Quiz: Mechanical Engineering

Exam: UPPSC AE/Lecturer

Topic: Engg. Mechanics

Each question carries 3 marks

Negative marking: 1 mark

Time: 8 Minutes

Q1. The parameters upon which the effect of a force on a body depends on

(a) Direction

(b) Magnitude

(c) Position

(d) All of these

Q2. The force of attraction, between the bodies of masses m_1 and m_2 situated at a distance ‘d’ apart, according to the Newton’s law of gravitation is given by

(a) F=(Gm_1 m_2)/d^2

(b) F=(m_1 m_2)/(Gd^2 )

(c) F=(Gm_1 m_2)/d

(d) F=(m_1 m_2)/Gd

Q3. Opening a Limca bottle is due to

(a) moment

(b) couple

(c) torque

(d) parallel forces

Q4. A body of weight 100 N is resting on a rough horizontal plane which has coefficient of friction μ, and can be just moved by a force of 40 N which is applied in horizontal direction. What will be the value of the coefficient of friction between horizontal plane and the body?

(a) 0.5

(b) 0.4

(c) 0.3

(d) none of the above

Q5. The linear velocity of a body rotating at 20 rad/s along a circular path of radius 10 m in m/sec is

(a) 200

(b) 1

(c) 400

(d) 2

Q6. An object drops from the top of a building. If it comes down half the height of building in 2 seconds, then the time taken by the object to reach the ground is

(a) 2.8 s

(b) 3.2 s

(c) 4.0 s

(d) 4.5 s

Solutions

S1. Ans. (d)

Sol. The parameters upon which the effect of a force on a body depends on

Direction,

Magnitude,

Position

Line of action.

S2. Ans. (a)

Sol. F=(Gm_1 m_2)/d^2

S3. Ans. (a)

Sol. Opening a Limca bottle is due to Moment applied on the bottle.

S4. Ans. (b) N

Sol.

μ F

F=μN

N=mg=W=100 N

μ=F/N=40/100=0.4

S5. Ans. (a)

Sol. We know that, linear velocity, V = ω.r

V=20×10=200 m/sec

S6. Ans. (a)

Let, the height of the building is x meter

So,

x/2=ut+1/2 gt^2

u=0 and t=2 sec

x/2=2g

x=4g

Also, we know that

v^2=u^2+2gx

v^2=0+2g×4g=8g^2

v=g√8

And,

v=u+gt

g√8=0+gt

t=(g√8)/g=2.83 sec