question_answer

Two light rods \[AB=a+b\]and \[CD=a-b\]symmetrically lying on a horizontal. There are kept intact by two strings AC and BD. The perpendicular distance between rods is a. The length of AC is given by

A) \[{{a}^{2}}+{{b}^{2}}\]                  

B) \[{{a}^{2}}-{{b}^{2}}\]

C) \[\sqrt{{{a}^{2}}-{{b}^{2}}}\]                      

D) \[\sqrt{{{a}^{2}}+{{b}^{2}}}\]

Correct Answer: D

Solution :

Since they are symmetrically lying on horizontal plane.

\[\therefore \]      \[AC=BD\]

\[\therefore \]      \[AE=BF=x\]

Now,     \[AB=(a-b)+2x\]

i.e.        \[a+b=a-b+2x\]\[\Rightarrow \]\[2b=2x\]

\[\therefore \]      \[x=b\]

Now, in \[\Delta ACE,\]\[{{x}^{2}}+{{a}^{2}}=A{{C}^{2}}\]

\[\Rightarrow \]   \[A{{C}^{2}}={{b}^{2}}+{{a}^{2}}\]

\[\therefore \]      \[AC=\sqrt{{{b}^{2}}+{{a}^{2}}}\]