Question

Find the number of coins 0.8 cm in radius and 0.4 cm thick to be melted to form a right circularcylinder of height 10 cm and 8 cm as radius.
(Hint. The shape of coin is cylindrical).​

Answer 1

Answer:

Coin,

R=0.5cm

H=0.4cm

volume of a coin= π *R² *H

= π *2.5 *0.4

volume of the cylinder = π *8²* 10

Let, the number of coins are= X

then, X*π*2.5*0.4 = π*64*20

» X*10 = 1280

»X = 128

So the number of coins are 128.

Please give me brainliest mark, I worked hard for this answer...............

Answer 2

$\sf \bf \huge {\boxed {\mathbb {QUESTION}}}$

$\tt Find\: the\: number\: of\: coins\: 0.8 \:cm \:in\: radius\\\tt and\: 0.4\: cm \:thick\: to \:be\: melted\: to\: form\: a\: right\\\tt circularcylinder\: of\: height \:10\: cm \:and \:8 \:cm\\\tt as\: radius.$

$\tt (Hint. \:The\: shape\: of\: coin\: is\: cylindrical).$

$\sf \bf \huge {\boxed {\mathbb {ANSWER}}}$

$\sf \bf {\boxed {\mathbb {GIVEN}}}$

• $\sf \bf Radius\:of\:the\:coin(r_1)=0.8\:cm$
• $\sf\bf Thickness\:(height)\:of \:the \:coin (h_1)=0.4\:cm$
• $\sf \bf Radius\:of\:the\: right\:circularcylinder(r_2)=8\:cm$
• $\sf \bf Height\:of\:the\:right\:circularcylinder(h_2)=10\:cm$

$\sf \bf {\boxed {\mathbb {TO\:SOLVE}}}$

• $\sf \bf Number\:of\:coins\:needed\:to\:make\:right\:circularcylinder$

$\sf \bf {\boxed {\mathbb {SOLUTION}}}$

$\tt {\underbrace {Number\:of\:coins\:be\:x}}$

${\boxed {\boxed {\color{blue} {\sf \bf Volume\:of\:cylinder=\pi {r}^{2} h}}}}$

• $\sf \bf \pi=\dfrac{22}{7}$
• $\sf \bf r=radius$
• $\sf \bf h=height$

$\sf \bf (Number\:of\:coins) \times (Volume\:of\:1\:coin)=(Volume\:of\:right\:circularcylinder)$

$\sf \bf x \times \pi {r_1}^{2} h_1=\pi {r_2}^{2} h_2$

$\tt \underline {Substitute\:the\:values}$

$\sf \bf\implies x \times \dfrac{22}{7} \times {(0.8)}^{2} \times 0.4 = \dfrac{22}{7} \times {(8)}^{2} \times 10$

$\sf \bf\implies x \times \dfrac{22}{7} \times 0.64 \times 0.4 = \dfrac{22}{7} \times 64 \times 10$

$\sf \bf\implies x \times \dfrac{22}{7} \times 0.256 = \dfrac{22}{7} \times 640$

$\sf \bf\implies x \times \dfrac{5.632}{\cancel {7}}= \dfrac{14080}{\cancel {7}}$

$\sf \bf \implies x \times 5.632=14080$

$\sf \bf \implies x= \dfrac{14080}{5.632}$

$\implies {\boxed {\color{green} {\sf \bf x=2500}}}$

$\sf \bf \therefore 2500\:coins\:are\:needed\:to\:form\:a\:right\:circularcylinder$

$\sf \bf \huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU}}}$

_________________________________________

$\sf \bf \huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

$\sf \bf Volume\:of\:cylinder=\pi {r}^{2} h$

$\sf \bf L.S.A\:of\:cylinder=2\pi r h$

$\sf \bf T.S.A\:of\:cylinder=2\pi r(r+h)$

${\mathbb{\colorbox {orange} {\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox {lime} {\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox {aqua} {@suraj5070}}}}}}}}}}}}}}}$