Question

12. The mass of a lead piece is 115 g. When it is
immersed into a measuring cylinder, the water
level rises from 20 mL mark 'to 30 mL mark,
Find :
(i) the volume of the lead piece,
(ii) the density of the lead in kg m-3.​

Answer 1

Given:-

• The mass of a lead piece = 115 g

• Volume of water (V¹)= 20 ml

• Volume of water (V²) = 30 ml

To Find:-

• The Volume of the lead piece = ?

• The density of the lead piece in kg m-³ = ?

Solution:-

• To calculate the volume and density of the lead piece at, 1st we have to find volume of lead piece as we know that volume of lead piece is (subtract V¹ from V²). After that we have to find Density of the lead piece. As given in the question the mass of a lead piece is 115 g. When it is immersed into a measuring cylinder, the water level rises from 20 mL mark 'to 30 mL mark. Simply by applying formula formula to calculate Density of the lead piece.

⟶ Volume of lead = Volume ² - Volume ¹

⟶ Volume of lead = 30 - 20

⟶ Volume of lead = 10 cm³

Let's calculate the density of lead:-

⟶ Density of lead = Mass / Volume

⟶ Mass of lead = 115 ⟶ Volume = 10 cm³

⟶ Density of lead = 115 / 10 cm³

⟶ Density of lead = 11.5 g cm-³

Now Find it's density in kg m-³ here:-

• We know that 1 g cm-³ = 1000 kg-³

⟶ Density of lead in kg m-³ = Density of lead × 1000

⟶ Density of lead in kg m-3 = 11.5 × 1000

⟶ Density of lead in kg m-³ = 11500 kg/ m-³

Hence,

• The volume of lead piece = 10 cm³

• The density of lead piece = 11500 kg/ m-³

Answer 2

Given :-

Mass  of a lead piece is 115 g.When it is  immersed into a measuring cylinder, the water  level rises from 20 mL mark 'to 30 mL mark,

To Find :-

Volume

Density

Solution :-

Volume of lead = Rise in level of water

Volume of lead = 30 - 20

Volume of lead = 10 cm³

For density

Density = Mass/Volume

Density = 115/10

Density = 11.5 gm/cm⁻³

Density = 11.5 × 1000 =  11500 kg/m⁻³