Question

A tank contains water of mercury as shown in the figure. A cubical block of side $10cm$ is in equilibrium inside (Relative density of the material of block $=7.2$, relative density of mercury $=13.6$).

Find the height of the block above the mercury level.

• A. $6cm$
• B. $4.7cm$
• C. $7cm$
• D. $8cm$

Answer 1

The correct option is B $4.7cm$

According to question.................

$\begin{array}{c}Theblockwillfloatinmercuryifitsweightisequaltotheweightof\\ \Rightarrow mercurydisplacedbyit.\\ \Rightarrow Densityofiron,{\rho }_i=7.2g/c{m}^3\\ \Rightarrow Densityofmercury,{\rho }_m=13.6g/c{m}^3\\ \Rightarrow Lenghtofeachsideoftheironblock=10cm\\ let,\\ \Rightarrow xbetheheightoftheblockabovethemercurylevelinthevessel.\\ then,\\ thevolumeoftheblockinsidethemercury.\\ \Rightarrow V_m=10\times 10\times (10-x)\\ Thisisalsoequaltothevolumeofmercurydisplayedbytheblock.\\ \Rightarrow Weightofthisvolumeofmercuryis,\\ \Rightarrow W=V_m{\rho }_{m}g=V_m\times 13.6\times 981=upthrustonblock\\ Totalvolumeoftheironblockis,\\ \Rightarrow V=10\times 10\times 10=1000c{m}^3\\ Hence,\\ W=V{\rho }_{i}g=V\times 7.2\times 981\\ then,\\ Theironblockfloats,therefore......\\ \Rightarrow Weightofmercurydisplaced=weightoftheironblock\\ \Rightarrow 100\times (10-x)\times 13.6\times 981=1000\times 7.2\times 981\\ \Rightarrow 100\times (10-x)\times 13.6=1000\times 7.2\\ \Rightarrow (10-x)\times 1360=7200\\ \Rightarrow 10-x=\frac{7200}{1360}\\ \Rightarrow 10-x=5.29\\ \Rightarrow x=10-5.29\\ \therefore x=4.71\\ sotheheightofhteblockabovemercurylevel=4.71cm\\ sothatthecorrectoptionisB.\end{array}$