Figure represents a light inextensible string ABCDE in which AB = BC = CD = DE and to which are attached masses M, m, and M at the points B, C, and D, respectively.
The system hangs freely in equilibrium with ends A and E of the string fixed in the same horizontal line. It is given that tan $α$ = 3/4 and tan $β$ = 12/5. Then the tension in the
string BC is

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2 mg

(13/10) mg

(3/10) mg

(20/11) mg

answer is B.

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Detailed Solution

$\begin{array}{l} \tan β = \frac{12}{5} ∴ \cos β = \frac{5}{13} \\ T_{1} \cos β + T_{2} \cos β = m g . . . . . . (i) \\ T_{1} \sin β = T_{2} \sin β . . . . . . (i i) \\ ∴ T_{1} = T_{2} = T \end{array}$

$∴ 2 T \cos β = m g \Rightarrow T = \frac{m g}{2 \cos β} \Rightarrow T = \frac{13}{10} m g$

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