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NHPC JE’21 ME: Daily Practices Quiz. 28-Oct-2021

Quiz: Mechanical Engineering
Exam: NHPC
Topic: Miscellaneous

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

Q1. A streamline is a line
(a) which is along path of the particle
(b) which is always parallel to the main direction of flow
(c) along which there is no flow
(d) on which tangent drawn at any point gives the direction of velocity

Q2. The loss of head due to sudden enlargement is attributed to
(a) Viscosity of fluid
(b) Generation of heat
(c) Roughness of pipe
(d) Production and dissipation of turbulent energy

Q3. Match List-I (Phenomena) with List-II (Causes) and select the correct answer using the codes given below the lists:
A. Shock wave
B. Flow separation
C. Capillary rise
D. Cavitation
1. Surface tension
2. Vapour pressure
3. Compressibility
4. Adverse pressure gradient
(a) 3 1 2 4
(b) 4 2 1 3
(c) 3 4 1 2
(d) 4 1 2 3

Q4. In reaction turbines, the draft tube is used
(a) for the safety of the turbine
(b) to convert the kinetic energy of flow by a gradual expansion of the flow cross-section
(c) to destroy the undesirable eddies
(d) for none of the above purposes

Q5. Hydraulic ram is a pump which works on the principle of
(a) water hammer
(b) centrifugal action
(c) reciprocating action
(d) hydraulic press

Q6. In a long cylindrical rod of radius R and a surface heat flux of q_0, the uniform internal heat generation rate is
(a) 2 q_0/R
(b) 2 q_0
(c) q_0/R
(d) (2〖 q〗_0)/R^2


S1. Ans (d)
Sol. A stream line is a line on which tangent drawn at any point gives the direction of velocity.

S2. Ans (d)
Sol. Loss of head due to sudden enlargement is ((V_1-V_2)²)/2g

S3. Ans (c)
Sol. Shock wave : compressibility
Flow separation : adverse pressure gradient
Capillary rise : surface tension
Cavitation : vapour pressure

S4. Ans (b)
Sol. The draft tube has two purposes as follows:
It permits a negative or suction head to be established at the runner exist, thus making it possible to install the turbine above the tail race level without loss of head.
It converts a large proportion of velocity energy rejected from the runner into useful pressure energy. i.e. acts as a recuperator of pressure energy.

S5. Ans (a)
Sol. The working of a hydraulic ram is based on the principle of water hammer or inertia pressure developed in the supply pipe.

S6. Ans (a)
T_w=(q ̅R)/2h+T_f
T_w-T_f=((q.) ̅R)/2h
Surface heat flux
q_0=h.(q ̅.R)/2h
q ̅=(2q_0)/R

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Published by VINAYAK KUMAR

Graduated from BIT Sindri, Postgraduated from IIT BHU Varanasi

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