Q1: Correction per chain length of 100 links along a slope having a rise of 1 unit in n horizontal units, is

ANS:A -

Given: gradient 1 in N.
Wkt gradient = H/L.
Chain length = 100links=20m chain
Gradient = H/L = 1/N.
H = L/N.
Correction slag = H^2/2L.
CS = (L/N)^2/(2*L).
Substitute L =20m.
We will get 100/n^2. or  We know 100 links =20 m chain.
Son horizontal =1 unit
20 m horizontal =20/n
i.e. h=20/n
L=20m.

Slope correction = h^2/2l
So (20/n)^2/2*20
= 100/n^2(ans) Correction per chain length of 100 links =100*α^2, here α is Radians. α=1/n(rise of 1 unit in n horizontal units).

Therefore the correction = 100/n^2.