A) 1.85
B) 1.5
C) 3
D) 1.37
Correct Answer: D
Solution :
If CM lies vertically below \[A\Rightarrow \]?as per choose coordinate axis in x-coordinate is equal to \[\frac{{{\ell }_{1}}}{2}\] \[{{X}_{cm}}=\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}\] \[\frac{{{\ell }_{1}}}{2}=\frac{({{\ell }_{1}})\left( \frac{{{\ell }_{1}}}{4} \right)+({{\ell }_{2}})\left( \frac{{{\ell }_{2}}}{2} \right)}{({{\ell }_{1}}+{{\ell }_{2}})}\] \[\frac{\ell _{1}^{2}}{2}+\frac{{{\ell }_{1}}{{\ell }_{2}}}{2}=\frac{\ell _{1}^{2}}{4}+\frac{\ell _{1}^{2}}{4}+\frac{\ell _{2}^{2}}{2}\] \[\frac{\ell _{1}^{2}}{4}+\frac{{{\ell }_{1}}+\ell & {{ & }_{2}}}{2}-\frac{\ell _{2}^{2}}{2}=0\] \[\ell _{1}^{2}+2{{\ell }_{1}}{{\ell }_{2}}-2\ell _{2}^{2}=0\] \[{{\ell }_{1}}=\frac{-2{{\ell }_{2}}\pm \sqrt{4\ell _{1}^{2}+4.1(+2\ell _{2}^{2})}}{2}\] \[{{\ell }_{1}}=\frac{-2{{\ell }_{2}}\pm \sqrt{12\ell _{2}^{2})}}{2}\] \[{{\ell }_{1}}=\frac{-2{{\ell }_{2}}\pm 2\sqrt{3}{{\ell }_{2}}}{2}\] \[{{\ell }_{1}}=\left( \sqrt{3}-1 \right){{\ell }_{2}}\] \[\frac{{{\ell }_{2}}}{\ell & & {{ & }_{1}}}=\frac{1}{\sqrt{3-1}}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}\] \[\frac{{{\ell }_{2}}}{\ell & & {{ & }_{1}}}=\frac{\sqrt{3}+1}{2}=\frac{2.732}{2}=1.366\] \[\simeq 1.37\]You need to login to perform this action.
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