Question

A solid sphere of radius 12cm is melted and recast into sphere ball of 1 cm radius find the number of such balls

Answer 1

Step-by-step explanation:

pls pls pls mark me as brainllist pls pls pls pls pls pls pls pls pls pls pls pls pls pls

Answer 2

Answer:

$❏ Question:-</p><p>@ How many balls of radius 1mm can be prepared by melting a sphere of radius 1cm.</p><p></p><p>❏ Solution:-</p><p>\begin{gathered}\bf{Given}\begin{cases}\text{Radius Of Large Sphere R=1 cm} \\ \text{Radius of Small Sphere r=1 mm =0.1 cm}\end{cases} \end{gathered} </p><p>Given{ </p><p>Radius Of Large Sphere R=1 cm</p><p>Radius of Small Sphere r=1 mm =0.1 cm</p><p> </p><p> </p><p> </p><p> </p><p></p><p>Let, the number of balls is = N pcs.</p><p></p><p>Now, The volume of the large sphere is</p><p></p><p>\sf\implies V_{large}= \frac{4}{3}\times\pi\times R^3⟹V </p><p>large</p><p> </p><p> = </p><p>3</p><p>4</p><p> </p><p> ×π×R </p><p>3</p><p> </p><p></p><p>\sf\implies V_{large}= \frac{4}{3}\times\pi\times 1^3\:cm^3⟹V </p><p>large</p><p> </p><p> = </p><p>3</p><p>4</p><p> </p><p> ×π×1 </p><p>3</p><p> cm </p><p>3</p><p> </p><p></p><p>\sf\implies \boxed{\red{V_{large}= (\frac{4\pi}{3})\:cm^3}}⟹ </p><p>V </p><p>large</p><p> </p><p> =( </p><p>3</p><p>4π</p><p> </p><p> )cm </p><p>3</p><p> </p><p> </p><p> </p><p></p><p>Now, for the Each of small sphere,</p><p></p><p>The volume of each small sphere is ,</p><p></p><p>\sf\implies V_{small}= \frac{4}{3}\times\pi\times r^3⟹V </p><p>small</p><p> </p><p> = </p><p>3</p><p>4</p><p> </p><p> ×π×r </p><p>3</p><p> </p><p></p><p>\sf\implies V_{small}= \frac{4}{3}\times\pi\times (0.1)^3\:cm^3⟹V </p><p>small</p><p> </p><p> = </p><p>3</p><p>4</p><p> </p><p> ×π×(0.1) </p><p>3</p><p> cm </p><p>3</p><p> </p><p></p><p>\sf\implies \boxed{\red{V_{small}= (\frac{4\pi}{3000})\:cm^3}}⟹ </p><p>V </p><p>small</p><p> </p><p> =( </p><p>3000</p><p>4π</p><p> </p><p> )cm </p><p>3</p><p> </p><p> </p><p> </p><p></p><p>Therefore, Number of small spherical balls is,</p><p></p><p>\sf\implies N= \frac{\cancel{\frac{4}{3}\pi} R^3}{\cancel{\frac{4}{3}\pi} r^3}⟹N= </p><p>3</p><p>4</p><p> </p><p> π</p><p> </p><p> r </p><p>3</p><p> </p><p>3</p><p>4</p><p> </p><p> π</p><p> </p><p> R </p><p>3</p><p> </p><p> </p><p> </p><p></p><p>\sf\implies N= \frac{ R^3}{r^3}⟹N= </p><p>r </p><p>3</p><p> </p><p>R </p><p>3</p><p> </p><p> </p><p> </p><p></p><p>\sf\implies N=(\frac{ R}{r})^3⟹N=( </p><p>r</p><p>R</p><p> </p><p> ) </p><p>3</p><p> </p><p></p><p>\sf\implies N=(\frac{ 1}{0.1})^3\:pcs⟹N=( </p><p>0.1</p><p>1</p><p> </p><p> ) </p><p>3</p><p> pcs</p><p></p><p>\sf\implies N=(10)^3\:pcs⟹N=(10) </p><p>3</p><p> pcs</p><p></p><p>\sf\implies\boxed{\large{ \red{N=1000\:pcs}}}⟹ </p><p>N=1000pcs</p><p> </p><p> </p><p>∴ Number of small balls is = 1000 pcs.$

Step-by-step explanation:

prefer the answer !!!!!!!!!