For the arrangement shown in the fig. the tension in the string is [Given: ${\tan }^{-1}(0.8) = {39}^0$]

If $\mathrm{\alpha }$ represents angle of repose, then, tan$\mathrm{\alpha }$ = 0.8
$\mathrm{\alpha }$ = ${\tan }^{-1}(0.8) = {39}^0$
The given angle of inclination is less is less than the angle of repose. So, the I kg block has no tendency to move. [Note that mg sin $\mathrm{\theta }$ is exactly balanced by the force of friction. So, T. = 0.]