Question

A cubical block of wood weighing 200 g has a lead piece fastened underneath. Find the mass of the lead piece which will just allow the block to float in water. The specific gravity of wood is 0.8 and that of lead is 11.3.

Answer 1

Let the mass of the lead piece be a kg.

Therefore, the total mass of the wood and the lead = 0.2 + a kg.

Density of lead = Density of water × specific gravity

= 1000 × 11.3 = 11300

Now, The volume of the lead piece = a/11300 m³    

When the wooden block is just allowed to float in water, it displaces a volume of water equal to its volume = 200/0.8 cm³

= 250 cm³  = 0.25 m²

Now, The Total volume of the water displaced = 0.25 + a/11300 cm³

Upthrust = Vρg

=  (250 + a/11300) × 1000 × 10

Now,

(0.2 + a) × 10 = (250 + X/11300)  × 1000 × 10

∴ a = 54.8 g.

Hope it helps.

Answer 2

Lead Density = $11.3 g/cm^3$

Wood density = $0.8 \times 1 g/cm^3 =0.8 cm^3$

The cumulative volume of the water displaced within the given condition = $(250 + x/11.3) cm^3$

Now while it's just allowed to float on the water, the wood weight plus the lead piece = 54.8 g

Explanation:

Step 1:

Let the lead component mass = $x g$  

Total Wood and lead weight = $200+ x g$  

Maximum Wood and Lead weight = $x/11.3cm^3$    

{ Since Lead Density = $11.3 g/cm^3$}  

Step 2:

When allowed to float in water, the wooden block displaces a volume of water equal to its volume  

$= 200/0.8cm^3 =250 cm^3$  

{ Because Wood density =$0.8 \times 1 g/cm^3 =0.8 cm^3$}  

The cumulative volume of the water displaced within the given condition =$(250 + x/11.3) cm^3$

Step 3:  

So the force of buoyancy

$=(250 + x/11.3) cm^3 \times 1 g/cm^3$

$=(250 + x/11.3) g$  

Step 4:

Now while it's just allowed to float on the water, the wood weight plus the lead piece = the buoyancy force.

$200+x = (250 + x/11.3)$    

$x-x/11.3=250-200$  

$(11.3x-x)/11.3=250-200$  

$10.3 x =50\times11.3$

$x = 50\times11.3/10.3$

$= 54.8 g$