Question

If 3 metallic spheres of radius 6 cm ,8 cm and 10 cm are melted to form a single sphere, the diameter of the sphere is

Answer 1

HeYa❤️...

Answer:

Given:

Radius(r1) of Sphere 1 = 6cm.

Radius(r2) of Sphere 2 = 8cm.

Radius(r3) of Sphere 3 = 10cm.

Formula Used:

$Volume \: of \: sphere \: = \frac{4}{3}\pi \: r_{1} {}^{3}$

Solution:

$Volume \: of \: sphere \: 1 = \frac{4}{3} \pi \: r_{1} {}^{3} \\ = \frac{4}{3} \times \frac{22}{7} \times 6 \times 6 \times 6 \: cm {}^{3} \\ = \frac{88}{21} \times 36 \times 6 \\ = 904.78 \: cm {}^{3}$

$volume \: of \: sphere \: 2 = \frac{4}{3} \pi \: r_{2} {}^{3} \\ = \frac{4}{3} \times \frac{22}{7} \times 8 \times 8 \times 8 \\ = \frac{88}{21} \times 64 \times 8 \\ = 2144.66 \: cm {}^{3}$

$volume \: of \: sphere = \frac{4}{3} \pi \: r_{3} {}^{3} \\ = \frac{4}{3} \times \frac{22}{7} \times 10 \times 10 \times 10 \\ = \frac{88}{21} \times 1000 \\ = 4188.79 \: cm {}^{3}$

$volume \: of \: single \: sphere = volume \: of \: sphere \: 1 + sphere \: 2 + sphere \: 3 \\ \frac{4}{3} \pi \: r {}^{3} = (904.78 + 2144.66 + 4188.79) \: cm {}^{3} \\ \frac{4}{3} \times \frac{22}{7} \times r {}^{3} =7238.23 \: cm {}^{3} \\ r {}^{3} = \frac{7238.23 \times 3 \times 7}{4 \times 22} \\ r{}^{3} = \frac{94305.96}{88} \\ r {}^{3} = 4039.95\\ r = \sqrt[3]{4039.95} \\ r = 15.9266cm$

Diameter of sphere = 2(radius of sphere)

= 2( 15.9266 ) cm

= 31.8532 cm

Hence, diameter of sphere is 31.8532 cm.

I hOpe It HeLpS Ya!

✌️✌️✌️✌️

Answer 2

Answer:

Diameter is 24cm.

Solution:

ATQ, Three spheres of radii 6cm, 8cm and 10cm respectively are melted to form another sphere, and we have to find its diameter.

Assuming that there is no waste created in the process of melting the spheres, the volume of the three spheres combined will be equal to the new single sphere.

Let the volume of the sphere with radius 6cm, 8cm and 10cm be V₁, V₂ and V₃ and their raddi be named r₁ , r₂ and r₃ respectively. Let the volume of the new sphere be V$\sf _t$ and radius be R.

$\boxed{\sf \ Volume \ of \ a \ sphere = \dfrac{4}{3} \pi r^3 \ }$

$\Longrightarrow \sf V_1 + V_2 + V_3 = V_t\\ \\ \\\Longrightarrow \sf \dfrac{4}{3}\pi{r_1}^3+\dfrac{4}{3}\pi{r_2}^3+\dfrac{4}{3}\pi{r_3}^3=\dfrac{4}{3}\pi{R}^3\\ \\ \\\Longrightarrow \sf \dfrac{4}{3}\pi\left({r_1}^3+{r_2}^3+{r_3}^3\right)=\dfrac{4}{3} \pi {R}^3\\ \\ \\\Longrightarrow \sf \left({r_1}^3 + {r_2}^3 + {r_3}^3\right) = {R}^3\\ \\ \\\Longrightarrow \sf \left({6}^3+{8}^3+{10}^3\right)={R}^3\\ \\ \\\Longrightarrow \sf 216+512+1000={R}^3\\ \\ \\\Longrightarrow \sf 1728=R^3$

$\Longrightarrow \sf R = \sf \sqrt[3]{1728}$

$\begin{array}{r | l}\sf 2 & \sf 1728 \\\cline{2-2} \sf 2 & \sf 864 \\\cline{2-2} \sf 2 & \sf 432 \\\cline{2-2} \sf 2 & \sf 216 \\\cline{2-2} \sf 2 & \sf 108 \\\cline{2-2} \sf 2 & \sf 54 \\\cline{2-2} \sf 3 & \sf 27 \\\cline{2-2} \sf 3 & \sf 9 \\\cline{2-2} \sf 3 & \sf 3 \\\cline{2-2} & \sf 1 \\\end{array}\\ \\ \\\sf 1728 = \sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3}\\ \\\sf \sqrt[3]{1728}=2\times2\times3\\ \\\sf\sqrt[3]{1728}=4\times3\\ \\\sf \sqrt[3]{1728}=12$

$\Longrightarrow \sf R = 12cm.$

Diameter = 2R

Diameter = 2(12)

Diameter = 24cm.

∴ The radius of the bigger sphere is 24cm.