This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
……………. conducted important research on the toxin responsible for gas gangrene and the antitoxin responsible for treating it.(a) Ida Bengston (b) Van Ermengem (c) Louis Pasteur (d) Alexander Fleming |
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Answer» Correct answer is: (a) Ida Bengston |
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| 2. |
How many different industries depend upon the Lactobacilli bacteria? |
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Answer» Industries like milk products, cider, cocoa, pickles, pharmaceuticals depend on Lactobacilli bacteria. |
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| 3. |
Match the pairs:‘A’ group‘B’ group1. Rhizobiuma. Food poisoning2. Clostridiumb. Nitrogen fixation3. Penicilliumc. Bakery products4. Yeastd. Production of antibiotics |
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| 4. |
Which among the following statements is incorrect for magnesium metal?(a) It burns in oxygen with a dazzling white flame (b) It reacts with cold water to form magnesium oxide and evolves hydrogen gas (c) It reacts with hot water to form magnesium hydroxide and evolves hydrogen gas (d) It reacts with steam to form magnesium hydroxide and evolves hydrogen gas |
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Answer» The answer is (b) It reacts with cold water to form magnesium oxide and evolves hydrogen gas. Magnessium reacts with cold water to form magnesium oxide and evolves hydrogen gas is a wrong statement because Magnessium Hydroxide is formed not Magnessium oxide. |
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| 5. |
………………… produce bottle-shaped endospores in adverse conditions.(a) Lactobacilli (b) Clostridium (c) Yeast (d) Rhizobium |
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Answer» Correct answer is: (b) Clostridium |
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| 6. |
Which types of cottage industries and factories can be started in areas with abundant milk production? |
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Answer» Cottage industries like the manufacture of milk products like ghee, cheese, paneer, curd, shrikhand, etc. and chocolate making can be started in areas with abundant milk production. |
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| 7. |
Which plant and animal diseases are caused by micro-organisms and what are the 1 measures to be taken against them? |
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Answer» Plant diseases:
Preventive Measures:
Animal diseases:
Preventive Measures:
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| 8. |
The rod-shaped bacteria found in milk or buttermilk are called ………(a) Rhizobium (b) Clostridium (c) Lactobacilli (d) Saccharomyces |
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Answer» Correct answer is: (c) Lactobacilli |
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| 9. |
A metal 'x' available in the form of ribbons burn in air with a dazzling white light, once heated to its ignition temperature. It leaves a white powder 'y' which dissolves in water. The solution so formed turns red litmus blue. Identify 'x' and 'y' and write the balanced chemical equation of the reaction. |
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Answer» Magnesium ribbon burns with dazzling white light, once treated to its ignition temperature. The white powder formed is Magnesium oxide which dissolves in water to give a basic solution which turns red litmus blue. Thus 'x' is Mg and 'y' is MgO. 2MgO+O2 → 2MgO MgO+H2O → Mg(OH)2 |
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| 10. |
Which of the following is not a double decomposition reaction ? A) Pb(NO3)2 + 2 KI → PbI2 + 2KNO3 B) NaCl + AgNO3 → AgCl + NaNO3 C) Na2SO4 + BaCl2 → BaSO4 + 2NaCl D) Fe + CuSO4 → FeSO4 + Cu |
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Answer» D) Fe + CuSO4 → FeSO4 + Cu |
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| 11. |
Find out the uses of fungi to plants and animals? |
Answer»
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| 12. |
What is decomposition reaction ? Explain it with suitable example. |
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Answer» A simple reactant breaks down to give simpler products and the process is known as decomposition reaction. Example: CaCO3(s) + Heat → CaO(s) +CO2(g) CaO(s)-Quick lime |
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| 13. |
Balance the following chemical equationMnO2 + HCI → MnCl2+ Cl2 + H2O |
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Answer» MnO2 + 4HCI → MnCl2+ 2H2O+Cl2 |
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| 14. |
What is the structure of lichen, a condiment? Where else is it used? |
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| 15. |
Which electronic shell is at a higher energy level K or L? |
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Answer» L – shell is at higher energy level, because it is far from nucleus than K shell. |
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| 16. |
An orbital is designated by …………. quantum numbers while an electron in an atom is designated by …………. quantum numbers.(A) two, three (B) three, two (C) four, two (D) three, four |
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Answer» Option : (D) three, four An orbital is designated by three quantum number (n , l , m) while an electron in an atom is designated by four quantam number (n , l , m and s) Option : (D) three, four |
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| 17. |
Two solutions X and Y are tested with universal indicator. Solution X turns orange whereas solution Y turns red. Which of the solutions is a stronger acid ? |
| Answer» Solution Y is a stronger acid. | |
| 18. |
Name the acid present in Tomato. |
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Answer» Tomato - Malic acid and citric acid |
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| 19. |
State the meaning of strong acids and weak acids. Give one example of each. |
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Answer» Strong acid: An acid which dissociates completely in water and produces a large amount of hydrogen ions. e.g., HCl. Weak acid: An acid which dissociates partially in water and produces a small amount of hydrogen ions. e.g., CH3COOH. |
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| 20. |
Write a short note on electron spin quantum number. |
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Answer» Electron spin quantum number (ms) :
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| 21. |
What are the values of n and l for 2p orbital? |
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Answer» n = 2 and l = 1 |
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| 22. |
The least electronegative element is A) Cs B) F C) CZ D) H |
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Answer» Correct option is A) Cs |
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| 23. |
An atom has two electrons in its 4s orbital. Write the values of the four quantum numbers for each of them. |
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Answer» For the 4s orbital, 4 stands for the principal quantum number n; s stands for the subshell s having the value of azimuthal quantum number, l = 0. In the ‘s’ subshell, there is only one orbital and has magnetic quantum number, ml = 0. The two electrons in this orbital have opposite spins. Thus, The four quantum numbers of two electrons in 4s orbital are :
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| 24. |
The bromine atom possesses 35 electrons. It has 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electron experiences the lowest effective nuclear charge? |
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Answer» Bromine (Atomic No. 35) = 1s2 2s2 2p6 3s2 2p6 3d10 4s2 4p5 2p orbital has 6 electrons. 3p orbital has 6 electrons and 4p orbital has 5 electrons. 4p electron experience lowest effective nuclear charge because of the maximum of screening or shielding effect. |
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| 25. |
The velocity associated with a proton moving in a potential difference of 1000 v is 4.37 x 105 ms-1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity. |
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Answer» v = 4.37 x 105 ms-1 m = 0.1 kg h = 6.626 x 10-34 Js λ = h/mv or, λ = {6.626 x 10-34}/{10-1 x 4.37 x 105} or, λ = 6.626/4.37 x 10-38 = 1.51 x 10-38 m. |
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| 26. |
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination (s) has/have the same energy lists.1. n = 4, l = 2, m1 = -2, ms = -1/22. n = 2, l = 2, m1 = 1, ms = +1/23. n = 4, l = 1, m1 = 0, ms = +1/24. n = 3, l = 2, m1 = -2, ms = -1/25. n = 3, l = 1, m1 = -1, ms = +1/26. n = 4, l = 1, m1 = 0, ms = +1/2 |
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Answer» (i) Represents 4d orbital (ii) Represents 3d orbital, (iii) Represents 4p orbital (iv) Represents 3d orbital (v) Represents 3p orbital (vi) represents 4p orbital 3p < 3d < 4p < 4d Combination (2) n = 3, l = 2, m1 = 1, ms = +1/2 and combination (4) n = 3, l = 2, m1 = -2 and ms = -1/2 have the same energy lists. Similarly combination (3) n = 4, l = 1, m1 = 0, and ms = 1/2 and combination (6) n = 4, l = 1, m1 = 0, ms = +1/2 have the same energy lists. Therefore, (v) < (iii) = (iv) < (vi) = (iii) < (i) |
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| 27. |
What are quantum number and Name them? |
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Answer» In order to define state energy and location of electron a set of four numbers is used. These numbers are called quantum numbers. • Principal quantum number (n) • Azimuthal quantum number (1) • Magnetic quantum number (m) • Spin quantum number (s). |
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| 28. |
When do you see an emission line? |
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Answer» The energy emitted by an electron is seen in the form of an electromagnetic energy and when the wavelength is. in the visible region it is visible as an emission line. |
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| 29. |
How would the velocity be effected if the position is known? |
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Answer» If the position of the electron is known with high degree of accuracy (Δx is small), then the velocity of the electron will be uncertain (Δ(Vx) is large.) |
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| 30. |
We don not see a car moving as a wave on the road why? |
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Answer» According to de Broglie’s relation, λ = h/mv ie λα1/m of the car is very large and its wavelength (λ) or wave character is negligible. Therefore, we do not see a car moving like a wave. |
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| 31. |
Define the term Angular nodes. |
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Answer» Angular nodes occur when the probability density wave function for the electron is zero along the directions specified by a particular angle. Number of angular nodes = l Total number of nodes = n – 1 |
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| 32. |
Which has more energy of election: (i) Last election of Cl− or 1/last electron of O2− (ii) n = 4,l = 3 or n = 5,l = 2 |
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Answer» Electronic Configuration: Cl− = 2, 8, 8 = 1s2 , 2s22p6 , 3s23p6 O2− = 2, 8 = 1s2 , 2s2sp6 Since last electron of Cl− occurs in 3p sub- shell and last electron of O2− occurs 2p subshell. As we know that, according to Aufbau principle, 3p Subshell has more energy than 2p subshell. Therefore, last electron of Cl− has more energy of electron. (ii) n = 4, l = 3 represent 4f n = 5, l = 2 represents 5p According to Aufbau principle, the energy of 4f is more than 5p. So, n = 4, l = 3 has more energy of electron. |
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| 33. |
Carbon with atomic number 6 and silicon with atomic number 14 belong to same group although carbon is non-metal and silicon is semi-metal. |
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Answer» Carbon with atomic number 6, shows electronic configuration 2, 4. Silicon with atomic number 14 shows electronic configuration 2, 8, 4. Both the elements have same valence electrons, hence they are placed in same group. |
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| 34. |
Explain giving reason, which of the following sets of quantum numbers are not possible:(i) n = 1,l = 1,ml = 0, ms = \(+\frac{1}{2}\)(ii) n = 0,l = 2, ml = -2, ms = \(+\frac{1}{2}\) |
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Answer» (i) When n = 1, value of l cannot be equal to 1. Thus it is not possible. (ii) Value of n i.e., shell starts from 1 and not from zero. Thus, it is not possible. |
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| 35. |
Define Radial nodes. |
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Answer» Radial nodes occur when the probability density of wave function for the electron is zero on a spherical surface of a particular radius. Number of radial nodes = n – l – 1. |
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| 36. |
What happens when an electron gains energy? |
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Answer» The electron moves to higher energy level called the excited state |
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| 37. |
The number of valence electrons in CL ion are:A. 7B. 8C. 1D. 6 |
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Answer» Distribution of electrons in chloride ion are: - Chlorine ion - K shell: 2 electrons L shell: 8 electrons M shell: 8 electrons Hence, the number of valence electrons in chloride ions are 8 as its outermost orbit |
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| 38. |
Does the electron retain the energy forever? |
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Answer» The electron loses the energy and comes back to its ground state. The energy emitted by the electron is seen in the form of electromagnetic energy. |
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| 39. |
Match the following :ElementNo.of Neutrona. \(^{40}_{18}Ar\)i. 7b. \(^{14}_{6}C\)ii. 21c. \(^{40}_{19}K\)iii. 8d. \(^{14}_{7}N\)iv. 22 |
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Answer» a – iv, b – iii, c – ii, d – i |
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| 40. |
The number of valence electrons in Na is -(A) 1 (B) 2 (C) 3 (D) 4 |
| Answer» (A) The number of valence electrons in Na is 1. | |
| 41. |
The frequency of light having wavelength 500 nm is a. 5 x 1015 Hzb. 5 x 1010 MHZ c. 2 x 10-15 Hzd. 6 x 1014 Hz |
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Answer» Correct option is d. 6 x 1014 Hz |
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| 42. |
If an element ‘X’ has mass number 11 and it has 6 neutrons, then write its representation. |
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Answer» The representation of the given element is \(^{11}_5X\). |
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| 43. |
The wave length of a wave is 100 nm. Find its frequency. |
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Answer» Wave length (λ) = 100 mm = 100 x 10-9 m = 10-7 m Speed of light (c) = 3 x 108 m/s We know c = vλ ⇒v = c/λ = 3 x 108/ 10-7 = 3 x 1015 Hertz. |
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| 44. |
How light can be characterized? |
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Answer» Light can be characterized by its wavelength and frequency (v) and these quantities related to the speed of light as c = vλ |
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| 45. |
Did Bohr’s model account for the splitting of line spectra of a hydrogen atom into finer lines? |
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Answer» No, Bohr’s model failed to account for splitting of line spectra |
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| 46. |
If lanthanoids and actinoids are inserted within the table, imagine how the table would be? |
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Answer» It looks very big in size, and it is difficult to identify, as these elements have similar properties. |
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| 47. |
What is spectrum? |
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Answer» A spectrum is a group of wavelengths. |
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| 48. |
Second ionization energy of an element is higher than its first ionization energy. Why? |
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Answer» 1. The energy required to remove an electron from unipositive ion is called second ionisation energy. 2. It is difficult to remove an electron from unipositive ion when compared with neutral atom due to an increase in nuclear attraction. 3. So always second ionisation energy is higher than first ionisation energy. |
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| 49. |
The ion of an element has 3 positive charges. Mass number of the atom is27 and the number of neutrons is 14. What is the number of electrons in the ion?A. 13B. 10C. 14D. 16 |
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Answer» Given that the Mass number of the atom is 27. ⇒ Number of protons + number of neutrons = 27 ⇒ Number of protons + 14 = 27 ⇒ Number of protons = 13 Therefore, number of electrons in a neutral atom = 13 The ion of the element has 3 positive charges. ⇒ Number of electrons = 13 – 3 = 10 Hence, option B is correct. |
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| 50. |
Define excitation energy, binding energy and ionization energy of an electron in an atom. |
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Answer» (1) Excitation energy of an electron in an atom : The energy required to transfer an electron from the ground state to an excited state (a state of higher energy) is called the excitation energy of the electron in that state. (2) Binding energy of an electron in an atom is defined as the minimum energy that should be provided to an orbital electron to remove it from the atom such that its total energy is zero. (3) Ionization energy of an electron in an atom is defined as the minimum energy required to remove the least strongly bound electron from a neutral atom such that its total energy is zero. |
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