After applying the correction for elevation and temperature, the runway length is 700 m. The corrected runway length (in m) for an effective gradient of 1% is __________ (round off to the nearest integer).

Question

cdquestions Admin · Accepted Answer

Given: Corrected runway length after applying elevation and temperature corrections: $$ l = 700 { m},\quad {Gradient} = 1\% $$ According to standards, for every 1% gradient, the runway length increases by 20%. $$ {Correction} = 700 \times \frac{20}{100} = 140 $$ $$ {Corrected length} = 700 + 140 = 840 { m} $$ $$ {Or simply: } 700 \times 1.2 = 840 { m} $$ Hence, the final corrected length is $ \boxed{840 { m}} $.

Answer

Answer

Answer

Answer

After applying the correction for elevation and temperature, the runway length is 700 m.
The corrected runway length (in m) for an effective gradient of 1% is __________ (round off to the nearest integer).

Show Hint

The Correct Option is A

Solution and Explanation

Top Questions on Design of flexible and rigid pavement using IRC codes

Questions Asked in GATE CE exam

After applying the correction for elevation and temperature, the runway length is 700 m.The corrected runway length (in m) for an effective gradient of 1% is __________ (round off to the nearest integer).