Question

Mass of the iron ball shown in the figure is 52 g. When it is put in the measuring cylinder following displacement is observed in the water level. Calculate the density of the iron ball in the C.G.S and S.I unit.

Ans:
Volume of the iron ball =................. Cubic centimeter.
Mass of the iron ball = .............. g
Density = ................ gcm-3
Density = ................. Kg m-3

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Volume\:of\:iron\:ball=20\:cm^{3}}}}$

$\green{\tt{\therefore{Density\:of\:iron\:ball=2.6\:g/cm^{3}}}}$

$\green{\tt{\therefore{Density\:of\:iron\:ball=2600\:kg/m^{3}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given: }} \\ \tt: \implies Mass \: of \: iron \: ball = 52 \: g \\ \\ \red{\underline \bold{To\:Find: }} \\ \tt: \implies Volume \: of \: iron \: ball =? \\ \\ \tt: \implies Density = ?$

• According to given question :

$\tt \circ \: \Delta \: V= V_{2} - V_{1} = 80 - 60 = 20 \: ml = 20 \: {cm}^{3} \ \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies Density = \frac{Mass}{Volume} \\ \\ \tt: \implies Density = \frac{52}{20} \\ \\ \green{\tt: \implies Density = 2.6 \: g/{cm}^{3} } \\ \\ \tt: \implies Density = 2.6 \times \frac{1000000}{1000} \\ \\ \green{\tt: \implies Density = 2600 \: kg /{m}^{3} }$

Answer 2

GIVEN:

• Mass of iron ball = 52g
• Volume 1 ( V₁ ) = 60 cm³
• Volume 2 ( V₂ ) = 80 cm³

TO FIND:

• Volume of iron ball = ?
• Density = ?

SOLUTION:

We know that

d = m/v

Where

• d : density = ?
• m : mass = 52g
• v : volume = 20 cm³

Finding volume

∆V = V₂ - V₁

→ 80 - 60

→ 20 cm³

→ Density = 52/20

→ Density = 2.6 g/cm³

Converting g/cm³ → kg/m³

→ Density = 2600 kg/m³

Hence, volume of iron ball = 20cm³ , density = 2600 kg/m³