A sphere of radius R made of a material of re | vedclass

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(A) According to the law of floatation,a body floats if the weight of the body is equal to the weight of the water displaced.$W_{	ext{body}} = W_{	e

Vedclass · Accepted Answer

$(\frac{\sigma}{\sigma - 1})^{1/3}$

Vedclass · Answer

(A) According to the law of floatation,a body floats if the weight of the body is equal to the weight of the water displaced.$W_{	ext{body}} = W_{	ext{water displaced}}$$V_{	ext{solid}} \cdot ho_{	ext{solid}} \cdot g = V_{	ext{displaced}} \cdot ho_{	ext{water}} \cdot g$Here,the volume of the solid material is $V_{	ext{solid}} = \frac{4}{3}\pi(R^3 - r^3)$ and the volume of water displaced is $V_{	ext{displaced}} = \frac{4}{3}\pi R^3$.Given relative density $\sigma = \frac{ho_{	ext{solid}}}{ho_{	ext{water}}}$,we have $ho_{	ext{solid}} = \sigma ho_{	ext{water}}$.Substituting these into the equation:$\frac{4}{3}\pi(R^3 - r^3) \cdot \sigma ho_{	ext{water}} \cdot g = \frac{4}{3}\pi R^3 \cdot ho_{	ext{water}} \cdot g$$\sigma(R^3 - r^3) = R^3$$\sigma R^3 - \sigma r^3 = R^3$$R^3(\sigma - 1) = \sigma r^3$$\frac{R^3}{r^3} = \frac{\sigma}{\sigma - 1}$$\frac{R}{r} = (\frac{\sigma}{\sigma - 1})^{1/3}$

$A$ sphere of radius $R$ made of a material of relative density $\sigma$ has a concentric cavity of radius $r$. It just floats when placed in a tank full of water. The value of the ratio $R/r$ will be:

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