Question

3. An ancient coin made of pure gold weighs 11.58g. Find its
volume when the density of gold is 19,300 kg/m^3.​

Answer 1

Answer :-

Volume of the coin is 0.6 cm³ .

Explanation :-

We have :-

→ Weight of the coin = 11.58 g

→ Density of gold = 19300 kg/m³

To find:-

→ Volume of the coin.

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Firstly, let's convert the unit of density of gold from kg/m³ to g/cm³ .

⇒ 1 kg/m³ = 0.001 g/cm³

⇒ 19300 kg/m³ = 19300(0.001)

⇒ 19.3 g/cm³

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Volume = Weight/Density

Substituting values, we get :-

⇒ Volume = 11.58/19.3

⇒ Volume = 0.6 cm³

Answer 2

Given : An ancient coin made of pure gold weighs 11.58g [ as , Mass ] & density of gold is 19,300 kg/m³ [ as , Density ] .

Need To Find : The Volume of coin .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀Given that ,

⠀⠀⠀⠀⠀⠀⠀▪︎⠀⠀An ancient coin made of pure gold weighs 11.58g [ as , Mass ]

⠀⠀⠀⠀⠀⠀⠀▪︎⠀⠀The density of gold is 19,300 kg/m³ [ as , Density ] .

⠀⠀⠀⠀⠀Converting Density from kg / m³ to g / cm³ [ kilo gram per cu. metre to gram per cu. centimetre ] :

$\qquad:\implies \sf Density \:\:=\:\: 19300 kg / m^3 \:$

$\qquad:\implies \sf Density \:\:=\:\: 19.30 g / cm^3 \:\qquad \bigg\lgroup \sf{ 1 kg/m^3 \:=\:0.0010 g/cm^3 }\bigg\rgroup$

$\qquad:\implies \bf Density \:\:=\:\: 19.30 g /c m^3 \:$

$\qquad \therefore \pmb{\underline{\purple{\bf \:Density \:\:=\:\: 19.30\: g /c m^3\: }} }\:\:\bigstar$

⠀⠀⠀⠀⠀Now ,

As , We know that ,

⠀⠀⠀⠀⠀⠀⠀▪︎⠀⠀Formula for VOLUME :

$\qquad \star \:\:\: \boxed { \pink{\pmb { \:\:\: Volume \:\:=\:\: \dfrac{\;\: Mass \:\:}{\:\: Density \:\:}\:\:\:\:}}}$

⠀⠀⠀⠀⠀Where,

• Mass is 11.58g & Density is 19.30 g/cm³ .

$\qquad \dashrightarrow \sf Volume \:\:=\:\: \dfrac{\;\: Mass \:\:}{\:\: Density \:\:}\:\:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf Volume \:\:=\:\: \dfrac{\;\: Mass \:\:}{\:\: Density \:\:}\:\:$

$\qquad \dashrightarrow \sf Volume \:\:=\:\: \dfrac{\;\: 11.58 \:\:}{\:\: 19.30 \:\:}\:\:$

$\qquad \dashrightarrow \sf Volume \:\:=\:\: \cancel {\dfrac{\;\: 11.58 \:\:}{\:\: 19.30 \:\:}}\:\:$

$\qquad \dashrightarrow \sf Volume \:\:=\:\: 0.6\:cm^3 \:$

$\qquad \therefore \pmb{\underline{\purple{\bf Volume \:\:=\:\: 0.6\:\:c m^3\: }} }\:\:\bigstar$

⠀⠀⠀⠀⠀⠀⠀▪︎⠀⠀Here , The volume is 0.6 cm³

$\therefore \underline {\sf Hence , \: Volume \: of \:coin \: is \:\bf \: 0.6\:\:c m^3\:} \:$