Right option is (a) 61.693 kW

For explanation I would say: Given : d1 = 300 mm or r1 = 150 mm ; d2 = 200 mm or r2 = 100 mm ; p = 0.1 N/mm^2 ; μ = 0.3 ; N = 2500 r.p.m. or ω = 2π × 2500/60 = 261.8 rad/s

Since the intensity of pressure ( p) is maximum at the inner radius (r2), therefore for uniform wear,

p.r2 = C or C = 0.1 × 100 = 10 N/mm

We know that the axial thrust,

W = 2 π C (r1 – r2) = 2 π × 10 (150 – 100) = 3142 N

and mean radius of the friction surfaces for uniform wear,

R = r1 + r2/2 = 150 + 100/2 = 125 mm = 0.125m

We know that torque transmitted,

T = n.μ.W.R = 2 × 0.3 × 3142 × 0.125 = 235.65 N-m …( n = 2,for both sides of plate effective)

∴ Power transmitted by a clutch,

P = T.ω = 235.65 × 261.8 = 61 693 W = 61.693 kW.