To find the density of a material of the body using moments

[leena19]

Homework Statement

"[URL=http://img193.imageshack.us/my.php?image=63378846.png[/URL]
Fig.A shows the balanced position of a light rod carrying an object X and a mass M .
Fig.B shows the balanced position of the same system when X is immersed in water.If the density of water is d,the density of the material made of X is given by,
1) [L[SUB]1[/SUB]/(L[SUB]1[/SUB]-L[SUB]2[/SUB])]d 2) (L[SUB]1[/SUB]/L[SUB]2[/SUB])d 3) [L[SUB]1[/SUB]/L[SUB]1[/SUB]+L[SUB]2[/SUB]]d 4) [(L[SUB]1[/SUB]-L[SUB]2[/SUB])/L1]d 5) (L[SUB]2[/SUB]/L[SUB]1[/SUB]) d

[h2]Homework Equations[/h2]
[h2]The Attempt at a Solution[/h2]
Taking moments about the pivot,
for A
mg*l = Mg*L[SUB]1[/SUB]
m=ML[SUB]1[/SUB]/l m=mass of X

for B
Mg*L[SUB]2[/SUB] = m[SUB]1[/SUB]g*l
m[SUB]1[/SUB]= ML[SUB]2[/SUB]/l m[SUB]1[/SUB]g = apparent weight of X in water

density of X= (real weight/apparent loss of weight in water) d
=(m/m-m[SUB]1[/SUB])*d
=(1-m/m[SUB]1[/SUB])d
=(1-L[SUB]1[/SUB]/L[SUB]2[/SUB])*d
gives me( L[SUB]2[/SUB]-L[SUB]1[/SUB]/L[SUB]2[/SUB])*d, which is not even among the choices given.
I can't figure out where I'm going wrong.

 

[alphysicist]

Hi leena19,

Homework Statement

"[URL=http://img193.imageshack.us/my.php?image=63378846.png[/URL]
Fig.A shows the balanced position of a light rod carrying an object X and a mass M .
Fig.B shows the balanced position of the same system when X is immersed in water.If the density of water is d,the density of the material made of X is given by,
1) [L[SUB]1[/SUB]/(L[SUB]1[/SUB]-L[SUB]2[/SUB])]d 2) (L[SUB]1[/SUB]/L[SUB]2[/SUB])d 3) [L[SUB]1[/SUB]/L[SUB]1[/SUB]+L[SUB]2[/SUB]]d 4) [(L[SUB]1[/SUB]-L[SUB]2[/SUB])/L1]d 5) (L[SUB]2[/SUB]/L[SUB]1[/SUB]) d

[h2]Homework Equations[/h2]



[h2]The Attempt at a Solution[/h2]
Taking moments about the pivot,
for A
mg*l = Mg*L[SUB]1[/SUB]
m=ML[SUB]1[/SUB]/l m=mass of X

for B
Mg*L[SUB]2[/SUB] = m[SUB]1[/SUB]g*l
m[SUB]1[/SUB]= ML[SUB]2[/SUB]/l m[SUB]1[/SUB]g = apparent weight of X in water

density of X= (real weight/apparent loss of weight in water) d
=(m/m-m[SUB]1[/SUB])*d
=(1-m/m[SUB]1[/SUB])d[/quote]

I have not looked over all of your work, but it looks like these last two lines are saying that:

$$\left(\frac{m}{m-m_1}\right) \to \left(1-\frac{m}{m_1}\right)$$

which is not true.

 

[leena19]

Thanks for replying!

I have not looked over all of your work, but it looks like these last two lines are saying that:

<br /> \left(\frac{m}{m-m_1}\right) \to \left(1-\frac{m}{m_1}\right)<br />

which is not true.

Oops!
after correcting it ,I get
[L₁/(L₁-L₂)]d which would be answer no.1,hopefully?


EDIT:
The answer is (1).
Thank you very much for the help!