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Physics Question on Relative Velocity

A solid sphere of mass 1 kg1 \,kg rolls without slipping on a plane surface Its kinetic energy is 7×10−3J7 \times 10^{-3} J. The speed of the centre of mass of the sphere is ___cms−1cm s ^{-1}.

Answer

12mv2+12mω2=7×10−3\frac1{2}mv^2 +\frac1{2}m{ω}^2 = 7 \times 10^{-3}

12mv2+12(25MR2)(VR)2=7×10−3\frac1{2}mv^2 +\frac1{2} (\frac2{5}MR^2)({\frac{V}{R}})^2 = 7 \times 10^{-3}

12MV2+[1+25]=7×10−3\frac1{2}MV^2 +[1+\frac2{5}] = 7 \times 10^{-3}

12(1)V2+75=7×10−3\frac1{2}(1)V^2 +\frac7{5} = 7 \times 10^{-3}

V2=10−2V^2 = 10^{-2}

V=10−1V = 10^{-1}

V=0.1 ms−1=10 cm−1V= 0.1\,ms^{-1} = 10\,cm^{-1}

So, The correct answer is 10.