Question

a piece of iron of density 7.8g/cm cube and volume 100 cm cube is fully immersed in water ( density of water 1g/cm cube ). Calculate -» weight of iron piece in air.
-» upthrust of water on iron piece.
-» apparent weight of iron piece in water. ( g = 10m/s square )
Please guys give me an appropriate answer as soon as possible.

Answer 1

Given,
Density = 7.8g/cm³
Volume = 100 cm³
We know, Mass = density × volume 
Then put values,
⇒7.8 × 100 = 780 g or 0.78kg
Then we know,
Weight  = mg = 0.78 × 9.8 = 7.64 N
So, weight = 7.64 N
It fully submerge in water. 
∴ Volume of displaced water = volume of iron peace = 100cm³
∴ Mass of displaced water = density × volume of displaced water
                                          = 1.0×100 [ density of water = 1.0cm³]
                                          = 100g or 0.1kg
weight of water = mg
                         = 0.1×9.8 
                        = 0.98 N or 1 N
Buoyant force = weight of displaced fluid [ Archimedes  Principal]
∴ Buoyant force = 0.98 N
Then, weight of iron piece in water = weight in air - buoyant force
                                                        = 7.64 - 0.98
                                                        = 6.66 N

Answer 2

Answer:

→ Weight in Air = V × d × g

= (0.0001 × 7800 × 10) kg

= 7.8 kg

= 7.8 N

→ Upthrust of water on iron piece= Buoyant Force

= V × d × g

= 0.0001 × 1000 × 10 N

= 1 N

→ Apparent weight of iron piece in water

= (7.8 - 1) N

= 6.8 N