NHPC JE'21 ME: Daily Practices Quiz. 02-Nov-2021 |_00.1
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NHPC JE’21 ME: Daily Practices Quiz. 02-Nov-2021

Quiz: Mechanical Engineering
Exam: NHPC JE
Topic: Miscellaneous

Each question carries 1 mark
Negative marking: 1/4 mark
Time: 6 Minutes

Q1. For an Oldham coupling used between two shafts, which among the following statements are correct?
I. Torsional load is transferred along shaft axis.
II. A velocity ratio of 1:2 between shafts is obtained without using gears
III. Bending load is transferred transverse to shaft axis.
IV. Rotation is transferred along shaft axis.
(a) I and II
(b) I and IV
(c) II and III
(d) II and IV

Q2. A metallic rod of 500 mm length and 50 mm diameter, when subjected to a tensile force of 100 KN at the ends, experiences an increase in its length by 0.5 mm and a reduction m its diameter by 0.015 mm. The Poisson’s ratio of the rod material is
A. 0.5
B. 0.2
C. 0.3
D. 0.5

Q3. The maximum theoretical work obtainable, when a system interacts to equilibrium with a reference environment, is called
(a) Entropy
(b) Enthalpy
(c) Exergy
(d) Rothalpy

Q4. The jobs arrive at a facility, for service, in a random manner. The probability distribution of number of arrivals of jobs in a fixed time interval is
(a) Normal
(b) Poisson
(c) Erlang
(d) Beta

Q5. A block R of mass 100 kg is placed on a block S of mass 150 kg as shown in the figure. Block R is tied to the wall by a massless and inextensible string PQ. If the coefficient of static friction for all surfaces is 0.4, the minimum force F (in KN) needed to move the block S is

(a) 0.69
(b) 0.88
(c) 0.98
(d) 1.37

Q6. Notch sensitivity varies between:
(a) 0–10
(b) 0.1 – 100
(c) 1 – 100
(d) 0–1

Solution

S1. Ans(b)
Sol. Oldham’s coupling is used to transmit power between two shafts having small angular misalignment. There is no bending load or reduction/increase of speed criterion. Hence only I and IV statements are correct.

S2. Ans(c)
Sol. Given that,
Length (l) = 500 mm
Elongation in length (∆l) = 0.5 mm
Diameter (d) = 50 mm
Reduction in diameter (∆d) = -0.015 mm
Poisson’s ratio = -(lateral strain)/(longitudinal strain)
Poisson’s ratio = -(∆d/d)/(∆l/l)
Poisson’s ratio = -((-0.015)/50)/(0.5/500)
Poisson’s ratio = 0.3

S3. Ans(c)
Sol. As per the definition of the Exergy (or) Available Energy, we know that the maximum portion of energy which could be converted into useful work by ideal processes which reduce the system to dead state is known as Exergy.

S4. Ans(b)
Sol. As we know, that the arrival rate depends upon the time factor. So, Poisson distribution can be chosen, but normal distribution expresses same result throughout.

S5. Ans(d)
Sol. Force required to move the block S will be;
F = Force required to move block S + Force required to move block R
Therefore; F = 0.4 × 100 × 9.81 + 0.4 × 250 × 9.81 = 1.37 KN

S6. Ans(d)
Sol. Notch sensitivity is defined as the ratio of increase of actual stress over nominal stress in fatigue to the increase stress over nominal stress in static loading.
q=(K_f-1)/(K_t-1)
where,
K_f = actual (or) fatigue stress concentration factor.
K_t = theoretical (or) static stress concentration factor.
So, it varies between 0 to 1.

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Published by VINAYAK KUMAR

Graduated from BIT Sindri, Postgraduated from IIT BHU Varanasi

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