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Ex 6.5

Ex 6.5, 1 (i)
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Ex 6.5, 1 (ii)

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Ex 6.5, 1 (iv)

Ex 6.5, 2 (i)

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Ex 6.5,24 Important You are here

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Ex 6.5, 27 (MCQ)

Ex 6.5,28 (MCQ) Important

Ex 6.5,29 (MCQ)

Last updated at April 15, 2021 by Teachoo

Ex 6.5, 24 Show that the right circular cone of least curved surface and given volume has an altitude equal to β2 time the radius of the baseLet r & h be the radius & height of a cone respectively And V & S be the volume & curved surface area of cone respectively Given volume of cone is constant Volume of cone = 1/3 π(πππππ’π )^2 (βπππβπ‘) V = 1/3 π(π)^2 β 3π/π=π^2 β h = (3π/π) 1/π^2 β= π/π^2 We need to show Curved surface Area is least & Altitude = β2 times of radius of base i.e. β=β2 π Now, Curved Surface Area of Cone = Ο ππ Putting π = β(β^2+π^2 ) S = Ο π β(β^2+π^2 ) S = Ο π β((π/π^2 )^2+π^2 ) Since V is constant β (3π/π) is also constant Let k = 3π/π S = Ο π β(π^2/π^4 +π^2 ) S = Ο π β((π^2 + π^6)/π^4 ) S = Ο π/π^2 β(π^2+π^6 ) S = π/π β(π^2+π^6 ) S = Ο [(β(π^2 + π^6 ) " " )/π] Since S has square root It will be difficult to differentiate So, we take Z = S2 Z = π^2 [(π^2 + π^6 " " )/π^2 ] Z = π^2 [(π^2 " " )/π^2 +π^4 ] Z = π^2 [π^2 π^(β2)+π^4 ] Since S is positive, S is minimum if A2 is minium So, we minimize Z = A2 Diff. Z w.r.t π πZ/ππ=π(π^2 [π^2 π^(β2)+π^4 ])/ππ πZ/ππ=π^2 [π^2 (β2π^(β2β1))+4π^3 ] πZ/ππ=π^2 [β2π^2 π^(β3)+4π^3 ] Putting π π/π π = 0 π^2 [β2π^2 π^(β3)+4π^3 ] = 0 β2π^2 π^(β3)+4π^3= 0 4π^3 = 2π^2 π^(β3) 4π^3 = (2π^2)/π^3 4π^3 Γ π^3 = 2π^2 4π^6 = 2π^2 π^6 = π^2/2 Finding (π ^π π)/(ππ^π ) πZ/ππ=π^2 [β2π^2 π^(β3)+4π^3 ] Diff w.r.t π₯ (π^2 Z)/(ππ^2 ) = π/ππ [π^2 [β2π^2 π^(β3)+4π^3 ]] (π^2 Z)/(ππ₯^2 ) = π^2 [β2π^2 (β3π^(β4))+4(3π^2)] (π^2 Z)/(ππ₯^2 ) = π^2 [6π^2 π^(β4)+12π^2 ] Since everything is positive, (π^2 Z)/(ππ₯^2 ) > 0 for π^6 = π^2/2 Thus, Surface area is minimum when π^6 = π^2/2 Now, β=π/π^2 βπ^2=π (βπ^2 )^2=π^2 β^2 π^4=π^2 Putting value of π^2=2π^6 β^2 π^4=2π^6 β^2=(2π^6)/π^4 β^2=2π^2 β=β(2π^2 ) π=πβπ Hence altitude equal to β2 times the radius of the base Hence proved