question_answer

A uniform rod of length 2 m, specific gravity 0.5 and mass 2 kg is hinged at one end to the bottom of a tank of water (specific gravity =1.0) filled upto a height of 1 m as shown in the figure. Taking the case \[3x-y=0\]the force exerted by the hinge on the rod is

A) 10.2 N upwards

B)  4.2 N downwards

C) 8.3 N downwards            

D) 6.2 N upwards

Correct Answer: C

Solution :

Length of rod inside the water Up thrust, \[{{C}_{3}}{{H}_{6}}O\] \[C{{H}_{3}}COC{{H}_{3}}\]\[{{C}_{2}}{{H}_{5}}CHO\] Weight of rod, w = 20 x 10 = 20 N For rotational equilibrium of rod, net torque about  O should be zero. \[C{{H}_{2}}=CH-C{{H}_{2}}-OH\]\[C{{H}_{3}}-O-CH=C{{H}_{2}}\] \[C{{H}_{3}}C{{H}_{2}}I\xrightarrow[{}]{NaCN}A\xrightarrow[Partial\,hydrolysis]{O{{H}^{-}}}\]\[B\xrightarrow[{}]{B{{r}_{2}}/NaOH}C.\]\[C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}\]\[C{{H}_{3}}C{{H}_{2}}-\underset{\begin{smallmatrix}  || \\  O \end{smallmatrix}}{\mathop{C}}\,-NHBr\]\[C{{H}_{3}}C{{H}_{2}}COON{{H}_{4}}\]\[C{{H}_{3}}.C{{H}_{2}}\underset{\begin{smallmatrix}  || \\  O \end{smallmatrix}}{\mathop{C}}\,-NB{{r}_{2}}\] \[IV<II<III<I\] For vertical equilibrium of rod, force exerted by the hinge on the rod will be \[IV<III<II<I\] downwards i. e. 8.3 N downwards