# Thread: I have two questions I'm stuck on.

1. ## I have two questions I'm stuck on.

An object made entirely from one material has a volume of 30m(3). It floats in a liquid of density 3g/cm(3) with only half its mass below the liquid's surface. The object is then placed in another liquid of the same volume but with a density of 1g/cm(3). If the final upthrust that will act on the object is expressed in the form a x 10(b)N give the values of a and b. (Assume g = 10 m/s(-2).)

The answer is a=b=3. I've no idea how to do it.

Also I have another question.

A study has claimed to show that abnormal blood levels of prostate specific antigen (PSA) may be an indicator of prostate cancer. Twenty percent of men with prostate cancer have normal PSA levels and at least 2 out of 3 men with abnormal PSA levels do not have prostate cancer. Tests were performed on a group of 2500 men and 1051 of them had normal blood PSA levels. What is the maximum possible number of men you would expect to have prostate cancer?

A- 200
B- 241
C- 483
D- 604
E- 724

Could anyone answer these questions for me and show the working. A huge thank you in advance.

2. In the 1st one you need to calculate the body's density.
When you do, you'll see it will sink in the 2nd liquid.

Then you just apply Archimedes buoyancy law/thing to calculate the force applied in the body.

The 2nd, you see there are 1449 men with abnormal PSA levels (its more than the 'normal' but ok). of those only a third is expected to have cancer.
1449/3=483

the text also says 20% of cancer people dont have abnormal values, so 483 is just 80%. so the final result is
$\frac{483}{80}*100=604$

3. Thank you for the answer to the second one (I made a really stupid mistake) but could you go in to more depth on the first one or give me a website explaining it?

4. In the 1st one you need to calculate the body's density.

The body is in equilibrium (floating) so the weight(P) of the body and the Boyauncy force (I) are equal
$P=\rho_{body} \times V_{body} \times g$
$I=\rho_{liquid} \times V_{underliquid} \times g$

so

$\rho_{body}=\frac{1}{2} \rho_{liquid}=1.5 g/cm^3$

it will sink in the 2nd liquid.

so the force from the 2nd liquid is
$I=\rho_{liquid} \times V_{underliquid} \times g$
$
I=1 \times 30000 \times 10 = 3 \times 10^5 N$