At a distance of 5 cm and 10 cm outward from the surface of a uniformly charged solid sphere, the potentials are 100 V and 75 V, respectively. Then

V=14πε0Q(R+d)R is radius, and d is distance from surface100=KQ(R+5)×10−2   .........(i)75=KQ(R+10)×10−2    ........(ii)From (i) and (ii), R = 10 cm and Q=503×10−10CPotential at surface is  V0=kQR=9×109×5010×10−2×3×10−10=150V Electric field on surface is E0=kQR2=1500Vm−1Potential at the center is VC=32V0=225V