Step-by-step Derivation:

1. Principle of Conservation of Mechanical Energy Since the object rolls up the curved surface without slipping, the static friction does no work. Assuming no other non-conservative forces like air resistance, the total mechanical energy of the object is conserved. We can equate the initial total energy ($E_i$) at the bottom to the final total energy ($E_f$) at the maximum height. $E_i = E_f$

2. Initial Energy ($E_i$) At the initial position (let's set the reference height $h=0$), the object has both translational and rotational kinetic energy. The potential energy is zero.

   • Translational Kinetic Energy: $KE_{trans} = \frac{1}{2}mv^2$
   • Rotational Kinetic Energy: $KE_{rot} = \frac{1}{2}I\omega^2$ The total initial energy is: $E_i = KE_{trans} + KE_{rot} + PE_i = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 + 0$

3. Final Energy ($E_f$) At the maximum height $h_{max}$, the object momentarily comes to rest. Therefore, its final translational and rotational velocities are zero, and its kinetic energy is zero. The energy is purely gravitational potential energy. $E_f = KE_f + PE_f = 0 + mgh_{max}$

4. Applying Energy Conservation Equating the initial and final energies: $\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh_{max}$

5. Using Rolling Condition and Moment of Inertia For an object rolling without slipping, the linear velocity $v$ and angular velocity $\omega$ are related by $v = R\omega$, where $R$ is the radius of the object. So, $\omega = v/R$. The moment of inertia $I$ for a body of mass $m$ and radius $R$ can be expressed in a general form as $I = k m R^2$, where $k$ is a dimensionless constant that depends on the shape of the object.

6. Solving for the Shape Factor (k) Substitute $I = k m R^2$ and $\omega = v/R$ into the energy conservation equation: $\frac{1}{2}mv^2 + \frac{1}{2}(k m R^2) \left( \frac{v}{R} \right)^2 = mgh_{max}$ $\frac{1}{2}mv^2 + \frac{1}{2}k m R^2 \frac{v^2}{R^2} = mgh_{max}$ $\frac{1}{2}mv^2 + \frac{1}{2}k mv^2 = mgh_{max}$ Factor out $\frac{1}{2}mv^2$ from the left side and cancel $m$ from both sides: $\frac{1}{2}v^2(1+k) = gh_{max}$

7. Using the Given Maximum Height The problem states that the maximum height reached is $h_{max} = \frac{3v^2}{4g}$. Substitute this value into the equation: $\frac{1}{2}v^2(1+k) = g \left( \frac{3v^2}{4g} \right)$ Now, we can cancel $v^2$ and $g$ from both sides: $\frac{1}{2}(1+k) = \frac{3}{4}$ Multiply both sides by 4 to solve for $k$: $2(1+k) = 3$ $2 + 2k = 3$ $2k = 1$ $k = \frac{1}{2}$

8. Identifying the Object Now we compare this value of $k$ with the known values for the objects listed in the options:

   • A: Ring: $I = mR^2 \implies k = 1$
   • B: Solid Sphere: $I = \frac{2}{5}mR^2 \implies k = 2/5$
   • C: Hollow Sphere: $I = \frac{2}{3}mR^2 \implies k = 2/3$
   • D: Disc: $I = \frac{1}{2}mR^2 \implies k = 1/2$ The calculated value $k = 1/2$ corresponds to a disc.

Therefore, the object is a disc.