Question

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? DONOT TAKE VOLUME AS 100 CENTIMETER CUBE AND I NEED EXPLANATION AND I WILL MARK YOU BRAINLIEST

Answer 1

Answer:

Answer ::-

Given :-

Density = 7.8 g/ cm^3

Volume = 100 cm^3

we know that ,

mass = density *volume

= 7.8 * 100 = 780 g

= 0.78 Kg

We know ,

weight = mg

= 0.78 * 10 = 7.8 N

It fully submerge in water .

Volume of displace water = density * volume of displace water = 1 * 100 = 100 g = 0.1 Kg

Weight of water = mg

= 0.1 * 10 = 1 N

Buoyant force = weight of displace 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

Answer 2

iron
​
=7.8×10
3
kg/m
3

ρ
Hg
​
=13.6×10
3
kg./m
3

Let volume of body=V
Weight of body = buoyant force
ρ
iron
​
Vg=ρ
Hg
​
V
′
g
7.8×10
3
×V=13.6×10
3
×V
′

V
′
=
13.6×10
3

7.8×10
3

​

V
′
=
136
78
​
V
V
′
=0.57V
Volume outside =V−V
′

V−0.57V
=0.43V