Question

3) A lead sinker of mass 225 grams and density of 11.3 g/cm3 is attached to the bottom of a wooden block of mass 25 grams and density 0.5 g/cm3. Calculate the apparent weight when both are submerged in water.

Answer 1

180.1 g

Step-by-step explanation:

Data provided in the question

Mass of lead sinker = 225 grams

Density = 11.3 g/cm^3

The Wooden block of mass = 25 grams

Density = 0.5/g cm^3

Based on the above information, the apparent weight is

Before that we need to do the following calculations

`V_1 = (m_1)/(D_1)`

`= (225)/(11.3)`

= 19.91 cm^3

`V_2 = (m_2)/(D_2)`

`= (25)/(0.5)`

= 50 cm^3

Now as we know that

V = V_1 + V_2

= 19.91 cm^3 + 50 cm^3

= 69.91 cm^3

Now the weight of dispacement of water is

`m = VD_(water)`

`= 69.91 cm^3 (1 (g)/(cm^3) )`

= 69.91 g

Therefore the apparent weight is

`W = m_1 + m_2 - m`

= 225 + 25 - 69.91 g

= 180.1 g