The tension at any point of the string AB will be same if

Options

A. `(sqrt(T_(1)^(2)+T_(2)^(2)))/(2)`
B. `(sqrt(T_(1)^(2)-T_(2)^(2))/(2))`
C. `(sqrt(T_(1)^(2)+T_(2)^(2))/(2))`
D. none of the above.

Answer

C

Explanation

Referring to the free budy diagrams of the bob we obtain,
`T_(1)-mg=ma`
`rArr T_(1)=m(g+a) " " ...(1)`
`mg-T_(2)=ma`
`rArr T_(2)=m(g-a)" " ...(2)`
When the cage moves horizontally with an acceleration a, the tension be T.

From the free body diagram, T `sin theta=ma`
And `T cos theta-mg =0`
`rArr (T sin theta)^(2)+(T-cos theta)^(2)=(ma)^(2)+(mg)^(2)`
`rArr T^(2)=m^(2)(g^(2)+a^(2)) " " ..(3)`
From (1) and (2)
`((T_(1))/(m))^(2)+((T_(2))/(m))^(2)=(g+a)^(2)+(g-a)^(2)`
`rArr (T_(1)^(2)+T_(2)^(2))/(2)=(g^(2)+a^(2))m^(2)" " ...(4)`
Equations (3) and (4), we obtain
`T^(2)=(1)/(2) (T_(1)^(2)+T_(2)^(2)) rArr T=sqrt((T_(1)^(2)+T_(2)^(2))/(2))`