This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 18651. |
What will be the output of the following Python code? def add (num1, num2): sum = num1 + num2 sum = add(20,30) print(sum) a. 50 b. 0 c. Null d. None |
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Answer» Answer is d. None |
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| 18652. |
Raghav is trying to write a tuple tup1 = (1,2,3,4,5) on a binary file test.bin. Consider the following code written by him. import pickle tup1 = (1,2,3,4,5) myfile = open("test.bin",'wb') pickle........ #Statement 1 myfile.close() Identify the missing code in Statement 1. a. dump(myfile,tup1) b. dump(tup1, myfile) c. write(tup1,myfile) d. load(myfile,tup1) |
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Answer» b. dump(tup1, myfile) |
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| 18653. |
A binary file employee.dat has following data Empno empname Salary101Anuj50000102Arijita40000103Hanika30000104Firoz60000105Vijaylakshmi40000def display(eno): f=open("employee.dat","rb") totSum=0 try: while True: R=pickle.load(f) if R[0]==eno: ......... #Line1 totSum=totSum+R[2] except: f.close() print(totSum) When the above mentioned function, display (103) is executed, the output displayed is 190000. Write appropriate jump statement from the following to obtain the above output.a. jump b. break c. continue d. return |
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Answer» Answer is c. continue |
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| 18654. |
The command used to give a heading to a graph is |
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Answer» The title() command is used to provide a title to the graph plotted in MATLAB. |
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| 18655. |
Logic gate diagram |
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Answer» Logic gates are the basic building blocks of any digital system. It is an electronic circuit having one or more than one input and only one output. The relationship between the input and the output is based on a certain logic. Based on this, logic gates are named as AND gate, OR gate, NOT gate etc. |
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| 18656. |
Which of the disease caused due to water pollution?(A) Eye problem (B) Respiratory problem(C) Dysentery (D) Deafness |
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Answer» The disease caused due to water pollution is Dysentery. |
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| 18657. |
Headquarter of North-Central railway is:-(A) Allahabad (B) Hajipur(C) Kolkata (D) Jabalpur |
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Answer» Headquarter of North-Central railway is Allahabad. |
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| 18658. |
Which one of the following is the major exported goods of India?(A) Engineering goods (B) Electronic goods(C) Chemical and related product (D) None of these |
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Answer» Engineering goods is the major exported goods of India. |
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| 18659. |
What is fermentation? |
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Answer» Fermentation is an anaerobic process in which energy can be released from glucose even though oxygen is not available. Fermentation occurs in yeast cells, and a form of fermentation takes place in bacteria and in the muscle cells of animals. 1.Yeast uses sugar for food. 2.Yeast grows and multiplies rapidly due to carbon compound in the sugar solution. 3.In the process of obtaining nutrition yeast cells convert the carbohydrates in the food into alcohol and carbon dioxide. 4.Also, the bacteria lactobacilli converts lactose, the sugar in milk into lactic acid. 5.This process is called fermentation. |
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| 18660. |
Hyderabad and Secunderabad city are referred as:-(A) Twins (B) Dark(C) Slum (D) None of these |
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Answer» Hyderabad and Secunderabad city are referred as Twins. |
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| 18661. |
The “Silicon Valley of India” refers to ______.1. Kulu valley2. Hyderabad3. Bangalore4. Kolar |
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Answer» Correct Answer - Option 3 : Bangalore The Correct Answer is Bangalore.
Hyderabad
Kulu valley
Kolar
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| 18662. |
Which of the following is known as the Diamond City of India? 1. Surat2. Panna3. Mumbai4. Jaipur |
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Answer» Correct Answer - Option 1 : Surat The Correct Answer is Surat.
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| 18663. |
Which of the following state has highest density of population in India?(A) Bihar (B) Uttar Pradesh(C) Kerala (D) Punjab |
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Answer» Bihar has highest density of population in India. The records of population density 2011 of India state that the density 2011 has increased from a figure of 324 to that of 382 per square kilometer. Bihar is the most thickly populated state |
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| 18664. |
The silk route is an early example of long distance trade which connecting:-(A) Spain-Norway (B) Seattle-Sanfrancisco(C) Kashmir-Kanyakumari (D) Rome-china |
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Answer» The silk route is an early example of long distance trade which connecting Rome-china. |
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| 18665. |
Which one of the following countries has the criterion of 300 population for urban classification?(A) Denmark (B) Finland(C) Iceland (D) Sweden |
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Answer» Iceland has the criterion of 300 population for urban classification. |
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| 18666. |
Consider the following minerals:1. Bentonite2. Chromite3. Kyanite4. SillimaniteIn India, which of the above is/are officially designated as major minerals?1. 1 and 2 only2. 4 only3. 1 and 3 only4. 2, 3 and 4 only |
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Answer» Correct Answer - Option 4 : 2, 3 and 4 only The correct answer is 2, 3, and 4 only.
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| 18667. |
Pin Valley National Park is located in which state of India?1. Himachal Pradesh2. Jammu & Kashmir3. Punjab4. Gujarat |
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Answer» Correct Answer - Option 1 : Himachal Pradesh The correct answer is Himachal Pradesh.
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| 18668. |
Transfer of words, messages, facts and thoughts is called?(A) Communication (B) Trade(C) Transportation (D) None of these |
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Answer» Transfer of words, messages, facts and thoughts is called Communication. |
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| 18669. |
Which of the following types of industry manufacturers raw material for other industry ?(A) Cottage industry (B) Basic industry(C) small Scale Industry (D) Footloose Industry |
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Answer» Basic industry manufacturers raw material for other industry. |
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| 18670. |
Identify and explain each of the parts of a computer system |
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Answer» There are five main hardware components in a computer system: Input, Processing, Storage, Output and Communication devices. Are devices used for entering data or instructions to the central processing unit. Are classifie according to the method they use to enter data. |
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| 18671. |
Aravalli ranges are an example of :A. stright mountainB. B block mountainC. C fold mountainD. D volcanic mountain |
| Answer» Correct Answer - C | |
| 18672. |
Which enzyme joins the Okazaki fragments? |
| Answer» The enzyme DNA ligase joins the Okazaki fragments. | |
| 18673. |
List the geographical information depicted using red colour. |
Answer»
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| 18674. |
Which is the official agency in India responsible for the preparation of topographical maps? Why are restrictions imposed on the use of topographical maps? |
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Answer» In India, the Survey of India is entrusted with the preparation of topographical maps. Certain restrictions have been imposed on the use of topographical maps of strategic regions owing to the national security concerns. |
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| 18675. |
How did Mount Everest, the highest peak in the Himalayas get that name? |
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Answer» The survey works in India began in 1802 under Col. William Lambton. Col. George Everest joined as an assistant to Lambton in 1818. This was the first survey that recorded the correct measurements of the Himalayan mountain ranges. As a tribute to George Everest who took charge of the survey after Lambton, the highest peak in the Himalayan mountain ranges was given the name Everest. |
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| 18676. |
Balance redox reaction by oxidation number and ion electron method |
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Answer» Ion electron Method Sno + \(\overset{+5}{NO}\)3- + H+ \(\rightarrow\) Sn+2 + \(\overset{-3}{NH}\)4+ + H2O first of all separating half cell reaction Sn \(\rightarrow\) Sn+2 +2e- (oxidation)....(i) No3- + 8e- \(\rightarrow\) NH4+ (reduction)....(ii) In equation (ii), adding 3H2o in the right side (for balancing oxygen) NO3- + 8e-\(\rightarrow\) NH4+ + 3H2O...(iii) Now, balancing atom, and charge adding 10H+ in the left side in equation (iii) we got NO3- + 8e- + 10H+ \(\rightarrow\) NH4\(\bigoplus\) + 3H2O....(iv) Now, multiply by '5' in equation (i) and then adding in equation (iv) we got 5Sn + NO3- + 10H+ \(\rightarrow\) 5Sn+2 + NH4+ + 3H2O....(v) equation (v) is a balance equation Ion electron Method \(\overset{0}{AS}\) + \(\overset{+5}{NO}\) 3- + H+ \(\rightarrow\) A\(\overset{+5}{SO}\)4-3 + \(\overset{+4}{NO}\)2 + H2O separating half cell reaction \(\overset{+5}{NO}\)3- + e- \(\rightarrow\) \(\overset{+4}{NO}\)2....(i) \(\overset{0}{AS}\) \(\rightarrow\) A \(\overset{+5}{SO}\)4-3 + 5e- ...(ii) For balancing oxygen, adding (i) we get NO3- + e- \(\rightarrow\) NO2 + H2O....(iii) For balancing Hydrogen and charge , adding 2H+ in the left side in the equation (iii) we got NO3- + 2H+ + e- \(\rightarrow\) NO2 + H2O....(iv) For equation (ii) For balancing oxygen, adding 4H2O in left side, ion equation (ii) we got AS- + 4H2O\(\rightarrow\) ASO4-3 + 5e-....... (v) For balancing Hydrogen, and change adding OH+ in the right side, in equation (v) we got, AS + 4H2O\(\rightarrow\) ASO4-3 + 8H+ + 5e-...(vi) Now equation (iv) Multiply by '5' then adding equation (vi) we got AS + 5NO3- + 2H+ \(\rightarrow\) ASO4-3 + 5NO2 + H2O.....(vii) equation (iv) is balance equation |
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| 18677. |
80. When \( Cl _{2} \) gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from(1) zero to \( +1 \) and zero to \( -5 \)(2) zero to \( -1 \) and zero to \( +5 \)(3) zero to \( -1 \) and zero to \( +3 \)(4) zero to \( +1 \) and zero to \( -3 \) |
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Answer» The reaction of chlorine gas with hot and concentrated sodium hydroxide solution is 3Cl2 + 6NaOH⟶NaClO3 + 5NaCl+3H2O Oxidation number of Cl is 0 in Cl2, −1 in NaCl and +5 in NaClO3 So the oxidation number of chlorine changes from Zero to -1 and Zero to +5. |
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| 18678. |
What are contour lines? |
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Answer» Contours are imaginary lines drawn connecting places having equal elevation from sea level. |
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| 18679. |
Where is the headquarters of Survey of India? |
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Answer» The headquarters of Survey of India Dehradun |
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| 18680. |
Assertion: Volume of a gas is inversely proportional to the number of moles of a gas. Reason: The ratio by volume of gaseous reactants and products is in agreement with their molar ratio.A. (a)If both assertion and reason are true and the reason is the correct explanation of the assertion.B. (b)If both assertion and reason are true and the reason is not the correct explanation of the assertion.C. (c )If assertion is true but reason is false.D. (d)If assertion is false but reason is true. |
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Answer» Correct Answer - D We know that from the reaction `H_(2)+Cl_(2) rarr 2HCl` that the ratio of the volume of gaseous reactants and products is in agreement with their molar ratio. The ratio of `H_(2):Cl_(2):HCl` volumes is `1:1:2` which is the same as their molar ratio. Thus volume of gas is directly related to the number of moles. Therefore, the assertion is false but reason is true. |
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| 18681. |
Isotopic number of `._(26)^(58)Fe` is: |
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Answer» Correct Answer - 6 6 Isotopic no `=n-p(58-26)-(26)=32-26=6` |
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| 18682. |
The number of electrons for `Zn^(2+)` cation that have the value of azimuthal quantum number `=0` is: |
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Answer» Correct Answer - 6 6 Azimuthal quantum number is zero for s-subshell. |
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| 18683. |
Which metal has a greater tendency to from metal oxide?A. `Cr`B. `Ca`C. `A1`D. Fe` |
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Answer» Correct Answer - B Because the change in free enegry, i.e., `DeltaG` is highly negative for this reaction. |
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| 18684. |
In a solid A, B, C are arranged as above, the formula of solid is :A. ABCB. `AB_(2)C_(2)`C. `A_(2)BC`D. `AB_(8)C_(2)` |
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Answer» Correct Answer - A `A=1` ltBrgt `B=(8)/(8)=1" "therefore ABC` `C=(2)/(2)=1` |
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| 18685. |
A 200 MVA and 1000 MVA synchronous machine has H1 = 2 MJ/MVA and H2 = 4 MJ/MVA respectively which are operating in parallel. Calculate the equivalent H constant for both relative to a 10 MVA base.1. 44 MJ/MVA2. 440 MJ/MVA3. 444 MJ/MVA4. 4400 MJ/MVA |
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Answer» Correct Answer - Option 2 : 440 MJ/MVA Concept: The KVA is inversely proportional to the inertia constant. \(S \propto \frac{1}{H}\) \(\Rightarrow \frac{{{S_{new}}}}{{{S_{old}}}} = \frac{{{H_{old}}}}{{{H_{new}}}}\) Calculation: Given that, S1 = 200 MVA, H1 = 2 MJ/MVA S2 = 1000 MVA, H2 = 4 MJ/MVA Common base = 10 MVA. We know that \(S \propto \frac{1}{H}\) \(\Rightarrow \frac{{{S_{new}}}}{{{S_{old}}}} = \frac{{{H_{old}}}}{{{H_{new}}}}\) For generator 1: \({H_{new}} = \frac{{200}}{{10}} \times 2 = 40\;MJ/MVA\) For generator 2: \({H_{new}} = \frac{{1000}}{{10}} \times 4 = 400\;MJ/MVA\) Equivalent inertia constant (H) = 400 + 40 = 440 MJ/MVA |
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| 18686. |
A cylindrical portion of radius r is removed from a solid sphere of radius R and uniform volume chargedensity `rho` in such a way that the axis of the hollow cylinder coincides with one of the diameters of thesphere. (r is negligible compared to R). Then the electric field intensity at point A isA. `(rhor)/(3epsilon_(0))hati`B. `-(rhor)/(3epsilon_(0))hati`C. `(rhor)/(6epsilon_(0))hati`D. `-(rhor)/(6epsilon_(0))hati` |
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Answer» Field at A Due at A due the cylinder of length 2R (which can assumed to be inifinite, since `r lt lt R`) `E_(2)=(2K(rhopir^(2)))/(r)(-hati)=-(rho)/(2epsilon_(0))rhati` `therefore` net field `E=E_(1)+E_(2)=(rhor)/(6epsilon_(0))hati` |
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| 18687. |
In the figure shown a massless spring of stiffness `k` and natural length `l_(0)` is rigidly attached to a block of mass `m` and is in vertical position. A wooden ball of mass `m` is released from rest to fall under gravity. Having fallen a height `h` the ball strikes the spring and gets stuck up in the spring at the top. What should be the minimum value of `h` so that the lower block will just lose contact with the ground later on ? Assume that `l_(0) gt gt (4mg)/(k)`. Neglect any loss of energy. (Given `k=mg//2`) |
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Answer» The minimum force needed to lift the lower block is equal to its weight. During upward motion the spring will get elongated. If elongation in the spring for just lifting the block is `x_(0)` then `kx_(0)=mg` `rArrx_(0)=(mg)/(k)`……`(i)` From `COE` `mg(l_(0)+h)=mg(l_(0)+x_(0))+(1)/(2)kx_(0)^(2)` `rArr mgh=mgx_(0)+(1)/(2)kx_(0)^(2)rArr mgh=((mg)^(2))/(k)+(1)/(2)(m^(2)g^(2))/(k)rArrh=(3mg)/(2k)` During down ward motion, suppose maximum compression in the spring is `x` From `COE` `mg(l_(0)+h)=mg(l_(0)-x)+(1)/(2)kx^(2)` `rArrmgh= -mgx+(1)/(2)kx^(2)rArrmg"(3mg)/(2k)= -mgx+(1)/(2)kx^(2)` `rArr 3(mg)^(2)= -2mgkx+k^(2)x^(2)` `rArr k^(2)x^(2)-2mgkx-3(mg)^(2)=0` `rArr x=(2mgk+-sqrt(4(mgk)^(2)+12k^(2)(mg)^(2)))/(2k^(2))=(2mgk+-4mgk)/(2k^(2))rArrx=(3mg)/(k)` |
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| 18688. |
Assertion `:-` The mechanical energy of earth-moon system remains same when a another heavenly body passes nearby the earth`-`moon system. Reason `:-` Force exerted by heavenly body on the earth`-`moon system is non`-`conservative.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertiion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are False |
| Answer» Correct Answer - D | |
| 18689. |
Find the ratio of stress present in wire-1 to the wire-2 shown in the diagram while both wires have same material `&` friction. A. 1B. 2C. 3D. 4 |
| Answer» Tension in wire 1 is two times the tension in wire 2. | |
| 18690. |
A uniform rod of length l is rotating in a gravity free region about an axis passing through its one end and perpendicular to it length. Choose graph between stress developed and distance the axis of roation.A. B. C. D. |
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Answer» `T=(m omega^(2))/(2l)(l^(2)-x^(2))` stress =T/A |
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| 18691. |
As per cooled water freezes spontaneously, its temperature rises to `0^(@)C. DeltaH` for the spontaneous process. `H_(2)O_((l))(-10^(@)C)toH_(2)O_((s))(0^(@)C)` is :A. zeroB. `+ve`C. `-ve`D. either of these |
| Answer» No energy is transferred to or from the system. The energy liberated in the freezing process warms the system to `0^(@)C`. Thus `DeltaH=0` | |
| 18692. |
Number of `Na^(+)` and `Cl^(-)` ions associated with each a unit cell of `NaCl` is: |
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Answer» `NaCl` has `fcc` structures `8` at the corners is shared by `8` unit cells. Each of `6 Na^(+)` ions at `fcc` is shared by `2` unit shell. `:.` No. of `Na^(+)` in a each unit cell `=8/8+6/2=4` `12 Cl^(-)` at the middle of each edge of cell is shared by `4` cell and `Cl^(-)` at the centre is wholly belong to it. `:.` No. of `Cl^(-)` in unit cell `=12/4+1=4` |
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| 18693. |
`1g` pure iron is dissolved in excess of `H_(2)SO_(4)`. The clear filtrate is made up `100 mL. 10mL` of this solution is treated with `0.1 M KMnO_(4)` solution till whole of the `Fe^(2+)` ions are oxidised to `Fe^(3+)` ions. Now `0.2 g Fe_(2)(SO_(4))_(3)` is dissolved in it. the solution is now treated with `Zn` and `H_(2)SO_(4)`. The amount of `K_(2)Cr_(2)O_(7)` to be dissolved to prepare `V mL` of `K_(2)Cr_(2)O_(7)`, which is just sufficient to completely oxidised `10 mL` of above `FeSO_(4)` solution ?A. `0.0875 g`B. `0.875 g`C. `8.75 g`D. `0.0087g` |
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Answer» Meq. Of `Kr_(2)Cr_(2)O_(7)` needed for `FeSO_(4)` =meq of `Fe` in `10 mL` `w/49xx1000/VxxV=1/56xx1000x10/100` `:. W=0.0875 g` |
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| 18694. |
`1g` pure iron is dissolved in excess of `H_(2)SO_(4)`. The clear filtrate is made up `100 mL. 10mL` of this solution is treated with `0.1 M KMnO_(4)` solution till whole of the `Fe^(2+)` ions are oxidised to `Fe^(3+)` ions. Now `0.2 g Fe_(2)(SO_(4))_(3)` is dissolved in it. the solution is now treated with `Zn` and `H_(2)SO_(4)`. The ratio of equivalent of `KMnO_(4)` and `K_(2)Cr_(2)O_(7)` used for reducing the solution is:A. `5//6`B. `6//5`C. `1`D. `2` |
| Answer» Eq. of `KMnO_(4)`=Eq. of `K_(2)Cr_(2)O_(7)`=Eq. of `Fe^(2+)` | |
| 18695. |
`1g` pure iron is dissolved in excess of `H_(2)SO_(4)`. The clear filtrate is made up `100 mL. 10mL` of this solution is treated with `0.1 M KMnO_(4)` solution till whole of the `Fe^(2+)` ions are oxidised to `Fe^(3+)` ions. Now `0.2 g Fe_(2)(SO_(4))_(3)` is dissolved in it. the solution is now treated with `Zn` and `H_(2)SO_(4)`. The volume of `0.1 M K_(2)Cr_(2)O_(7)` used after reducing the solution mixture with `Zn+H_(2)SO_(4)` is :A. `4.64 mL`B. `5.46 mL`C. `3.46 mL`D. `2.64 mL` |
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Answer» Meq. Of `Kr_(2)Cr_(2)O_(7)=Meq. Of Fe^(2+)` [from `Fe+` from `Fe_(2)(SO_(4))_(3)`] `V=4.64 mL` |
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| 18696. |
`1g` pure iron is dissolved in excess of `H_(2)SO_(4)`. The clear filtrate is made up `100 mL. 10mL` of this solution is treated with `0.1 M KMnO_(4)` solution till whole of the `Fe^(2+)` ions are oxidised to `Fe^(3+)` ions. Now `0.2 g Fe_(2)(SO_(4))_(3)` is dissolved in it. the solution is now treated with `Zn` and `H_(2)SO_(4)`. The volume of `0.1 M KMnO_(4)` used after reducing the solution mixture with `Zn+H_(2)SO_(4)` is:A. `5.572 mL`B. `3.572 mL`C. `4.572 mL`D. `6.572 mL` |
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Answer» Meq. of `Fe^(2+)` [form `Fe`]+Meq. of `Fe^(2+)` [from `Fe_(2)(SO_(4))_(3))`]=Meq of `KMnO_(4)` `1/56xx1000xx10/100+0.2/(400/2)xx1000=0.1xx5xxV` `:. V=([1.786+1])/0.5=5.572 mL` |
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| 18697. |
`1g` pure iron is dissolved in excess of `H_(2)SO_(4)`. The clear filtrate is made up `100 mL. 10mL` of this solution is treated with `0.1 M KMnO_(4)` solution till whole of the `Fe^(2+)` ions are oxidised to `Fe^(3+)` ions. Now `0.2 g Fe_(2)(SO_(4))_(3)` is dissolved in it. the solution is now treated with `Zn` and `H_(2)SO_(4)`. Select the correct statement. `1.` The `Fe^(3+)` ions present in solution are reduced by `Zn` and `H_(2)SO_(4)` `2. H_(2)` gas formed by the action of `Zn` and `H_(2)SO_(4)` is reducing agent. `3.` Atomic form of `H` formed by the action of `Zn` and `H_(2)SO_(4)` is reducing agent `4.` Nascent form of `H` formed by the action of `Zn` and `H_(2)SO_(4)` is reducing agentA. `1,2`B. `1,3`C. `1,4`D. `1,2,3` |
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Answer» `Zn+H_(2)SO_(4)toZnSO_(4)+underset("Nascent H")(2H)` `Fe^(3+)+HtoFe^(2+)+H^(+)` |
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| 18698. |
`1g` pure iron is dissolved in excess of `H_(2)SO_(4)`. The clear filtrate is made up `100 mL. 10mL` of this solution is treated with `0.1 M KMnO_(4)` solution till whole of the `Fe^(2+)` ions are oxidised to `Fe^(3+)` ions. Now `0.2 g Fe_(2)(SO_(4))_(3)` is dissolved in it. the solution is now treated with `Zn` and `H_(2)SO_(4)`. The volume of `KMnO_(4)` needed to convert `Fe^(2+)` ions to `Fe^(3+)` ions in `100 mL` original solution is:A. `71 mL`B. `142 mL`C. `35.7 mL`D. `80 mL` |
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Answer» Meq. Of `Fe^(2+)`=Meq. Of `KMnO_(4)` (for `100 mL` solution) `1/56xx1000=0.1xx5xxV` `:. V=35.7 mL` |
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| 18699. |
A charged particle (charge `q`) is moving in a circle of radius `R` with unifrom speed `v`. The associated magnetic moment `mu` is given byA. `(qvR)/(2)`B. `qvR^(2)`C. `(qvR^(2))/(2)`D. `qvR` |
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Answer» Correct Answer - A As revolving charge is equivalent to a current, so `I = af = q xx (omega)/(2pi)` But `omega = (v)/(R )` where `R` is radius of circle and `v` is unifrom speed of charged particle. Therefore, `I = (qv)/(2pi R)` Now, magetic moment associated with charged particles is given by `mu = IA = I xx pi R^(2)` or `mu = (qv)/(2piR) xx piR^(2) = (1)/(2)qvR` |
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| 18700. |
The maximum number of atoms that may lie in the same plane in `(CH_(3))_(2)C=SF_(2)(CH_(3))(2)` is …… |
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Answer» Correct Answer - 8 Fact |
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