Approach

The total downward shift of Disc A is the sum of the compression in the spring ($x_s$) and the compression in the rubber pad ($x_p$). Since the components are in series, the force is the same through the spring and the pad. We have the characteristic curve for the pad (Force vs $x_p$) and a linear equation for the spring. We can solve for the equilibrium state by finding the intersection of these two relationships.

1. Force and Displacement Relations

When equilibrium is established:

• The force exerted by the load $D$ (mass $M$) is transmitted through the spring to the rubber pad.
• $F = Mg$
• This force causes a compression $x_p$ in the pad (governed by the graph) and a compression $x_s$ in the spring ($F = k x_s$).

The total downward shift of Disc A is given as 10 cm. This shift is the sum of the compressions: $$ x_{\text{total}} = x_s + x_p = 10 \text{ cm} $$ $$ x_s = 10 – x_p \quad (\text{units in cm}) $$

2. Graphical Solution

We have two equations for the Force $F$: 1. From the spring: $F = k x_s$. Given $k = 100 \text{ N/m} = 1 \text{ N/cm}$. Substituting $x_s$: $$ F = 1 \cdot (10 – x_p) = 10 – x_p $$ 2. From the pad: $F$ follows the curve given in the graph.

We plot the line $F = 10 – x$ on the provided graph (a straight line connecting $(0, 10)$ on the y-axis and $(10, 0)$ on the x-axis) and find where it intersects the curve.

Visual inspection of the intersection point on the grid:

• The line intersects the curve at $x_p \approx 7$ cm.
• At this x-value, the Force $F \approx 3$ N.

Checking consistency: If $x_p = 7$, then $F = 10 – 7 = 3$ N. This point lies on the characteristic curve shown.

3. Calculating Mass

The equilibrium force is $F = 3$ N. Since $F = Mg$: $$ 3 = M(10) $$ $$ M = 0.3 \text{ kg} = 300 \text{ g} $$