Given, the coefficient of friction between A and B,µ1=0.35 The coefficient of friction between B and surface, µ2=0.5 Mass of block A,ma=100‌kg Mass of block B,mb=300‌kg Frictional force between blocks A and B, f1=µ1m2g =0.35×100×9.8 =0.35×980 = 343 N Frictional force between block B and surface, f2=µ2(mα+mb)g =0.5(100+300)9.8 =0.5×4×980=1960N Minimum force required to move block B, F=f1+f2=343+1960 ⇒ = 2303 N Which is nearest to 2350N. Hence, option (c) is correct.