RD Sharma Solutions for Class 9 Maths Chapter 19 Surface Area and Volume of a Right Circular Cylinder
Chapter 19 RD Sharma Solutions for Class 9 Maths introduces students to the concept of calculating the surface area and volume of a right circular cylinder. The RD Sharma solutions for this chapter are crafted to simplify learning for students with varying levels of understanding, ensuring each concept is explained clearly and thoroughly.
These solutions aim to strengthen conceptual knowledge and are especially helpful for exam preparation. With step-by-step explanations, students can effectively grasp the techniques used to solve problems based on curved surface area, total surface area, and volume of cylinders.
As Class 9 marks the introduction of several core mathematical concepts, mastering these topics is essential. The solutions provided here are prepared by subject experts, following the latest CBSE syllabus for 2023–24. This ensures that students are learning in sync with their curriculum and are well-prepared to score high marks in their annual examinations.
RD Sharma Solution for Class 8 Chapter 19 - Download PDF
Here are the RD Sharma Solutions Class 9 Maths Chapter19 Surface Area and Volume of a Right Circular Cylinder Solutions, designed to help students prepare effectively for their exams. By referring to these solutions and practicing the problems, students can boost their confidence and improve their scores.
RD Sharma Solution for Class 9 Chapter 19 - Question with Answers
Curved Surface Area (CSA) of a Cylinder
Q1: Find the curved surface area of a cylinder with radius 7 cm and height 10 cm.
Ans: CSA = 2πrh = 2 × (22/7) × 7 × 10 = 440 cm²
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Q2: A cylinder has height 15 cm and diameter 14 cm. Find its CSA.
Ans: Radius = 7 cm → CSA = 2πrh = 2 × (22/7) × 7 × 15 = 660 cm²
Q3: If the CSA of a cylinder is 352 cm² and height is 8 cm, find the radius.
Ans: 2πrh = 352 → 2 × (22/7) × r × 8 = 352 → r = 7 cm
Q4: The CSA of a cylinder is 264 cm² and radius is 6 cm. Find its height.
Ans: h = CSA / (2πr) = 264 / (2 × (22/7) × 6) = 7 cm
Q5: A hollow pipe is 35 cm long with inner radius 3 cm. Find its CSA.
Ans: CSA = 2πrh = 2 × (22/7) × 3 × 35 = 660 cm²
Total Surface Area (TSA) of a Cylinder
Q6: Find the total surface area of a cylinder with radius 5 cm and height 12 cm.
Ans: TSA = 2πr(h + r) = 2 × (22/7) × 5 × (12 + 5) = 1705/7 = ~243.57 cm²
Q7: TSA of a cylinder is 462 cm² and height is 7 cm. Find the radius.
Ans: TSA = 2πr(h + r) → Plug and solve → r = 3.5 cm
Q8: A cylinder has a TSA of 616 cm² and radius of 7 cm. Find its height.
Ans: TSA = 2πr(h + r) → 616 = 2 × (22/7) × 7(h + 7) → h = 7 cm
Q9: Find TSA of a cylinder with diameter 10 cm and height 20 cm.
Ans: Radius = 5 cm → TSA = 2πr(h + r) = 2 × (22/7) × 5 × 25 = 785.71 cm²
Q10: TSA of a cylinder is 462 cm², radius 3.5 cm. Find height.
Ans: Use formula → TSA = 2πr(h + r) → h = 7 cm
Volume of a Cylinder
Q11: Find the volume of a cylinder with radius 7 cm and height 10 cm.
Ans: V = πr²h = (22/7) × 7² × 10 = 1540 cm³
Q12: A cylinder has diameter 14 cm and height 20 cm. Find its volume.
Ans: Radius = 7 cm → V = πr²h = (22/7) × 49 × 20 = 3080 cm³
Q13: The volume of a cylinder is 1760 cm³ and height is 10 cm. Find the radius.
Ans: πr²h = 1760 → r² = 56 → r = ~7.48 cm
Q14: Find the volume of a cylindrical water tank with radius 2 m and height 3.5 m.
Ans: V = πr²h = (22/7) × 4 × 3.5 = 44 m³
Q15: A cylindrical jar has radius 4 cm and height 12 cm. Find its volume in litres.
Ans: V = πr²h = (22/7) × 16 × 12 = 603.43 cm³ = 0.603 litres
Application-Based Problems
Q16: How much sheet is required to make the curved surface of a cylinder of radius 10 cm and height 21 cm?
Ans: CSA = 2πrh = 2 × (22/7) × 10 × 21 = 1320 cm²
Q17: A cylinder has radius 5 cm. If the height is doubled, what happens to the volume?
Ans: Volume also doubles.
Q18: Two cylinders have the same volume. One has radius 3 cm, the other has radius 6 cm. Compare their heights.
Ans: Since V = πr²h, the height becomes 1/4 in the second case.
Q19: A cylindrical can is open at the top. Radius = 7 cm, height = 10 cm. Find the surface area.
Ans: Area = πr² + 2πrh = (22/7) × 49 + 2 × (22/7) × 7 × 10 = 214.57 + 440 = 654.57 cm²
Q20: A cylinder has the same CSA and TSA. What is the relation between radius and height?
Ans: 2πrh = 2πr(h + r) → h = 0 or r = 0, which is not possible. Only true if CSA = TSA − Area of two ends.
Real-life Conversion Questions
Q21: A cylindrical pipe is 1 m long and 10 cm in diameter. Find its volume in litres.
Ans: Radius = 5 cm = 0.05 m → V = πr²h = 3.14 × 0.0025 × 1 = 0.00785 m³ = 7.85 litres
Q22: Water flows through a cylindrical pipe of radius 2 cm at a speed of 5 m/min. How much water flows in 1 minute?
Ans: Volume = πr²h = π × 4 × 500 = 6283.2 cm³ or 6.283 litres
Q23: A cylindrical glass holds 500 ml of water. Radius = 4 cm. Find its height.
Ans: 500 ml = 500 cm³ → πr²h = 500 → h = 500 / (3.14 × 16) = ~9.95 cm
Q24: Find the cost to paint the outer surface of a cylinder with r = 10 cm, h = 15 cm at ₹5/cm².
Ans: TSA = 2πr(h + r) = 2 × 3.14 × 10 × 25 = 1570 cm² → Cost = 1570 × 5 = ₹7850
Q25: A hollow cylindrical pipe has inner radius 4 cm, outer radius 5 cm, and length 10 cm. Find its volume.
Ans: V = πh(R² − r²) = 3.14 × 10(25 − 16) = 3.14 × 90 = 282.6 cm³
Conceptual Questions
Q26: What is the formula for CSA of a cylinder?
Ans: CSA = 2πrh
Q27: What is the formula for total surface area of a closed cylinder?
Ans: TSA = 2πr(h + r)
Q28: Write the formula for the volume of a cylinder.
Ans: V = πr²h
Q29: If the radius of a cylinder is doubled, how does the volume change?
Ans: Volume becomes 4 times (since r² term).
Q30: A cylinder is cut vertically. What shape is seen in cross-section?
Ans: A rectangle
Challenge Questions
Q31: Volume of a cylinder is 1760 cm³, radius = 7 cm. Find height.
Ans: V = πr²h → 1760 = 22/7 × 49 × h → h = 8 cm
Q32: CSA of a cylinder = 440 cm², height = 10 cm. Find radius.
Ans: 2πrh = 440 → r = 7 cm
Q33: Height of a cylinder = radius. If volume is 500 cm³, find height.
Ans: Let h = r → V = πr²r = πr³ = 500 → r = ∛(500/π) ≈ 5.3 cm
Q34: Which formula is used when a cylinder is open at one end?
Ans: TSA = πr² + 2πrh
Q35: Can two cylinders have same volume but different surface areas?
Ans: Yes, if radius and height vary accordingly.
FAQs on RD Sharma Solutions for Class 9 Maths Chapter 19
What does Chapter 19 of RD Sharma Class 9 Maths cover?
Chapter 19 focuses on the surface area (curved and total) and volume of a right circular cylinder. It includes the derivation of formulas, conceptual questions, numerical problems, and application-based exercises.
What is the formula for curved surface area of a cylinder?
The curved surface area (CSA) of a right circular cylinder is given by:
CSA = 2πrh, where r is the radius and h is the height.
How is the total surface area (TSA) of a cylinder calculated?
Total Surface Area = 2πr(h + r), which includes both the curved surface and the two circular ends of the cylinder.
What is the volume formula of a right circular cylinder?
Volume of a cylinder = πr²h, where r is the radius and h is the height.
Are RD Sharma solutions for Chapter 19 aligned with the latest CBSE syllabus?
Yes, RD Sharma solutions for Class 9 Chapter 19 are updated as per the CBSE 2023–24 syllabus and follow the current academic guidelines.
How do these solutions help students in exam preparation?
These solutions offer step-by-step explanations, solved examples, and a clear understanding of formulas and concepts, making it easier for students to solve similar questions confidently during exams.