Dear student, here the angular acceleration of both the shafts will be same

so, α1 = α2

=> ω1*r1 = ω2*r2

where  r1  is the radius of pulley fitted on main spindle and  r2  is the radius of  pulley fitted on machine spindle.

=> 250*(d/2) = 25*(20/2)

=> d = 25*20/250 = 2 c.m.

Therefor diameter of the pulley to be fitted on main spindle is 2c.m.

All the best

Hope this helps

Hiii amaresh ....the answer to your question is.....

Here the angular acceleration of both the shafts will be same

so, α1 = α2
=> ω1*r1 = ω2*r2
=> 250*(d/2) = 125*(20/2)
=> d = 125*20/250 = 10cm

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