A block has 30% of it's volume immersed in water when the container at rest. What will be the % volume immersed when the block is accelerated upwards with acceleration a ?

When you equate the V×d(solid)×g with V(submerged in water ×d(water) ×g, you get d(solid)=3d(water)/10. When you take the container upwards with an acceleration a the net acceleration will be (g+a) but it will be for both water and the object, so it will cancel. I think it doesn't matter for any value of a, what do you say?

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TopNewestWe will solve for the percentage immersed at rest when we have the densities of liquid and block. Let the percentage be P.

We have \({\rho}_{block}Vg=P{\rho}_{liquid}Vg \)

Solving we have \( P = \dfrac{{\rho}_{block}}{{\rho}_{liqiud}} \)

It is worthy to note that the answer is independent of \(g\).

So when the container is accelerated only the effective gravity changes, hence the answer is independent of \(a\).

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Comment deleted Dec 29, 2015

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When you equate the V×d(solid)×g with V(submerged in water ×d(water) ×g, you get d(solid)=3d(water)/10. When you take the container upwards with an acceleration a the net acceleration will be (g+a) but it will be for both water and the object, so it will cancel. I think it doesn't matter for any value of a, what do you say?

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Me too same. But I'll wait till 'The Great Ronak Agarwal' answers.

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\( \dfrac{V_{in}}{V_{total}} = \dfrac{d_{body}}{d_{liquid}} \) which is independent of the effective acceleration as it cancels out on both sides.

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