Let us draw a free body diagram of the block with respect to the disc (with pseudo force ma )N1​=mg= normal reaction on the block from groundN2​=masinθ= Normal reaction on the block from AB side of grooveNow, ar​= acceleration of block w.r.to disc=mmacosθ−f1​−f2​​(f1​=μN1​ and f2​=μN2​)∴ar​=mmacosθ−μN1​−μN2​​​Substituting the values of N1​ and N2​, we getar​​=acosθ−μg−μasinθ=25×54​−52​×10−52​×25×53​=10 m/s2​