Relative density of ice in cylinder

[leena19]

Homework Statement

A measuring cylinder contains 60cm³ of oil at 0 degrees celsius .When a piece of ice is dropped into the cylinder,it sank completely in oil and the oil level rose to 90 cm³ mark.When the ice melted the oil level came down to 87cm³ mark.The relative density of ice is
(1)0.80 (2) 0.85 (3) 0.90 (4)0.95 (5)0.98

Homework Equations

The Attempt at a Solution

I guess the volume of ice is 90-60=30cm³
and the relative density of ice can be given as = real weight of ice(mg) / apparent loss in weight in oil(mg-m'g)?

mg would be 30*d*g,where d is the density of oil ,but how do I find its apparent weight in oil.I think it has something to do with the 87cm³ but I can't figure out what.

Hope someone can help.
Thanks in advance.

 

[Cyosis]

To calculate the relative density you need to have a reference density. For this problem I guess you're expected to calculate the relative density of ice with respect to water.

Then RD=\rho_{ice} / \rho_{water}. Now try to find expressions for both densities.

 

[leena19]

Thanx.
so \rho_{ice} = m/30
& \rho_{water}= m/27

therefore RD = (m/30)*(27/m) = 0.9,which would be answer no.3?
but I have a slight problem,in some questions I've used the equations
RD of the material of a body= weight of body/weight of an equal volume of water
and to find the RD of a liquid=weight of a given volume of a liquid/weight of an equal vol.of water
or upthrust in liquid/upthrust in water
I can't undersatand why we can't solve the above problem using these equations

 

[Cyosis]

but I have a slight problem,in some questions I've used the equations
RD of the material of a body= weight of body/weight of an equal volume of water

Let's use this definition then. The weight of the body of ice is W_i=m_ig=\rho_i V_i g.
The weight of an equal volume of water is W_w=m_w g=\rho_w V_i g. Putting this in your definition of RD yields:

<br /> RD=\frac{\rho_i V_i g}{\rho_w V_i g}=\frac{\rho_i}{\rho_w}<br />

So as you can see these two equations, while perhaps looking different, really are just the same.

 

[leena19]

Oh yes!They are the same.
Thanks so much,Cyosis.