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A solid metal cylinder of 10 cm height and 14 cm diameter is melted and re-cast into two cones in the proportion of 3 : 4 (volume), keeping the height 10 cm. What would be the percentage change in the flat surface area before and after?
- a)9 %
- b)16 %
- c)25%
- d)50 %
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A solid metal cylinder of 10 cm height and 14 cm diameter is melted an...
Change in flat surface area = 21k – 14k = 7k
% change in flat surface area 7k/14k * 100 = 50%.
% change in flat surface area 7k/14k * 100 = 50%.
Most Upvoted Answer
A solid metal cylinder of 10 cm height and 14 cm diameter is melted an...
Given data:
Height of cylinder, h = 10 cm
Diameter of cylinder, d = 14 cm
Radius of cylinder, r = d/2 = 7 cm
Volume of cylinder = πr²h
= π(7)²(10)
= 490π cm³
Let the volumes of two cones be 3x and 4x respectively.
Total volume of cones = 3x + 4x = 7x
As the height of cones is given as 10 cm, we can find the radius of each cone as follows:
Let the radius of cone with volume 3x be r1 and the radius of cone with volume 4x be r2.
As the volume of a cone is (1/3)πr²h, we have:
(1/3)πr1²(10) = 3x
r1² = (9x/π)×(3/10)
r1 = √(27x/10π)
(1/3)πr2²(10) = 4x
r2² = (16x/π)×(3/10)
r2 = √(48x/10π)
As the metal is conserved in the process, we have:
Volume of cylinder = Volume of cones
490π = (1/3)πr1²(10) × 3 + (1/3)πr2²(10) × 4
490 = (1/3)(27x/10) × 30 + (1/3)(48x/10) × 40
490 = 27x + 64x
x = 5
Therefore, the volumes of the two cones are 15π cm³ and 20π cm³ respectively.
Flat surface area of the cylinder = 2πr² + 2πrh
= 2π(7)² + 2π(7)(10)
= 98π + 140π
= 238π cm²
Flat surface area of the two cones can be found as follows:
Base radius of cone with volume 3x, b1 = r1/√2
Base radius of cone with volume 4x, b2 = r2/√2
Slant height of cone with volume 3x, l1 = √(h² + b1²)
Slant height of cone with volume 4x, l2 = √(h² + b2²)
Flat surface area of cone with volume 3x = πb1l1
= π(r1/√2)√(h² + b1²)
= π(√(27x/10π)/√2)√(100 + 27x/10)
= π√(27x/20)(10 + 27x/10)
Flat surface area of cone with volume 4x = πb2l2
= π(r2/√2)√(h² + b2²)
= π(√(48x/10π)/√2)√(100 + 48x/10)
= π√(12x/5)(10 + 24x/5)
Total flat surface area of the two
Height of cylinder, h = 10 cm
Diameter of cylinder, d = 14 cm
Radius of cylinder, r = d/2 = 7 cm
Volume of cylinder = πr²h
= π(7)²(10)
= 490π cm³
Let the volumes of two cones be 3x and 4x respectively.
Total volume of cones = 3x + 4x = 7x
As the height of cones is given as 10 cm, we can find the radius of each cone as follows:
Let the radius of cone with volume 3x be r1 and the radius of cone with volume 4x be r2.
As the volume of a cone is (1/3)πr²h, we have:
(1/3)πr1²(10) = 3x
r1² = (9x/π)×(3/10)
r1 = √(27x/10π)
(1/3)πr2²(10) = 4x
r2² = (16x/π)×(3/10)
r2 = √(48x/10π)
As the metal is conserved in the process, we have:
Volume of cylinder = Volume of cones
490π = (1/3)πr1²(10) × 3 + (1/3)πr2²(10) × 4
490 = (1/3)(27x/10) × 30 + (1/3)(48x/10) × 40
490 = 27x + 64x
x = 5
Therefore, the volumes of the two cones are 15π cm³ and 20π cm³ respectively.
Flat surface area of the cylinder = 2πr² + 2πrh
= 2π(7)² + 2π(7)(10)
= 98π + 140π
= 238π cm²
Flat surface area of the two cones can be found as follows:
Base radius of cone with volume 3x, b1 = r1/√2
Base radius of cone with volume 4x, b2 = r2/√2
Slant height of cone with volume 3x, l1 = √(h² + b1²)
Slant height of cone with volume 4x, l2 = √(h² + b2²)
Flat surface area of cone with volume 3x = πb1l1
= π(r1/√2)√(h² + b1²)
= π(√(27x/10π)/√2)√(100 + 27x/10)
= π√(27x/20)(10 + 27x/10)
Flat surface area of cone with volume 4x = πb2l2
= π(r2/√2)√(h² + b2²)
= π(√(48x/10π)/√2)√(100 + 48x/10)
= π√(12x/5)(10 + 24x/5)
Total flat surface area of the two
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A solid metal cylinder of 10 cm height and 14 cm diameter is melted and re-cast into two cones in the proportion of 3 : 4 (volume), keeping the height 10 cm. What would be the percentage change in the flat surface area before and after?a)9 %b)16 %c)25%d)50 %Correct answer is option 'D'. Can you explain this answer?
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A solid metal cylinder of 10 cm height and 14 cm diameter is melted and re-cast into two cones in the proportion of 3 : 4 (volume), keeping the height 10 cm. What would be the percentage change in the flat surface area before and after?a)9 %b)16 %c)25%d)50 %Correct answer is option 'D'. Can you explain this answer? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A solid metal cylinder of 10 cm height and 14 cm diameter is melted and re-cast into two cones in the proportion of 3 : 4 (volume), keeping the height 10 cm. What would be the percentage change in the flat surface area before and after?a)9 %b)16 %c)25%d)50 %Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solid metal cylinder of 10 cm height and 14 cm diameter is melted and re-cast into two cones in the proportion of 3 : 4 (volume), keeping the height 10 cm. What would be the percentage change in the flat surface area before and after?a)9 %b)16 %c)25%d)50 %Correct answer is option 'D'. Can you explain this answer?.
A solid metal cylinder of 10 cm height and 14 cm diameter is melted and re-cast into two cones in the proportion of 3 : 4 (volume), keeping the height 10 cm. What would be the percentage change in the flat surface area before and after?a)9 %b)16 %c)25%d)50 %Correct answer is option 'D'. Can you explain this answer? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A solid metal cylinder of 10 cm height and 14 cm diameter is melted and re-cast into two cones in the proportion of 3 : 4 (volume), keeping the height 10 cm. What would be the percentage change in the flat surface area before and after?a)9 %b)16 %c)25%d)50 %Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solid metal cylinder of 10 cm height and 14 cm diameter is melted and re-cast into two cones in the proportion of 3 : 4 (volume), keeping the height 10 cm. What would be the percentage change in the flat surface area before and after?a)9 %b)16 %c)25%d)50 %Correct answer is option 'D'. Can you explain this answer?.
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