Correct option is (c) 615 N

To explain: Diameter of effort wheel, D = 400 mm

Diameter of longer axle, d1 = 2 x 150 = 300 mm

Diameter of the smaller axle, d2 = 2 x 100 = 200 mm

Diameter of the rope, dr = 10 mm

therefore, V.R. = 2(D + dr )/(d1 + dr ) – (d2 + dr )

= 2(400 + 10)/ (300 + 10) – (200 + 10)

= 820/100 = 8.2

Effort, P = 100 N

ȵ = 75%

Let W = load which can be lifted by the machine

ȵ = M.A./V.R.

0.75 = W/P x 8.2

W = 0.75 x 100 x 8.2 = 615 N.