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CBSE Class 12-science Answered

Show that the volume of largest cone that can be inscribed in a sphere of radius R is 8 over 27 of the volume of the sphere.
Asked by Topperlearning User | 21 Aug, 2014, 01:22: PM
answered-by-expert Expert Answer

Let space straight r space be space the space radius space of space the space inscribed space cone space and space straight h space be space the space height. Let space straight R space be space the space radius space of space the space sphere. Using space pythagoras space theorem left parenthesis straight h minus straight R right parenthesis squared plus straight r squared equals straight R squared rightwards double arrow straight h squared minus 2 hR plus straight R squared plus straight r squared equals straight R squared rightwards double arrow straight h squared minus 2 hR plus straight r squared equals 0 rightwards double arrow straight r squared equals 2 hR minus straight h squared Let space straight V subscript straight s space be space the space volume space of space sphere space and space straight V subscript straight c space be space the space volume space of space the space cone. straight V subscript straight s equals 4 over 3 πR cubed Also comma space straight V subscript straight c equals 1 third πr squared straight h straight V subscript straight c equals 1 third straight pi left parenthesis 2 hR minus straight h squared right parenthesis straight h space space space space equals space 1 third straight pi left parenthesis 2 straight h squared straight R minus straight h cubed right parenthesis Differenetiating space straight w. straight r. straight t. space straight h comma space we space get comma dV subscript straight c over dh equals 1 third straight pi left parenthesis 4 hR minus 3 straight h squared right parenthesis Now comma dV subscript straight c over dh equals 0 rightwards double arrow 1 third straight pi left parenthesis 4 hR minus 3 straight h squared right parenthesis equals 0 rightwards double arrow straight h equals fraction numerator 4 straight R over denominator 3 end fraction For space maxima space or space minima comma fraction numerator straight d squared straight V subscript straight c over denominator dh squared end fraction equals 1 third straight pi left parenthesis 4 straight R minus 6 straight h right parenthesis space space space space space space space space space space space space equals 1 third straight pi left parenthesis 4 straight R minus 6 cross times fraction numerator 4 straight R over denominator 3 end fraction right parenthesis space space space space space space space space space space space space equals fraction numerator negative 4 straight R over denominator 3 end fraction straight pi less than 0 therefore Volume space is space maximum. straight r squared equals 2 hR minus straight h squared space space space equals 2 open parentheses fraction numerator 4 straight R over denominator 3 end fraction close parentheses straight R minus open parentheses fraction numerator 4 straight R over denominator 3 end fraction close parentheses squared space space space equals fraction numerator 8 straight R squared over denominator 3 end fraction minus fraction numerator 16 straight R squared over denominator 9 end fraction space space space equals fraction numerator 8 straight R squared over denominator 9 end fraction straight V subscript straight c equals 1 third straight pi left parenthesis 2 straight h squared straight R minus straight h cubed right parenthesis space space space space space equals 1 third straight pi open parentheses 2 open parentheses fraction numerator 4 straight R over denominator 3 end fraction close parentheses squared straight R minus open parentheses fraction numerator 4 straight R over denominator 3 end fraction close parentheses cubed close parentheses space space space space space equals 1 third straight pi open parentheses fraction numerator 32 straight R cubed over denominator 9 end fraction minus fraction numerator 64 straight R cubed over denominator 27 end fraction close parentheses straight V subscript straight c space equals 1 third straight pi fraction numerator 32 straight R cubed over denominator 27 end fraction....... left parenthesis straight i right parenthesis straight V subscript straight s equals 4 over 3 πR cubed....... left parenthesis ii right parenthesis left parenthesis straight i right parenthesis divided by left parenthesis ii right parenthesis comma straight V subscript straight c over straight V subscript straight s equals 8 over 27 rightwards double arrow straight V subscript straight c equals 8 over 27 straight V subscript straight s Hence space Proved

Answered by Renu Varma | 21 Aug, 2014, 03:22: PM

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