \times 21 = \frac{22}{7} \times 196 \times 21 \). Simplifying: \( 22 \times 28 \times 21 = 12936 \text{ cm}^3 \). Each sphere has radius 3.5 cm and volume \( \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (3.5)^3 = \frac{4}{3} \times \frac{22}{7} \times 42.875 = \frac{297}{3} \text{ cm}^3 \). The number of spheres is \( \frac{12936}{\frac{297}{3}} = 12936 \times \frac{3}{297} = 72 \).
In simple words: Work out the cylinder's volume and the sphere's volume, then divide one by the other to get the count.

Exam Tip: Take care when calculating cubic values and use cancellation in fractions to simplify before dividing.

Question 18. A solid cone of radius 5 cm and height 9 cm is melted and made into small cylinders of radius 0.5 cm and height 1.5 cm. Find the number of cylinders so formed.
Answer: The solid cone has volume \( \frac{1}{3} \pi R^2 H = \frac{1}{3} \times \frac{22}{7} \times (5)^2 \times 9 = \frac{1}{3} \times \frac{22}{7} \times 25 \times 9 \). Simplifying: \( \frac{1}{3} \times \frac{22}{7} \times 225 = 75 \text{ cm}^3 \). Each small cylinder has volume \( \pi r^2 h = \frac{22}{7} \times (0.5)^2 \times 1.5 = \frac{22}{7} \times 0.25 \times 1.5 = \frac{22}{7} \times 0.375 \). Cancelling the \( \frac{22}{7} \) terms: number of cylinders is \( \frac{25 \times 9}{3 \times 0.25 \times 1.5} = \frac{75}{0.375} = 200 \).
In simple words: Since both the cone and cylinder use \( \pi \), those cancel out when dividing. Find the cone's volume and divide by one cylinder's volume.

Exam Tip: When two solids share \( \pi \), you can cancel it out early to make arithmetic simpler.

Question 19. A metallic sphere of radius 10.5 cm is melted and then recast into small cones, each of radius 3.5 cm and height 3 cm. Find the number of cones thus obtained.
Answer: The metallic sphere has volume \( \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (10.5)^3 \). Computing: \( (10.5)^3 = 1157.625 \), so volume is \( \frac{4}{3} \times \frac{22}{7} \times 1157.625 = 4630.5 \text{ cm}^3 \). Each cone has volume \( \frac{1}{3} \pi R^2 h = \frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times 3 = \frac{1}{3} \times \frac{22}{7} \times 12.25 \times 3 = 36.75 \text{ cm}^3 \). The number of cones is \( \frac{4630.5}{36.75} = 126 \).
In simple words: Divide the sphere's volume by one cone's volume to see how many cones you can make.

Exam Tip: Double-check your calculation of cubic values before dividing - a small error there makes the final count wrong.

Question 20. A certain number of metallic cones each of radius 2 cm and height 3 cm are melted and recast in a solid sphere of radius 6 cm. Find the number of cones.
Answer: Each cone has volume \( \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (2)^2 \times 3 = \frac{1}{3} \times \frac{22}{7} \times 4 \times 3 = \frac{1}{3} \times \frac{22}{7} \times 12 \). The sphere has volume \( \frac{4}{3} \pi R^3 = \frac{4}{3} \times \frac{22}{7} \times (6)^3 = \frac{4}{3} \times \frac{22}{7} \times 216 \). Since both have the common factor \( \frac{22}{7} \), dividing gives: \( n = \frac{4 \times 216}{\frac{1}{3} \times 12} = \frac{864}{4} = 216 \). Correcting: \( n \times r^2 h = 4R^3 \), so \( n \times 4 \times 3 = 4 \times 216 \), giving \( n = \frac{864}{12} = 72 \).
In simple words: Set up an equation where the total cone volume equals the sphere volume, then solve for the number of cones.

Exam Tip: Carefully cancel common factors like \( \pi \) and coefficients to reduce the equation before solving.

Question 21. A vessel is in the form of an inverted cone. Its height is 11 cm and the radius of its top, which is open, is 2.5 cm. It is filled with water upto the rim. When some lead shots, each of which is a sphere of radius 0.25 cm, are dropped into the vessel, 2/5 of the water flows out. Find the number of lead shots dropped into the vessel.
Answer: The cone's radius is 2.5 cm and its height is 11 cm. Its total volume is \( \frac{1}{3} \pi R^2 H = \frac{1}{3} \times \frac{22}{7} \times (2.5)^2 \times 11 = \frac{137.5\pi}{15} \text{ cm}^3 \). The water that flows out is \( \frac{2}{5} \) of this: \( V = \frac{2}{5} \times \frac{137.5\pi}{15} = \frac{137.5\pi}{37.5} \). Each lead shot is a sphere with radius 0.25 cm and volume \( \frac{4}{3} \pi (0.25)^3 = \frac{4\pi \times 0.015625}{3} \text{ cm}^3 \). Setting volume of water flown out equal to total volume of spheres: \( \frac{137.5\pi}{37.5} = n \times \frac{4\pi \times 0.015625}{3} \), which gives \( n = 440 \).
In simple words: Find what fraction of the cone's water spills out, then divide by the volume of one lead shot.

Exam Tip: Remember that the fraction of water displaced equals the total volume of the submerged objects.

Question 22. The surface area of a solid metallic sphere is 616 cm². It is melted and recast into smaller spheres of diameter 3.5 cm. How many such spheres can be obtained?
Answer: Given surface area \( 4\pi R^2 = 616 \), we solve for the large sphere's radius: \( R^2 = \frac{616}{4\pi} = \frac{616}{4 \times \frac{22}{7}} = \frac{616 \times 7}{4 \times 22} = \frac{4312}{88} = 49 \), so \( R = 7 \) cm. The smaller spheres have diameter 3.5 cm, so radius \( r = 1.75 \) cm. Volume of the large sphere is \( \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (7)^3 = \frac{1372\pi}{3} \text{ cm}^3 \). Volume of each small sphere is \( \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (1.75)^3 = \frac{4}{3}\pi \times 5.359375 \). Setting up the ratio: \( n = \frac{(7)^3}{(1.75)^3} = \frac{343}{5.359375} = 64 \).
In simple words: Use the surface area to find the large sphere's radius, then divide its volume by one small sphere's volume.

Exam Tip: When working with spheres and surface area, solve for the radius first before finding volume.

Question 23. The surface area of a solid metallic sphere is 1256 cm². It is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate (i) the radius of the solid sphere. (ii) the number of cones recast. (Use \( \pi = 3.14 \)).
Answer: (i) Given \( 4\pi R^2 = 1256 \), we find \( R^2 = \frac{1256}{4 \times 3.14} = \frac{1256}{12.56} = 100 \), so \( R = 10 \) cm. (ii) The sphere's volume is \( \frac{4}{3}\pi R^3 = \frac{4}{3} \times 3.14 \times (10)^3 = \frac{4}{3} \times 3.14 \times 1000 = 4186.67 \text{ cm}^3 \). Each cone has volume \( \frac{1}{3} \pi r^2 h = \frac{1}{3} \times 3.14 \times (2.5)^2 \times 8 = \frac{1}{3} \times 3.14 \times 6.25 \times 8 = 52.33 \text{ cm}^3 \). The number of cones is \( \frac{4186.67}{52.33} \approx 80 \).
In simple words: First use surface area to get the sphere's radius. Then find how many cones fit in the sphere by dividing volumes.

Exam Tip: When \( \pi \) is given as a specific value like 3.14, use it consistently throughout both parts of the question.

Question 24. A cylindrical can whose base is horizontal and of radius 3.5 cm contains sufficient water so that when a sphere is placed in the can, the water just covers the sphere. Given that the sphere just fits into the can, calculate: (i) the total surface area of the can in contact with water when the sphere is in it. (ii) the depth of the water in the can before the sphere was put into the can. Given your answer as proper fractions.
Answer: (i) Since the sphere just fits inside the cylinder, its radius equals the cylinder's radius: \( R = r = 3.5 \) cm. When the water just covers the sphere, the water height is \( h = 2R = 7 \) cm. The surface area in contact with water includes the curved side and the base: \( T = 2\pi rh + \pi r^2 = \pi r(2h + r) = \frac{22}{7} \times 3.5 \times (2 \times 7 + 3.5) = 22 \times 0.5 \times 17.5 = 11 \times 17.5 = 192.5 \text{ cm}^2 \). (ii) The sphere's volume is \( \frac{4}{3}\pi (3.5)^3 \). The volume of water before the sphere was placed equals the volume of water after the sphere is added. Initial water depth \( d \) satisfies: \( \pi r^2 d = \pi r^2 \times 7 - \frac{4}{3}\pi r^3 \). Simplifying: \( d = 7 - \frac{4}{3} \times 3.5 = 7 - 4.667 = 2.333 = \frac{7}{3} \) cm.
In simple words: The sphere fits snugly in the cylinder, so both have the same radius. The water height when the sphere is in equals twice the sphere's radius.

Exam Tip: For this problem, recognise that the submerged sphere displaces a volume of water equal to its own volume, and use this to find the original depth.

Question 1. The volume of a conical tent is 462 m³ and the area of the base 154 m². The height of the conical tent is:
(a) 15 m
(b) 12 m
(c) 9 m
(d) 24 m
Answer: (c) 9 m
In simple words: Use the cone volume formula. Substitute the known area value to find the height directly.

Exam Tip: Remember that the base area is already given - avoid computing it again from the radius. Substitute directly into the volume formula.

Question 2. The radius of a roller 100 cm long is 14 cm. The curved surface area of the roller is
(a) 13200 cm²
(b) 15400 cm²
(c) 4400 cm²
(d) 8800 cm²
Answer: (d) 8800 cm²
In simple words: A roller is a cylinder. Apply the curved surface area formula using the radius and height (length) provided.

Exam Tip: Always check whether you need curved surface area or total surface area - the question clearly asks for curved only, so exclude the two circular bases.

Question 3. If two cylinders of the same lateral surface have their radii in the ratio 4:9, then the ratio of their heights is
(a) 2:3
(b) 3:2
(c) 4:9
(d) 9:4
Answer: (d) 9:4
In simple words: When lateral surface areas are equal, the height ratio is the reciprocal of the radius ratio.

Exam Tip: Set the two lateral surface area expressions equal and simplify - the \( 2\pi \) terms cancel out cleanly.

Question 4. The radii of two cylinders are in the ratio 2:3 and their heights are in the ratio 5:3. The ratio of their volumes is
(a) 10:17
(b) 20:27
(c) 17:27
(d) 20:37
Answer: (b) 20:27
In simple words: The volume of a cylinder depends on both radius squared and height. Multiply the squared radius ratio by the height ratio.

Exam Tip: Volume ratios involve the radius squared - forgetting to square the radius ratio is a common error that will drop your marks.

Question 5. The total surface area of a cone whose radius is \( \frac{r}{2} \) and slant height 2l is
(a) \( 2\pi r(l + r) \)
(b) \( \pi r\left(l + \frac{r}{4}\right) \)
(c) \( \pi r(l + r) \)
(d) \( 2\pi rl \)
Answer: (b) \( \pi r\left(l + \frac{r}{4}\right) \)
In simple words: Apply the total surface area formula for a cone using the given radius and slant height, then simplify.

Exam Tip: Total surface area includes both the curved lateral area and the circular base - do not forget either component.

Question 6. If the diameter of the base of cone is 10 cm and its height is 12 cm, then its curved surface area is
(a) \( 60\pi \) cm²
(b) \( 65\pi \) cm²
(c) \( 90\pi \) cm²
(d) \( 120\pi \) cm²
Answer: (b) \( 65\pi \) cm²
In simple words: First find the radius by halving the diameter. Then use the Pythagorean theorem to get the slant height. Finally, use the curved surface area formula.

Exam Tip: The slant height is not the same as the vertical height - you must calculate it using \( l = \sqrt{r^2 + h^2} \).

Question 7. If the radius of a hemisphere is 5 cm, then its volume is
(a) \( \frac{250\pi}{3} \) cm³
(b) \( \frac{500\pi}{3} \) cm³
(c) \( 75\pi \) cm³
(d) \( \frac{125\pi}{3} \) cm³
Answer: (a) \( \frac{250\pi}{3} \) cm³
In simple words: A hemisphere is half a sphere. Use the hemisphere volume formula with the given radius.

Exam Tip: Remember the factor \( \frac{2}{3} \) in the hemisphere volume formula - it comes from half of the sphere volume formula.

Question 8. If the ratio of the diameters of the two spheres is 3:5, then the ratio of their surface areas is
(a) 3:5
(b) 5:3
(c) 27:125
(d) 9:25
Answer: (d) 9:25
In simple words: Surface area of a sphere depends on radius squared. Since diameter is 2 times the radius, the surface area ratio is the diameter ratio squared.

Exam Tip: Surface area involves the square of the radius (or diameter), so a 3:5 diameter ratio gives a 9:25 surface area ratio.

Question 9. The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratio of the surface areas of the balloon in the two cases is
(a) 1:4
(b) 1:3
(c) 2:3
(d) 2:1
Answer: (a) 1:4
In simple words: The hemisphere surface area increases with the square of the radius. When radius doubles, surface area becomes 4 times larger.

Exam Tip: Surface area grows as the square of linear dimensions - doubling the radius means the surface area multiplies by 4, not 2.

Question 10. If two solid hemispheres of same base radius r are joined together along with their bases, then the curved surface of this new solid is
(a) \( 4\pi r^2 \)
(b) \( 6\pi r^2 \)
(c) \( 3\pi r^2 \)
(d) \( 8\pi r^2 \)
Answer: (a) \( 4\pi r^2 \)
In simple words: Two hemispheres joined at their bases form a complete sphere. The curved surface area is that of one full sphere.

Exam Tip: When the hemispheres are joined, their flat circular bases touch and are not part of the curved surface - only count the two curved surfaces.

Question 11. If a solid of one shape is converted to another, then the surface area of the new solid
(a) remains same
(b) increases
(c) decreases
(d) can't say
Answer: (d) can't say
In simple words: The surface area after reshaping depends on the final shape - it can increase, decrease, or stay the same.

Exam Tip: Volume stays constant when reshaping, but surface area changes unpredictably - without knowing the final shape, you cannot determine the outcome.

Question 12. The volume of the largest right circular cone that can be carved out from a cube of edge 4.2 cm is
(a) 9.7 cm³
(b) 77.6 cm³
(c) 58.2 cm³
(d) 19.4 cm³
Answer: (d) 19.4 cm³
In simple words: The largest cone fits inside the cube with its circular base on one face and its apex at the opposite face. The radius is half the edge length and the height equals the edge length.

Exam Tip: Visualize how a cone sits inside a cube - its base diameter equals the cube edge, and its height equals the cube edge.

Question 13. The volume of the greatest sphere cut off from a circular cylindrical wood of base radius 1 cm and height 6 cm is
(a) \( 288\pi \) cm³
(b) \( \frac{4\pi}{3} \) cm³
(c) \( 6\pi \) cm³
(d) \( 4\pi \) cm³
Answer: (b) \( \frac{4\pi}{3} \) cm³
In simple words: The largest sphere that fits inside the cylinder has a radius equal to the cylinder's radius, which is 1 cm.

Exam Tip: The sphere must fit entirely inside the cylinder - its radius cannot exceed the cylinder's base radius, even though the cylinder is taller.

Question 14. The volumes of two spheres are in the ratio 64:27. The ratio of their surface areas is
(a) 3:4
(b) 4:3
(c) 9:16
(d) 16:9
Answer: (d) 16:9
In simple words: From the volume ratio, find the radius ratio by taking the cube root. Then square this radius ratio to get the surface area ratio.

Exam Tip: Volume involves radius cubed, surface area involves radius squared - use the relationship between these two exponents to link the ratios.

Question 15. If a cone, a hemisphere and a cylinder have equal bases and have same height, then the ratio of their volumes is
(a) 1:3:2
(b) 2:3:1
(c) 2:1:3
(d) 1:2:3
Answer: (a) 1:3:2
In simple words: Write the volume formula for each shape using the same base radius r and height h. The cone volume is \( \frac{1}{3}\pi r^2 h \), the hemisphere volume is \( \frac{2}{3}\pi r^3 \), and the cylinder volume is \( \pi r^2 h \). When the base and height are equal, compute their ratio.

Exam Tip: For a hemisphere with height equal to cylinder height, the hemisphere radius equals that height - apply the formulas carefully and simplify the ratio step by step.

Question 16. If a sphere and a cube have equal surface areas, then the ratio of the diameter of the sphere to the edge of the cube is
(a) 1 : 2
(b) 2 : 1
(c) \( \sqrt{\pi} : \sqrt{6} \)
(d) \( \sqrt{6} : \sqrt{\pi} \)
Answer: (d) \( \sqrt{6} : \sqrt{\pi} \)
In simple words: When a sphere and a cube have the same outer surface area, the ratio of the sphere's width (diameter) to the cube's side length equals \( \sqrt{6} : \sqrt{\pi} \).

Exam Tip: Equate the surface area formulas \( 4\pi r^2 = 6a^2 \) and solve for the ratio \( d : a \) by taking square roots of both sides.

Question 17. A solid piece of iron in the form of a cuboid of dimensions 49 cm × 33 cm × 24 cm is molded to form a sphere. The radius of the sphere is
(a) 21 cm
(b) 23 cm
(c) 25 cm
(d) 19 cm
Answer: (a) 21 cm
In simple words: When a rectangular block is melted and reformed into a ball, the volume stays the same. Use this fact to find the ball's radius.

Exam Tip: Always equate volumes: volume of cuboid = \( \frac{4}{3}\pi r^3 \). Substitute values carefully and simplify using \( \pi = \frac{22}{7} \) to avoid arithmetic errors.

Question 18. If a solid right circular cone of height 24 cm and base radius 6 cm is melted and recast in the shape of sphere, then the radius of the sphere is
(a) 4 cm
(b) 6 cm
(c) 8 cm
(d) 12 cm
Answer: (b) 6 cm
In simple words: A cone is melted down and reshaped into a ball. The amount of material (volume) does not change, so you can find the ball's radius using this.

Exam Tip: Set cone volume equal to sphere volume: \( \frac{1}{3}\pi R^2 h = \frac{4}{3}\pi r^3 \). Cancel \( \pi \) and multiply by 3 to simplify before solving for r.

Question 19. If a solid circular cylinder of iron whose diameter is 15 cm and height 10 cm is melted and recasted into a sphere, then the radius of the sphere is
(a) 15 cm
(b) 10 cm
(c) 7.5 cm
(d) 5 cm
Answer: (c) 7.5 cm
In simple words: A cylinder is melted and poured into a sphere shape. The total amount of material stays the same, so calculate the sphere's radius from this fact.

Exam Tip: Remember that cylinder radius = diameter ÷ 2 = 7.5 cm. Equate volumes and solve: \( \pi r^2 h = \frac{4}{3}\pi R^3 \) gives \( R^3 = (7.5)^3 \).

Question 20. The number of balls of radius 1 cm that can be made from a sphere of radius 10 cm is
(a) 100
(b) 1000
(c) 10000
(d) 100000
Answer: (b) 1000
In simple words: A large ball is cut up into many tiny balls. Divide the big ball's volume by one small ball's volume to find how many fit.

Exam Tip: Use the formula \( n = \frac{R^3}{r^3} = \frac{10^3}{1^3} = 1000 \). This works because both shapes are spheres with the same density.

Question 21. A metallic spherical shell of internal and external diameters 4 cm and 8 cm respectively is melted and recast into the form of a cone of base diameter 8 cm. The height of the cone is
(a) 12 cm
(b) 14 cm
(c) 15 cm
(d) 18 cm
Answer: (b) 14 cm
In simple words: A hollow metal shell (like a thick-walled ball) is melted and reshaped into a cone. The metal's amount stays the same, so calculate the height using equal volumes.

Exam Tip: The volume of a hollow shell = volume of outer sphere - volume of inner sphere = \( \frac{4}{3}\pi(R^3 - r^3) \). Set this equal to the cone's volume and solve for height.

Question 22. A cubical ice cream brick of edge 22 cm is to be distributed among some children by filling ice cream cones of radius 2 cm and height 7 cm up to its brim. The number of children who will get the ice cream cones is
(a) 163
(b) 263
(c) 363
(d) 463
Answer: (c) 363
In simple words: A cube of ice cream is divided equally into cone-shaped containers. Find how many cones can be filled by dividing the cube's volume by one cone's volume.

Exam Tip: Calculate the cube's volume \( 22^3 = 10648 \) cm³, then one cone's volume \( \frac{1}{3}\pi r^2 h = \frac{88}{3} \) cm³. Divide to get the number of cones: \( \frac{10648 \times 3}{88} = 363 \).

Question 23. Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is
(a) 4 cm
(b) 3 cm
(c) 2 cm
(d) 6 cm
Answer: (c) 2 cm
In simple words: A cylinder is melted and divided equally into 12 identical balls. Use the equal volumes to find each ball's size.

Exam Tip: Set cylinder volume = 12 × (one sphere's volume). This gives \( \pi r^2 h = 12 \times \frac{4}{3}\pi R^3 \). Substitute \( r = 1 \), \( h = 16 \), and solve for R.

Question 24. A rectangular sheet of paper of size 11 cm × 7 cm is first rotated about the side 11 cm and then about the side 7 cm to form a cylinder, as shown in the diagram. The ratio of their curved surface areas is:
(a) 1 : 1
(b) 7 : 11
(c) 11 : 7
(d) \( \frac{11\pi}{7} : \frac{7\pi}{11} \)
Answer: (a) 1 : 1
In simple words: A flat piece of paper is rolled in two different ways to make two cylinders. Both cylinders have the same curved (side) surface area even though they look different.

Exam Tip: When rotating about the 11 cm side, the height is 7 cm and the radius is \( \frac{11}{2\pi} \). When rotating about the 7 cm side, the height is 11 cm and the radius is \( \frac{7}{2\pi} \). The curved surface area formula \( 2\pi rh \) gives the same result both times.

Question 25. A right angle triangle shaped piece of hard board is rotated completely about its hypotenuse, as shown in the diagram. The solid so formed is always:
(i) a single cone
(ii) a double cone
Which of the statement is valid?
(a) only (i)
(b) only (ii)
(c) both (i) and (ii)
(d) neither (i) nor (ii)
Answer: (b) only (ii)
In simple words: When you spin a right-angled triangle around its longest side (the hypotenuse), the shape you get is two cones joined point-to-point, not just one cone.

Exam Tip: Visualise rotating a right triangle about its hypotenuse - both the other two sides sweep out cone shapes, one above the hypotenuse and one below, creating a double cone (or bicone) shape.

Question 26. A solid sphere is cut into identical hemispheres.
Statement 1: The total volume of two hemispheres is equal to the volume of the original sphere.
Statement 2: The total surface area of two hemispheres together is equal to the surface area of the original sphere.
Which of the following is valid?
(a) Both the statements are true
(b) Both the statements are false
(c) Statement 1 is true, and statement 2 is false
(d) Statement 1 is false, and statement 2 is true
Answer: (c) Statement 1 is true, and statement 2 is false
In simple words: When you cut a ball in half, the two half-balls have the same total volume as the whole ball. However, their total outer surface area is larger than the original ball's surface because each half now has a flat circular face.

Exam Tip: Volume statement: \( 2 \times \frac{2}{3}\pi r^3 = \frac{4}{3}\pi r^3 \) - TRUE. Surface area statement: each hemisphere has curved surface \( 2\pi r^2 \) plus flat base \( \pi r^2 \), so total is \( 2(3\pi r^2) = 6\pi r^2 \), which is greater than \( 4\pi r^2 \) - FALSE.

Question. Assertion Reason Type Questions

(Note: The source document ends with this heading but provides no assertion-reason questions following it. No further content to process.)

Question. Assertion (A): The surface area of largest sphere that can be inscribed in a hollow cube of side 'a' cm is πa² cm².
Answer: When a sphere is inscribed inside a hollow cube, the sphere's diameter equals the cube's side length (a). This means the radius is a/2. The surface area formula for a sphere is 4πr². Substituting r = a/2 gives: 4π(a/2)² = 4π(a²/4) = πa². The assertion is correct.
In simple words: A sphere fit inside a cube touches all six sides. Its width across is the same as the cube's side. This makes the surface area exactly πa².

Exam Tip: Remember that when a sphere is inscribed in a cube, the diameter equals the side length - this is the key relationship. The reason given in the question (about πr³) is a volume formula, not a surface area formula, so it cannot be the correct reason.

Question. Assertion (A): From a solid wooden cylinder of height 15 cm and diameter 14 cm, a conical cavity of same height and same base diameter is hollowed out. The volume of the cone is 770 cm³.
Answer: The cylinder has a height of 15 cm and diameter of 14 cm, so its radius is 7 cm. The cone shares the same dimensions: height 15 cm and radius 7 cm. Using the cone volume formula V = (1/3)πr²h:

\( V = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 15 = \frac{1}{3} \times 22 \times 7 \times 15 = 22 \times 7 \times 5 = 770 \text{ cm}^3 \)

The assertion is true. The reason states that a cylinder's volume is πr²h, which is a correct formula, making this a valid reason for the assertion.
In simple words: A cone inside a cylinder with the same base and height holds one-third of the cylinder's volume. With these measurements, that equals 770 cm³.

Exam Tip: Both the assertion and reason are correct AND the reason explains why the assertion is true - this is the key distinction. The cone's volume is exactly one-third of what the cylinder would hold with identical base and height.

Question. Assertion (A): From a solid wooden cylinder of height 15 cm and diameter 14 cm, a hemispherical depression of same base diameter is carved out. The volume of remaining wood is 1591⅓ cm³.
Answer: The cylinder has radius 7 cm and height 15 cm. The hemisphere also has radius 7 cm. The volume of remaining wood is found by subtracting the hemisphere's volume from the cylinder's volume:

Volume of cylinder = πr²h = (22/7) × 7² × 15 = 22 × 7 × 15 = 2310 cm³

Volume of hemisphere = (2/3)πr³ = (2/3) × (22/7) × 7³ = (2/3) × 22 × 49 = 2156/3 cm³

Remaining volume = 2310 - 2156/3 = (6930 - 2156)/3 = 4774/3 = 1591⅓ cm³

The assertion is true. The reason correctly identifies both formulas (πr²h for cylinder and (2/3)πr³ for hemisphere), making it the correct explanation.
In simple words: Start with a solid cylinder. Scoop out a hemisphere from the top. What's left is the cylinder volume minus the hemisphere volume, which gives 1591⅓ cm³.

Exam Tip: This is a composite solid problem. Always identify each component shape separately, calculate each volume, then add or subtract as required. Both statements are correct and logically connected.

Question 1. A cylindrical container is to be made of tin sheet. The height of the container is 1 m and its diameter is 70 cm. If the container is open at the top and the tin sheet costs Rs. 300 per m², find the cost of the tin for making the container.
Answer: Because the container is open at the top, the total surface area includes only the curved surface plus the base:

Radius = 70/2 = 35 cm = 0.35 m

Surface area = 2πrh + πr²
\( = 2 \times \frac{22}{7} \times 0.35 \times 1 + \frac{22}{7} \times 0.35 \times 0.35 \)
\( = 2 \times 22 \times 0.05 + 22 \times 0.05 \times 0.35 \)
\( = 2.2 + 0.385 = 2.585 \text{ m}^2 \)

Total cost = 2.585 × 300 = Rs. 775.50
In simple words: Find the curved side area and the flat bottom area. Add them together. Multiply by the price per square meter to get the total cost.

Exam Tip: Always note whether the top is open or closed - this changes the surface area calculation. For an open-top cylinder, exclude the top circle from the total surface area.

Question 2. A cylinder of maximum volume is cut out from a wooden cuboid of length 30 cm and cross section of square of side 14 cm. Find the volume of the cylinder and the volume of wood wasted.
Answer: The largest cylinder that fits inside the cuboid has a diameter equal to the side of the square cross section (14 cm), giving a radius of 7 cm. The height of the cylinder equals the cuboid's length (30 cm).

Volume of cylinder = πr²h = (22/7) × 7² × 30 = 22 × 7 × 30 = 4620 cm³

Volume of cuboid = length × width × height = 30 × 14 × 14 = 5880 cm³

Volume of wood wasted = 5880 - 4620 = 1260 cm³
In simple words: The cylinder takes up as much space as possible inside the rectangular block. What's left over is the wasted wood. The larger block volume minus the cylinder volume gives the waste.

Exam Tip: When finding maximum volume of one shape inside another, the inscribed shape's dimensions are constrained by the outer shape's dimensions. Here, the cylinder's diameter cannot exceed the square's side.

Question 3. Find the volume and the total surface area of a cone having slant height 17 cm and base diameter 30 cm. Take π = 3.14
Answer: Given: slant height l = 17 cm, diameter = 30 cm, so radius r = 15 cm

Using the relationship l² = r² + h²:
\( 17^2 = 15^2 + h^2 \)
\( 289 = 225 + h^2 \)
\( h = 8 \text{ cm} \)

Volume of cone = (1/3)πr²h = (1/3) × 3.14 × 15² × 8 = (1/3) × 3.14 × 225 × 8 = 1884 cm³

Total surface area = πr(l + r) = 3.14 × 15 × (17 + 15) = 3.14 × 15 × 32 = 1507.2 cm²
In simple words: First find the height using the Pythagorean theorem. Then use that height to calculate the volume. For surface area, add the base area and the curved side area together.

Exam Tip: Always verify that you have the height before calculating volume. Use the three-dimensional Pythagorean theorem (l² = r² + h²) when slant height is given instead of vertical height.

Question 4. Find the volume of a cone given that its height is 8 cm and the area of base 156 cm².
Answer: Given: height h = 8 cm and base area = 156 cm²

The volume formula for a cone is: V = (1/3) × base area × height

V = (1/3) × 156 × 8 = 1248/3 = 416 cm³
In simple words: When you already know the base area, you don't need to calculate it from the radius. Just multiply the base area by the height, then divide by 3.

Exam Tip: This alternative formula V = (1/3) × base area × height is useful when the base area is given directly, avoiding the need to find the radius first.

Question 5. The circumference of the edge of a hemispherical bowl is 132 cm. Find the capacity of the bowl.
Answer: The edge of the hemispherical bowl forms a circle with circumference 132 cm:

\( 2\pi r = 132 \)
\( 2 \times \frac{22}{7} \times r = 132 \)
\( r = \frac{132 \times 7}{22 \times 2} = \frac{924}{44} = 21 \text{ cm} \)

Volume of hemisphere = (2/3)πr³ = (2/3) × (22/7) × 21³ = (2/3) × 22 × 21² = 2 × 22 × 441 = 19404 cm³
In simple words: Find the radius from the circumference. Then use the hemisphere volume formula with that radius.

Exam Tip: The capacity (volume) of a hemispherical bowl is half the volume of a complete sphere. Start by finding the radius from the given circumference, then proceed with the volume calculation.

Question 6. The volume of a hemisphere is 2425½ cm³. Find its curved surface area.
Answer: Given: volume of hemisphere = 2425½ = 4851/2 cm³

Using the volume formula (2/3)πr³ = 4851/2:
\( \frac{2}{3} \times \frac{22}{7} \times r^3 = \frac{4851}{2} \)
\( r^3 = \frac{4851 \times 3 \times 7}{2 \times 22 \times 2} = \frac{101871}{88} = \frac{9261}{8} = \left(\frac{21}{2}\right)^3 \)
\( r = \frac{21}{2} \text{ cm} \)

Curved surface area = 2πr² = 2 × (22/7) × (21/2)² = 2 × (22/7) × (441/4) = 44 × 441/28 = 693 cm²
In simple words: Work backwards from the volume to find the radius. Then use that radius to find the curved surface area.

Exam Tip: When working backwards from volume to radius, simplify the cubic root carefully. The curved surface area of a hemisphere is 2πr² (half of a full sphere's surface area).

Question 7. A solid wooden toy is in the shape of a right circular cone mounted on a hemisphere. If the radius of the hemisphere is 4.2 cm and the total height of the toy is 10.2 cm, find the volume of the toy.
Answer: The toy consists of a hemisphere (radius 4.2 cm) with a cone mounted on top. Since the hemisphere contributes a height of 4.2 cm to the total, the cone's height is:

\( h_{\text{cone}} = 10.2 - 4.2 = 6 \text{ cm} \)

Total volume = Volume of cone + Volume of hemisphere
\( = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 \)
\( = \pi r^2\left(\frac{1}{3}h + \frac{2}{3}r\right) \)
\( = \frac{1}{3} \times \frac{22}{7} \times (4.2)^2 \times (6 + 2 \times 4.2) \)
\( = \frac{22 \times 17.64 \times 14.4}{21} = \frac{5588.352}{21} = 266.112 \text{ cm}^3 \)
In simple words: Find how tall the cone part is by subtracting the hemisphere's height from the total. Then add the cone's volume and hemisphere's volume together.

Exam Tip: For composite solids, identify each component clearly and calculate their volumes separately before adding. The hemisphere's height equals its radius, which is crucial for finding the cone's height.

Question 8. A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Find the volume and the total surface area of the solid.
Answer: The solid has two hemispherical ends (which together form a complete sphere) and a cylindrical middle section.

Radius r = 7/2 cm
Height of cylinder = Total height - (2 × radius) = 19 - 7 = 12 cm

Total volume = 2 × (hemisphere volume) + cylinder volume
\( = \frac{2}{3}\pi r^3 + \pi r^2 h = \pi r^2\left(\frac{4r}{3} + h\right) \)
\( = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \left(\frac{4 \times 7}{3 \times 2} + 12\right) \)
\( = \frac{77}{2} \times \left(\frac{14}{3} + 12\right) = \frac{77}{2} \times \frac{50}{3} = \frac{3850}{6} = 641\frac{2}{3} \text{ cm}^3 \)

Total surface area = 2 × (hemisphere surface) + cylinder curved surface
\( = 4\pi r^2 + 2\pi rh = \pi r(4r + 2h) \)
\( = \frac{22}{7} \times \frac{7}{2} \times \left(4 \times \frac{7}{2} + 2 \times 12\right) \)
\( = 11 \times (14 + 24) = 11 \times 38 = 418 \text{ cm}^2 \)
In simple words: A sphere in the middle, wrapped by a cylinder. Calculate the sphere volume and cylinder volume separately, then add them. For surface area, add the sphere's outer layer and the cylinder's curved side.

Exam Tip: The two hemispheres together make one complete sphere. The cylinder height is found by subtracting twice the radius (the combined height of both hemispheres) from the total height.

Question 9. The radius and height of a right circular cone are in the ratio 5 : 12. If its volume is 2512 cm³, find its slant height. (Take π = 3.14)
Answer: Let radius r = 5k and height h = 12k for some constant k.

Using the volume formula:
\( \frac{1}{3}\pi r^2 h = 2512 \)
\( \frac{1}{3} \times 3.14 \times (5k)^2 \times 12k = 2512 \)
\( \frac{1}{3} \times 3.14 \times 25k^2 \times 12k = 2512 \)
\( \frac{1}{3} \times 3.14 \times 300k^3 = 2512 \)
\( 314k^3 = 2512 \)
\( k^3 = 8 \)
\( k = 2 \)

Therefore: r = 5 × 2 = 10 cm and h = 12 × 2 = 24 cm

Slant height: \( l = \sqrt{r^2 + h^2} = \sqrt{100 + 576} = \sqrt{676} = 26 \text{ cm} \)
In simple words: When two measurements are in a ratio, use a variable times that ratio for each. Substitute into the volume formula to find that variable. Then calculate the slant height using the Pythagorean theorem.

Exam Tip: Ratio problems are solved by setting the quantities as multiples of a common variable. Once you find that variable using the given constraint (volume), the actual dimensions follow immediately. Always use the Pythagorean theorem for slant height.

Question 10. A cone and a cylinder are of the same height. If diameters of their bases are in the ratio 3 : 2, find the ratio of their volumes.
Answer: Let the height of cone and cylinder be h. The diameter of the cone's base is 3a, so its radius r₁ = 3a/2. The diameter of the cylinder's base is 2a, so its radius r₂ = 2a/2 = a.

Volume of cone (V₁) = (1/3)πr₁²h = (1/3)π(3a/2)²h

Volume of cylinder (V₂) = πr₂²h = π(a)²h

The ratio V₁/V₂ = [(1/3)π(3a/2)²h] / [πa²h] = (1/3) × (9a²/4) / a² = (1/3) × (9/4) = 3/4

Therefore, the ratio of their volumes is 3 : 4.
In simple words: When you divide the cone's volume by the cylinder's volume, you get 3 divided by 4.

Exam Tip: Always set up the ratio by dividing cone volume by cylinder volume in the order asked. Simplify carefully by cancelling π and h before computing the numerical ratio.

Question 11. A solid cone of base radius 9 cm and height 10 cm is lowered into a cylindrical jar of radius 10 cm, which contains water sufficient to submerge the cone completely. Find the rise in water level in the jar.
Answer: Radius of cone (r) = 9 cm, Height of cone (h) = 10 cm. Volume of the cone = (1/3)πr²h = (1/3)π(9)² × 10 = (1/3)π × 81 × 10 = 270π cm³

Let h₁ be the rise in water height in the jar. Radius of jar (r₁) = 10 cm. Since the cone is fully submerged, the volume of water that rises equals the cone's volume.

Volume of water rise = π(r₁)²h₁ = (1/3)πr²h

π(10)² × h₁ = (1/3)π(9)² × 10

100h₁ = (1/3) × 81 × 10 = 270

h₁ = 270/100 = 2.7 cm
In simple words: The cone pushes the water up. The amount the water rises is found by dividing the cone's volume by the jar's base area.

Exam Tip: The key is recognizing that the volume displaced by the submerged cone equals the volume of water that rises in the cylinder. Use the formula πr₁²h₁ = (1/3)πr²h and solve for h₁.

Question 12. An iron pillar has some part in the form of a right circular cylinder and the remaining in the form of a right circular cone. The radius of the base of each of cone and cylinder is 8 cm. The cylindrical part is 240 cm high and the conical part is 36 cm high. Find the weight of the pillar if one cu. cm of iron weighs 7.8 grams.
Answer: Radius of base of cone (r) = 8 cm, Radius of cylinder (r) = 8 cm. Height of cylindrical part (h₁) = 240 cm, Height of conical part (h₂) = 36 cm.

Total volume = Volume of cylinder + Volume of cone = πr²h₁ + (1/3)πr²h₂ = πr²(h₁ + h₂/3) = π(8)² × (240 + 36/3) = π × 64 × (240 + 12) = π × 64 × 252

Using π = 22/7:
V = (22/7) × 64 × 252 = 22 × 64 × 36 = 50,688 cm³

Weight of 1 cm³ of iron = 7.8 g
Weight of 50,688 cm³ = 50,688 × 7.8 = 395,366.4 g = 395.3664 kg

Hence, the weight of the pillar is 395.3664 kg.
In simple words: Find the total volume by adding the cylinder and cone volumes. Then multiply by the weight per unit volume to get the final weight.

Exam Tip: Factor out πr² from both volume terms to simplify the calculation. Remember to convert grams to kilograms by dividing by 1000 at the end.

Question 13. A circus tent is made of canvas and is in the form of right circular cylinder and a right circular cone above it. The diameter and height of cylindrical part of tent are 126 m and 5 m respectively. The total height of the tent is 21 m. Find the total cost of tent if canvas used costs Rs. 36 per square metre.
Answer: Diameter of cylindrical part = 126 m, so radius r = 63 m. Height of cylindrical part (h₁) = 5 m. Total height of tent = 21 m, so height of conical part (h) = 21 - 5 = 16 m.

Slant height of cone (l) = √(r² + h²) = √(63² + 16²) = √(3969 + 256) = √4225 = 65 m

Surface area of tent = Lateral surface area of cylinder + Lateral surface area of cone = 2πrh₁ + πrl = πr(2h₁ + l) = (22/7) × 63 × (2 × 5 + 65) = 22 × 9 × (10 + 65) = 198 × 75 = 14,850 m²

Cost per m² = Rs. 36
Total cost = 14,850 × 36 = Rs. 534,600

Hence, the total cost of tent is Rs. 534,600.
In simple words: Calculate the surface area by finding the curved surface of both shapes. Multiply the total area by the cost per unit area to get the final price.

Exam Tip: Remember that the tent covers only the lateral (curved) surfaces, not the base. The slant height must be calculated correctly before finding the cone's lateral surface area.

Question 14. The entire surface of a solid cone of base radius 3 cm and height 4 cm is equal to entire surface of a solid right circular cylinder of diameter 4 cm. Find the ratio of their (i) curved surfaces (ii) volumes.
Answer: Radius of base of cone (r₁) = 3 cm, Height of cone (h₁) = 4 cm. Slant height of cone (l) = √(3² + 4²) = √(9 + 16) = √25 = 5 cm.

Let height of cylinder be h₂ cm and radius be r₂ cm. From diameter 4 cm, r₂ = 2 cm. Total surface area of cone = πr₁(l + r₁) = π × 3 × (5 + 3) = 24π. Total surface area of cylinder = 2πr₂(r₂ + h₂) = 2π × 2 × (2 + h₂) = 4π(2 + h₂).

Given: 24π = 4π(2 + h₂)
24 = 4(2 + h₂)
6 = 2 + h₂
h₂ = 4 cm

(i) Ratio of curved surfaces = (πr₁l) / (2πr₂h₂) = (π × 3 × 5) / (2π × 2 × 4) = 15π / 16π = 15 : 16

(ii) Ratio of volumes = [(1/3)πr₁²h₁] / [πr₂²h₂] = (r₁²h₁) / (3r₂²h₂) = (3² × 4) / (3 × 2² × 4) = (9 × 4) / (3 × 4 × 4) = 36 / 48 = 3 : 4

Hence, the ratio of curved surfaces is 15 : 16 and the ratio of volumes is 3 : 4.
In simple words: First find the cylinder's height using the equal surface area condition. Then compute the two ratios by dividing corresponding formulas.

Exam Tip: Set up the equal surface area equation first to find the unknown height. For ratios, cancel π and common factors before simplifying to lowest terms.

Question 15. A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. Find the radius of the sphere.
Answer: Radius of base of cone (r) = 2.1 cm and height (h) = 8.4 cm. Let radius of sphere be R cm. Since the cone is recast into a sphere, their volumes are equal.

Volume of cone = Volume of sphere
(1/3)πr²h = (4/3)πR³

Cancelling π/3 from both sides:
r²h = 4R³
R³ = (r²h) / 4 = (2.1)² × 8.4 / 4 = 4.41 × 8.4 / 4

Notice that 8.4 = 4 × 2.1, so:
R³ = (2.1)² × 4 × 2.1 / 4 = (2.1)³
R = 2.1 cm

Hence, the radius of the sphere is 2.1 cm.
In simple words: Set the volumes equal and solve for R. When you substitute the numbers, you find that R³ equals 2.1 cubed, so R is 2.1 cm.

Exam Tip: Look for numerical relationships (like 8.4 = 4 × 2.1) that simplify the algebra. This avoids unnecessary computation and reduces chance of error.

Question 16. Find the least number of coins of diameter 2.5 cm and height 3 mm which are to be melted to form a solid cylinder of radius 3 cm and height 5 cm.
Answer: Radius of coin (r) = 2.5/2 = 1.25 cm. Height of coin (h) = 3 mm = 0.3 cm. Radius of cylinder (R) = 3 cm. Height of cylinder (H) = 5 cm.

Let n be the number of coins needed. The total volume of all coins must equal the cylinder's volume:
n × (Volume of each coin) = Volume of cylinder
n × πr²h = πR²H
n = (R²H) / (r²h) = (3² × 5) / (1.25² × 0.3) = 45 / (1.5625 × 0.3) = 45 / 0.46875 = 96

Hence, 96 coins are required to form a solid cylinder.
In simple words: Divide the cylinder's volume by each coin's volume to find how many coins you need.

Exam Tip: Convert all measurements to the same unit (cm in this case) before calculating volumes. Use the volume ratio formula to find n, then verify your answer makes sense.

Question 17. A hemisphere of lead of radius 8 cm is cast into a right circular cone of base radius 6 cm. Determine the height of the cone correct to 2 places of decimal.
Answer: Radius of hemisphere (r) = 8 cm. Radius of cone (R) = 6 cm. Let height of cone be h cm. Since the hemisphere is cast into a cone, their volumes are equal.

Volume of hemisphere = Volume of cone
(2/3)πr³ = (1/3)πR²h
2r³ = R²h
h = (2r³) / R² = (2 × 8³) / 6² = (2 × 512) / 36 = 1024 / 36 = 28.44 cm

Hence, the height of the cone is 28.44 cm.
In simple words: Set hemisphere volume equal to cone volume, then solve for the cone's height using algebra.

Exam Tip: When setting up the volume equation, remember the hemisphere formula is (2/3)πr³. Simplify by cancelling π/3 before substituting numbers.

Question 18. A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of water in the cylinder.
Answer: Radius of hemispherical bowl (r) = 6 cm. Radius of cylinder (R) = 4 cm. Let h be the height of water in the cylinder. The water from the hemisphere fills the cylinder, so their volumes are equal.

Volume of hemisphere = Volume of water in cylinder
(2/3)πr³ = πR²h
(2/3) × 6³ = 4² × h
(2/3) × 216 = 16h
144 = 16h
h = 9 cm

Hence, the height of water in the cylinder is 9 cm.
In simple words: The water volume stays the same when poured from the bowl into the cylinder. Use this to find how high the water reaches in the cylinder.

Exam Tip: Set the two volumes equal and cancel π from both sides. Substitute the radius values and solve for the unknown height algebraically.

Question 19. A sphere of diameter 6 cm is dropped into a right circular cylinder vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely

Question 19. A sphere of diameter 12 cm is submerged in water. By how much will the water level rise in the cylindrical vessel?
Answer: The sphere has a radius of 3 cm. The cylinder has a radius of 6 cm. When the sphere is fully submerged, it displaces water equal to its own volume. Setting the volume of the sphere equal to the volume of water displaced in the cylinder:

\( \frac{4}{3}\pi r^3 = \pi R^2 h \)

\( \frac{4}{3} \times 3^3 = 6^2 \times h \)

\( \frac{4}{3} \times 27 = 36h \)

\( 36 = 36h \)

\( h = 1 \text{ cm} \)

The water level rises by 1 cm.
In simple words: When you put a ball in water, the water pushes up. How much it pushes up depends on how big the ball is compared to the container.

Exam Tip: Always equate the volume of the solid being submerged to the volume of liquid displaced - this is the key principle for water displacement problems. Check your final answer by substituting back into the original equation.

Question 20. A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, find the uniform thickness of the cylinder.
Answer: The volume of the solid sphere must equal the volume of the hollow cylinder.

Volume of sphere:
\( V = \frac{4}{3}\pi r_1^3 = \frac{4}{3}\pi \times 6^3 = 4\pi \times 2 \times 36 = 288\pi \text{ cm}^3 \)

For the hollow cylinder with external radius R = 5 cm, internal radius r, and height h = 32 cm:
\( V = \pi(R^2 - r^2)h \)

\( 288\pi = \pi(5^2 - r^2) \times 32 \)

\( 288 = 32(25 - r^2) \)

\( \frac{288}{32} = 25 - r^2 \)

\( 9 = 25 - r^2 \)

\( r^2 = 16 \)

\( r = 4 \text{ cm} \)

Thickness = External radius - Internal radius = 5 - 4 = 1 cm
In simple words: The metal from the sphere is stretched into a hollow tube. The space between the outer edge and inner hole is the thickness.

Exam Tip: For hollow cylinders, remember to use \( \pi(R^2 - r^2)h \) where R is the outer radius and r is the inner radius. Set volumes equal and solve for the unknown radius first, then find thickness as the difference.

Question 21. In the adjoining diagram, a tilted right circular cylindrical vessel with base diameter 7 cm contains a liquid. When placed vertically, the height of the liquid in the vessel is the mean of two heights shown in the diagram. Find the area of wet surface, when the cylinder is placed vertically on a horizontal surface. (Use \( \pi = \frac{22}{7} \))
Answer: When the tilted cylinder is placed upright, the liquid height becomes the average of the two heights shown. From the diagram, the two heights are 1 cm and 6 cm.

Height of liquid: \( h = \frac{1 + 6}{2} = \frac{7}{2} \text{ cm} \)

Diameter of base = 7 cm, so radius: \( r = \frac{7}{2} \text{ cm} \)

The wet surface includes the circular base and the curved side that touches the liquid:
\( \text{Area of wet surface} = \pi r^2 + 2\pi rh = \pi r(r + 2h) \)

\( = \frac{22}{7} \times \frac{7}{2} \left( \frac{7}{2} + 2 \times \frac{7}{2} \right) \)

\( = \frac{22}{7} \times \frac{7}{2} \left( \frac{7}{2} + 7 \right) \)

\( = 11 \times \left( 3.5 + 7 \right) \)

\( = 11 \times 10.5 \)

\( = 115.5 \text{ cm}^2 \)
In simple words: The wet area is the bottom circle plus the side of the cylinder that gets wet. Add the area of the circle to the curved wall area that the liquid touches.

Exam Tip: Wet surface includes BOTH the base and the lateral (side) surface in contact with liquid - do not forget either part. Use the formula \( \pi r(r + 2h) \) as a shortcut to combine both areas efficiently.

Question 22. A manufacturing company prepares spherical ball bearings, each of radius 7 mm and mass 4 g. These ball bearings are packed into boxes. Each box can have maximum of 2156 cm³ of ball bearings. Find the:
(a) maximum number of ball bearings that each box can have.
(b) mass of each box of ball bearings in kg.
Answer:
(a) Given: radius of each ball bearing = 7 mm, volume of box = 2156 cm³ = 2156 × 10³ mm³

Volume of one ball bearing:
\( V_{\text{ball}} = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 7^3 = \frac{88}{3} \times 49 \text{ mm}^3 \)

Maximum number of ball bearings:
\( N = \frac{\text{Volume of box}}{\text{Volume of each ball}} = \frac{2156 \times 10^3}{\frac{4}{3} \times \frac{22}{7} \times 7^3} \)

\( = \frac{2156 \times 10^3}{\frac{88}{3} \times 49} = \frac{2156 \times 3 \times 1000}{88 \times 49} = \frac{6468000}{4312} = 1500 \)

Maximum number of ball bearings = 1500

(b) Mass of each box = Number of balls × Mass of each ball
\( = 1500 \times 4 \text{ g} = 6000 \text{ g} = 6 \text{ kg} \)

Mass of each box = 6 kg
In simple words: Divide the box volume by each ball's volume to get how many fit. Then multiply the number of balls by the weight of one ball to get the total weight.

Exam Tip: Always convert units consistently before dividing volumes - both should be in the same unit (either all cm³ or all mm³). In part (b), remember to convert grams to kilograms at the end by dividing by 1000.