An object has a weight of 4 N in air, 3 N in water and 2.8 N in a salt solution. If the density of water is 1000 kg m^{-3}, the density of the salt solution is

A. 830 kg m^{-3}B. 1070 kg m^{-3}C. 1200 kg m^{-3}D. 1430 kg m^{-3}

__Solution__Show solution >>

This is a problem that requires the application of Archimedes’ principle.

The object has a weight of 4 N in air. Because the upthrust on the object when it is in air is negligible, we can say that the weight of the object is 4 N.

When the object is (completely immersed) in water, its weight is 3 N, so it experiences an apparent loss of weight of 1 N due to the upthrust acting on the object.

Upthrust = Apparent loss of weight

and

Upthrust = *Vρ*_{water}*g*

Where*V* – volume of the object

ρ

ρ

_{water}

*– density of water*

*– the acceleration due to gravity*

g

g

Now the object is immersed in a salt solution with density *ρ*_{salt} and its weight becomes 2.8 N. It experiences an apparent loss of weight of 1.2 N, therefore

Upthrust = *V**ρ*_{salt}*g *= 1.2 N - (2)

Because the volume of object remains the same, we substitute equation 1 into equation 2.

Hide solution <<

**Question source:**Malaysia National Physics Competition Panel

## No comments:

Post a Comment