question_answer

A uniform rod is suspended horizontally from its mid-point. A piece of metal whose weight is w is suspended at a distance \[l\] from the mid- point. Another weight \[{{W}_{1}}\]is suspended on the other side at a distance \[{{l}_{1}}\]the mid-point to bring the rod to a horizontal position. When \[\omega \]is completely immersed in water, \[{{\omega }_{1}}\]needs to be kept at a distance \[{{l}_{2}}\] from the mid-point to get the rod back into horizontal position. The specific gravity of the metal piece is

A)  \[\frac{w}{{{w}_{1}}}\]

B)  \[\frac{w{{l}_{1}}}{wl-{{w}_{1}}{{l}_{2}}}\]

C)  \[\frac{{{l}_{1}}}{{{l}_{1}}-{{l}_{2}}}\]

D)  \[\frac{{{l}_{1}}}{{{l}_{2}}}\]

Correct Answer: C

Solution :

  \[Wl={{W}_{1}}{{l}_{1}}\] ?(i) where \[\rho =\] specific gravity \[Wl\left( 1-\frac{1}{\rho } \right)={{W}_{1}}{{l}_{2}}\] \[\left( 1-\frac{1}{\rho } \right)=\frac{{{W}_{1}}{{l}_{2}}}{Wl}\] [from Eq. (ii)] \[\left( 1-\frac{1}{\rho } \right)=\frac{{{W}_{1}}{{l}_{2}}}{{{W}_{1}}{{l}_{1}}}\] \[1-\frac{1}{\rho }=\frac{{{l}_{2}}}{{{l}_{1}}}\Rightarrow \frac{1}{\rho }=1-\frac{{{l}_{2}}}{{{l}_{1}}}\] \[\frac{1}{\rho }=\frac{{{l}_{1}}-{{l}_{2}}}{{{l}_{1}}}\]or \[\rho =\frac{{{l}_{1}}}{{{l}_{1}}-{{l}_{2}}}\]