# Quiz: Mechanical Engineering 12th May

Quiz: Mechanical Engineering
Exam: GATE
Topic: Miscellaneous

Each question carries 2 mark
Negative marking: 1/3 mark
Time: 20 Minutes

Q1. A two phase liquid vapour mixture of a pure substance has a pressure of 120 bar and occupies 0.3 m^3. If the mass of started liquid present in mixture is 2.1 kg and the mixture specific volume is 0.050 m^3/kg, what is the mass of vapour present in the mixture?
(a) 2.1 kg
(b) 3.1 kg
(c) 2.9 kg
(d) 3.9 kg

Q2. A motor car is going due north at a speed of 50 km/h. It makes a 90° left turn without changing the speed. What is the change in velocity of car?
(a) 50 km/h towards west
(b) 71 km/h south west
(c) 71 km/h north west
(d) zero

Q3. A round disk of 150 mm diameter is to be blanked from strip of 3.2 mm, half hard cold rolled steel whose shear strength = 310 MPa. Then ratio of die opening diameter to punch diameter will be, if the radial clearance is governed by e = 0.075 (t) mm where ’t’ thickness of strip
(a) < 1

(b) > 1
(c) = 1
(d) None of these

Q4. A steel component has a theoretical stress concentration factor 1.6 and notch sensitivity of 0.7 then its effect on endurance limit will be:
(a) 70.42% increases
(b) 70.72% decreases
(c) 29.58% increases
(d) 29.58% decreases

Q5. For the element shown in ﬁgure, principal stresses are 80 MPa and — 20 MPa. What is the value of stresses σ_x and σ_y?

(a) σ_x = – 10 MPa, σ_y = 70 MPa
(b) σ_x = 70 MPa, σ_y = 10 MPa
(c) σ_x = 70 MPa, σ_y = – 10 MPa
(d) σ_x = -70 MPa, σ_y = -10 MPa

Q6. At what height from water surface, a centrifugal pump may be installed in the following case to avoid cavitation: atmospheric pressure = 101 kPa, vapour pressure 2.34 kPa, inlet and other losses in suction pipe = 1.55 m, effective head of pump 52.5 m and cavitation parameter σ = 0.118.
(a) 2 m
(b) 2.2 m
(c) 2.3 m
(d) 2.5 m

Q7. A body of mass 10 kg falls from a height of 5 m and is stopped within (1/10)second on the ground. What is the force of interaction?
(a) 100 N
(b) 0 N
(c) 1000 N
(d) 1100 N

Q8. In a slotted lever quick return mechanism, the coriolis component of acceleration of slider becomes zero for how many times in one complete rotation of driving crank?
(a) 1
(b) 2
(c) 3
(d) 4

Q9. What is the main objective of spheroidise annealing?
(a) For improving the machinability
(b) For producing hard surface and tough core
(c) For relieving the internal stresses
(d) For localized grain reﬁnement

Q10. A liquid is compressed in a cylinder from a volume of 0.0113 m^3 at 6.87 × 10^6 N/m^2 pressure to 0.0112 m^3 at 13.73 × 10^6 N/m^2. The compressibility of the liquid is
(a) 1.29 × 10^(-9) m^2/N
(b) 7.75 × 10^8 N/m^2
(c) 2.4 × 10^(-9) m^2/N
(d) 8.4 × 10^8 N/m^2

Solutions

S1. Ans.(d)
Sol. V = 0.3 m^3
v_mixture = 0.050 m^3/kg
m_f = 2.1 kg
m_g = ?
v_mixture = V_mixture/(m_f + m_g )
0.050 = (0.3 )/(2.1 + m_g )
0.1505 + 0.050 m_g = 0.3
m_g = 3.9 kg

S2. Ans.(b)
Sol.

Change in velocity = Final velocity – initial velocity
= –50i ̂ – 50j ̂
= – 50 (i ̂+ j ̂)
That is in third quadrant. So direction in south-west.
Magnitude of change in velocity = √((-50)^2+(-50)^2 ) = 50√2
Change in magnitude of velocity = 70.7 m/s

S3. Ans.(b)
Sol. In blanking and punching operation, always size of die opening > size of punch, so the ratio of die opening diameter to punch diameter is greater than 1.

S4. Ans.(d)
Sol. as we know
k_t = 1.6
q = 0.7
k_f = 1 + q (k_t – 1)
k_f = 1 + 0.7 (1.6 – 1)
k_f = 1.42
1/k_f = 0.7042
% reduction in endurance limit = (1 – 0.7042) × 100 = 29.58%

S5. Ans.(c)
Sol. τxy= 30 MPa σ_1,2 = ((σ_x + σ_y)/2) ±√(((σ_x-σ_y)/2)^2+(τ_xy )^2 ) ((σ_x+σ_y)/2) = a √(((σ_x-σ_y)/2)^2+τ_xy^2 ) = b σ(1,2) = a ± b
Now, σ_1 = a + b
80 = a + b
σ_2 = a – b
– 20 = a – b
Now, solving eqs. (1) and (2)
60 = 2a
a = 30
b = 50
(σ_x+σ_y)/2 = 30
σ_x + σ_y = 60
√(((σ_x-σ_y)/2)^2+τ_xy^2 ) = 50
((σ_x-σ_y)/2)^2 + 30^2 = 50^2
(σ_x-σ_y ) = 80
By using equation (3) and (4)
σ_x – σ_y = 80
σ_x + σ_y = 60
σ_x = 70 MPa on solving
σ_y = – 10 MPa

S6. Ans.(c)
Sol. σ = (P_a/ρg – P_v/ρg – h_s – h_fs)/H_m
P_a = 101 kPa = 101000 N/m^2
P_v = 2.34 kPa = 2340 N/m^2
H_m = 52.5 m, h_(f_s ) = 1.55 m, σ = 0.118
Substituting all the values
0.118 = ((101000 –2340)/9810 – h_s – 1.55 )/52.5
h_s = 2.312 m

S7. Ans.(d)
Sol.

Velocity when block reaches the ground = √2gh
= √(2×10×5) = 10 m/s
By momentum conservation :
(F) × dt = Momentum just after striking the ground – momentum just before striking the ground
(N – mg) × dt = m × 0 – (– m × 10)
(N – mg) = (m×10)/dt
N = (10 × 10)/((1/10)) + 10 × 10
Force of interaction, N = 1100 N

S8. Ans.(d)
Sol. Coriolis component of acceleration will only be zero if either angular velocity of slotted lever is zero or the velocity of slider is zero. The possible 4 conditions are
• Two at the extremes of slotted lever.

Two when the driving crank and slotted lever are vertical because at that position, velocity of slider will be zero.

S9. Ans.(a)
Sol. Spheroide annealing is used for improving the machinability of medium and high carbon steel. Material is heated near the lower critical temperature and cooled extremely slowly around 20°C/hour in the furnace. The reason of machinability improvement is spheroidise formation.

S10. Ans.(a)
Sol. The compressibility of a liquid is defined as
β = 1/K = ((dV/V))/(dp )
β = -(((0.0112-0.0113)/0.0113))/((13.73×10^6-6.87×10^6))
β = 1.29 × 10^(-9) m^2/N

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