Correct Answer - A::B::D
Initially, the spring has an elongation, `DeltaL=mg//k`, where m is the mass of hanging block and k is the force constant of the spring. It means, initially spring has some strain energy. When the boy pushes the block upwards, the potential energy of the block increases and work is done by the boy during lifting of the block but spring becomes strain free. It means, initially, strain energy stored in spring is lost. Hence, the work done by the boy will be equal to increase in gravitational potential energy of the block minus loss in strain energy stored in the spring. Hence, only (c) is correct, i.e., all other options (a), (b), (d) and (e) are not correct.