OP Malhotra Class 10 Maths Solutions Chapter 15 Three Dimensional Solids Exercise 15 (A)

Question 1. Find the area of curved surface and total surface area of the cylinders whose heights and radii are given below:
(i) h = 12 cm, r = 7 cm
(ii) h = 10 cm, r = 7 cm
(iii) h = 5 cm, r = 21 cm
(iv) h = 20 cm, r = 14 cm
(v) h = 16 m, r = 10.5 m
Answer:
(i) For a cylinder with height \( h = 12 \) cm and radius \( r = 7 \) cm:
Curved surface area \( = 2\pi rh \)
\( = 2 \times \frac{22}{7} \times 7 \times 12 \)
\( = 44 \times 12 = 528 \) cm\(^2 \)
Total surface area \( = 2\pi r(h + r) \)
\( = 2 \times \frac{22}{7} \times 7(12 + 7) \)
\( = 44 \times 19 = 836 \) cm\(^2 \)

(ii) For a cylinder with height \( h = 10 \) cm and radius \( r = 7 \) cm:
Curved surface area \( = 2\pi rh \)
\( = 2 \times \frac{22}{7} \times 7 \times 10 \)
\( = 44 \times 10 = 440 \) cm\(^2 \)
Total surface area \( = 2\pi r(h + r) \)
\( = 2 \times \frac{22}{7} \times 7(10 + 7) \)
\( = 44 \times 17 = 748 \) cm\(^2 \)

(iii) For a cylinder with height \( h = 5 \) cm and radius \( r = 21 \) cm:
Curved surface area \( = 2\pi rh \)
\( = 2 \times \frac{22}{7} \times 21 \times 5 \)
\( = 2 \times 22 \times 3 \times 5 = 660 \) cm\(^2 \)
Total surface area \( = 2\pi r(h + r) \)
\( = 2 \times \frac{22}{7} \times 21(5 + 21) \)
\( = 2 \times 22 \times 3 \times 26 \)
\( = 132 \times 26 = 3432 \) cm\(^2 \)

(iv) For a cylinder with height \( h = 20 \) cm and radius \( r = 14 \) cm:
Curved surface area \( = 2\pi rh \)
\( = 2 \times \frac{22}{7} \times 14 \times 20 \)
\( = 2 \times 22 \times 2 \times 20 = 1760 \) cm\(^2 \)
Total surface area \( = 2\pi r(h + r) \)
\( = 2 \times \frac{22}{7} \times 14(20 + 14) \)
\( = 2 \times 22 \times 2 \times 34 \)
\( = 88 \times 34 = 2992 \) cm\(^2 \)

(v) For a cylinder with height \( h = 16 \) m and radius \( r = 10.5 \) m:
We can write \( r = 10.5 = \frac{21}{2} \) m.
Curved surface area \( = 2\pi rh \)
\( = 2 \times \frac{22}{7} \times \frac{21}{2} \times 16 \)
\( = 22 \times 3 \times 16 = 1056 \) m\(^2 \)
Total surface area \( = 2\pi r(h + r) \)
\( = 2 \times \frac{22}{7} \times \frac{21}{2} \left(16 + \frac{21}{2}\right) \)
\( = 66 \left(16 + 10.5\right) \)
\( = 66 \times 26.5 = 1749 \) m\(^2 \).
In simple words: To find the curved area, we multiply 2 by pi, by the radius, and by the height. For the total area, we add the area of the top and bottom circles to the curved area. Remember to use the correct units like cm² or m².

🎯 Exam Tip: Always pay attention to the units given in the question (cm or m) and ensure your final answer also reflects those units correctly. Double-check your calculations, especially when dealing with fractions or decimals.

Question 2. A cylindrical tank 7 m in diameter, contains water to a depth of 4 m. Find the total area of the wet surface. (π = 3.14)
Answer:
The diameter of the base of the tank is \( 7 \) m.
So, the radius \( (r) = \frac{7}{2} = 3.5 \) m.
The depth of the water \( (h) = 4 \) m.
The wet surface includes the curved surface area and the area of the base (since it's a tank containing water).
Total wet surface area \( = 2\pi rh + \pi r^2 = \pi r(2h + r) \)
Given \( \pi = 3.14 \).
\( = 3.14 \times 3.5 (2 \times 4 + 3.5) \)
\( = 3.14 \times 3.5 (8 + 3.5) \)
\( = 3.14 \times 3.5 \times 11.5 \)
\( = 126.385 \) m\(^2 \).
Rounding to two decimal places, the total wet surface area is \( 126.39 \) m\(^2 \). This value represents the total inner area of the tank that is in contact with water.
In simple words: We need to find the area of the tank's sides and its bottom that are touching the water. We use the formula for the curved surface area plus the area of one circle (the base) since the water fills the bottom.

🎯 Exam Tip: For problems involving liquids in a container, the "wet surface area" typically includes the curved surface and the base, but not the top unless specified. Always read carefully to see if \( \pi \) has a specific value (like 3.14 or \( \frac{22}{7} \)) to use.

Question 3. Find the whole surface of a hollow cylinder open at both ends, whose external diameter is 14 cm, thickness 2 cm and height 20 cm.
Answer:
The external diameter of the hollow cylinder is \( 14 \) cm.
So, the external radius \( (R) = \frac{14}{2} = 7 \) cm.
The thickness of the cylinder is \( 2 \) cm.
The inner radius \( (r) = \text{External radius} - \text{Thickness} = 7 - 2 = 5 \) cm.
The height of the cylinder \( (h) = 20 \) cm.

The whole surface area of a hollow cylinder open at both ends consists of:
1. Curved surface area of the outer cylinder \( (2\pi Rh) \).
2. Curved surface area of the inner cylinder \( (2\pi rh) \).
3. Area of the two rings at the top and bottom \( (2 \times (\pi R^2 - \pi r^2)) \).

Total surface area \( = 2\pi Rh + 2\pi rh + 2\pi (R^2 - r^2) \)
\( = 2\pi h(R + r) + 2\pi (R^2 - r^2) \)
\( = 2\pi (20)(7 + 5) + 2\pi (7^2 - 5^2) \)
\( = 2\pi (20)(12) + 2\pi (49 - 25) \)
\( = 40\pi (12) + 2\pi (24) \)
\( = 480\pi + 48\pi \)
\( = 528\pi \) cm\(^2 \).
This calculation represents the entire exposed surface area of the hollow cylinder, including its inner and outer walls and the top and bottom ring edges.
In simple words: Imagine a pipe that is open at both ends. Its total surface area is the area of its outer side, its inner side, and the two circular edges at the top and bottom. We find these three areas and add them up.

🎯 Exam Tip: For hollow cylinders, remember to calculate both inner and outer curved surface areas. If the cylinder is open, there are also two annular (ring) areas at the ends, which are \( \pi(R^2 - r^2) \) each, so \( 2\pi(R^2 - r^2) \) for both ends.

Question 4. Find the radius of the cylinder if area of its curved surface is 110 cm², and height 5 cm. (Take \( \pi=\frac{22}{7} \))
Answer:
Given the curved surface area of the cylinder \( = 110 \) cm\(^2 \).
The height \( (h) = 5 \) cm.
The formula for the curved surface area is \( 2\pi rh \).
So, \( 2\pi rh = 110 \)
Substitute the given values and \( \pi = \frac{22}{7} \):
\( 2 \times \frac{22}{7} \times r \times 5 = 110 \)
\( \frac{220}{7} \times r = 110 \)
To find \( r \), we rearrange the equation:
\( r = \frac{110 \times 7}{220} \)
\( r = \frac{7}{2} \)
\( r = 3.5 \) cm.
This means the cylinder has a radius of 3.5 cm, which creates the given curved surface area with its specific height.
In simple words: We know the curved area of a cylinder and its height. We use the formula for curved area and work backward to find the missing radius.

🎯 Exam Tip: When a problem asks to find a missing dimension (like radius or height) given the surface area, always write down the formula first, substitute the known values, and then solve for the unknown variable using algebraic manipulation.

Question 5. Find the height of the cylinder if area of its curved surface is 13.2 cm², and radius 6 cm.
Answer:
Given the curved surface area of the cylinder \( = 13.2 \) cm\(^2 \).
The radius \( (r) = 6 \) cm.
The formula for the curved surface area is \( 2\pi rh \).
So, \( 2\pi rh = 13.2 \)
We will use \( \pi = \frac{22}{7} \).
\( 2 \times \frac{22}{7} \times 6 \times h = 13.2 \)
\( \frac{264}{7} \times h = 13.2 \)
To find \( h \), we rearrange the equation:
\( h = \frac{13.2 \times 7}{264} \)
\( h = \frac{92.4}{264} \)
\( h = 0.35 \) cm.
Since \( 1 \) cm \( = 10 \) mm, we can convert the height to millimeters:
\( h = 0.35 \times 10 = 3.5 \) mm. *Correction: The OCR calculation `0.35 cm = 35 mm` seems to have a typo, `0.35 cm` should be `3.5 mm`. I will correct the final output accordingly. However, the calculation itself leads to 0.35 cm, and 0.35 cm is 3.5 mm. The source output `35 mm` has an extra zero. I will output `3.5 mm`.*
The height of the cylinder is 0.35 cm or 3.5 mm, which is quite small, indicating a wide and short cylinder.
In simple words: We are given the curved surface area and the radius of a cylinder. We use the formula for curved surface area to calculate its height. Remember to convert centimeters to millimeters if asked.

🎯 Exam Tip: Be careful with decimal calculations and unit conversions. Always ensure the units of your final answer match what is expected or are clearly stated if a conversion is made.

Question 6. A garden roller is 75 cm in diameter and 105 cm in width. What area does it cover in 14 revolutions?
Answer:
The diameter of the garden roller is \( 75 \) cm.
So, the radius \( (r) = \frac{75}{2} \) cm.
The width of the roller acts as the height \( (h) = 105 \) cm.

The area covered in one revolution is equal to the curved surface area of the roller.
Curved surface area \( = 2\pi rh \)
\( = 2 \times \frac{22}{7} \times \frac{75}{2} \times 105 \)
\( = 22 \times 75 \times 15 \) (since \( \frac{105}{7} = 15 \))
\( = 24750 \) cm\(^2 \).

Area covered in \( 14 \) revolutions:
\( = 24750 \times 14 \)
\( = 346500 \) cm\(^2 \).

To convert this area to square meters, we know that \( 1 \) m \( = 100 \) cm, so \( 1 \) m\(^2 = 100 \times 100 = 10000 \) cm\(^2 \).
Area in m\(^2 = \frac{346500}{10000} \)
\( = 34.65 \) m\(^2 \).
This garden roller covers a significant area with just 14 rotations, showing its efficiency in leveling.
In simple words: The roller covers an area equal to its curved surface in one turn. To find the area covered in many turns, we multiply the area of one turn by the number of turns. Then, we change the answer from square centimeters to square meters.

🎯 Exam Tip: When dealing with rollers, the distance covered in one revolution is always the curved surface area. Remember the conversion factor for area: \( 1 \) m\(^2 = 10000 \) cm\(^2 \).

Question 7. 10 cylindrical pillars of a building have to be painted. If the diameter of each pillar is 50 cm and the height 4 m, what will be the cost of painting these at the rate of 50 paise per m²? (Use \( \pi = 3.14 \))
Answer:
For each cylindrical pillar:
Diameter \( = 50 \) cm.
Radius \( (r) = \frac{50}{2} = 25 \) cm.
To match the painting rate, convert radius to meters: \( r = \frac{25}{100} = 0.25 \) m.
Height \( (h) = 4 \) m.

The area to be painted is the curved surface area of one pillar (assuming top and bottom are not painted).
Curved surface area of one pillar \( = 2\pi rh \)
\( = 2 \times 3.14 \times 0.25 \times 4 \)
\( = 2 \times 3.14 \times 1 \) (since \( 0.25 \times 4 = 1 \))
\( = 6.28 \) m\(^2 \).

Total curved surface area for \( 10 \) pillars:
\( = 10 \times 6.28 \)
\( = 62.8 \) m\(^2 \).

Rate of painting \( = 50 \) paise per m\(^2 \).
Convert paise to Rupees: \( 50 \) paise \( = \frac{50}{100} = 0.50 \) Rs. per m\(^2 \).

Total cost of painting \( = \text{Total area} \times \text{Rate per unit area} \)
\( = 62.8 \times 0.50 \)
\( = 31.40 \) Rs.
The total cost for painting all 10 pillars is Rs. 31.40, which is a very reasonable price.
In simple words: First, find the curved area of one pillar. Then, multiply by 10 for all pillars. Finally, multiply this total area by the cost per square meter to get the final painting cost. Remember to use consistent units and convert paise to Rupees.

🎯 Exam Tip: Always ensure all measurements are in consistent units (e.g., all meters or all centimeters) before starting calculations. Also, pay attention to the value of \( \pi \) to be used and the currency conversions for cost problems.

Question 8. The diameter of a roller 120 cm long is 84 cm. If it takes 500 complete revolutions to level a playground, determine the cost of levelling at the rate of 30 paise per m². (Use \( \pi=\frac{22}{7} \))
Answer:
The length of the roller is its height \( (h) = 120 \) cm.
The diameter of the roller is \( 84 \) cm.
So, the radius \( (r) = \frac{84}{2} = 42 \) cm.

The area covered in one revolution is the curved surface area of the roller.
Curved surface area \( = 2\pi rh \)
\( = 2 \times \frac{22}{7} \times 42 \times 120 \)
\( = 2 \times 22 \times 6 \times 120 \) (since \( \frac{42}{7} = 6 \))
\( = 44 \times 720 \)
\( = 31680 \) cm\(^2 \).

Area of the playground leveled by \( 500 \) revolutions:
\( = 31680 \times 500 \)
\( = 15840000 \) cm\(^2 \).

To convert this area to square meters, divide by \( 10000 \) (since \( 1 \) m\(^2 = 100 \times 100 \) cm\(^2 \)).
Area in m\(^2 = \frac{15840000}{10000} \)
\( = 1584 \) m\(^2 \).

Rate of leveling the ground \( = 30 \) paise per m\(^2 \).
Convert paise to Rupees: \( 30 \) paise \( = \frac{30}{100} = 0.30 \) Rs. per m\(^2 \).

Total cost of leveling \( = \text{Total area} \times \text{Rate per unit area} \)
\( = 1584 \times 0.30 \)
\( = 475.20 \) Rs.
The total cost to level the playground is Rs. 475.20, which is reasonable for a large area.
In simple words: First, find the area the roller covers in one spin. Then, multiply this by 500 spins to get the total area of the playground. Convert this area to square meters, and finally, multiply by the cost per square meter to find the total cost.

🎯 Exam Tip: Be meticulous with unit conversions, especially when changing from cm\(^2 \) to m\(^2 \). A common mistake is to divide by 100 instead of 10000. Also, ensure the cost calculation uses the correct currency unit.

Question 9. The curved surface area of a cylinder is 1000 cm². A wire of diameter 5 mm is wound round it so as to cover it completely. Find the length of the wire.
Answer:
Given the curved surface area of the cylinder \( = 1000 \) cm\(^2 \).
The diameter of the wire \( = 5 \) mm.
To convert the wire diameter to centimeters: \( 5 \) mm \( = \frac{5}{10} = 0.5 \) cm.

When the wire is wound around the cylinder to cover its curved surface completely, the total length of the wire can be found. Each turn of the wire will cover a height equal to its diameter.

Let \( h \) be the height of the cylinder and \( r \) be its radius.
The curved surface area \( = 2\pi rh = 1000 \) cm\(^2 \).

The number of coils (turns) of the wire will be the height of the cylinder divided by the diameter of the wire.
Number of coils \( = \frac{\text{Height of cylinder}}{\text{Diameter of wire}} = \frac{h}{0.5} \)

The total length of the wire is the number of coils multiplied by the circumference of the cylinder.
Length of wire \( = (\text{Number of coils}) \times (2\pi r) \)
\( = \frac{h}{0.5} \times (2\pi r) \)
\( = \frac{2\pi rh}{0.5} \)
We know that \( 2\pi rh = 1000 \) cm\(^2 \).
So, Length of wire \( = \frac{1000}{0.5} \)
\( = 2000 \) cm.
To convert to meters: \( 2000 \) cm \( = \frac{2000}{100} = 20 \) m.
This length of wire is substantial, highlighting that even thin wires add up quickly when coiled many times.
In simple words: The curved area of the cylinder is made up of many circular turns of wire. Each turn is a circle, and the thickness of the wire tells us how many turns fit vertically. We use the cylinder's curved area and the wire's thickness to find the total length of the wire needed.

🎯 Exam Tip: For problems involving winding wire, remember that the "height" covered by each turn of wire is its diameter, and the total length of wire is the number of turns multiplied by the circumference of the object it's wound around.

Question 10. An iron pipe 20 cm long has exterior diameter equal to 25 cm. If the thickness of the pipe is 1 cm, find the whole surface area of the pipe.
Answer:
The length of the iron pipe is its height \( (h) = 20 \) cm.
The exterior diameter \( = 25 \) cm.
So, the exterior radius \( (R) = \frac{25}{2} = 12.5 \) cm.
The thickness of the pipe \( = 1 \) cm.
The inner radius \( (r) = \text{Exterior radius} - \text{Thickness} = 12.5 - 1 = 11.5 \) cm.
We can write \( R = \frac{25}{2} \) cm and \( r = \frac{23}{2} \) cm.

The whole surface area of this closed iron pipe includes:
1. Outer curved surface area \( (2\pi Rh) \).
2. Inner curved surface area \( (2\pi rh) \).
3. Area of the two annular (ring) ends \( (2 \times (\pi R^2 - \pi r^2)) \).

Total surface area \( = 2\pi Rh + 2\pi rh + 2\pi (R^2 - r^2) \)
\( = 2\pi h(R + r) + 2\pi (R^2 - r^2) \)
\( = 2\pi (20) \left(\frac{25}{2} + \frac{23}{2}\right) + 2\pi \left(\left(\frac{25}{2}\right)^2 - \left(\frac{23}{2}\right)^2\right) \)
\( = 40\pi \left(\frac{48}{2}\right) + 2\pi \left(\frac{625}{4} - \frac{529}{4}\right) \)
\( = 40\pi (24) + 2\pi \left(\frac{96}{4}\right) \)
\( = 960\pi + 2\pi (24) \)
\( = 960\pi + 48\pi \)
\( = 1008\pi \) cm\(^2 \).
Using \( \pi = \frac{22}{7} \):
\( = 1008 \times \frac{22}{7} \)
\( = 144 \times 22 = 3168 \) cm\(^2 \).
This calculation shows the total amount of surface, both inside and out, that would need to be painted or coated.
In simple words: For a closed hollow pipe, we need to calculate the area of its outer surface, its inner surface, and the two ring-shaped ends. We add all these parts together to get the total surface area.

🎯 Exam Tip: Always draw a simple diagram for hollow cylinder problems to visualize the different surfaces (inner, outer, and ends). Remember that the ends are rings, so their area involves the difference between the outer and inner circle areas.

Question 11. 50 circular plates, each of radius 7 cm and thickness 0.5 cm, are placed one above the other to form a right circular cylinder. Find Its total surface area.
Answer:
Each circular plate has a radius \( (r) = 7 \) cm.
Each plate has a thickness \( = 0.5 \) cm.
When \( 50 \) such plates are stacked one above the other, they form a cylinder.
The height of this cylinder \( (h) = \text{Number of plates} \times \text{Thickness of each plate} \)
\( = 50 \times 0.5 = 25 \) cm.

Now we have a cylinder with radius \( r = 7 \) cm and height \( h = 25 \) cm.
The total surface area of this cylinder \( = 2\pi r(h + r) \)
\( = 2 \times \frac{22}{7} \times 7 (25 + 7) \)
\( = 2 \times 22 \times 32 \)
\( = 44 \times 32 = 1408 \) cm\(^2 \).
Alternatively, we can calculate the curved surface area and the area of the top and bottom circles separately:
Curved surface area \( = 2\pi rh \)
\( = 2 \times \frac{22}{7} \times 7 \times 25 \)
\( = 44 \times 25 = 1100 \) cm\(^2 \).
Area of top and bottom (two circles) \( = 2\pi r^2 \)
\( = 2 \times \frac{22}{7} \times 7^2 \)
\( = 2 \times \frac{22}{7} \times 49 \)
\( = 2 \times 22 \times 7 = 308 \) cm\(^2 \).
Total surface area \( = 1100 + 308 = 1408 \) cm\(^2 \).
This cylinder made from stacked plates illustrates how individual components combine to form a larger geometric shape.
In simple words: When many plates are stacked, they form a cylinder. The total thickness of all plates becomes the height of the cylinder. Then, we use the formula for the total surface area of a cylinder with its radius and this new height.

🎯 Exam Tip: Recognize that stacking identical circular objects creates a cylinder. The height of this new cylinder is the sum of the thicknesses, and its radius is the same as the individual objects.