A small roller of diameter 20 cm has an axle of diameter 10 cm (see figure below on the left). It is on a horizontal floor and a meter scale is positioned horizontally on its axle with one edge of the scale on top of the axle (see figure on the right). The scale is now pushed slowly on the axle so that it moves without slipping on the axle, and the roller starts rolling without slipping. After the roller has moved 50 cm, the position of the scale will look like (figures are schematic and not drawn to scale)

Let

outer radius of the roller $$R = 10 \text{ cm},$$
radius of the axle              $$r = 5 \text{ cm}.$$

When the roller rolls without slipping on the floor, the distance moved by its centre is related to its angular displacement $$\theta$$ by the rolling‐condition

$$s = R\,\theta \qquad -(1)$$

The scale is pushed on the axle without slipping, so the length through which the scale slides over (or “unwinds” from) the axle equals the arc length traced at the axle surface:

$$\ell = r\,\theta \qquad -(2)$$

The roller is stated to have moved $$s = 50 \text{ cm}.$$ Substituting this in $$(1)$$ gives the common angular displacement:

$$\theta = \frac{s}{R} = \frac{50}{10} = 5 \text{ rad}.$$

From $$(2)$$, the relative slide of the scale over the axle is

$$\ell = r\,\theta = 5 \times 5 = 25 \text{ cm}.$$

Initial arrangement (at $$t = 0$$): the leftedge (0-cm mark) of the metre-scale is exactly on the topmost point of the axle. Hence the axle top touches the 0-cm mark.

Final arrangement (after the motion): during the 50 cm travel, the axle top has come in contact with the 25-cm mark of the scale (because the scale has slipped over the axle by 25 cm). Therefore, relative to the axle,

• the left edge of the scale is now $$25 \text{ cm}$$ to the left of the axle centre, and
• the right edge is $$100 - 25 = 75 \text{ cm}$$ to the right of the axle centre.

Since the axle centre itself has shifted 50 cm to the right, the absolute positions measured from the initial axle position are

left edge  : $$50 - 25 = 25 \text{ cm},$$
right edge : $$50 + 75 = 125 \text{ cm}.$$

Thus, after the roller has travelled 50 cm:

• about one-quarter (25 cm) of the scale projects behind the roller, and
• about three-quarters (75 cm) project ahead of it, with the axle situated 25 cm from the left end of the scale.

The only figure that shows the axle a quarter of the scale-length from the left end and three-quarters from the right end is Option E.

Option E which depicts 25 cm of the scale behind the roller and 75 cm in front of it is therefore the correct choice.