250+ TOP MCQs on Vehicles and Powertrain and Answers

Automobile Engineering Multiple Choice Questions on “Vehicles and Powertrain”.

1. What is running resistance of the vehicle?
a) Rolling resistance
b) Aerodynamic resistance
c) Sum of rolling and aerodynamic resistance
d) Traction force
Answer: c
Clarification: Running resistance is the sum of rolling and aerodynamic resistance. F_running = F_rolling + F _aerodynamic = (μ_k * m_vehicle * g) + (0.5 * ρ * C_drag * A_front * V²). It is the resistance that will affect when the vehicle is running. Running resistance is directly proportional to the mass of the vehicle, frontal area, and velocity. Increasing any of the above parameters will increase running resistance. Tractive force is the effort from the powertrain. Rolling resistance is the frictional resistance that wheels have to overcome. Aerodynamic resistance is the drag force acting against the motion of the vehicle.

2. Suppose the road gradient is G = 10%. What is the corresponding angle (approximate value in radians)?
a) 5 rad
b) 0.1 rad
c) 1 rad
d) 10 rad
Answer: b
Clarification: G = tan (α). For small angles α, tan (α) = α. Hence α = 10% = 0.1 rad. The corresponding angle for road gradient G = 10% is 0.1 rad.

3. If the vehicle mass is 800 kg, what is the gradient force (approximate value in N) caused by the road gradient 10%?
a) 500 N
b) 600 N
c) 700 N
d) 800 N
Answer: d
Clarification: F_gradient = m_vehicle * g * sin (α). For small angle, sin (α) = α. Hence, α = 10% = 0.1 rad. Taking g = 10 m/s², F_gradient = 800*10*0.1 = 800 N.

4. What will happen if the traction force is negative?
a) The vehicle will accelerate
b) The vehicle will decelerate
c) The vehicle will first accelerate and then decelerate
d) The vehicle will run at a constant speed
Answer: b
Clarification: F_net = m_vehicle* a_vehicle = F_traction – F_rolling – F_aerodynamic. If the F_tractionis negative, then F_net will be more negative ⇒ a_vehicle is negative. Therefore the vehicle will decelerate.

5. Suppose the vehicle is running at a constant speed on the flat road with rolling resistance = 100 N and aerodynamic resistance = 100 N, what is the traction force required?
a) 100 N
b) 0 N
c) 200 N
d) 50 N
Answer: c
Clarification: F_net = F_traction – F_rolling – F_aerodynamic – F_gradient = 0 N (constant speed). Flat road ⇒ F_gradient = 0. Hence F_traction = F_rolling + F_aerodynamic + F_gradient = 100 + 100 + 0 = 200 N.

6. Which one of the curves below represents a road gradient of 10%, if the mass of the vehicle is 1 ton?

a) B
b) A
c) D
d) C
Answer: a
Clarification: F_gradient = m_vehicle * g * sin (α). For small angles α, sin (α) = α = G. At G = 10%, sin (α) = 0.1 = α. Hence F_gradient = 1000 * 9.81 * 0.1 = 981 N. Using G = 0% curve as base curve and at speed = 0 Kmph, curve B is higher than the G = 0% curve across the full speed range.

7. What will happen if the vehicle is made 50% heavier, but all other parameters remain the same?
a) The running resistance will decrease
b) The running resistance will increase
c) The running resistance will remain the same
d) The aerodynamic resistance will increase by a factor of 1.5 squared
Answer: b
Clarification: F_running = F_rolling + F_aerodynamic = (μ_k * m_vehicle * g) + (0.5 * ρ * C_drag * A_front * V²). The running resistance is the sum of rolling and aerodynamic resistance. Running resistance is directly proportional to the mass of the vehicle. As the mass of vehicle increases rolling resistance increases and thus, the running resistance increases. The aerodynamic resistance is independent of the vehicle mass.

8. If F_rolling = 150 N, F_aerodynamic = 400 N, F_traction = 600 N, F_gradient = 0 N, what is the net force acting on the vehicle?
a) 100 N
b) 50 N
c) 550 N
d) 750 N
Answer: b
Clarification: F_net = F_traction – F_rolling – F_aerodynamic – F_gradient = 600 – 150 – 400 – 0 = 50 N.

9. If the traction resistance is equal to the total running resistance, then which of the following will happen?
a) The vehicle will accelerate
b) The vehicle will decelerate
c) The vehicle will run at a constant velocity
d) The vehicle will come to rest
Answer: c
Clarification: F_traction = F_running (given). F_running = F_rolling + F_aerodynamic and F_net = F_traction – (F_rolling + F_aerodynamic). F_net = F_traction – F_running = 0 N = m_vehicle * a_vehicle ⇒ a_vehicle = 0 ⇒ V_vehicle = constant. Therefore the vehicle will run at a constant velocity.

10. If the vehicle is running on the road having 10% gradient at a constant speed, then on flat road it will accelerate at approximately 1 m/s2 (All the parameters remain the same).
a) True
b) False
Answer: a
Clarification: F net = F_traction – F_rolling – F_aerodynamic – F_gradient. In case 1: Due to constant speed ⇒ a_vehicle = 0 ⇒ F_net = 0 N and F_traction = F_rolling + F_aerodynamic + F _gradient. In case 2: Due to flat road, F_gradient = 0 N and F_net = F_traction – F_rolling – F_aerodynamic. As all parameter remain the same ⇒ F_traction will be the same. Therefore F_net = F_gradient. F_net = m_vehicle * a_vehicle = m_vehicle * g * sin (α) = F_grad. For small values, sin (α) = α. Therefore a_vehicle = g * α = 9.81 * 0.1 = 0.981 m/s² which is approximately 1 m/s².

Author engineerPosted on May 26, 2024May 26, 2024Categories Objective Questions