# The required slope correction for a length of 60 m along a gradient of 1 in 20 is:

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The required slope correction for a length of 60 m along a gradient of 1 in 20 is:
1. 7.5 m
2. 7.5 cm
3. 0.75 cm
4. 5.75 cm

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Correct Answer - Option 2 : 7.5 cm

Explanation

Given,

Length = 60 m

$\tan {\rm{\theta }} = \frac{1}{{20}} = \frac{h}{{60}}$

⇒ h = 3 m

We know slope correction is given by

${\rm{Slope\;correction}} = \frac{{{{\rm{h}}^2}}}{{2{\rm{L}}}}$

${{\rm{C}}_{\rm{S}}} = \frac{{{{\left( {3} \right)}^2}}}{{2 \times 60}} = 0.075{\rm{\;m}}$

CS = 0.075 m = 7.5 cm (Negative)

Note:

Correction due to slope will always be negative because slope distance is greater than corresponding horizontal distance.