Given, m1​=10 kg,m2​=5 kg,ω=10rad/sr=0.3 m,r=0.124 m∴r2​=r−r1​=0.176 m​(a) Masses m1​ and m2​ are at rest with respect to rotating table. Let f be the friction between mass m1​ and table.Free body diagram of m1​ and m2​ with respect to ground is shown below.T=m2​r2​ω2…(i)Since, m2​r2​ω2Tand friction on m1​ will be inward (toward centre)f+T=m1​rω2…(ii)From Eqs. (i) and (ii), we getf=m1​r1​ω2−m2​r2​ω2…(iii)=(m1​r1​−m2​r2​)ω2=(10×0.124−5×0.176)(10)2 N=36 N​Therefore, frictional force on m1​ is 36 N (inwards)(b) From Eq. (iii)f=(m1​r1​−m2​r2​)ω2Masses will start slipping when this force is greater than fmax​ or(m1​r1​−m2​r2​)ω2>fmax​>μm1​g∴ Minimum value of ω isωmin​=m1​Γ1​−m2​r2​μm1​g​​=10×0.124−5×0.1760.5×10×9.8​​or ωmin​=11.67rad/s(c) From Eq. (iii), frictional force f=0when, m1​r1​=m2​r2​orr2​r1​​=m1​m2​​=105​=21​and r=r1​+r2​=0.3 m∴r1​=0.1 m and r2​=0.2 mi.e. mass m2​ should placed at 0.2 m and m1​ at 0.1 m from the centre O.