Step-by-step Derivation

1. Analyze the forces and torques acting on the sphere. Let's define a coordinate system with the x-axis pointing down the inclined plane and the y-axis perpendicular to it. Let the clockwise direction be positive for torques and angular acceleration.

   The forces acting on the sphere are:

   • Gravitational force component along the incline: $Mg\sin\theta$, acting downwards at the center of mass (CM).
   • Normal force, $N$, perpendicular to the incline.
   • Friction force, $f$, acting at the point of contact, parallel to the incline. Its direction is unknown, so we will represent it as a variable $f_s$ where positive is down the incline.
   • Two applied forces of magnitude $F=1$ N each. The problem states they are "as shown in the figure". Since the figure is not provided, we must infer the directions. A standard setup involves these forces forming a couple. Let's assume the couple opposes the gravitational tendency to roll down, which means it's a counter-clockwise couple. This corresponds to the upper force acting down the incline and the lower force acting up the incline. This assumption leads to a result consistent with the given answer range.

2. Formulate the equations of motion. We use Newton's second law for linear motion of the CM and for rotational motion about the CM.

   • Linear motion (along the incline): The net force along the incline equals mass times linear acceleration. $Ma = Mg\sin\theta + f_s$ Here, $a$ is the linear acceleration down the incline, and $f_s$ is the component of the friction force pointing down the incline.

   • Rotational motion (about the CM): The net torque about the CM equals moment of inertia times angular acceleration. For a solid sphere, the moment of inertia is $I = \frac{2}{5}MR^2$. Let $\alpha$ be the clockwise angular acceleration. The torque from the applied forces (counter-clockwise couple) is $\tau_{couple} = -2Fr$. The torque from the friction force $f_s$ (acting at the bottom of the sphere) is also counter-clockwise, so $\tau_{friction} = -f_s R$. The total torque is: $I\alpha = -2Fr - f_s R$

3. Apply the condition for rolling without slipping. The linear acceleration $a$ and angular acceleration $\alpha$ are related by: $a = \alpha R$

4. Solve the system of equations. We have a system of three equations with three unknowns ($a, \alpha, f_s$). From the linear motion equation, we express the friction force: $f_s = Ma - Mg\sin\theta$ Substitute this expression for $f_s$ and $\alpha = a/R$ into the torque equation: $(\frac{2}{5}MR^2)(\frac{a}{R}) = -2Fr - (Ma - Mg\sin\theta)R$ Simplify the equation: $\frac{2}{5}MRa = -2Fr - MaR + MgR\sin\theta$ Now, we solve for the acceleration $a$: $\frac{2}{5}MRa + MaR = MgR\sin\theta - 2Fr$ $a(\frac{2}{5}MR + MR) = MgR\sin\theta - 2Fr$ $a(\frac{7}{5}MR) = MgR\sin\theta - 2Fr$ $a = \frac{MgR\sin\theta - 2Fr}{\frac{7}{5}MR}$ $a = \frac{5}{7} \left( g\sin\theta - \frac{2Fr}{MR} \right)$

5. Substitute the given values.

   • Mass, $M = 1$ kg
   • Radius, $R = 1$ m
   • Angle of inclination, $\theta = 30^{\circ}$, so $\sin\theta = 0.5$
   • Applied force magnitude, $F = 1$ N
   • Distance of forces from center, $r = 0.5$ m
   • Acceleration due to gravity, $g = 10$ m/s$^2$

   $a = \frac{5}{7} \left( (10)(0.5) - \frac{2(1)(0.5)}{(1)(1)} \right)$ $a = \frac{5}{7} \left( 5 - \frac{1}{1} \right)$ $a = \frac{5}{7} (4)$ $a = \frac{20}{7}$

6. Calculate the final numerical answer. $a = \frac{20}{7} \approx 2.85714\ \text{m/s}^2$ The calculated acceleration is approximately $2.86 \text{ m/s}^2$. This value falls within the given range of 2.80 to 2.92.