TGPSC HMWSSB Manager 2015 Paper 2 · Question 16

Question 16 of the TSPSC Manager (Engineering), HMWSSB, 2015: Paper 2 (Civil Engineering) (1 November 2015). Try it first, then open the answer and explanation.

1. Manager 2015 Paper 2 (Civil) · Q16 · Transportation Engineering · Highway Engineering

   Find out the rate of change of centrifugal acceleration for a design speed of 75 kmph, using IRC recommended formula.

   1. 0.63 m/sec³
   2. 0.53 m/sec³
   3. 0.48 m/sec³
   4. 0.73 m/sec³
   Show answer

   Answer: (2) 0.53 m/sec³

   IRC formula: C = 80/(75 + V) = 80/(75 + 75) = 0.53 m/sec³. This lies within the IRC limits of 0.5 to 0.8 m/sec³, so it is adopted.

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