question_answer

A mass less rod is suspended by two identical strings AB and CD of equal length. A block of mass m is suspended from point 0 such that BO is equal to 'y'. Further, it is observed that the frequency of 1st harmonic (fundamental frequency) in AB is equal to 2nd harmonic frequency in CD. Then, length of BO is

A)  L/5                                        

B)  4L/5

C)  3L/4                     

D)  L/4

Correct Answer: A

Solution :

\[{{S}_{4}}O_{6}^{2-}\]                                  \[C{{l}_{2}}{{O}_{7}}\] For rotational equilibrium,\[CC{{l}_{4}}\]\[CC{{l}_{4}}\]\[\text{(a}\,\text{=}\,\text{20}{{\text{L}}^{\text{2}}}\text{mo}{{\text{l}}^{\text{-2}}}\text{atm}\,\text{and}\,\text{b}\ \text{=}\,\text{0}\text{.14}\,\text{L}\,\text{mo}{{\text{l}}^{\text{-1}}}\text{)}\]