Physics, asked by jayanthbabu9207, 11 months ago

A piece of iron of mass 156g and density 7.8gm per c.c floats on mercury of density 13.6gm per c.c what is the minimum force required to submerge it

Answers

Answered by paritoshprasad077
2

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

Answered by bharatjoshi94247
0

Answer:

Explanation:

Final answer is 6.8 N

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