question_answer

In the figure shown, ABC is a uniform wire. If centre of mass of wire lies vertically below point A, then \[\frac{BC}{AB}\] is close to :

A) 1.85                 

B)        1.5    

C) 1.37                 

D)         3

Correct Answer: C

Solution :

[c] Centre of mass \[{{x}_{cm}}=\frac{x}{2}\frac{(\rho x)\left( \frac{x}{2} \right)\frac{1}{2}+\rho y\left( \frac{y}{2} \right)}{\rho (x+y)}\] \[\Rightarrow \]\[\frac{1}{2}+\frac{y}{x}=\frac{{{y}^{2}}}{{{x}^{2}}}\] \[\therefore \] \[\frac{BC}{AB}=\frac{y}{x}=\frac{1+\sqrt{3}}{2}=1.37\]