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. |
A heat engine works between a source and a sink maintaned at constant temperatures `T_(1)` and `T_(2)`. For the efficiency to be greatestA. `T_(1)`m and `T_(2)`. Should be highB. `T_(1)` and `T_(2)` should be lowC. `T_(1)` should be low and `T_(2)` should be highD. `T_(1)` should be high and `T_(2)` should be low |
| Answer» `eta=1-(T_(2))/(T_(1))` | |
| 2. |
Which is the triploid tissue in a fertilised ovule? How is the triploid condition achieved? |
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Answer» The triploid tissue in the ovule is the endosperm. Its triploid condition is attained due to the fusion of two polar nuclei and one nucleus of male gamete (also referred to as triple fusion). |
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| 3. |
Hardy-Weinberg principle is as (a) Allele frequencies in a population are not stable(b)Allele frequencies in a population are stable and is constant from generation to generation. (c) Allele frequencies in a population are stable and is not constant from generation to generation. (d) All of the above |
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Answer» Hardy-Weinberg principle is as Allele frequencies in a population are stable and is constant from generation to generation. |
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| 4. |
Sieve tubes in angiosperms are associated with specialised parenchyma cells. Name those cells. How do they help sieve-tube members? |
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Answer» Companion cells, help in physiological working of sieve-tube members. |
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| 5. |
Identify the cell of xylem that helps in the upward conduction of water in the plant is :-(A) Sieve tube (B) Xylem parenchyma (C) Companion cell (D) Tracheids |
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Answer» Answer: Xylem parenchyma |
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| 6. |
Front Office night auditor needs to report to the: a) Head of Accounts department b) Head of Front Office department c) Heads of both Front office and Accounts department d) Heads of both HR and Accounts department |
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Answer» Correct option: c) Heads of both Front office and Accounts department |
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| 7. |
Reserpine is used as medicine for snake bite. It is extracted from roots and bark of Rauwolfia serpentina. |
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Answer» Phenotypic ratio is 3 : 1 Genotypic ratio is 1 : 2 : 1 |
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| 8. |
How the contact resistance can increase? |
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Answer» Humidity + salty air, dust, poor contact pressure |
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| 9. |
Can we file the pitted contacts of contactor? |
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Answer» No, use emery paper and etc. |
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| 10. |
What material used for contactor? |
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Answer» Silver-nickel for <100A Silver-cadmium oxide for large currents. |
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| 11. |
Can we use ac coils in dc circuit? |
| Answer» Yes with economy resistor in series. | |
| 12. |
Why pure silver is not used in contactor? |
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Answer» Affected by sulphur, mechanical or arcing damages (adv. Lower voltage drop) |
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| 13. |
Why copper contact are not used in contactor? |
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Answer» Because corrosion rate increases. Poor surface property , large closing force required. |
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| 14. |
Why shading rings provided in armature core of ac contactor. Why not for dc contactor? |
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Answer» Because the force developed is not steady in ac therefore contacts will chatter but if shading ring is used force developed becomes steady due to splitting of phases of flux, therefore contacts becomes bounce free and humming sound reduces. |
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| 15. |
Do all eukaryotic cells have nucleus and only one nucleus? |
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Answer» There are eukaryotic cells without a nucleus and others with more than one nucleus. Osteoclasts, the cells responsible for the resorption of the osseous matrix, for example, are multinucleate cells; striated muscle fibres are multinucleate too. Red blood cells are an example of enucleated specialized cells. |
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| 16. |
What are cells with a delimited nucleus called ? What are the main elements of the nucleus? |
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Answer» Cells with the delimited nucleus are called eukaryotic cells. Organisms composed of one or more eukaryotic cells are called eukaryotes. The mains elements of the nucleus are the chromatin (made of DNA molecules), the nucleolus, the karyolymph, or nucleoplasm, and the nuclear membrane (or karyotheca). |
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| 17. |
What are the primary and the secondary constrictions of a chromosome? What is the other name given to the secondary constriction? |
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Answer» Primary constriction is the narrower region of a condensed chromosome where the centromere, the structure that unites identical chromatids, is located. Secondary constriction is a region similar to the primary constriction, narrower than the normal the thickness of the chromosome too, and in general, it is related to genes that coordinate the formation of the nucleolus and control the ribosomic RNA (rRNA) synthesis. For this reason the secondary constrictions (that can be one or more in chromosome) is called nucleolus organizer region (NOR). |
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| 18. |
Insoinic acid is the biological precursor of (A) Cytosine and Uric acid (B) Adenylve acid and Glucine floc acid (C) Orotic acid and Uridylic acid (D) Adenosine acid Thymidine |
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Answer» (B) Adenylve acid and Glucine floc acid |
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| 19. |
The most common substrate used in distilleries for the production of ethanol is ________. (a) Soyameal (b) Groundgram (c) Molasses (d) Corn meal |
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Answer» The most common substrate used in distilleries for the production of ethanol is Molasses. |
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| 20. |
Pills have to taken daily for a period of ...a... days starting preferably within the first ...b... days of menstrual cycle. After a gap of ...... days during which ...d... occurs, it has to be repeated in the same pattern.A. a-28/29, b-7, c-5, d- ovulation.B. a-28/29, b 7, c-5, d-menstruationC. a-21, b-7, c-5, d-menstruationD. a-21, 6-5, c-7, d-menstruation |
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Answer» Correct Answer - D (D) Pills have to taken daily for a period of 21 days starting preferably within the first 5 days of menstrual cycle. After a gap of 7 days during which menstruation occurs, it has to be repeated in the same pattern. |
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| 21. |
Read the following lines and answer question on the basis of the same:The presence of dragonflies can reveal changes in the water ecosystems more quickly than studying other animals or plants. In fact, from the nymph to the adult stage, the dragonfly has a significant, positive ecological impact. Dragonfly eggs are laid and hatched in or near water, so their lives impact both water and land ecosystems. Once hatched, dragonfly nymphs can breathe underwater which enables them to eat mosquito larvae, other aquatic insects and worms, and even small aquatic vertebrates like tadpoles and small fish and in the air. Adult dragonflies capture and eat adult mosquitoes. Community wide mosquito control programs that spray insecticides to kill adult mosquitoes also kill dragonflies(i) The approach to biological control includes: a. Import and release of an insect pest to a new area to provide hosts for natural enemies b. Import and release of natural enemies from the native home of an alien insect pest that has invaded a new area c. Preservation of natural enemies (predators & parasitoids) that are already established in an area d. Use of insecticides to reduce alien insect pests to establish new equilibrium position.(ii) Two diseases less likely to occur in a region with plenty of dragonflies are_____a. Yellow fever and amoebic dysentery b. Malaria and Yellow fever c. Anthrax and typhoid d. Cholera and typhoid(iii) Dragonflies indicate positive ecological impact as-a. The presence of dragonflies indicates polluted water. b. Dragonfly nymphs selectively eat mosquito larvae. c. They help to decrease the probability of diseases spread by vectors. d. Dragonfly do not cause any harm to beneficial species.(iv) The most effective stages in the life cycle of dragonfly that eradicate mosquitoes are-a. Larvae and Adult b. Caterpillar and Adult c. Nymph and Adult d. Pupa and Adult(v) Assertion: Releasing dragonflies in areas where there is an outbreak of malarial diseases can be an environment friendly method of control. Reason: Dragon flies are dominant species and will not allow mosquitoes to reproducea. Both assertion and reason are true, and the reason is the correct explanation of the assertion.b. Both assertion and reason are true, but the reason is not the correct explanation of the assertion.c. Assertion is true but reason is false.d. Both assertion and reason are false |
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Answer» (i) a. Preservation of natural enemies (predators & parasitoids) vectors. (ii) b. Malaria and Yellow fever (iii) c. They help to decrease the probability of diseases spread by mosquitoes, horseflies and deer flies. (iv) c. Nymph and Adult (v) c. Assertion is true statement, but reason is false. |
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| 22. |
Assertion: In Ophrys one petal of the flower bears an uncanny resemblance to the female bee. Reason: Two closely related species competing for the same resource can coexist simultaneously. (a) Both assertion and reason are true, and the reason is the correct explanation of the assertion. (b) Both assertion and reason are true, but the reason is not the correct explanation of the assertion. (c) Assertion is true but reason is false. (d) Both assertion and reason are false. |
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Answer» (c) Assertion is true statement, but reason is false. |
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| 23. |
What colouration do roans possess? Why? |
| Answer» Roans possess the mixture of red and white colour side by side due to codominant alleles for red and white traits. | |
| 24. |
Assertion: A community with more species is more stable than that with less species.Reason: More the number of species, lesser the variation in the total biomass production year after year. (a) Both assertion and reason are true, and the reason is the correct explanation of the assertion. (b) Both assertion and reason are true, but the reason is not the correct explanation of the assertion. (c) Assertion is true but reason is false. (d) Both assertion and reason are false |
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Answer» (a) Both assertion and reason are true, and the reason is the correct explanation of the assertion. |
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| 25. |
The most likely lethal mutation is (A) Substitution of adenine for cytosine (B) Insertion of one nucleotide (C) Deletion of three nucleotides (D) Substitution of cytosine for guanine |
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Answer» (B) Insertion of one nucleotide |
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| 26. |
Assertion: E. coli having pBR322 with DNA insert at BamHI site cannot grow in medium containing tetracycline. Reason: Recognition site for Bam HI is present in tetR region of pBR322. (a) Both assertion and reason are true, and the reason is the correct explanation of the assertion. (b) Both assertion and reason are true, but the reason is not the correct explanation of the assertion. (c) Assertion is true but reason is false. (d) Both assertion and reason are false. |
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Answer» (a) Both assertion and reason are true, and the reason is the correct explanation of the assertion |
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| 27. |
Assertion: An organism with lethal mutation may not even develop beyond the zygote stage. Reason: All types of gene mutations are lethal. (a) Both assertion and reason are true, and the reason is the correct explanation of the assertion. (b) Both assertion and reason are true, but the reason is not the correct explanation of the assertion. (c) Assertion is true but reason is false. (d) Both assertion and reason are false |
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Answer» (b) Both Assertion and Reason are true but Reason, is not the correct explanation of Assertion. |
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| 28. |
Give any two examples for sacred grooves in India. |
Answer»
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| 29. |
Mention the reasons for population explosion. |
Answer»
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| 30. |
Write the Karyotype and mention any two symptoms of Klinefelter’s syndrome. |
Answer»
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| 31. |
What do you mean by chromosomal disorder? Describe the cause and related abnormalities in the following genetic disorders-(i) Down's syndrome(ii) Klinefelter's syndrome(iii) Turner's syndrome |
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Answer» Chromosomal disorder: A chromosome abnormality disorder, anomaly aberration or mutation is a missing, extra, or irregual portion of chromosomal DNA. It can be from an atypical number of chromosomes or a Structural abnormality in one or more chromosomes. Chromosome mutation was formerly used in a strict sense to mean a change in a chromosomal segment, involving more than one gone. A karyotype refers to a full set of chromosomes from an individual that can be compared to a "normal" karyotype for the species via genetic testing. A chromosome anomaly may be detected or confirmed in this manner. Chromosome anomalies usually occur when there is an error in cell division following meiosis or mitosis. There are many types of chromosome anomalies. They can be organized into two basic groups, numerical and structural anomalies. (i) Down syndrome is a chromosomal condition that is associated with intellectual disability, a characteristic facial appearance, and weak muscle tone (hypotonia) in infancy. All affected individuals experience cognitive delays, but the intellectual disability is usually mild to moderate. People with Down syndrome may have a variety of birth defects. About half of all afected children are born with a heart defect. Digestive abnormalities, such as a blockage of the instestine, are less common. Individuals with Down syndrome have an increased risk of developing serveral medical conditions. These include gastroesophageal reflux, which is a backflow of acidic stomach contents into the esophagus, and celiac disease, which is an intolerance of a wheat protein called qluten. About 15 percent of people with Down syndrome have an underactive thyroid gland (hypothyroidism). The thyroid gland is a butterfly-shaped organ in lower neck that produces hormones. Individuals with Down syndrome also have an increased risk of hearing and vision problems. Additionally, a small percentage of children with Down syndrome develop cancer of blood-forming cells (leukemia). (ii) Klinefelter syndrome is a chromosomal condition that affects male physical and cognitive development. Its signs and symptoms vary among affected individuals. Affected individuals typically have small testes that do not produce as much testosterone as usual. Testosterone is the hormone that directs male sexual development before birth and during puberty. A shortage of testosterone can Iead to delayed or incomplete puberty, breast enlargement (gynecomastia), reduced facial and body hair, and an inability to have biological children (infertility). Some affected indivdiduals also have genital differences including undescended testes (cryptorchidism) the opening of the urethra on the underside of the penis (hypospadias), or an unusually small pensis (micropenis). (iii) Turner syndrome is a chromosomal condition that affects development in females. The most common feature of Turner syndrome is short stature, which becomes evident by about age 5. An early loss of ovarian function (ovearian hypofunction or premature ovarian failure) is also very common. The ovaries develop normally at first, but egg cells (oocytes) usually die prematurely and most ovarian tissue degenerates before birth. Many affected girls do not undergo puberty unless they receive hormone therapy, and most are unable to conceeive (infertile). A small percentage of females with.Turner syndrome retain normal ovarian function through young adulthood. |
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| 32. |
Assertion : Down syndrome, Klinfelter syndrome and Turner syndrome are chromosomal disorders.Reason : In Klinfelter syndrome females are sterile.(1) If both assertion and reason are true and reason is the correct explanation of assertion.(2) If both assertion and reason are true but reason is not the correct explanation of assertion.(3) If assertion is true but reason is false.(4) If both assertion and reason are false |
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Answer» (3) If assertion is true but reason is false. |
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| 33. |
The figure below shows a vaccum tube containing electrodes made of different metals, 1 and 2 whose work functions are `phi_(1)` and `phi_(2)` The electrodes are illuminated simultaneously. The maximum kinetic energy of photoelectrons reaching plate 2 is 1 eV and maximum kinetic energy of photoelectrons reaching plate 1 is 3 eV. Assume that photoelectron emitted from either plate do not interact with each other `phi_(1)=1.5` eV and `phi_(2)=0.7` eV. Find wavelength (in nm) of the electromagnetic wave used |
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Answer» Correct Answer - `400` `K_(2max)=(hc)/lambda-phi_(1)-ev_(0)=1` `K_(1max)=(hc)/lambdaphi_(2)+ev_(0)=3` `(2hc)/lambda(phi_(1)+phi_(2))=4` `(2xx1240)/(4+2.2)=lambda` `lambda=400nm` |
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| 34. |
The voltage applied to an X-ray tube is 18 kV. The maximum mass of phton emitted by the X-ray tube will beA. `2xx10^(-13) kg `B. `3.2xx10^(-36) kg`C. `3.2xx10^(-32) kg`D. `9.1xx10^(-31) kg` |
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Answer» Energy of photon is given by `mc^(2)` now the maximum energy of photon is equal to the maximum energy of electorn =e V hence, `mc^(2) =ev implies m=(eV)/(c^(2))=(1.6xx10^(-19)xx18xx10^(3))/((3xx10^(8))^(2))=3.2xx10^(-32) kg` |
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| 35. |
If the de-Broglie wavelength of an electron beam is `5xx10^(-10)` m, then what is the retarding potential (in volt) necessary to completely stop the beam. (Round off to nearest integer) |
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Answer» Correct Answer - `6` `p=h/lambda=(6.6xx10^(-34))/(5xx10^(-10)) = 1.32xx10^(-24)` `E=p^(2)/(2m)=qV` `((1.32xx10^(-24)1)^(2))/(2xx9xx10^(-31)xx1.6xx10^(-19))=V` `=(1.32xx1.32xx10^(-48))/(18xx1.60xx10^(-50))=(132xx132)/(18xx160)=121/20=6.05 V approx 6V` |
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| 36. |
Assertion : Soft and hard `X`-rays differ in frequency as well as velocity. Reason : The penetrating power of hard `X`- rays is more than the penetrating power of soft `X`-rays.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 Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion `&` Reason are False. |
| Answer» Correct Answer - B | |
| 37. |
X-rays of high penetrating power are called hard X-ray. Hard X-rays have energy of the order of `10^5 eV`. The minimum potential difference through which the electrons should be accelerated in an X-ray tube to obtain X-ray of energy `10^5eV` is:A. `2xx10^(5) V`B. 50 kVC. 40 kVD. `10^(5) V` |
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Answer» If electrons are accelerated through a potential difference V, the maximum energy of emitted photon could be `E_(max)=eV :. 10^(5) eV=eV` `implies V= 10^(5)V.` |
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| 38. |
The energy that should be added to an electron, to reduce its de-Broglie wavelength from `2xx10^(-9) m` to `0.5xx10^(-9)m` will be:A. 1.1 MeVB. 0.56 MeVC. 0.56KeVD. 5.6 eV |
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Answer» `lambda=h/p=h/(sqrt(2mE))` `:. E=(h^(2))/(2m lambda^(2))` `Delta E=(h^(2))/(2m)(1/(lambda_(1)^(2))-1/(lambda_(2)^(2)))` Put `lambda_(1)=0/5xx10^(-9) m` & `lambda_(2)=2xx10^(-9) m` and solve. |
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| 39. |
A H–atom in ground state has time period T = 1.6 × 10–16 sec. find the frequency of electron in first excited state(1) 7.8 × 1014 (2) 7.8 × 1016 (3) 3.7 × 1014 (4) 3.7 × 1016 |
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Answer» Answer is (1) T ∝ r/v ∝ n2/z ∝ n/z ∝ n3/z2 T1/T2 = n31/n32 = 1/8 T2 = 8T1 = 8 × 1.6 × 10–16 = 12.8 × 10–16 f2 = {1}/{12.8 x 10-16} ≈ 7.8 x 1014 |
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| 40. |
In the diagram shown above n1, n2 and n3 are refractive indices of the media 1, 2 and 3 respectively. Which one of the following is true in this (a) n1 - n2 (b) n1 > n2 (c) n2 > n3(d) n3 > n1 |
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Answer» Correct option is (d) n3 > n1 |
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| 41. |
The pitch of a screw gauge is `0.55mm` and there are 100 divisions on its circular scale. The instrument reads `+2` divisions when nothing is put in between its jaws. In measuring the diameter of a wire, there are 8 divisions on the main scale and `83^(rd)` division coincides with the reference. Then the diameter of the wire isA. `4.05mm`B. `4.405mm`C. `3.05mm`D. `1.25mm` |
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Answer» `Deltal = 0.5mm` `N = 100 divisions` zero correction = 2 divisions Reading =Measured value + zero correction . `=(8 xx 0.5)mm + (83 -2) xx (0.5)/(100)` `4mm + 81 xx (0.5)/(100)mm` `=4.405mm` . |
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| 42. |
Breakdown does not destroy a zener diode provided the zener current is less than the a. Breakdown voltage b. Zener test current c. Maximum zener current rating d. Banier potential |
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Answer» (c) Maximum zener current rating |
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| 43. |
An input transducer converts a. Voltage to current b. Current to voltage c. An electrical quantity to a nonelectrical quantity d. A nonelectrical quantity to an electrical quantity |
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Answer» (d) A nonelectrical quantity to an electrical quantity |
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| 44. |
`CuSO_4` crystallise in rock salt structure.Its cell parameter can be determined by various experimental methods like electrical conductivity measurement, colligative properties measurement, pH measurement etc. A cubic crystal of `CuSO_4` of edge length 17.1 mm, is dissolved in water to make 500 ml solution of pH 5.`root3(1.5)=1.14)` Given : `Cu(H_2O)_6^(2+)+H_2O hArr [Cu(H_2O)_5(OH)]^(+)+H_3O^(+) , K=10^(-5)` In the given solution is made 1 M with respect to `[Cu^(2+)]` becomes `10^(-15)` M therefore `K_f` for the formation of `Cu(NH_3)_4^(2+)` isA. `10^10`B. `10^8`C. `2xx10^8`D. `2xx10^10` |
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Answer» Correct Answer - D `{:(Cu(H_2O)_6^(2+)+H_2OhArr,[Ca(H_2O)_5(OH)]^(+),+H_3O^(+)),(C, , ),(C-x,x,x):}` `10^(-5)=x^2/(C-x)=((10^(-5))^2)/(C-10^(-5))` `C=2xx10^(-5)` mol/lt `:.` moles of `CuSO_4` dissolved =`2xx10^(-5)xx0.5 =10^(-5)` Number of units cells `=(10^(-5)xxN_A)/4=6/4xx10^(18)=1.5xx10^(18)` number of unit cells along one edge of the cube =`root3(1.5xx10^(18))=1.14xx10^8` If edge length of F C C unit cell is a Now`{:(Cu^(2+)+,4NH_3hArr,Cu(NH_3)_4^(2+)),(2xx10^(-5), , ),(10^(-15),1,2xx10^(-5)):}` `K_f=([Cu(NH_3)_4^(2+)])/([Cu^(2+)][NH_3]^(4))=(2xx10^(-5))/(10^(-15)xx1)=2xx10^10` |
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| 45. |
1 ppm is equivalent to a. 0.1% b. 0.01% c. 0.001% d. 0.0001% |
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Answer» The correct answer is: (d) 0.0001% |
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| 46. |
The shortest wavelength of transition in Paschen series of `He^(+)` ion in nanometer (nm) is `(1/(R_(H))=91.12nm)`. |
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Answer» Correct Answer - `205.02` For shortest wavelength `n_(2)=oo` `1/(lamda_("min"))=4xxR_(H)[1/(3^(2))-1/(oo^(2))]` ltrbgt `1/(lamda_("min"))=4/9 R_(H)` `lamda_("min")=9/4xx1/(R_(H))nm` `=205.02nm` |
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| 47. |
The difference between the longest wavelength line of the Balmer series and shortest wavelength lion of the Lyman series for a hydrogenic atom (Atomic number Z) equal to `Delta lambda`. The value of the Rydberg constant for the given atom isA. `(5)/(31)(1)/(Delta lambda.Z^(2))`B. `(5)/(36)(z^(2))/(Delta lambda)`C. `(31)/(5)(1)/(Delta lambda. Z^(2))`D. none |
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Answer» `Deltalambda =lambda_(1)-lambda_(2)=(1)/(R[(1)/((2)^(2))-(1)/((3)^(2))]z^(2))-(1)/(R[a-0]z^(2))` On solving, `R=(31)/(5)(1)/(Delta lambda.Z^(2))` |
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| 48. |
By 2010, scientists surely ______ a cure for cancer. A) are discovering B) have been discovered C) will have discovered D) had discovered |
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Answer» Correct option is C) will have discovered |
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| 49. |
I ______ what the problem is by tomorrow. A) will be discovering B) have discovered C) will have discovered D) will be discovered |
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Answer» Correct option is C) will have discovered |
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| 50. |
Once I discovered the techniques that worked for me, my confidence grew and I began to see challenges as opportunities to move forward and succeed. Now I enjoy meeting people and my relationships are much better. I have more friends and can rely on them when I feel myself slipping back to where I was before.I am achieving my goals with ease because I have developed my self-confidence.I no longer miss the opportunities that life offers. I am more optimistic and my mind no longer limits me. Life is finally good !(1) What benefits does the writer get after discovering the techniques?(2) What change comes over the writer ?(3) What is the result of his developing self-confidence ?(4) What is the ultimate truth that the writer discovers ?(5) Pick out the words from the passage which mean:(i) ahead(ii) one who thinks positively |
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Answer» (1) After discovering the techniques that works for the writer, the writer’s confidence grows and he begins to see challenges as opportunities to move forward and succeed. (2) Now the writer can enjoy meeting people and his relationships are much better. (3) As a result of his developing self-confidence, the writer is achieving his goals with ease. He no longer misses the opportunities that life offers. He is more optimistic and his mind no longer limits him. (4) The ultimate truth that the writer discovers is ‘Life is finally good’. (5) These words are: |
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