Answer:
true
Explanation:
Shear strains and direct strains are independent components in a strain tensor at a point. Also the poisson ratio is defined as
μ = - [tex]\frac{lateral strain}{longitudinal strain}[/tex]
Which is independent of shear strains Thus this affect does not apply on shear strains
A wooden block (SG = 0.6) floats in oil (GS = 0.8). What fraction of the volume of the block is submerged in oil?
Answer:
3/4 th fraction of the volume of block is submerged in oil.
Explanation:
We know that
density of the block, ρ[tex]_{b}[/tex] =SG[tex]\times[/tex]density of water
= 0.6[tex]\times[/tex] 1000
= 600 kg/[tex]m^{3}[/tex]
density of the oil, ρ[tex]_{o}[/tex] =SG [tex]\times[/tex] denity of water
= 0.8[tex]\times[/tex] 1000
= 800 kg/[tex]m^{3}[/tex]
Let acceleration due to gravity, g = 9.81 m/[tex]s^{2}[/tex]
Volume of block submerged in oil is [tex]V_{1}[/tex]
Volume of block above the oil surface is [tex]V_{2}[/tex]
The total volume of the block is V = [tex]V_{1}[/tex]+[tex]V_{2}[/tex]
Therefore for a partially submerged body, we know that
Buoyant force = total weight of the block
and
total weight of the block = Weight of the fluid displaced by the block
ρ[tex]_{b}[/tex]\timesV\timesg = ρ[tex]_{o}[/tex]\times[tex]V_{1}[/tex]\times g
600([tex]V_{1}[/tex]+[tex]V_{2}[/tex]) = 800 [tex]V_{1}[/tex]
600[tex]V_{1}[/tex] + 600[tex]V_{2}[/tex] = 800 [tex]V_{1}[/tex]
600[tex]V_{2}[/tex] = 800 [tex]V_{1}[/tex] - 600[tex]V_{1}[/tex]
600[tex]V_{2}[/tex] = 200[tex]V_{1}[/tex]
[tex]\frac{V_{1}}{V_{2}}[/tex] = [tex]\frac{600}{200}[/tex]
Thus [tex]V_{1}[/tex] = 600
[tex]V_{2}[/tex] = 200
V = 600+200
= 800
Therefore fraction of volume of the block submerged in oil is,
[tex]\frac{V_{1}}{V}[/tex] = [tex]\frac{600}{800}[/tex]
[tex]\frac{V_{1}}{V}[/tex] = [tex]\frac{3}{4}[/tex]
Which is/are not a mechanism commonly associated with tool wear (mark all that apply)?a. Adhesion b. Attrition c. Abrasion d. Coercion
Answer: d)Coercion
Explanation:Tool wear is defined as the situation when the cutting tool is subjected to the regular process of cutting metal then they tend to wear because of the continuous action of cutting and facing stresses and pressure . The mechanism that does not happen during this process are coercion that means the process of exerting forces on any material forcefully against the will or need. Therefore, adhesion,attrition and abrasion are the process of tool wear .So the correct option is (d)
An alternating current E(t) =120 sin(12t) has been running through a simple circuit for a long time. The circuit has an inductance of L = 0.2 henrys, a resistor of R = 5 ohms and a capacitor of capcitance C = 0.043 farads. What is the amplitude of the current I?
Answer:
Explanation:
we have given E(t)=120 sin(12t)
R=5 ohm
L=0.2 H
ω=12 ( from expression of E)
[tex]X_L=0.2\times 12=2.4[/tex] ohm
[tex]X_C=\frac{1}{\omega \times C}=\frac{1}{12\times 0.043}=1.9379\ ohm[/tex]
[tex]Z=\sqrt{R^2+\left ( \omega L-\frac{1}{\omega C} \right )^2}[/tex]
[tex]Z=\sqrt{5^2+\left ( \2.4-1.9379 )^2}[/tex]
=5.021 ohm
so amplitude of current = [tex]\frac{v}{z}=\frac{120}{5.021}=23.89[/tex]
Why the inviscid, incompressible, and irrotational fields are governed by Laplace's equation?
Answer: Laplace equation provides a linear solution and helps in obtaining other solutions by being added to various solution of a particular equation as well.
Inviscid , incompressible and irrotational field have and basic solution ans so they can be governed by the Laplace equation to obtain a interesting and non-common solution .The analysis of such solution in a flow of Laplace equation is termed as potential flow.
The heat rate is essentially the reciprocal of the thermal efficiency. a)- True b)- False
Answer:
a). TRUE
Explanation:
Thermal efficiency of a system is the defined as the ratio of the net work done to the total heat input to the system. It is a dimensionless quantity.
Mathematically, thermal efficiency is
η = net work done / heat input
While heat rate is the reciprocal of efficiency. It is defined as the ratio of heat supplied to the system to the useful work done.
Mathematically, heat rate is
Heat rate = heat input / net work done
Thus from above we can see that heat rate is the reciprocal of thermal efficiency.
Thus, Heat rate is reciprocal of thermal efficiency.
To measure the voltage drop across a resistor, a _______________ must be placed in _______________ with the resistor. Ammeter; Series Ammeter; Parallel Voltmeter; Series Voltmeter; Parallel
Answer:
The given blanks can be filled as given below
Voltmeter must be connected in parallel
Explanation:
A voltmeter is connected in parallel to measure the voltage drop across a resistor this is because in parallel connection, current is divided in each parallel branch and voltage remains same in parallel connections.
Therefore, in order to measure the same voltage across the voltmeter as that of the voltage drop across resistor, voltmeter must be connected in parallel.
What is the Thermodynamic (Absolute) temperature scale?
Answer:
0 K
Explanation:
The thermodynamic absolute temperature is that temperature at which there is an infinite cooling and so there is no movement of any molecules or particles it is given by kelvin. 0 K is that temperature at which there is no movement of molecule so 0 K or -273°C in degree Celsius is the absolute temperature in thermodynamic temperature scale.
The Molybdenum with an atomic radius 0.1363 nm and atomic weight 95.95 has a BCC unit cell structure. Calculate its theoretical density (g/cm^3).
Solution:
Given :
atomic radius, r = 0.1363nm = 0.1363×10⁻⁹m
atomic wieght, M = 95.96
Cell structure is BCC (Body Centred Cubic)
For BCC, we know that no. of atoms per unit cell, z = 2
and atomic radius, r =[tex]\frac{a\sqrt{3} }{4}[/tex]
so, a = [tex]\frac{4r}{\sqrt{3}}[/tex]
m = mass of each atom in a unit cell
mass of an atom = [tex]\frac{M}{N_{A} }[/tex],
where, [tex]N_{A}[/tex] is Avagadro Number = 6.02×10^{23}
volume of unit cell = a^{3}
density, ρ = [tex]\frac{mass of unit cell}{volume of unit cell}[/tex]
density, ρ = [tex]\frac{z\times M}{a^{3}\times N_{A}}[/tex]
ρ = [tex]\frac{2\times 95.95}{(\frac{4\times 0.1363\times 10^{-9}}{\sqrt{3}})^{3}\times (6.23\times 10^{23})}[/tex]
ρ = 10.215gm/[tex]cm^{3}[/tex]
Which of the following are not related to a materials structure? (Mark all that apply) a)- Atomic bonding b)- Crystal structure c)- Atomic number d)- Microstructure
The answer is c) atomic number
Water flows at the rate of 200 I/s upwards through a tapered vertical pipe. The diameter at Marks(3) CLO5) the bottom is 240 mm and at the top 200 mm and the length is 5m. The pressure at the bottom is 8 bar, and the pressure at the topside is 7.3 bar. Determine the head loss through the pipe. Express it as a function of exit velocity head.
Answer: 5.35m
Explanation:
By using energy equation:
[tex]\frac{P_1}{\gamma}+z_1+\frac{v_1^{2} }{2g} =\frac{P_2}{\gamma}+z_2+\frac{v_2^{2} }{2g}+h_{L}[/tex]
[tex]\gamma=specific weight[/tex]
[tex]v_{1} =\frac{Q_1}{A_1} =\frac{0.2}{\frac{\pi }{4} \times 0.24^2} =4.42 m/s\\v_{2} =\frac{Q_2}{A_2} =\frac{0.2}{\frac{\pi }{4} \times 0.2^2} =6.37 m/s[/tex]
[tex]h_{L}=\frac{P_1-P_2}{\gamma}+z_1-z_2+\frac{v_1^{2}-v_2^{2} }{2g}[/tex]
[tex]h_{L}=\frac{(8-7.3)\times 100 }{9.81} +0+5+\frac{4.42^2-6.37^2}{2\times 9.81}[/tex]
[tex]h_L=7.135+3.927\\h_L=11.062m[/tex]
exit velocity head = [tex]\frac{v_2^{2} }{2g}[/tex]=2.068m
head loss as a function of exit velocity head is=[tex]\frac{11.062}{2.068}[/tex]
[tex]h_L=K\times V_e[/tex]
head loss as a function of exit velocity head =5.35m
Primary Creep: slope (creep rate) decreases with time
Answer:
true
Explanation:
Creep is known as the time dependent deformation of structure due to constant load acting on the body.
Creep is generally seen at high temperature.
Due to creep the length of the structure increases which is not fit for serviceability purpose.
When time passes structure gain strength as the structure strength increases with time so creep tends to decrease.
When we talk about Creep rate for new structure the creep will be more than the old structure i.e. the creep rate decreases with time.
I need help on Problem 2.5
I would like an explanation of how you got your answer. Thank you!!
Answer:
0.424
Explanation:
The electrical energy is the electrical power times time:
E = Pt
E = (IV)t
E = (1.5 A) (110 V) (300 s)
E = 49,500 J
The heat absorbed by the cookie dough is the mass times specific heat capacity times increase in temperature:
Q = mCΔT
Q = (1 kg) (4200 J/kg/K) (5 K)
Q = 21,000 J
So the fraction of electrical energy converted to internal energy is:
Q / E = 21,000 / 49,500
Q / E = 0.424
The remainder of the electrical energy is used to do work.
Why degree of crystallinity affects polymer properties?
Answer:
Degree of crystallinity affects polymer properties as increasing in crystallinity means higher the thermal stability and harder the material. Basically, crystallinity strongly affects polymer properties as it defines the degree of long range order in a material. Crystallinity is also determined by the size as well as molecular chain orientation.
A 12-ft circular steel rod with a diameter of 1.5-in is in tension due to a pulling force of 70-lb. Calculate the stress in the rod
Answer:
The stress in the rod is 39.11 psi.
Explanation:
The stress due to a pulling force is obtained dividing the pulling force by the the area of the cross section of the rod. The respective area for a cylinder is:
[tex] A=\pi*D^2/4 [/tex]
Replacing the diameter the area results:
[tex] A= 17.76 in^2 [/tex]
Therefore the the stress results:
[tex] σ = 70/17.76 lb/in^2 = 39.11 lb/in^2= 39.11 psi [/tex]
Design a solid steel shaft to transmit 14 hp at a speed of 2400rpm f the allowable shearing stress is given as 3.5 ksi.
Answer:
Is required a 0.8 inches diameter steel shaft.
Explanation:
With the power P and the rotating speed n (RPM), we can find the torque applied:
T = P/N
Before calculating the torque, we convert the power and rotating speed units:
[tex]P = 14\ HP * 550\ \frac{\frac{lb.ft}{s}}{HP} *\frac{12\ in}{ft} = 92400\ \frac{lb.in}{s} [/tex]
[tex]n=2400\ RPM .\frac{2\pi/60\frac{rad}{s}}{RPM}= 251\frac{rad}{s}[/tex]
Replacing the values, the torque obtained is:
[tex]T = \frac{92400\ lb.in/s}{251\ rad/s} = 368\ \ lb.in[/tex]
Then the maximum shearing stress will be located at the edge of the shat at the Maximus radius:
[tex]Smax =\ \frac{T.R}{J}[/tex]
Here J is the moment of inertia and R a radius. For a solid shaft, it is calculated by:
[tex]J =\frac{\pi.D^4}{32}[/tex]
Where D is shaft's diameter. Replacing the expression of J in
[tex]Smax =\frac{T.R}{\frac{\pi.D^4}{32}}[/tex]
As the radius is half of the diameter:
[tex]Smax =\frac{T.D}{\frac{2*\pi.D^4}{32\\} } = \frac{16T}{\pi.D^3}[/tex]
For the maximum stress of 3.5 ksi (3500 psi = 3500\ lb/in^2) and the calculated torque:
[tex]Smax = \frac{16.368\ lb.in}{3500\ lb/in^2*\pi.D^3}[/tex]
Solving for D:
[tex]D =\sqrt[3]{16.368\ lb.in / (3500\pi\ lb/in^2)}} = 0.8\ in[/tex]
Why is a Screw Pump a quiet operating pump?
Answer:
Screw pumps like those that are used in viscous fluids transportation have lubricating properties. The fluid flows in a line axially without any disturbence. Since, the motion of fluid is free from any sort of rotation even at high speeds there is no turbulence and motion of the pump is quite. Thus screw Pumps provides smooth operation with extremely low pulsation, lower noise levels and higher efficiency.
The contact angle between the mercury surface and capillary tube wall is______ A) Less than 90 B) Equal to 90 C) Greater than 90 D) Varying with mercury level
Answer:
The Answer to the question is :
Explanation:
The contact angle between the mercury surface and capillary tube wall is Greater than 90.
If the surface of the solid is hydrophobic, the contact angle will be greater than 90 °. On very hydrophobic surfaces the angle can be greater than 150º and even close to 180º.
What is the output of a system having the transfer function G = 2/[(s + 3) x(s + 4)] and subject to a unit impulse?
Answer:
output=[tex]\frac{2}{(s+3)(s+4)}[/tex]
Explanation:
output =transfer function ×input
here transfer function G=[tex]\frac{2}{(S+3)(S+4)} {}[/tex]
input = unit impulse
in S domain unit impulse =1
so output =[tex]\frac{2}{(S+3)(S+4)} {}[/tex]
=[tex]\frac{2}{(s+3)(s+4)}[/tex]
Which of the following components of a PID controlled accumulates the error over time and responds to system error after the error has been accumulated? a)- Proportional b)- Derivative c)- Integral d)- ON/OFF.
Answer:
D Is the answer babe....
Answer:
d)- ON/OFF.
Explanation:
ON/OFF components of a PID controlled accumulates the error over time and responds to system error after the error has been accumulated.
Describe the design experiments method for product and process optimization
Answer:
Design of Experiment also known as DoE is a systematic methodology to understand how any process or parameters of any product affects the response variables like physical properties or performance of any product, etc. It serves the purpose of making the job easier.
It is technique to generate valuable information required with minimum experimentation with the use of these:
Statistical methodologyMathematical analysis to predict the output within the limits of experiments at any point.Experimental extremities
Process Optimization:
It includes the following:
Product prototype should be used in designingOptimization of significant value adding activitiesDetection and minimization of errorStrong built concept for the designTesting and validation of the process for further improvement in efficiencyTime, quality, environmental, manufacting and overall cost, safety and operational constraints should be duly noted.For a fluid with a Prandtl Number of 1000.0, the hydrodynamic layer is thinner than the thermal boundary layers. a) True b) False
Answer:
(b)False
Explanation:
Given:
Prandtl number(Pr) =1000.
We know that [tex]Pr=\dfrac{\nu }{\alpha }[/tex]
Where [tex]\nu[/tex] is the molecular diffusivity of momentum
[tex]\alpha[/tex] is the molecular diffusivity of heat.
Prandtl number(Pr) can also be defined as
[tex]Pr=\left (\dfrac{\delta }{\delta _t}\right )^3[/tex]
Where [tex]\delta[/tex] is the hydrodynamic boundary layer thickness and [tex]\delta_t[/tex] is the thermal boundary layer thickness.
So if Pr>1 then hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.
In given question Pr>1 so hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.
So hydrodynamic layer will be thicker than the thermal boundary layer.
A disk-shaped part is to be cast out of aluminum. The diameter 500 mm and thickness = 20 mm. If the mold constant = 3.0 s/mm2 in Chvorinov's Rule, how long will it take the casting to solidify, in minutes?
Answer:
t =253.8s
Explanation:
Chvorinov's Rule can be written as:
[tex]t=B(\frac{V}{A} )^{n}[/tex]
where t is the solidification time,
V is the volume of the casting,
A is the surface area of the casting that contacts the mold,
n is a constant
B is the mold constant
The S.I. units of the mold constant B are s/m2.
According to Askeland, the constant n is usually 2.
[tex]V=\frac{\pi D^{2}h }{4} = 3.9*10^6[/tex]mm3
[tex]As=\pi D h+2\frac{\pi }{4} D^{2} =0.424*10^6[/tex] mm2
[tex]V/A=9.198[/tex]mm
[tex]t = 3.0*9.198^2[/tex] =253.8s
Answer:
Chvorinov's Rule with Askeland Method: t = 4.286694102 minutes
Chvorinov's Rule with Degarmo Method:
Minimum time required at constant n = 1.5 : t = 1.408751434 minutesMaximum time required at constant n = 2.0 : t = 4.286694102 minutesExplanation:
Data:
Aluminum disc
Diameter (D) = 500 mm
Thickness = Height (h) = 20 mm
Mold Constant (C) = 3.0 sec / [tex]mm^{2}[/tex]
Required:
Solidification time (t) in minutes = ?
Formula:
The solidification time can be found by using the Chvorinov's Rule:
[tex]t = C (\frac {V}{A})^{n}[/tex]
Where;
t = solidification time
C = mold constant
V = Volume of disc
A = Surface area of disc
n = constant
Note: According to Askeland n = 2.0 and According to Degarmo n varies 1.5 to 2.0 therefore , we will do for both method and by Degarmo method we can predict maximum and minimum solidification time.
Solution:
First, we will find the volume of the disc
disc = cylinder
therefore, Volume of cylinder is given by:
[tex]V = \frac{\pi }{4} * D^{2} * H[/tex]
Where:
V = Volume of Cylinder
H = Height of disc
D = Diameter of disc
Now, putting dimensional values in above equation
[tex]V = \frac{\pi }{4} * 500^{2} *20[/tex]
V = 3926990.817 [tex]mm^{3}[/tex]
Second, we will find the surface area of the disc
Therefore, surface area of cylinder is given by:
[tex]A = (\pi * D * H) + (2 * \frac{\pi }{4} * D^{2} )[/tex]
Where:
A = Surface area of disc
D = Diameter of disc
H = Height of disc
Now, putting dimensional values in above equation
[tex]A = (\pi * 500 * 20) + (2 * \frac{\pi }{4} * 500^{2} )[/tex]
A = 424115.0082 [tex]mm^{2}[/tex]
Finally, Moving towards the final solution
Chvorinov's Rule with Askeland Method n = 2:Rewriting the equation:
[tex]t = C (\frac {V}{A})^{2}[/tex]
Putting the dimensional and constants values in the equation
[tex]t = 3.0 (\frac {3926990.817}{424115.0082})^{2}[/tex]
t = 257.2016461 seconds
Converting to minutes
t = 4.286694102 minutes
Chvorinov's Rule with Degarmo Method n = 1.5 (Minimum Solidification Time)Rewriting the equation:
[tex]t = C (\frac {V}{A})^{2}[/tex]
Putting the dimensional and constants values in the equation
[tex]t = 3.0 (\frac {3926990.817}{424115.0082})^{1.5}[/tex]
t = 84.52508604 seconds
Converting to minutes
t = 1.408751434 minutes
Chvorinov's Rule with Degarmo Method n = 2.0 (Maximum Solidification Time)
Rewriting the equation:
[tex]t = C (\frac {V}{A})^{2}[/tex]
Putting the dimensional and constants values in the equation
[tex]t = 3.0 (\frac {3926990.817}{424115.0082})^{2}[/tex]
t = 257.2016461 seconds
Converting to minutes
t = 4.286694102 minutes
The melting point of Pb (lead) is 327°C, is the processing at 20°C hot working or cold working?
Answer:
Explained
Explanation:
Cold working: It is plastic deformation of material at temperature below recrystallization temperature. whereas hot working is deforming material above the recrystallization temperature.
Given melting point temp of lead is 327° C and lead recrystallizes at about
0.3 to 0.5 times melting temperature which will be higher that 20°C. Hence we can conclude that at 20°C lead will under go cold working only.
Which of these is not applicable to tuning for some Overshoot on Start-up? Select one: 1)-No overshoot during normal modulating control only 2)-KC-0.2 KU 3)- KC 0.33 KU 4)- KC 0.59 KU 5)- KC-0.78 KU
Answer: 5) KC-0.78 KU
Explanation:
KC-0.78 KU is not defined for tuning of overshoot on the Start-up as, this method is only applicable for there is some turning constants and also there is no overshoot during the normally modulating control. But some overshoot at start up are applicable. For the continuous cycling method, some overshoot is same as for closed loop tuning.
In a horizontal pipeline a 150 mm diameter pipe is connected to a 250 mm diameter pipe. The flow rate in the pipeline is 0.15 m^3/s. Take the connection as a sudden enlargement and determine: (a) The pressure head loss vhen the vater flovs from the large pipe to the smaller pipe (take Cc=0.64) (b) The pressure head loss when vater flovs from the small pipe to the larger pipe (c) The loss of power in both cases
Answer:
a) [tex]h_L=1.17m[/tex]
b)[tex]h_L=1.52m[/tex]
c)[tex]P_1=1.721 kN[/tex]
[tex]P_2=2.236 kN[/tex]
Explanation:
velocities of the pipe;
velocity of small dia pipe
[tex]v_{small}=\frac{Q}{A_{small}} =\frac{0.15}{\frac{\pi}{4}\times d^2 }= \frac{0.15}{\frac{\pi}{4}\times 0.15^2 }=8.52m/s[/tex]
velocity of larger dia pipe
[tex]v_{large}=\frac{Q}{A_{large}} =\frac{0.15}{\frac{\pi}{4}\times d^2 }= \frac{0.15}{\frac{\pi}{4}\times 0.25^2 }=3.05m/s[/tex]
a) pressure head loss when water is flowing from large to smaller pipe
[tex]h_L=(\frac{1}{C_c} -1)^2 \times \frac{v^2}{2g}[/tex]
[tex]h_L=(\frac{1}{0.64} -1)^2 \times \frac{8.52^2}{2\times 9.81} = 1.17m[/tex]
b) pressure head loss when water flow from small pipe to large pipe
[tex]h_L= \frac{v^2}{2g}(1-\frac{A_{small}}{A_{large}} )^2[/tex]
[tex]h_L=\frac{8.52^2}{2\times9.81}(1-\frac{\frac{\pi}{4}\times 0.15^2}{\frac{\pi}{4}\times 0.25^2}} )^2= 1.52 m[/tex]
c) power loss in both cases are
[tex]P_1= \rho g Q h_{L1}= 1000 \times 9.81\times 0.15\times 1.17= 1.721 kN[/tex]
[tex]P_2= \rho g Q h_{L2}= 1000 \times 9.81\times 0.15\times 1.52= 2.236 kN[/tex]
An automobile having a mass of 884 kg initially moves along a level highway at 68 km/h relative to the highway. It then climbs a hill whose crest is 69 m above the level highway and parks at a rest area located there. For the automobile, determine its changes in kinetic energy, in kJ
Answer:
ΔK.E. = - 142.72 kJ
Explanation:
mass = 884 kg
initial velocity = 68 km/h = 68 \times \frac {5}{18} = 18.89 m/s
final velocity = 0 m/s
height = 69 m
change in kinetic energy :
ΔK.E. = [tex]\dfrac{1}{2}m(v_f^2-v_i^2)[/tex]
ΔK.E. =[tex]\dfrac{1}{2}\times 884 \times (0^2-18.89^2)[/tex]
ΔK.E. =-142,716.05 J
ΔK.E. =-142.72 kJ
hence change in kinetic energy of the automobile is -142.72 kJ
In closed forging, why is flash needed and what is the effect with the overall pressure?
Explanation:
Closed die forging:
This is also known as impression die forging.By using the two die ,impression are made to produce desired forge product.A small gap provides between these two dies ,this small gap is called flash gutter.By the help of flash gutter excess ,material can flow and forms flash.
This flash plays an important role in the closed die forging.Generally friction is high in the flash gutter due to high ratio of length to thickness.So material feels high pressure and confirm the filling of die cavity.
Which atom bond in atomic interaction combines electrons, filling its valence zone a) Van der Vaals bond; b) a covalent bond; c) ionic bond; d) metallic bond.
Answer: Covalent bond
Explanation: Covalent bond is the bond that gets created when there is a sharing of electrons among atoms and hence creating atomic interaction. The bond formed is from the shared pair because they allow the atoms or ions to achieve stability by completely filling the outer shell of the electron and thus form the covalent bond .Therefore, the correct option is the option(b) .
An important material for advanced electronic technologies is the pure silicon.a)-True b)-False
Answer: False
Explanation:
Pure silicon is a good semiconductor, which conducts electricity when is mixed with some other component and can also work as an insulator.It is found abundantly on the earth's crust area. It is widely used in mixed form in the advanced electronic technologies but not in pure silicon form . So silicon is the important material in the advanced electronic technologies but not the pure silicon form.
Which of the following is/are not a common crystal structure in metals (mark all that apply)? a. Face-centered cubic (FCC) b. Face-centered orthorhombic (FCOR) C. Body-centered cubic (BCC) (HCP)
Answer: b) FCOR( Face-centered orthorhombic)
Explanation: Face centered orthorhombic lattice is the lattice that has eight lattice corner points and they also have each face with center lattice point.This lattice structure is usually not common in metals.Whereas the face centered lattice structure , body structure lattice and hexagonal close packed lattice are common in metal.Thus option (b) is the correct option.