6. On an urban road, the free mean speed was measured as 70 kmph and the average spacing between the vehicles under jam condition as 7.0 m. The speed-flow-density equation is given by
   

   U = Usf		1 -	k		and q = UK
   			k1

   
   where, U → space-mean speed (kmph); U_sf → free mean speed (kmph); k → density (veh/km); k_j → jam density (veh/ km); q → flow (veh/hr). The maximum flow (veh/hr) per lane for this condition is equal to
   

1. 1. 2000
      

   
   
   

   2. 2500
      

   
   
   

   3. 3000
      

   
   
   

   4. None of these

   
   
   

---

1. View Hint View Answer Discuss in Forum

   u = usf		1 -	k
   			kj

   
   q = uk
   

   q	= usf		1 -	k
   k				kj

   
   

   ⇒ q = usf		1 -	k		k
   			kj

   
   Diff. w.r.t.k,
   

   dq	= 0 ⇒ usf		1 -	2k		= 0
   dk				kj

   
   

   ⇒ k =	kj
   	2

   
   

   ∴ qmax = usf =		1 -	1		×	kj	= usf ×	kj
   			2			2		4

   
   

   = 70 ×	1000	= 2500
   	7 × 4

   
   

   Correct Option: B

   u = usf		1 -	k
   			kj

   
   q = uk
   

   q	= usf		1 -	k
   k				kj

   
   

   ⇒ q = usf		1 -	k		k
   			kj

   
   Diff. w.r.t.k,
   

   dq	= 0 ⇒ usf		1 -	2k		= 0
   dk				kj

   
   

   ⇒ k =	kj
   	2

   
   

   ∴ qmax = usf =		1 -	1		×	kj	= usf ×	kj
   			2			2		4

   
   

   = 70 ×	1000	= 2500
   	7 × 4

   
   

---

7. In signal design as per Indian Roads Congress specifications, if the sum of the ratios of normal flows to saturation flow of two directional traffic flow is 0.50 and the total lost time per cycle is 10 seconds, the optimum cycle length in seconds is
   

1. 1. 100
      

   
   
   

   2. 80
      

   
   
   

   3. 60
      

   
   
   

   4. 40

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Optimum cycle length =	1.5L + 5
   	1 - 4

   
   

   =	1.5 × 10 + 5	= 40 s
   	1 - 0.5

   Correct Option: D

   Optimum cycle length =	1.5L + 5
   	1 - 4

   
   

   =	1.5 × 10 + 5	= 40 s
   	1 - 0.5

---

---

8. If a two-lane national highway and a two-lane state highway intersect at right angles, the number of potential conflict points at the intersection, assuming that both the roads are two-way is
   

1. 1. 11
      

   
   
   

   2. 17
      

   
   
   

   3. 24
      

   
   
   

   4. 32

   
   
   

---

1. View Hint View Answer Discuss in Forum

   No. of conflict points in 2 way 2 lane = 24

   Correct Option: C

   No. of conflict points in 2 way 2 lane = 24

---

9. If the jam density is given as k_j and the free flow speed is given as u_f, the maximum flow for a linear traffic speed density model is given by which of the following options?
   

1. 1. 	1	kj × uf
      	4

      
      

   
   
   

   2. 	1	kj × uf
      	3

      
      

   
   
   

   3. 	3	kj × uf
      	5

      
      

   
   
   

   4. 	2	kj × uf
      	3

   
   
   

---

1. View Hint View Answer Discuss in Forum

   For max. flow, k =	kj
   	2

   
   

   u =	uf
   	2

   
   q = k × u
   

   qmax =	kj	×	uf	=	1	.kj.uf
   	2		2		4

   
   

   Correct Option: A

   For max. flow, k =	kj
   	2

   
   

   u =	uf
   	2

   
   q = k × u
   

   qmax =	kj	×	uf	=	1	.kj.uf
   	2		2		4

   
   

---

---

10. For designing a 2-phase fixed type signal at an intersection having North-South and East-West road where only straight ahead traffic permitted, the following data are available.
    
    Total time lost per cycle is 12 seconds. The cycle length (seconds) as per Webster’s approach is
    

1. 1. 67
      

   
   
   

   2. 77
      

   
   
   

   3. 87
      

   
   
   

   4. 91

   
   
   

---

1. View Hint View Answer Discuss in Forum

   In N-S direction : Higher flow = 1000, saturation flow = 2500
   

   ∴ y1 =	1000	= 0.4
   	2500

   
   In E-W direction : Higher flow = 900, saturation flow = 3000
   

   y2 =	900	= 0.3
   	3000

   
   Y = y₁ + y₂ = 0.7
   Lost time per cycle, L = 2n + R = 12 s
   

   Optimum cycle time =	1.5L + S
   	1 - 4

   
   

   =	1.5 × 12 + 5	= 77
   	1 - 0.7

   Correct Option: B

   In N-S direction : Higher flow = 1000, saturation flow = 2500
   

   ∴ y1 =	1000	= 0.4
   	2500

   
   In E-W direction : Higher flow = 900, saturation flow = 3000
   

   y2 =	900	= 0.3
   	3000

   
   Y = y₁ + y₂ = 0.7
   Lost time per cycle, L = 2n + R = 12 s
   

   Optimum cycle time =	1.5L + S
   	1 - 4

   
   

   =	1.5 × 12 + 5	= 77
   	1 - 0.7

---