A piece of iron of mass 156 g and density 7.8 g/3
floats on mercury of density
13.6 g/3
. What is the minimum force required to submerge it?
Answers
Answer:
6.8 N
Explanation:
Given :-
Density = 7.8 g/ cm^3
Volume = 100 cm^3
mass = density × volume
⇒ 7.8 * 100 = 780 g
⇒ 0.78 Kg
weight = mass × gravitation
⇒ 0.78 * 10 = 7.8 N
It fully submerges in water.
Volume of displace water = density × volume of displace water
⇒ 1 × 100 = 100 g = 0.1 Kg
Weight of water = mass × gravitation
⇒ 0.1 * 10 = 1 N
Buoyant force = weight of displaced fluid
∴ Buoyant force = 1 N
(weight of iron piece of water) = (weight in air) - (buoyant force)
⇒ ( 7.8 - 1 ) N
⇒ 6.8 N
(Hope it helped you)
Answer:
Given :-
Density=7.8 g/ cm^3
Volume=100 cm^3
we know that,
mass = density*volume
= 7.8*100=780g
= 0.78 kg
we know,
weight = mg
= 0.78*10=7.8 N
it fully submerge in water.
volume of display water= density*
volume of display water=1*100=g=0.1 kg
weight of water = mg
=0.1*10=1 N
buoyant force= weight of display fluid
Therefore,
buoyant force =1 N
the weight of iron piece of water = weight in air- buoyant force=(7.8-1) N
=6.8 N