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 point 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 alpha=3//4` and `tan beta=12//5`. Then the tension in the string BC is

Options

A. `2mg`
B. `(13//10)mg`
C. `(3//10)mg`
D. `(20//11)mg`

Answer

B

Explanation

`tan beta=12/5 :.cos beta = 5/13`
`T_(1) cos beta + T_(2) cos beta =mg` …(i)
`T_(1) sin beta =T_(2) sin beta` …(ii)
`:. T_(1)=T_(2)=T`
`:. 2T cos beta =mg implies T=(mg)/(2 cos beta)implies T=(13)/(10) mg`.