Chapter 4 - Force and Motion - Learning Path Questions and Exercises - Exercises - Page 137: 50

$20.5N$

Vertically, $\frac{1}{2}mg=T_{2}cos30^{\circ}=T_{1}sin45^{\circ}$ $T_{2}=\frac{mg}{2cos30^{\circ}}=\frac{10\times9.8}{2cos30^{\circ}}=57N$ $T_{AB}=T_{1}cos45^{\circ}-T_{2}sin30^{\circ}$ $T_{1}cos45^{\circ}=\frac{mg}{2}=49$ Therefore, $T_{AB}=49-28.5=20.5N$