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:

0.6cm³ is the required volume of gold coin.

Explanation:

According to the Question

It is given that

• Weight(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

Question :-

• An ancient coin made of pure gold weighs 11.58g. Find its volume when the density of gold is 19300 kg/m³.

Answer :-

• Volume of gold coin is 0.6 cm³.

Explanation :-

Given :-

• Weight of gold coin (Mass) = 11.58 g
• Density of gold = 19300 kg/m³

To Find :-

• Volume of gold coin?

Solution :-

❏ Converting units of density :-

We know that,

✪ $\bf 1\:kg/m^3 = 0.001\:g/cm^3$ ✪

Hence,

↦ $\sf Density \:in \:g/cm^3 = 19300\:\times\:0.001$

↦ $\sf Density \:in \:g/cm^3 = 19300\:\times\:\dfrac{1}{1000}$

↦ $\sf Density \:in \:g/cm^3 = \dfrac{19300\:\times\:1}{1000}$

↦ $\sf Density \:in \:g/cm^3 = \dfrac{193\cancel{00}}{10\cancel{00}}$

↦ $\sf Density \:in \:g/cm^3 = {\cancel{\dfrac{193}{10}}}$

➦ $\bf \red{Density\: in \:g/cm^3 = 19.3 \:g/cm^3}$

∴ Density of gold is 19.3 g/cm³.

• Now, we have mass of gold coin as well as density of gold. So, we can easily find volume of gold coin by using well known formula i.e, formula of volume.

We know that,

✪ $\bf Volume = \dfrac{Mass}{Density}$ ✪

Putting values in formula we get,

⇒ $\sf Volume = {\cancel{\dfrac{11.58}{19.3}}}$

➠ $\bf \purple{Volume = 0.6\:cm^3}$

∴ Volume of gold coin is 0.6 cm³.

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