Question

Three metallic spheres of radii 6 cm.,
8 cm. and 10 cm. respectively are melted
together to form a single solid sphere.
Find the radius of the resulting sphere.
(AS)
1..​

Answer 1

Answer:

12 cm

Step-by-step explanation:

Given that,

Three metallic spheres are melted to form a single solid sphere.

Also, the radii are respectively:-

• 6 cm
• 8 cm
• 10 cm

To find the radius of resulting sphere.

Let the required radius be 'r' cm.

Clearly, the sum of all the volumes of all these three spheres will be equal to the volume of resulting sphere.

As we know that,

Volume of a sphere = 4/3 π r^3

Where, r is the radius.

Therefore, we will get,

$= > \frac{4}{3} \pi {(6)}^{3} + \frac{4}{3} \pi {(8) }^{3} + \frac{4}{3} \pi {(10)}^{3} = \frac{4}{3} \pi {r}^{3} \\ \\ = > \frac{4}{3} \pi( {6}^{3} + {8}^{3} + {10}^{3} ) = \frac{4}{3} \pi {r}^{3} \\ \\ = > {r}^{3} = {6}^{3} + {8}^{3} + {10}^{3} \\ \\ = > {r}^{3} = 216 + 512 + 1000 \\ \\ = > {r}^{3} = 1728 \\ \\ = > {r}^{3} = {(12)}^{3} \\ \\ = > r = 12$

Hence, Radius of resulting sphere is 12 cm.

Answer 2

Answer:

⋆ DIAGRAM :

$\setlength{\unitlength}{1.5pt}\begin{picture}(150,100)\thicklines\put(0,0){\circle{70}}\put(0,0){\circle*{1.5}}\put(-1,0){\line(1,0){14}}\put(-0.5,1.5){\scriptsize\sf6 cm}\put(40,0){\circle{70}}\put(40,0){\circle*{1.5}}\put(40,0){\line(1,0){14}}\put(40,1.5){\scriptsize\sf8 cm}\put(80,0){\circle{70}}\put(80,0){\circle*{1.5}}\put(80,0){\line(1,0){14}}\put(80,1.5){\scriptsize\sf10 cm}\put(120,0){\circle{70}}\put(120,0){\circle*{1.5}}\put(120,0){\line(1,0){14}}\put(120,1.5){\scriptsize\sf R cm}\put(17,0){\sf+}\put(58,0){\sf+}\put(98,0){\sf=}\end{picture}$

• $\rm r_1 = 6\:cm$
• $\rm r_2 = 8\:cm$
• $\rm r_3 = 10\:cm$

Let the Radius of Resulting Sphere be R.

$\rule{110}{1}$

$\underline{\bigstar\:\boldsymbol{According\:to\:the\:Question :}}$

$:\implies\sf Volume_{(Sphere_1 + Sphere_2 + Sphere_3)} =Volume_{(Sphere)}\\\\\\:\implies\sf \dfrac{4}{3}\pi(r_1)^3+\dfrac{4}{3}\pi(r_2)^3+\dfrac{4}{3}\pi(r_3)^3=\dfrac{4}{3}\pi(R)^3\\\\\\:\implies\sf \dfrac{4}{3}\pi\bigg\lgroup(6)^3+(8)^3+(10)^3\bigg\rgroup=\dfrac{4}{3}\pi(R)^3\\\\\\:\implies\sf (6)^3+(8)^3+(10)^3 =(R)^3\\\\\\:\implies\sf 216 + 215 + 1000 = (R)^3\\\\\\:\implies\sf 1728 = (R)^3\\\\\\:\implies\sf \sqrt[3]{1728} = R\\\\\\:\implies\sf \sqrt[3]{12 \times 12 \times 12} = R\\\\\\:\implies\underline{\boxed{\sf R = 12 \:cm}}$

⠀

$\therefore\:\underline{\textsf{Hence, Radius of resulting sphere is \textbf{12 cm}}}.$

$\rule{170}{2}$

$\bigstar\:\sf Squares\:\&\:Cubes:\\\begin{tabular}{|c|c|c|}\cline{1-3}\sf Number&\sf Square&\sf Cube\\\cline{1-3}\sf1&\sf1&\sf1\\\cline{1-3}\sf2&\sf4&\sf8\\\cline{1-3}\sf3&\sf9&\sf27\\\cline{1-3}\sf4&\sf16&\sf64\\\cline{1-3}\sf5&\sf25&\sf125\\\cline{1-3}\sf6&\sf36&\sf216\\\cline{1-3}\sf7&\sf49&\sf343\\\cline{1-3}\sf8&\sf64&\sf512\\\cline{1-3}\sf9&\sf81&\sf729\\\cline{1-3}\sf10&\sf100&\sf1000\\\cline{1-3}\sf11&\sf121&\sf1331\\\cline{1-3}\sf12&\sf144&\sf1728\\\cline{1-3}\sf13&\sf169&\sf2197\\\cline{1-3}\sf14&\sf196&\sf2744\\\cline{1-3}\sf15&\sf225&\sf3375\\\cline{1-3}\sf16&\sf256&\sf4096\\\cline{1-3}\sf17&\sf289&\sf4913\\\cline{1-3}\sf18&\sf324&\sf5832\\\cline{1-3}\sf19&\sf361&\sf6859\\\cline{1-3}\sf20&\sf400&\sf8000\\\cline{1-3}\end{tabular}$