Quiz: Mechanical Engineering

Exam: GATE

Topic: Miscellaneous

Each question carries 2 mark

Negative marking: 1/3 mark

Time: 20 Minutes

Q1. A bar having a cross-section area of 700 mm² is subjected to axial loads at the positions indicated the value of stress in the segment QR is

(a) 40 MPa

(b) 50 MPa

(c) 70 MPa

(d) 120 MPa

Q2. The strain energy stored in the beam with flexural rigidity EI and loaded as shown in the figure is

(a) (P^2 L^3)/3EI

(b) (2 P^2 L^3)/3EI

(c) (4 P^2 L^3)/3EI

(d) (8 P^2 L^3)/3EI

Q3. A quick return mechanism is shown below. The crank OS is driven at 2 rev/s in counter clockwise direction.

If the quick return ratio is 1: 2, then the length of the crank in mm is

(a) 250

(b) 250 √3

(c) 500

(d) 500 √3

Q4. Considering massless rigid rod small oscillations, the natural frequency (in rad/s) of vibration of the system shown in the figure is

(a) √(400/1)

(b) √(400/2)

(c) √(400/3)

(d) √(400/4)

Q5. A solid circular shaft needs to be designed to transmit a torque of 50 N.m. if the allowable shear stress of the material is 140 MPa, assuming a factor of safety of 2, minimum allowable design diameter in mm is

(a) 8

(b) 16

(c) 24

(d) 34

Q6. The velocity triangles at the inlet and exit of the rotor of a turbomachine are shown. V denotes the absolute velocity of the fluid, W denotes the relative velocity of the fluid and U denotes the blade velocity. Subscripts 1 and 2 refer to inlet and outlet respectively if V_2=W_1 and V_1=W_2 then the degree of reaction is

(a) 0

(b) 1

(c) 0.5

(d) 0.25

Q7. An incompressible fluid flows over a flat plate with zero pressure gradient. The boundary layer thickness is 1 mm at a location where the Reynolds number is 1000. If the velocity of the fluid alone is increased by a factor of 4, then the boundary layer thickness at the same location, in mm will be

(a) 4

(b) 2

(c) 0.5

(d) 0.25

Q8. Which of the following statements are TRUE, when the cavitation parameter σ = 0?

(i) the local pressure is reduced to vapor pressure

(ii) cavitation starts

(iii) boiling of liquid starts

(iv) cavitation stops

(a) i, ii and iv

(b) only ii and iii

(c) only i and iii

(d) i, ii and iii

Q9. Consider a steam power plant using a reheat cycle as shown. Steam leaves the boiler and enters the turbine at 4MPa, 350°C(h_3=3095 kJ\/kg), the steam is reheated to 350°C (h_5=3170 kj\/kg), and then expanded in a low pressure turbine to 10 kPa (h_6=2165 kj\/kg) the specific volume of liquid handled by the pump can be assumed to be

The thermal efficiency of the plant neglecting pump work is

(a) 15.8%

(b) 41.1%

(c) 48.5%

(d) 58.6%

Q10. The enthalpy at the pump discharge (h_2 ) is

(a) 0.33 kJ/kg

(b) 3.33 kJ/kg

(c) 4.0 kJ/kg

(d) 33.3 kJ/kg

Solutions

S1. Ans.(a)

Sol. At face Q 35 kN fore is acting towards right hence

( x > y )

So, x-y = 35 kN

At face R 49 kN force is acting towards right hence (y > Z)

X = 63 kN

If

x = 63 kN

x-y =35 kN

Y = 28 kN

y-z = 49 kN

z = -21 kN

σ_QR=(28×〖10〗^3 N)/(700 mm^2 )=40 MPa

S2. Ans.(c)

Sol. total strain energy stored is given by

2∫_O^L▒〖((Px)^2 dx)/2EI+((PL)^2.2L)/2EI〗

=(2P^2)/2EI ∫_O^L▒〖x^2 dx+(P^2 L^3)/EI〗

=P^2/EI×L^3/3+(P^2 L^3)/EI

=(P^2 L^3)/3EI+(P^2 L^3)/EI=4/3 (P^2 L^3)/EI

S3. Ans.(a)

Sol. Given: 2θ/(2π-2θ )=1/2

∵θ=π/3

In triangle OMP

Cos θ = r/OP,

Cos (π/3)=r/OP

Cos(π/3)=r/500

OR r=250 mm

S4. Ans.(d)

Sol. θ → change in angular positions of rod in anticlockwise

direction. Restoring torque

=(krθ)r will at

In clockwise direction.

So,

τ=K r^2 θ

Iα= -kr^2 θ

α=(kr^2)/I θ

ω_n=√((kr^2)/I)

I= m(2r^2 )=1 (4r^2 )

I= √((400×r^2)/(1×4x^2 ))=√(400/4)

ω_n=10 rad\/sec

S5. Ans.(b)

Sol. T = 50 N.m = 140 MPa, N = 2

τ_permissible=140/N=70 MPa

∵τ_per=16T/(πd^3 )

d≥[16T/(π.τ_per )]^(1\/3)≥[(16×30)/(π×70×〖10〗^6 )]^(1\/3)

d≥0.0153 m

d≥15.37 mm

d≈16 mm

S6. Ans.(c)

Sol. Degree reaction,

R=(Static pressure change in the runner)/(Total energy transfer in the runner)

When static pressure change,

=(W_2^2-W_1^2)/2+(U_1^2-U_2^2)/2

=(W_2^2-W_1^2)/2∵U_1=U_2

Total energy transfer,

=(V_1^2-V_2^2)/2+(W_2^2-W_1^2)/2+(U_1^2-U_2^2)/2

=(W_2^2-W_1^2)/2 +(W_2^2-W_1^2)/2

= W_2^2-W_1^2

R= (W_2^2-W_1^2 \/2)/(W_2^2-W_1^2 )

R=1/2=0.5

S7. Ans.(c)

Sol. boundary layer thickness,

δ ∝1/(〖R_ex〗^(1/2) ) for laminar flow

and R_ex=Vx/ν

i.e.R_ex ∝ V

∵δ∝1/V^(1\/2)

δ_1 V_1^(1\/2)=δ_2 V_2^(1\/2)

V^(1\/2)=2δ_2 V^(1\/2)

δ_2=1\/2

δ_2=0.5 mm

S8. Ans.(d)

Sol. Cavitation parameter, σ = 0 if means:

the local pressure is reduce to vapor pressure

cavitation starts

boiling of liquid starts.

S9. Ans.(b)

Sol. Given data: h_3=3095 kJ\/kg;h_4=2604 kJ\/kg; h_5=3170 kJ\/kg ;h_6=2165 kJ/kg;〖 h〗_1=29.3 kJ\/kg

Network output per kg,

W_net=(h_3-h_4 )+(h_5-h_6 )

=(3095-2609)+(3170-2165)

=1491 kJ\/kg

Net heat supplied

q_s=(h_3-h_2 )+(h_2-h_4 )

=(h_3-h_1 )+(h_5-h_4 )

=3095-29.3+3170-2609

=3626.7 kJ\/kg

Thermal efficiency,

η=W_net/q_s =1491/3626.7

=0.4111

=41.11%

S10. Ans.(d)

Sol. Enthalpy at exit of pump must be greater than enthalpy at inlet of pump i.e. h_2 must be greater than h_1=29.3 kJ\/kg. Among the four give option is greater than h_1=29.3 kJ\/kg which is 33.3 kJ/kg. Hence option (d) is correct.