Answer
The Tension in the cord $ = 3.7 \times 10^{-3}$ $N$
Work Step by Step
To find the tension in cord ($ T$), we balance the forces in vertical direction.
So, the forces in vertical direction are:
i) Vertical component of tension (upwards) $= Tcos37^{\circ} = \frac{4}{5}T$
ii) Weight (downwards) $=mg$
Assuming upward direction positive, writing the force equation with proper sign convention gives:
$\frac{4}{5}T-mg=ma$
Since the sphere is at rest, net acceleration is zero.
So the above equation can becomes:
$\frac{4}{5}T-mg=0$
$\frac{4}{5}T=mg$
Substituting known values in the equation and solving gives :
$\frac{4}{5}T =[3\times 10^{-4} kg]\times [9.8$ $m/s^{2}]$
$T=3.675\times 10^{-3} N$
$T \approx 3.7 \times 10^{-3}$ $N$
