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Standard IX

Physics

Archimedes' Principle

If a body flo...

Question

If a body floats with ${(\frac{p}{q})}^{t h}$ of its volume above the surface of water, find the relative density of the body.

\frac{q + p}{q}

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1 - \frac{p}{q}

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\frac{1 - p}{q}

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\frac{p}{q}

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Solution

The correct option is B $1 - \frac{p}{q}$
Volume of body immersed = $(1 - \frac{p}{q})$ Volume
Hence Relative density = $1 - \frac{p}{q}$

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If a body floats with (pq)th of its volume above the surface of water, find the relative density of the body.

The correct option is B 1−pqVolume of body immersed = (1−pq) VolumeHence Relative density = 1−pq

If a body floats with ${(\frac{p}{q})}^{t h}$ of its volume above the surface of water, find the relative density of the body.

The correct option is B $1 - \frac{p}{q}$
Volume of body immersed = $(1 - \frac{p}{q})$ Volume
Hence Relative density = $1 - \frac{p}{q}$