Two partical`A` and `B` each of mass `m` are kept stationary by applying a horizontal force `F = mg` on partical `B` as show in figure .Then

Options

A. `tan beta = 2 tan alpha`
B. `2T_(1) = 5T_(2)`
C. `sqrt2 T_(1) = sqrt5 T_(2)`
D. `alpha = beta`

Answer

A, C

Explanation

Resultant of `mg` and `mg` is `sqrt(2) mg`.

Therefore `T_(2)` should be equal and opposite of this .
or `T_(2) = sqrt(2) mg`…(i)
Further, `T_(2) cos beta = mg` …(ii)
and `T_(2) cos beta = mg` …(iii)
or `sin beta = cos beta rArr beta = 45^(@)`
`T_(1) cos alpha = mg + T_(2) cos beta`
`= mg + sqrt(2) mg ((1)/(sqrt(2)))`
or `T_(1) cos alpha = 2 mg` ...(iv)
`T_(1) sin alpha = T_(2) sin beta = sqrt(2) mg ((1)/(sqrt(2)))`
`T_(1) sin alpha = mg` ...(v)

From Eqs. (iv) and (v), we get
`tan alpha = (1)/(2)` and `T_(1) =sqrt(5) mg`
`tan beta = tan 45^(@) = 1` and `T_(2) =sqrt(2) mg`
`:. tan beta = 2 tan alpha` and `sqrt(2) T_(1) =sqrt(5) T_(2)`