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  •  In the figure shown ABC is a uniform wire. If centre of mass of wire lies vertically below point A, then <img alt="\frac{BC}{AB}" src="https://entrancecorner.oncodecogs.com/gif.latex?%

 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 :

 

Option: 1

1.85     


Option: 2

1.37


Option: 3

1.5


Option: 4

 3

 


Answers (1)

best_answer

\\ \text{If CM liesvertically below A} \Rightarrow \text{ as per choose axis in x coordinate is equal to} \ \frac{l_{1}}{2} \\ \\ X_{c m}=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}}= \frac{l_{1}}{2}\\ \\ \\

\begin{aligned} &\frac{ {l}_1}{2}=\frac{\left(\lambda l_1\right)\left[\frac{l_1}{4}\right]+\left(\lambda l_2\right)\left[\frac{l_2}{2}\right]}{\lambda\left(\mathrm{l}_1+\mathrm{l}_2\right)}\\ &\frac{l_1^2}{2}+\frac{l_1 l_2}{2}=\frac{l_1^2}{4}+\frac{l_2^2}{2}\\ &\frac{l_1^2}{4}+\frac{l_1 l_2}{2}-\frac{l_2^2}{2}=0\\ &l_1^2+2 l_1 l_2-2 l_2^2=0 \end{aligned}

\begin{aligned} &l_{1}=\frac{-2 l_{2} \pm \sqrt{4 l_{1}^{2}+4.1\left(+2 l_{2}^{2}\right)}}{2}\\ &l_{1}=\frac{-2 l_{2} \pm 2 \sqrt{12 l_{2}^{2}}}{2}\\ &l_{1}=\frac{-2 l_{2} \pm 2 \sqrt{3} l_{2}}{2}\\ &l_{1}=(\sqrt{3}-1) l_{2}\\ &\frac{l_{2}}{l_{1}}=\frac{1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}\\ &\frac{l_{2}}{l_{1}}=\frac{\sqrt{3}+1}{2}\\ &\frac{l_{2}}{l_{1}}=1.365 \end{aligned}

 

Posted by

Ritika Jonwal

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