Step 1: Understand what the graph is showing.
The lower part of the figure plots flow rate against density for two conditions: the "Unconstrained flow" curve (applies to a wide section with no bottleneck) and the "Constrained flow" curve (applies to a narrower, lower-capacity section, like S2). Both curves rise to a peak (the maximum possible flow, called capacity) and then fall as density keeps increasing and speed drops.
A single horizontal dashed line passes through the four labeled points 1, 2, 3, 4. Since all four lie on this SAME horizontal line, they all represent the SAME flow rate, just reached at different densities on the two different curves.
Step 2: Use the given state of S1.
Section-1 (wide, unconstrained) is given to be at density \(D_2\), on the ascending, uncongested branch of the unconstrained curve (point 2). This fixes the flow rate carried by the whole road, since vehicles are conserved along the road and this same flow rate must, in steady conditions, also pass through every section downstream.
Step 3: Work out the state at the bottleneck, S2.
S2 is explicitly stated to be capacity constrained relative to the flow from S1, meaning S2 physically cannot carry this flow rate while staying on the free-flowing (low density) branch of ITS OWN (constrained) curve.
So traffic backs up as it approaches S2, forming a queue, and S2 is forced onto the congested (high density, slow speed) branch of the constrained curve, at density \(D_4\) (point 4), rather than the free-flow branch \(D_1\). Vehicles get through S2 only by packing closer together and moving more slowly.
Step 4: Work out the state right after the bottleneck, S3.
As vehicles squeeze through S2 and enter S3, they leave the bottleneck at the MAXIMUM rate S2 can discharge, which is the critical (capacity) flow value, right at the peak of the flow-density curve.
So immediately downstream, S3 is observed at the critical density \(D_3\) (point 3, the peak), the density at which flow is maximized.
Step 5: Work out the state further downstream, S4.
S4 is unconstrained and located further from the pinch point, giving the discharged traffic room to spread out and speed back up. Carrying the same flow rate but now on the free-flowing branch of the unconstrained curve, S4 settles at the lowest density, \(D_1\) (point 1).
Step 6: Assemble the answer and rule out other options.
So S2 is at \(D_4\), S3 is at \(D_3\), and S4 is at \(D_1\), matching option (D).
Option (A) wrongly keeps S4 at \(D_2\), ignoring the further recovery in speed and drop in density downstream. Option (B) wrongly puts S2 at \(D_3\) (the peak, as if S2 were running exactly at capacity rather than congested beyond it) and reverses S3 and S4. Option (C) swaps S3 and S4.
Final Answer:
S2 - D4 ; S3 - D3 ; S4 - D1, option (D).