Q1: Offsets are measured with an accuracy of 1 in 40. If the point on the paper from both sources of error (due to angular and measurement errors) is not to exceed 0.05 cm on a scale of 1 cm = 20 m, the maximum length of offset should be limited to

ANS:B - 28.28 m

Actually, the maximum length of offset 1/40 cm it means 0.025mm.

But the above question's maximum length of offset 0 .05cm. It means 0.5 mm or 1/20 cm
So, the maximum length of offset=s/20*r/√2.
S = 20
r = 40.
So, 20/20*40/√2.
= 28.32m. Formula : Limiting error = (L/sin@) * (1/r) * (1/S).

∠ = 45' for max limiting offset.
1/r = 1:40 = accuracy of measurement.
1/S= 1:20 = scale.
Limiting error = 0.05 cm.

L = 20*(2)^0.5.
L= 28.28 m. The displacement due to angular error =Lsin.
and it should be equal to displacement due to Linear error = L/r.
Corresponding displacement on paper=√2(L/r)(1/S)= √2(Lsin-/S).
Corresponding if a limit of accuracy in plotting is 0.05cm then,
√2(L/r)(1/S)=.05,
L= (.05x40x20)/√2.
= 28.28.
Where S=1 in accuracy.
r=1 in scale. The center-to-center dimension for a 45-degree bend is equal to the desired size of the offset times the cosecant 1.414.

A cosecant is used to determine the distance between the centers of the two bends used to make an offset. A 45-degree angle has a cosecant of 1.414.