Specific Gravity Question...

[TicAL]

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I'm having trouble attacking these specific gravity questions, can anyone here explain a good way to go about answering these things. Here's an example problem:

A 25 kg object has an apparent weight of 200 N when placed in a fluid with a specific gravity of 0.6. What is the specific gravity of the object?

 

[aamartin81]

I'm having trouble attacking these specific gravity questions, can anyone here explain a good way to go about answering these things. Here's an example problem:

A 25 kg object has an apparent weight of 200 N when placed in a fluid with a specific gravity of 0.6. What is the specific gravity of the object?

This helps for me:

Weight/Bouyant Force = denisty (specific gravity) multiplier

250N/50N = 5

5 x 0.6 = 3

Good luck, let me know if you need more help!

Adam

 

[BaylorGuy]

The best way i've figured it out would be the specific gravity of the object equals the weight of the object over the bouyant force of the object.

Thus for the above problem:
Specific Gravity=(25kg*10m/s^2)/((25kg*10m/s^2)-200N)

thus the specific gravity would equal 5.

There was a specific gravity problem in AAMC 5 question 77 in PS that i got this solution from. Its a really easy way to remember it.

 

[medworm]

yep. don't forget to factor in the specific gravity of the liquid (0.6).

normally in water, you don't have to worry about that b/c
the s.g. of water = 1, but in this case it's 0.6 for the liquid.

so s.g. = 5 x .6 = 3.

 

[SilvrGrey330]

whats the 5 X .6 for? and where do we use .6, is it necessary?

 

[SilvrGrey330]

we posted at same time, thanks...problem solved

 

[place]

All you need is 2 equations:

1) F(boyancy)=density(fluid) x Volume(submerged) x gravity

2) s.g. = density (substance in questions)/ density H20

density of H2O = 1

if you knew the volume of the object, then you could easily find the s.g. by using the second equation... density = mass/volume... but we dont know this so we need to find it by using the first equation... The weight of the object when put in the fluid is 400 N.. the weight of the object when its in the air would be 25kg x g = ~250N .. Therefore, the F(boyancy) must be the difference (50N). Plug that into equation 1. we can find the density of the fluid by using the second equation (.6 = x/1) density = .6
now find the volume(submerged) by doing algebra
50N=.6 x V x 10
V=8.3L
Thats it!!! now use the mass of the object and divide it by the volume we just found to get the density of the object. That = 3Kg/L
now put that into equation 2 to find the answer...

s.g. = 3

 

[place]

sorry, made a mistake... Then weight of the object when put in the fluid is, as you said, 200N .. Not 400 as i put... the rest of the calculations are right

 

[cfdavid]

All you need is 2 equations:

1) F(boyancy)=density(fluid) x Volume(submerged) x gravity

2) s.g. = density (substance in questions)/ density H20

density of H2O = 1

if you knew the volume of the object, then you could easily find the s.g. by using the second equation... density = mass/volume... but we dont know this so we need to find it by using the first equation... The weight of the object when put in the fluid is 400 N.. the weight of the object when its in the air would be 25kg x g = ~250N .. Therefore, the F(boyancy) must be the difference (50N). Plug that into equation 1. we can find the density of the fluid by using the second equation (.6 = x/1) density = .6
now find the volume(submerged) by doing algebra
50N=.6 x V x 10
V=8.3L
Thats it!!! now use the mass of the object and divide it by the volume we just found to get the density of the object. That = 3Kg/L
now put that into equation 2 to find the answer...

s.g. = 3

I did it this way too, but it seems kind of long.
But, it's intuitive, so I think I'll stay the course. I don't want to memorize an equation I've never seen before out of concern of not remembering it at all during the test.

 

[TicAL]

Awesome, I think I've got it. Thanks guys. 🙂

 

[place]

Yes, it is long.. there are not many questions on the MCAT that require this much calculation. If there is, it will only be a couple. Most questions can be deduced by little or no calculations, but its good to know how to do these "longer" problems so you can figure out the ones they ask on the real MCAT