A uniform solid sphere of mass m and radius R is projected with speed v0 along a rough horizontal floor with no initial spin. The coefficient of…

On the floor, friction acts at the contact point, so angular momentum about any point on the floor line is conserved: $m v_0 R=m v R+\tfrac25 mR^2\,\omega$ with $v=\omega R$ at the end of slipping. This gives $v_0=\tfrac75 v$, so $v=\tfrac57 v_0$ (independent of $\mu$); the rest of the kinetic energy, $\tfrac27$ of the initial value, is lost as heat during slipping. On the incline rolling is without slipping, so mechanical energy is conserved: $\tfrac12 mv^2+\tfrac12\cdot\tfrac25 mR^2\omega^2=\tfrac{7}{10}mv^2=mgh$. Hence $h=\dfrac{7v^2}{10g}=\dfrac{7}{10g}\cdot\dfrac{25v_0^2}{49}=\dfrac{5v_0^2}{14g}$. Option (A) assumes no loss of energy during slipping and ignores rotation, (B) ignores the rotational energy on the incline, and (C) assumes no energy loss at all during slipping. The correct option is (D).