Two particles A and B , each of mass m, are kept stationary by applying a horizontal force F = mg on particle B as shown in the figure. Then

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Two particles A and B , each of mass m, are kept stationary by applying a horizontal force F = mg on particle B as shown in the figure. Then

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$2 \tan β = \tan α$

$2 T_{1} = 5 T_{2}$

$T_{1} \sqrt{2} = T_{2} \sqrt{5}$

None of these

answer is C.

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Detailed Solution

From the figure $T_{2} \cos β = m g, T_{2} \sin β = m g$

$\begin{array}{l} T_{1} \sin α = T_{2} \sin β = m g \\ T_{1} \cos α = m g + T_{2} \cos β = 2 m g \\ \Rightarrow \frac{\cos α}{\sin α} = \frac{2 m g}{m g} \\ \tan α = \frac{1}{2} \Rightarrow \cos α = \frac{2}{\sqrt{5}} \\ T_{1} \frac{2}{\sqrt{5}} = 2 m g \Rightarrow T_{1} = m g \sqrt{5} \\ \frac{T_{1}}{T_{2}} = \frac{\sqrt{5}}{\sqrt{2}} \Rightarrow \sqrt{2} T_{1} = \sqrt{5} T_{2} \end{array}$