Mechanical Engineering Test

Result

Mechanical Engineering Test - 2

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
2 / -0.66

The Taylor’s tool life exponent n is 0.4 and tool changing time is 2 minutes. For maximum production rate the tool life in minutes is

A
3

B
1

C
2

D
6

Solution

Concept:

Expression for tool life for maximum production rate
Question 2
2 / -0.66

The hydrostatic stress for the state of stress represented by σ_ij given below is _______ MPa

A
100

B
101

C
102

D
105

Solution

Concept:

Any state of stress can be decomposed into a hydrostatic (or mean) stress σ_mI and a deviatoric stress s, according to:
Question 3
2 / -0.66

What will be the force exerted by a jet of water of diameter 100 mm on a stationary flat plate when the jet strikes the plate normally with the velocity of 20 m/s?

[assume, Density of water = 1000 kg/m³, π = 3.14]

A
3.14 kN

B
3140 kN

C
31.4 kN

D
0.314 kN

Solution

Concept:

Force exerted by a jet of water on a flat plate:
Question 4
2 / -0.66

An external force of 1 N is applied on the block of 1 kg as shown in the figure. The magnitude of the friction force F_s is (where, μ = 0.3, g = 10 m/s²):

A
0.3 N

B
0.1 N

C
3 N

D
1 N

Solution

Concept:

To determine the magnitude of the friction force (F_s) when an external force of 1 N is applied to a 1 kg block, we need to use the formula for frictional force:

Given:

Since the frictional force is 3 N, and the applied force is 1 N, the block does not move because the applied force is less than the maximum static friction force.

Hence, the actual friction force will be equal to the applied force (since the block is not moving):
Question 5
2 / -0.66

A turbine develops 9000 kW when running at 100 rpm. The head on the turbine is 25 m. If the head on the turbine is reduced to 16 m. Determine the speed

A
84 rpm

B
87 rpm

C
80 rpm

D
90 rpm

Solution
Question 6
2 / -0.66

The shape factor in the case of radiation exchange between a hemispherical surface and plane surface is:

A
0.8

B
0.5

C
0.3

D
0.1

Solution

Concept:

Shape Factor is defined as the fraction of radiation energy leaving a surface that reaches another surface, for example, F12 means the fraction of energy leaving the surface 1 reaches surface 2.

Summation Rule: F₁₁ + F₁₂ = 1 and F₂₁ + F₂₂ = 1

Reciprocity theorem: A₁ F₁₂ = A₂ F₂₁ (A is the surface area)

Calculation:

Let us consider the circular part of the hemisphere as one and the spherical part as surface 2

From summation rule F₁₁ + F₁₂ = 1

Radiation energy leaving the surface 1 will not strike the surface again, therefore, F₁₁ = 0

∴ F₁₂ = 1

Now, from reciprocity theorem

A₁F₁₂ = A₂F₂₁

πR² = F₂₁ × 2πR²

F₂₁ = 0.5

From F₂₁ + F₂₂ = 1 ⇒ F₂₂ = 0.5
Question 7
2 / -0.66

Let P and Q be two real matrices of size 4 × 6 and 5 × 4, respectively. If rank(Q) = 4 and rank(QP) = 2, then rank(P) is equal to ________

A
1

B
2

C
3

D
4

Solution

Concept:

ρ(QP) = 2, ρ(Q) = 4

From the properties of rank, we know that:

ρ(QP) ≤ mon [ρ(Q), ρ(P)]

2 ≤ min (4, ρ(P))

The possible values of ρ(P) are 2, 3, 4

on the other hand, by Sylvester's ranks inequality

ρ(Q) + ρ(P) - 4 ≤ ρ(QP)

4 + ρ(P) - 4 ≤ 2

ρ(P) ≤ 2

Thus we conclude that ρ(P) = 2

Alternate Method

From the Properties of Rank:

Let A and B be the matrices of size m × p and n × p respectively, then

(i) If rank(A) = p, then rank(AB) = rank(B)

(ii) If rank(B) = p, then rank(AB) = rank(A)

By applying the above property, rank(P) = 2.
Question 8
2 / -0.66

Find the total horizontal force acting on a horizontal cylinder floating water.

ρ = density of fluid, D = diameter of cylinder, w = length of cylinder

A

B

C

D

Solution

Question 9

2 / -0.66

Match the following NC codes with their options

List - I	List - II
A. M 03	1. Coolant supply ON
B. M 04	2. Dwell
C. G 04	3. Spindle rotation (CW)
D. M 08	4. Spindle rotation (CCW)

A – 3, B – 4, C – 2, D – 1

A – 4, B – 3, C – 2, D – 1

A – 4, B – 3, C – 1, D – 2

A – 2, B – 4, C – 1, D – 3

Solution

Important G codes

G00 – Rapid Transverse

G01 – Linear Interpolation

G02 – Circular Interpolation (CW)

G03 – Circular Interpolation (ACW)

G04 – Dwell

G – 97 – Spindle Speed

Important M codes

M 00 – Program Stop

M 03 – Spindle (CW)

M 04 – Spindle (CCW)

M 05 – Spindle Stop

M 08 – Coolant on

M 09 – Coolant off

M 10 – Clamp on

M 11 – Clamp off

M 02 or M 30 – Program stop, reset to start.

Question 10
2 / -0.66

A schematic diagram of the poring basin and the sprue of a gating system is shown in the figure. The depth of molten metal in the pouring basin is 100 mm and the height of the sprue is 1,500 mm.

Considering the cross-section of the sprue is circular, the ratio d₁:d₂ to avoid aspiration is

A
3 : 2

B
5 : 6

C
15 : 16

D
1 : 2

Solution

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Mechanical Engineering Test - 2

Result

Mechanical Engineering Test - 2

Score

-

Rank

-

SHARING IS CARING

Question 1

3

1

2

6

Solution

Question 2

100

101

102

105

Solution

Question 3

3.14 kN

3140 kN

31.4 kN

0.314 kN

Solution

Question 4

0.3 N

0.1 N

3 N

1 N

Solution

Question 5

84 rpm

87 rpm

80 rpm

90 rpm

Solution

Question 6

0.8

0.5

0.3

0.1

Solution

Question 7

1

2

3

4

Solution

Question 8

Solution

Question 9

A – 3, B – 4, C – 2, D – 1

A – 4, B – 3, C – 2, D – 1

A – 4, B – 3, C – 1, D – 2

A – 2, B – 4, C – 1, D – 3

Solution

Question 10

3 : 2

5 : 6

15 : 16

1 : 2

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.