Question

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:

The required volume of gold coin will be 0.6cm³

Explanation:

Given:-

According to the Question

• Mass of gold coin = 11.58g
• Density of gold ,d = 19300kg/m³

we have to calculate the volume of the gold coin.

Firstly we change the unit here .

Coversion of density into g/cm³

As we know that

➽ 1kg/m³ = 1/1000g/cm³

➽ 1kg/m³ = 0.001g/cm³

➽ 19300kg/m³ = 19300*0.001 g/cm³

➽ 19300kg/m³ = 19.3 g/cm³

Now, calculating the volume of gold coin.

As we know the relationship between density ,Volume and mass .

• Volume = Mass/Density

substitute the value we get

➽ volume of gold coin = 11.58/19.3

➽ volume of gold coin = 0.6cm³

• Hence, the volume of gold coin is 0.6cm³

Answer 2

$\huge \rm {Answer:-}$

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$\sf \red {\underline{Given:-}}$

★Weight of Coin=11.58g

★Density of gold = 19,300 kg/$\bf {m^{3}}$

★ Converting M.K.S into C.G.S

$\to \bf{1kg/m^{3}=\frac{1}{1000}g/cm^{3}}$

$\to \bf{19300kg/m^{2}=\frac{19300}{1000}g/cm^{3}}$

$\to \bf {Density=19.3g/cm^{3}}$

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$\sf \blue {\underline{To\: Find:-}}$

★Volume of Coin=?

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$\sf \pink {\underline{We\: know,}}$

$\large \implies \bf {Volume=\frac{Weight}{Density}}$

★Supplanting the given values we get,

$\to \bf {Volume=\frac{11.58g}{19.3g/cm^{3}}}$

★Adjusting the decimals,

$\to \bf {Volume=\frac{1158\times{\cancel{10}}\cancel{g}}{193\cancel{g}/cm^{3}\times10\cancel{0}}}$

$\implies \bf \green {Volume=0.6cm^{3}}$

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$\sf \purple {\underline{Henceforth,}}$

★Volume of Coin=${0.6cm^{3}}$