1. Given dimensions

   • Roller diameter $= 20\text{ cm}$, so outer radius $R = 10\text{ cm}$
   • Axle diameter $= 10\text{ cm}$, so axle radius $r = 5\text{ cm}$

2. Rolling condition for the roller

   The roller rolls on the floor without slipping.

   If the center of the roller moves with speed $V$ and angular speed is $\omega$, then for pure rolling on the ground, $V = R\omega$

3. Condition for the scale on the axle

   The meter scale is pushed slowly and moves on the top of the axle without slipping.

   The speed of the top point of the axle relative to ground is the translational speed of the center plus the rotational speed contribution: $v_{\text{top axle}} = V + r\omega$

   Using $V = R\omega$, $v_{\text{scale}} = R\omega + r\omega = (R+r)\omega$

4. Relating scale speed and roller speed

   Hence, $\frac{v_{\text{scale}}}{V} = \frac{(R+r)\omega}{R\omega} = \frac{R+r}{R}$

   Substituting $R=10$ cm and $r=5$ cm, $\frac{v_{\text{scale}}}{V} = \frac{10+5}{10} = \frac{3}{2}$

   So, $v_{\text{scale}} = \frac{3}{2}V$

5. Displacement relation

   Since the motion is slow and the ratio remains constant, displacement obeys the same ratio: $s_{\text{scale}} = \frac{3}{2}s_{\text{roller}}$

   The roller moves $50$ cm, so $s_{\text{scale}} = \frac{3}{2}\times 50 = 75\text{ cm}$

6. Relative position of the scale with respect to roller

   The roller center (and axle) moves forward by $50$ cm, while the scale itself moves forward by $75$ cm.

   Therefore, relative to the axle, the scale shifts ahead by $75 - 50 = 25\text{ cm}$

   So after the roller has moved $50$ cm, the meter scale will have advanced 25 cm forward relative to the axle.

7. Matching with the figure

   The correct schematic must show the scale displaced by $25$ cm ahead relative to the axle.

   This corresponds to Option C.

Comparison with stored answer: Stored correct answer is C, which matches the derived result.