An offset is measured with an accuracy of 1 in 40. If the scale of plotting is 1 cm=20 m. find the length of the offset so that the displacement of the point on the paper from both sources of error may not exceed 0.25 mm.

Correct option is C

​$$\text{Given:},\text{Accuracy of offset} = 1 \text{ in } 40$$
$$\text{Scale} = 1 \text{ cm} = 20 \text{ m}$$
$$\Rightarrow 1 \text{ mm} = 2 \text{ m}$$
$$\text{Let length of offset} = L \text{ m}$$
$$\text{Error in measurement on ground} = \frac{L}{40}$$
$$\text{Error on paper due to measurement} = \frac{L/40}{2} = \frac{L}{80} \text{ mm}$$
$$\text{Error due to plotting} = \frac{L}{80} \text{ mm}$$
$$\text{Resultant displacement on paper} =$$ $$\sqrt{\left(\frac{L}{80}\right)^2 + \left(\frac{L}{80}\right)^2}= \frac{L\sqrt{2}}{80}$$
$$\text{Given maximum allowable displacement} = 0.25 \text{ mm}$$
​$$\frac{L\sqrt{2}}{80} = 0.25$$​​
$$L = \frac{0.25 \times 80}{\sqrt{2}} = \frac{20}{\sqrt{2}} \text{ m}$$​​​​​​​​​​​