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Answer» <html><body><p></p><p>x+1=0</p><p>x=−1</p><p>8×1+4+5=17</p><p>Sometimes referred to as the <a href="https://interviewquestions.tuteehub.com/tag/incompressibility-1040188" style="font-weight:bold;" target="_blank" title="Click to know more about INCOMPRESSIBILITY">INCOMPRESSIBILITY</a>, the bulk modulus is a measure of the ability of a substance to withstand changes in <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> when under compression on all sides. ... It is equal to the quotient of the applied <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> divided by the relative deformation.Given:</p><p>A <a href="https://interviewquestions.tuteehub.com/tag/wooden-744658" style="font-weight:bold;" target="_blank" title="Click to know more about WOODEN">WOODEN</a> block of mass 400 gram float vertically in a liquid of density 0.8 gram/cm cube. The volume of the block is 625 cm³.</p><p></p><p>To find:</p><p>Volume of block above liquid surface ?</p><p></p><p>Calculation:</p><p>In case of floating, the weight of the block is being balanced by the buoyant force:</p><p></p><p>\therefore \: F_{b} = mg∴F </p><p>b</p><p> </p><p> =mg</p><p></p><p>\implies \: (V_{inside}) \rho g = (V_{total}) \sigma g⟹(V </p><p>inside</p><p> </p><p> )ρg=(V </p><p>total</p><p> </p><p> )σg</p><p></p><p>\implies \: \dfrac{V_{inside}}{V_{total}} = \dfrac{ \sigma}{ \rho}⟹ </p><p>V </p><p>total</p><p> </p><p> </p><p>V </p><p>inside</p><p> </p><p> </p><p> </p><p> = </p><p>ρ</p><p>σ</p><p> </p><p> </p><p></p><p>\implies \: \dfrac{V_{inside}}{625} = \dfrac{ (\<a href="https://interviewquestions.tuteehub.com/tag/frac-2657407" style="font-weight:bold;" target="_blank" title="Click to know more about FRAC">FRAC</a>{400}{625} )}{ 0.8}⟹ </p><p>625</p><p>V </p><p>inside</p><p> </p><p> </p><p> </p><p> = </p><p>0.8</p><p>( </p><p>625</p><p>400</p><p> </p><p> )</p><p> </p><p> </p><p></p><p>\implies \: \dfrac{V_{inside}}{625} = \dfrac{ 0.64}{ 0.8}⟹ </p><p>625</p><p>V </p><p>inside</p><p> </p><p> </p><p> </p><p> = </p><p>0.8</p><p>0.64</p><p> </p><p> </p><p></p><p>\implies \: \dfrac{V_{inside}}{625} = 0.8⟹ </p><p>625</p><p>V </p><p>inside</p><p> </p><p> </p><p> </p><p> =0.8</p><p></p><p>\implies \: V_{inside} = 500 \: {cm}^{3}⟹V </p><p>inside</p><p> </p><p> =500cm </p><p>3</p><p> </p><p></p><p>\implies \: V_{outside} =(625 - 500 )\: {cm}^{3}⟹V </p><p>outside</p><p> </p><p> =(625−500)cm </p><p>3</p><p> </p><p></p><p>\implies \: V_{outside} =125 \: {cm}^{3}⟹V </p><p>outside</p><p> </p><p> =125cm </p><p>3</p></body></html>


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