For the arrangment shown in the fig. the tension in the string is [Given: `tan^(-1)(0.8)=39^(@)`].

Options

A. 6N
B. `6.4N`
C. `0.4N`
D. zero

Answer

D

Explanation

If alpha represent angle of repose, then, `tanalpha=0.8`
`alpha=tan^(-1)(0.8)=39^(@)`
The given angle of inclination is less is less than the angle of repose. So, the `1kg` block has no tendency to move. [note that `mgsintheta` is exactly balanced by the force of friction. So, `T=0` .]