Correct Answer - A::C
In this case, three forces are acting on the object:
1. tension (T)
2. weight (mg) and
3. applied force (F)
Using work-enegy theorem.
`W _ (n et)=DeltaKE`
or `W_(T) + W_(mg) + W_(F)=0` ...(i)
as `DeltaKE=0`
because `K_(i) = K_(f) = 0`
Further, `W_T=0,` . as tension is always perpendicular to displacement.
`W_(mg) =-mgh`
or `w_(mg)=-mgl(1-costheta)`
Substituting these values in Eq. (i), we get
`W_(F) = mgl(1-costheta)`.