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

\frac{q + p}{q}

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1 - \frac{p}{q}

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\frac{q - p}{q}

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\frac{p}{q}

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Solution

The correct options are
B $1 - \frac{p}{q}$
C $\frac{q - p}{q}$
Weight of body $=$ upthrust
$v ρ_{b} g = (1 - \frac{p}{q}) v ρ_{w} g$
$∴ \frac{ρ_{b}}{ρ_{w}} = 1 - \frac{p}{q}$
i.e., R.D. of body $= (1 - \frac{p}{q})$ or $(\frac{q - p}{q})$

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If a body floats with (pq)th of its volume above the surface of the water, then the relative density of the body is :

The correct options are B 1−pq C q−pqWeight of body = upthrustvρbg=(1−pq)vρwg∴ρbρw=1−pqi.e., R.D. of body =(1−pq) or (q−pq)

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

The correct options are
B $1 - \frac{p}{q}$
C $\frac{q - p}{q}$
Weight of body $=$ upthrust
$v ρ_{b} g = (1 - \frac{p}{q}) v ρ_{w} g$
$∴ \frac{ρ_{b}}{ρ_{w}} = 1 - \frac{p}{q}$
i.e., R.D. of body $= (1 - \frac{p}{q})$ or $(\frac{q - p}{q})$