Answer
$\triangle T \approx -0.01648$ mg
The tension is negative, decreased.
Work Step by Step
Write the differential form such as:
$dT=\dfrac{\partial T}{\partial r} dr + \dfrac{\partial T}{\partial R} dR=[\dfrac{-4mgRr}{(2r^2+R^2)^2} ]dr + [\dfrac{mg(2R^2-r^2)}{(2r^2+R^2)^2} ] dR$
$\triangle T \approx [\dfrac{-4mg (Rr)}{(2r^2+R^2)^2} ] \times \triangle r + [\dfrac{mg(2R^2-r^2)}{(2r^2+R^2)^2} ] \times \triangle R$
After plugging the given value.
$\triangle T \approx [\dfrac{-4mg \times (3) \times (0.7)}{(2(0.7)^2+(3)^2)^2} ] \times (0.1) + [\dfrac{mg \times (2(0.7)^2-(3)^2)}{(2(0.7)^2+(3)^2)^2} ] \times (0.1) \approx -0.01648$ mg
This means that the tension is negative and thus, , decreased.
