Question

Two mtallic right circular cones hacing heights 4.1 cm and 4.3 cm. The radii of their bases 2.1 cm each have been melted togeter to form a sphere. Find the diameter of the sphere. ​

Answer 1

Answer :

Diameter of sphere is 4.2 cm

step by step Explanation

r1 = 2.1 cm , h1 = 4.1 cm

r2 = 2.1 cm , h2 = 4.3 cm

V1 = (1/3) ×pi×r^2×h = (1/3)×3.14×(2.1)^2×4.1 = 18.92 cm^2

V2 = (1/3) ×pi×r^2×h = (1/3)×3.14×(2.1)^2×4.3= 19.83 cm^2

V1+V2 = volume of sphere = 18.92 + 19.83 cm^2 = 38.92 cm^2

=> (4/3)×(pi)×r^3 = 38.92 cm^2

=> (4/3) × 3.14 × r^3 = 38.92

=> r = 2.1 cm

d = 2r = 2(2.1) = 4.2 cm^2

Answer 2

GIVEN :-

• Heights of the two metallic right circular cones are 4.1 cm and 4.3 cm
• Radius of the those two right circular cones is 2.1 cm
• The two metallic right circular cones melted and recast into a sphere.

TO FIND :-

• The diameter of the sphere

SOLUTION :-

Volume of a cone is given by ,

$\\ \star \: {\boxed{\purple{\sf{Volume = \dfrac{1}{3}\pi r^2 h}}}}$

Height of the first cone (h₁) = 4.1 cm

Radius of the first cone (r₁) = 2.1 cm

Now , Let us find the volume of the first right circular cone.

By substituting the values we have ,

$\\ : \implies \sf \: v_1 = \frac{1}{3} \times 3.14 \times {(2.1)}^{2} \times (4.1)$

$\\ : \implies \sf \: v_1 = \frac{1}{3} \times 3.14\times4.41 \times 4.1$

$\\ : \implies \sf \: v_1 = \frac{1}{3} \times 56.77$

$\\ : \implies \boxed{\red{\mathfrak{v_1 = 18.92\: cm {}^{2} }}}$

Now

Height of the second cone (h₂) = 4.3 cm

Radius of the second cone (r₂) = 2.1 cm

Now , Volume of second cone is ,

$\\ : \implies \sf \: v_2 = \frac{1}{3} \times 3.14 \times {(2.1)}^{2} \times 4.3$

$\\ : \implies \sf \: v_2 = \frac{1}{3} \times 3.14 \times 4.41 \times 4.3$

$\\ : \implies \sf \: v_2 = \frac{1}{3} \times 59.51$

$\\ : \implies \boxed{\red{\mathfrak { \: v_2 =19.83 \: {cm}^{2} }}}$

The sum of the Volumes of first cone (v₁) and second cone (v₂) becomes the volume of the sphere (vₛ)

Volume of the sphere is given by ,

$\\ \star\boxed{\purple{\sf{Volume = \dfrac{4}{3}\pi r^3}}}$

$\\ : \implies \sf \: v_1 + v_2 = \frac{4}{3}\pi {r}^{3}$

$\\ : \implies \sf \: 18.92 + 19.83 = \frac{4}{3} \times 3.14 \times {r}^{3}$

$\\ : \implies \sf \: 38.75 = 1.33 \times 3.14 \times {r}^{3}$

$\\ : \implies \sf \: 38.92 = 4.17 \times {r}^{3}$

$\\ : \implies \sf \: \frac{38.92}{4.17} = {r}^{3}$

$\\ : \implies \sf \: {r}^{3} = 9.33$

$\\ : \implies \sf \: r = \sqrt[3]{9.33}$

$\\ : \implies \boxed{\red{\mathfrak{r = 2.10\:cm}}}$

Now the relation between radius and diameter is given by ,

$\\ \star\boxed{\purple{\sf{diameter = 2(radius)}}}$

$\\ : \implies \sf \: d = 2(2.10) \: cm$

$\\ : \implies \boxed{\pink{\sf {\: d = 4.2 \: cm}}} \: \bigstar$

$\\ \underline {\sf{Hence\:, \:The\:diameter\:of\:the\:sphere\:is\: \bold{4.2\:cm}}}$