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Question 21 (Or 2nd) - CBSE Class 10 Sample Paper for 2019 Boards - Solutions of Sample Papers for Class 10 Boards

Last updated at Oct. 1, 2019 by

Question 21 (OR 2 nd Β  question)

A solid sphere of radius 3 cm is melted and then recast into small spherical balls each of diameter 0.6 cm. Find the number of balls

A solid sphere of radius 3 cm is melted and then recast into small

Question 21 (Or 2nd) - CBSE Class 10 Sample Paper for 2019 Boards - Part 2
Question 21 (Or 2nd) - CBSE Class 10 Sample Paper for 2019 Boards - Part 3
Question 21 (Or 2nd) - CBSE Class 10 Sample Paper for 2019 Boards - Part 4

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Question 21 (OR 2nd question) A solid sphere of radius 3 cm is melted and then recast into small spherical balls each of diameter 0.6 cm. Find the number of balls Since the big sphere is melted to make balls Volume of big sphere = Volume of balls Volume of big sphere = Number of balls Γ— Volume of 1 ball ∴ Number of balls = (π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ 𝑏𝑖𝑔 π‘ π‘β„Žπ‘’π‘Ÿπ‘’)/(π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ 1 π‘π‘Žπ‘™π‘™) Volume of big sphere Radius = 3 cm Volume of sphere = 4/3 πœ‹(π‘Ÿ)3 = 4/3Γ—πœ‹Γ—(3)^3 = 4/3Γ—πœ‹Γ—27 Volume of 1 ball Diameter = 0.6 cm Radius = 0.6/2 = 0.3 cm = 3/10 cm Volume of 1 ball = 4/3 πœ‹(π‘Ÿ)3 = 4/3Γ—πœ‹Γ—(3/10)^3 = 4/3Γ—πœ‹Γ—27/1000 Now, Number of balls = (π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ 𝑏𝑖𝑔 π‘ π‘β„Žπ‘’π‘Ÿπ‘’)/(π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ 1 π‘π‘Žπ‘™π‘™) = (4/3 Γ— πœ‹ Γ— 27" " )/(4/3 Γ— πœ‹ Γ— 27/1000) = 27" " /( 27/1000) = 27" " /27 Γ— 1000 = 1000 ∴ Number of balls is 1000

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.