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1.	If the car is moving on the level road at a speed of 50 km/h has a wheelbase 4 m, a distance of C.G. from ground 800 mm and distance from rear wheels is 1.2 m. Coefficient of friction is 0.8. What is the retardation if the brakes are applied to all wheels?(a) 7.8 m/s^2(b) 6.8 m/s^2(c) 5.8 m/s^2(d) 4.8 m/s^2This question was posed to me during an interview for a job.Question is from Braking System topic in chapter Braking System of Automobile Engineering
Answer» Correct answer is (a) 7.8 m/s^2 The BEST I can explain: When the brakes are applied on all wheels and the vehicle is moving on the level road, a = μgwhere ‘μ’ is the coefficient of FRICTION, ‘L’ is wheelbase, ‘x’ is a distance of C.G. from rear wheels, and ‘h’ is the distance of the C.G. from the SURFACE of the road. THEREFORE a = 0.8 9.81 = 7.8 m/s^2.

2.	If the car is moving on the level road at a speed of 50 km/h has a wheelbase 4 m, a distance of C.G. from ground 800 mm and distance from rear wheels is 1.2 m. Coefficient of friction is 0.8. What is the retardation if the brakes are applied to the front wheels?(a) 3.4 m/s^2(b) 2.4 m/s^2(c) 1.4 m/s^2(d) 4.4 m/s^2I have been asked this question in unit test.The question is from Braking System topic in portion Braking System of Automobile Engineering
Answer» Correct OPTION is (b) 2.4 m/s^2 The best explanation: When the brakes are applied on the front wheels and the VEHICLE is MOVING on the LEVEL road, a = \(\frac{\mu * g * x}{L-\mu* h}\)where ‘μ’ is the coefficient of friction, ‘L’ is wheelbase, ‘x’ is a DISTANCE of C.G. from rear wheels, and ‘h’ is the distance of the C.G. from the surface of the road. Therefore a = \(\frac{0.7 * 9.81 * 1.2}{4-(0.8 * 0.8)}\) = 2.4 m/s^2.

3.	On what principle does the braking system in the car work?(a) Frictional force(b) Gravitational force(c) Magnetic force(d) Electric forceI had been asked this question by my school teacher while I was bunking the class.My query is from Braking System in portion Braking System of Automobile Engineering
Answer» Correct choice is (a) Frictional force To ELABORATE: Frictional force is the main force which acts and plays an important role in the braking system. The disc pads get to RUB against the drum and which creates the FRICTION and OPPOSES the MOTION of the rotational motion of the drum.

4.	In a disc brake, which component provides the pad-to-disc adjustment?(a) Bleed screw(b) Piston(c) Caliper(d) Piston sealI have been asked this question by my college director while I was bunking the class.I'm obligated to ask this question of Braking System in section Braking System of Automobile Engineering
Answer» RIGHT OPTION is (d) PISTON seal To elaborate: Piston seal provides the pad-to-disc ADJUSTMENT in a disc brake. When the brake is applied, the piston seal DEFORMS.

5.	The metering valve is used to proportion the braking effect between the front and the rear axle.(a) True(b) FalseI have been asked this question in unit test.The above asked question is from Braking System in section Braking System of Automobile Engineering
Answer» Right CHOICE is (b) False The best explanation: The PROPORTIONING valve is used to proportion the BRAKING effect between the front and the rear axle.

6.	The car is moving down on the 14° inclined road with horizontal at 36 km/h which is having wheelbase 1.4 m. The C.G. of the car is 0.9 m above the road. The coefficient of friction is 0.75. What is the retardation of the car if the brakes are applied to all the four wheels?(a) 4.76 m/s^2(b) 5.76 m/s^2(c) 6.76 m/s^2(d) 7.76 m/s^2I got this question in unit test.My question is based upon Braking System in chapter Braking System of Automobile Engineering
Answer» The correct answer is (a) 4.76 m/s^2 To EXPLAIN I would say: α is the inclination angle in degree = 14° μ is the coefficient of friction = 0.75. When the vehicle is MOVING down on the inclined road and BRAKES are APPLIED to all FOUR wheels than the retardation a = g(μcosα – sinα) = 9.81(0.75cos14° – sin14°) = 4.76 m/s^2.

7.	The car is moving up on the 12o inclined road with horizontal at 36 km/h which is having wheelbase 1.4 m. The C.G. of the car is 0.9 m above the road. The coefficient of friction is 0.7. What is the retardation of the car if the brakes are applied to all the four wheels?(a) 8.75 m/s^2(b) 8.81 m/s^2(c) 7.71 m/s^2(d) 6.81 m/s^2This question was posed to me in an international level competition.I would like to ask this question from Braking System topic in section Braking System of Automobile Engineering
Answer» Right CHOICE is (a) 8.75 m/s^2 Easy explanation: α is the INCLINATION ANGLE in degree = 12°, μ is the coefficient of FRICTION = 0.7. When the vehicle is moving up on the INCLINED road and brakes are applied to all four wheels than the retardation a = g(μcosα + sinα) = 9.81(0.7cos12° + sin12°) = 8.75 m/s^2.

8.	Generally which brakes are on the front wheels?(a) Drum brake(b) Disk brake(c) Shoe brake(d) Double shoe brakeThe question was asked in examination.I would like to ask this question from Braking System topic in chapter Braking System of Automobile Engineering
Answer» Right answer is (b) Disk brake To explain: Disk brakes are used USUALLY on the FRONT wheels. When the brakes are APPLIED the 70% of the weight is transferred to the front wheels.Because of which the brakes on the front wear out FASTER.

9.	If the car is moving on the level road at a speed of 50 km/h has a wheelbase 3 m, a distance of C.G. from ground 700 mm and distance from rear wheels is 1.1 m. Coefficient of friction is 0.7. What is the retardation if the brakes are applied to the rear wheels?(a) 3.7 m/s^2(b) 2.7 m/s^2(c) 1.7 m/s^2(d) 4.7 m/s^2The question was asked during an online exam.Enquiry is from Braking System topic in chapter Braking System of Automobile Engineering
Answer» The correct ANSWER is (a) 3.7 m/s^2 Best explanation: When the brakes are APPLIED on the rear wheels and the vehicle is MOVING on the level road, a = \(\frac{\mu * G * (L-x)}{L+μ * h}\) where ‘μ’ is the coefficient of friction, ‘L’ is wheelbase, ‘x’ is a distance of C.G. from rear wheels, and ‘h’ is the distance of the C.G. from the surface of the road. THEREFORE a = \(\frac{0.7 * 9.81 * (3-1.1)}{3+(0.7 * 0.7)}\) = 3.7 m/s^2.

10.	What is the braking torque at leading shoe if resultant frictional force acts at a distance of 250 mm from the brake drum center,coefficient of friction between the shoe and the drum as 0.5, the free ends of the two shoes are pushed apart with a force of 300 N which is acting at a distance of 320 mm from anchor, and two shoes are anchored together 170 mm away from the brake drum center?(a) 276.6 Nm(b) 256.6 Nm(c) 266.6 Nm(d) 246.6 NmI got this question during an internship interview.This intriguing question comes from Braking System topic in section Braking System of Automobile Engineering
Answer» The correct option is (c) 266.6 Nm Explanation: TL =\(\frac{W * L * \mu * R}{M-(\mu * R)} = \frac{320 * 300 * 0.5 * 250}{170-(0.5 * 250)}\) = 266666.66 NMM = 266.6 Nm where L is the DISTANCE at which the force acts from the ANCHOR, W is the force from the anchor, μ is the coefficient of friction, R is the distance of RESULTING friction force from brake drum center, M is the distance between the two anchors.