Q1: The slope correction for a length of 30 m along a gradient of 1 in 20, is

ANS:A - 3.75 cm

Hypotenusal allowance.

Slope correction will be 75/(n^2)links (for 30 m chain) , where 1:n is the slope.
Cs = 75/(20 ^2) links,
= 0.1875 links of 30 m chain,
= 0.1875*20 (one links =20cm),
= 3.75 cm => Answer. Slope Correction Formula=h^2/2l ---> (1)
Slope=1:20 =1/20 = 0.05.

h= slope * l= (1/20)(30) = 1.5m.

Now eq(1) becomes,
(1.5)^2/2(30) = 0.0375m.
1m = 100cm.
So,
0.0375(100cm) = 3.75cm. For 20 m chain slope correction is 1m.
Therefore 1/20 * 30 = 1.5,
So (1.5 * 1.5)/2 * 30 = 0.0375m,
ie 0. 0.375 * 100 = 3.75. Slope correction =h^2/2L.
So, 1:20.
20m=1m.
1m=1/20m.
30m=1/20*30=1.5.
H= 1.5.
(1.5) ^2/2*30=0.0375m.
= 3.75 cm Answer. h=1 in 20 or 0.5 in 10.
for l = 30 m. h=1.5.
So, that = h2/2l (slope correction formula)= (1.5)2/2*30 = 0.0375m. x 100 = 3.75cm.