(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

An aluminum cylinder weighs 1.03 N. When this same cylinder is completely submerged in alcohol, the volume of the displaced alcohol is 3.89 multiplied by 10-5 m3. If the cylinder is suspended from a scale while submerged in the alcohol, the scale reading is 0.732 N. What is the specific gravity of the alcohol?

2. Relevant equations

F_{scale}+ F_{Buoyancy}= mg

F_{Bouyancy}= DVg

Specific gravity = D_{object}/ D_{medium}

3. The attempt at a solution

First I used F_{scale}+ F_{Buoyancy}= mg

(.732)(F_{B}) = 1.03

F_{B}= .298 N

Then I used F_{B}= DVg

.298 = D_{alcohol}(3.89e-5)(9.8)

D_{alcohol}= .298 / [(3.89e-5)(9.8)]

D_{alcohol}= 781.7009 kg.m^{3}

Then I used Specific gravity = D_{object}/ D_{medium}, where D = mv.

Specific gravity = (m/v) / D_{alcohol}

Specific gravity = [(1.02/9.8)/3.85e-5] / (781.7009)

Specific gravity = 3.458

I tried this answer and it was wrong. Can anyone help out? Thank you!

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# Homework Help: Specific Gravity of Alcohol.

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