Step 1: Formula for Intermediate Sight Distance (ISD).
The intermediate sight distance (ISD) is taken as twice the stopping sight distance (SSD).
\[
ISD = 2 \times SSD
\]
Step 2: Substitution.
Given that SSD = 80 m, we have:
\[
ISD = 2 \times 80 = 160 \, \text{m}.
\]
Step 3: Correction with respect to Passing Sight Distance.
As per IRC recommendations, ISD is the average of SSD and PSD when both are provided:
\[
ISD = \frac{SSD + PSD}{2} = \frac{80 + 300}{2} = \frac{380}{2} = 190 \, \text{m}.
\]
Step 4: Conclusion.
Hence, the intermediate sight distance is 190 m, so the correct answer is (A).
Match LIST-I with LIST-II (adopting standard notations):\[\begin{array}{|c|c|} \hline \textbf{LIST-I (Parameter)} & \textbf{LIST-II (Formula)} \\ \hline \\ \text{A. Cubic parabola equation} & \text{IV. $\dfrac{X^3}{6RL}$} \\ \\ \hline \\ \text{B. Shift in transition curve} & \text{II. $\dfrac{L^2}{24R}$} \\ \\ \hline \\ \text{C. Length of valley curve} & \text{III. $\dfrac{N S^2}{(1.50 + 0.035S)}$} \\ \\ \hline \\ \text{D. Length of summit curve} & \text{I. $\dfrac{N S^2}{4.4}$} \\ \\ \hline \end{array}\] Choose the most appropriate match from the options given below:
Which of the following parameters are required for the design of a transition curve for a highway system?
(A) Rate of change of grade
(B) Rate of change of radial acceleration
(C) Rate of change of super elevation
(D) Rate of change of curvature
Choose the most appropriate answer from the options given below:
A weight of $500\,$N is held on a smooth plane inclined at $30^\circ$ to the horizontal by a force $P$ acting at $30^\circ$ to the inclined plane as shown. Then the value of force $P$ is:
A steel wire of $20$ mm diameter is bent into a circular shape of $10$ m radius. If modulus of elasticity of wire is $2\times10^{5}\ \text{N/mm}^2$, then the maximum bending stress induced in wire is: