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. |
Intravenous infusions of adrenaline and noradrenaline have similar effects on A. Skeletal muscle blood flow. B. Renal blood flow. C. Skin blood flow. D. Diastolic arterial pressure. E. Heart rate. |
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Answer» A. False Adrenaline increases, and noradrenaline reduces skeletal muscle blood flow. B. True Both decrease renal blood flow. C. True Both cause cutaneous vasoconstriction. D. False Noradrenaline raises diastolic pressure; adrenaline lowers it. E. False Adrenaline increases heart rate; noradrenaline raises mean arterial pressure and causes reflex cardiac slowing. |
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| 2. |
Figure shows some relationships between lung volume (increasing upward) and oesophageal pressure (increasing to the right) during normal tidal breathing. In this diagram: a. The intra-oesophageal pressure is equal to atmospheric pressure at point A. b. The changes during the respiratory cycle follow the path ABDC. c. The slope of the line AD increases when lung compliance increases. d. The width of the loop CB increases when airway resistance increases. e. AD increases in length during exercise. |
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Answer» a. False Intra-oesophageal pressure is similar to intrapleural pressure which is negative with respect to atmospheric pressure at the beginning of a normal inspiration. b. True In both inspiration (ABD) and expiration (DCA) volume changes lag pressure changes, thus the relationship is a hysteresis loop, rather than the straight line AD. c. True Since compliance is volume change per unit pressure change. d. True The greater the airway resistance, the more does air flow lag behind pressure changes, hence the greater the hysteresis. e. True During exercise, both pressure changes and lung volume changes increase so that tidal volume increases. |
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| 3. |
What is natality? |
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Answer» Number of birth rate during a given period in the population. |
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| 4. |
Figure shows the effects of intravenous adrenaline and noradrenaline infusions on some cardiovascular variables. For each event or trace a–e below, select the best option from the following list. 1. Arterial pressure. 2. Heart rate. 3. Cardiac output. 4. Peripheral resistance. 5. The period of adrenaline infusion. 6. The period of noradrenaline infusion. a. Event A. b. Trace C. c. Trace D. d. Trace E. e. Trace F.(After Barcroft and Swan (1953) Sympathetic Control of Human Blood Vessels, Edward Arnold, London.) |
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Answer» a. Event A: Option 5 Period of adrenaline infusion. This corresponds to a period of increased heart rate and widening of the pulse pressure with a fall in diastolic pressure, whereas at the second event heart rate decreased and both systolic and diastolic pressure increased, indicating that noradrenaline was infused during period B. b. Trace C: Option 2 Heart rate. Adrenaline accelerates the heart but noradrenaline causes reflex slowing produced by the steep rise in mean arterial pressure. This is the only scale giving an initial value (around 80) corresponding to a normal (slightly apprehensive) heart rate. c. Trace D: Option 1 Arterial pressure. Adrenaline lowers the diastolic pressure but noradrenaline raises it. This is the only dual trace corresponding to systolic and diastolic pressures; the initial value 130/80 corresponds to normal blood pressure. d. Trace E: Option 3 Cardiac output. Adrenaline raises cardiac output but noradrenaline reduces it because of the reflex depression of cardiac activity. Again the scale corresponds to an initially normal/slightly raised cardiac output around 6–7 litres/minute. e. Trace F: Option 4 Peripheral resistance. Noradrenaline raises total peripheral resistance because of its predominant effect on alpha adrenoceptors which mediate vasoconstriction; adrenaline lowers it because of its predominant effect on beta adrenergic receptors which mediate vasodilation. These units are appropriate for peripheral resistance: from the equation Mean arterial pressure = Cardiac output x Peripheral resistance we can derive that Peripheral resistance = Mean arterial pressure/Cardiac output For the initial state this equals approximately 100/7 = 14–15 as in F (the units are mmHg/litre/minute). Note that for this question it was necessary to consider all aspects of the diagram together, rather than consecutively. |
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| 5. |
Figure shows some of the changes that occur throughout the phases of a menstrual cycle. For each of the physiological variables a–e below, select the most appropriate option from the following list of traces. 1. Trace A. 2. Trace B. 3. Trace C. 4. Trace D. a. Oestradiol level. b. Core temperature. c. Luteinizing hormone level. d. Progesterone level. e. Inhibin level. |
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Answer» a. Option 3 Trace C. Oestrogens produced by the follicular cells of the developing ovum dominate the follicular phase of the cycle and peak about the time of ovulation; the level is still somewhat raised during the luteal phase. b. Option 1 Trace A. Core temperature tends to rise about the time of ovulation and then falls again coming up to the following menstruation. c. Option 4 Trace D. Luteinizing hormone secretion peaks at about the time of ovulation (in a similar manner to follicle-stimulating hormone, but its concentration tends to be greater). d. Option 2 Trace B. Progestogens formed by the corpus luteum dominate the luteal phase of the cycle and cause the endometrium to enter the secretory phase. e. Option 2 Trace B. Inhibin is a hormone produced by the ovary that inhibits secretion of follicle-stimulating hormone by the anterior pituitary gland. It peaks in the luteal phase of the cycle with a time course similar to the progestogen level. |
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| 6. |
Name the stain used to visualize DNA fragments in gel electrophoresis. |
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Answer» Ethidium Bromide |
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| 7. |
Figure shows four points on the pressure–volume diagram of a left ventricle. For each of the descriptions a–e below, select the best option from the following list of points and lines. 1. D 2. C 3. A 4. B 5. DC 6. DA 7. AB 8. CB 9. CD 10. AD 11. BA 12. BC 13. ABCD 14. DCBA 15. DCAB 16. ADCB a. The beginning of diastole. b. The end of systole. c. Isometric contraction. d. The segment between two points where the trace would depart maximally from a straight line. e. One cardiac cycle starting with the onset of diastolic filling. |
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Answer» a. Option 1 D. At the beginning of diastole ventricular volume is around 50 ml and the pressure is around 120 mmHg (above atmospheric) because the isometric phase of diastole has not yet occurred. b. Option 1 D. At the end of systole, ventricular volume has returned to 50 ml and the pressure is around 120 mmHg, having fallen from a maximal value around 140 mmHg as the aortic valve closes. c. Option 12 BC. By definition volume is unchanged during isometric contraction, while pressure rises from about zero to a level which will open the aortic valve at arterial diastolic pressure, around 90 mmHg; thus isometric contraction is represented by the vertical line BC. d. Option 9 CD. The two segments BC and DA represent isometric systole and isometric diastole, so are completely straight. During ventricular filling, AB, there is a small (around 5 mmHg) rise in pressure during atrial systole. In contrast, during the ejection phase of ventricular systole, CD, the pressure rises from arterial diastolic to arterial systolic (from around 90 to 140 mmHg) and then falls back to around 120 mmHg as the aortic valve closes. e. Option 13 ABCD. Diastolic filling starts when ventricular volume is around 50 ml and its pressure close to atmospheric. The pressure–volume trace then moves to B where filling is complete, and continues in an anti-clockwise direction. |
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| 8. |
The by products formed during the oxidation of food |
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Answer» 1. carbon-di-oxide 2. water |
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| 9. |
For each of the intracellular organelles A–E, select the best option from the following list of descriptions. 1. Sites of protein synthesis rich in RNA. 2. Intracellular membrane-bound structures containing enzymes that can destroy most cellular structures. 3. Granules in a layer produced by high-speed centrifugation of cells. 4. Structures lying close to the nucleus responsible for organizing the microtubular systems. 5. Membrane-bound organelles associated with numerous enzymes that catalyse a variety of anabolic and catabolic reactions. A. Lysosomes. B. Centrosomes. C. Peroxisomes. D. Microsomes. E. Ribosomes. |
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Answer» A. Option 2 Intracellular membrane-bound structures containing enzymes that can destroy most cellular structures. The postmortem breakdown of the lysosomal membranes releases lysosomal enzymes that autolyse (cause self-destruction of) the cell. B. Option 4 Structures lying close to the nucleus responsible for organizing the microtubular systems. Centrosomes are made up of two centrioles. At mitotic division the centrosomes are duplicated and one goes to each end of the mitotic spindle. The microtubules they control allow movement within the cell. C. Option 5 Membrane-bound organelles associated with numerous enzymes that catalyse a variety of anabolic and catabolic reactions. They are involved in the oxidation of some long chain fatty acids. Drugs that can modify peroxisome behaviour are being used in the attempt to lower lipid levels in the blood. D. Option 3 Granules in a layer produced by high-speed centrifugation of cells. This is the generic name for the cellular organelles brought down by high-speed centrifugation. E. Option 1 Sites of protein synthesis rich in RNA. Ribosomes can be attached to the endoplasmic reticulum where they synthesize proteins such as hormones. |
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| 10. |
Sudden application of cold water to the A. Hand causes local vasoconstriction. B. Hand causes vasodilation in the opposite hand. C. Oesophagus can cause changes in the electrocardiogram. D. External auditory meatus causes nausea and nystagmus. E. Whole body by immersion causes apnoea. |
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Answer» A. True Due to the direct effect of cold on vascular smooth muscle. B. False Immersion of a hand in cold water provokes general vasoconstriction and a rise in blood pressure – the cold pressor response. C. True By cooling the myocardium. D. True By inducing currents in endolymph which stimulate vestibular receptors. E. False By stimulating uncontrollable gasping under water, it may result in drowning. |
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| 11. |
Normal healthy young adults can tolerate loss of half of their A. Renal tissue without developing renal failure. B. Pulmonary tissue without developing respiratory failure. C. Circulating platelets without developing a haemorrhagic tendency. D. Seminal sperm count without suffering from infertility. E. Liver without developing hepatic failure. |
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Answer» A. True Normal function can be maintained with one kidney. B. True Though maximum exercise tolerance is reduced, considerable exertion is possible. C. True The platelet count must fall by more than 75 per cent before bleeding problems arise. D. True Infertility is unlikely unless the count falls to around a quarter of the normal value. E. True In short, most body functions carry at least a 50 per cent reserve in the young adult. |
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| 12. |
Why is r-HUEPO preferred over blood transfusion in such cases where a person has excessive blood loss due to accidents? OR Differentiate between primary and secondary animal cell cultures. |
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Answer» No donor is required for transfusion, no transfusion facilities, no risk of transfusion related infection (any two) OR The maintenance of growth of cells under laboratory conditions in suitable culture medium is known as primary cell culture. The primary cell culture is sub-cultured in fresh growth media to develop secondary cultures. |
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| 13. |
I liked to read books about TRAVELS in my holiday. A) rests B) walks C) plays D) journeysE) balls |
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Answer» Correct option is D) journeys |
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| 14. |
Samuel told me that they were PLANNING to see the gallery in a few days. A) staying B) playing C) going D) starting E) coming |
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Answer» Correct option is C) going |
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| 15. |
Mention the length and breadth of the side gallery of the Badmintion Court? |
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Answer» The side and back gallery shall be of 212, and 112, respectively. |
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| 16. |
In which site of fallopian tube does fertilisation take place? |
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Answer» Ampullary – Isthmic junction. |
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| 17. |
Which tissue is used as an explants to obtain virus free plants in tissue culture? |
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Answer» Meristem is used as an explants to obtain virus free plants in tissue culture. shoot meristem is the tissue that is used as an explant to obtain virus free plants in virus culture. |
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| 18. |
Mention the weight of the shuttle? |
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Answer» The weight of the shuttle is from 73 to 85 grams. |
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| 19. |
Mention the role of Alpha interferon in treatment of cancer. |
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Answer» Activates the immune system of cancer patient and helps in destroying the tumor. |
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| 20. |
Which of the following is not a method of ex situ conservation?(1) National Parks(2) Micropropagation(3) Cryopreservation(4) In vitro fertilization |
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Answer» Correct option is (1) National Parks |
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| 21. |
What is a shuttle vector? |
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Answer» Vectors used in eukaryotic cells which are constructed in such a way so that they can exist both in eukaryotic cells and E. coli. |
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| 22. |
Interferon β is used for the treatment of A. Hepatitis C B. Hepatitis B C. Multiple Sclerosis D. Chronic Granulomatous disease |
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Answer» C. Multiple Sclerosis |
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| 23. |
Name the amino acids involved in the catalytic triad that regulates charge -relay system in the enzyme Chymotrypsin? |
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Answer» His-57, Asp-102 and Ser-195 |
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| 24. |
In-situ activation of chymotrypsin takes place in the A. jejunum. B. duodenum. C. ileum. D. pancreas. |
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Answer» Answer is B. Duodenum |
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| 25. |
In the enzyme chymotrypsin, why does Ser 195 develop a negative charge on its -OH group? A. Negatively charged Asp COOresidue pulls the Ser–OH proton through His B. Negatively charged Ser COOresidue pulls the Asp–OH proton through His C. Positively charged Asp COOresidue pulls the Ser–OH proton through His D. Positively charged Ser COOresidue pulls the Asp–OH proton through His |
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Answer» A. Negatively charged Asp COOresidue pulls the Ser–OH proton through His |
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| 26. |
Which technique is used for improving laundry detergent subtilisin? A. SDS-PAGE B. Mass Spectrometry C. Site directed Mutagenesis D. Protein fingerprinting |
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Answer» C. Site directed Mutagenesis |
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| 27. |
Define refraction of light. What is the cause of refraction? Give two everyday examples of refraction of light. |
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Answer» Change in the path of light ray when it passes from one medium to another medium is called refraction of Speed of light is different in different media. The speed of light changes when it passes from one medium to another is the main causes of refraction. Examples of refraction of light: 1. A pencil partially immersed in water appears to be bent because of the refraction of light coming from the part of pencil that is under water. 2. A lemon kept in water in a glass tumbler appears to be bigger than its actual size, when viewed from the sides. |
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| 28. |
The goal of mass spectrometric analysis is to create A. gas phase ions from polar charged molecules. B. polar charged molecules from gas phase ions. C. liquid ions from non-polar molecules. D. non-polar molecules from liquid ions. |
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Answer» A. Create gas phase ions from polar charged molecules |
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| 29. |
State the laws of reflection. What is the nature of image formed by a plane mirror? |
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Answer» Laws of reflection: 1. The angle of incidence is equal to the angle of reflection. 2. The incident ray, the normal to the mirror at the point of incidence and the reflected ray, all lie in the same plane. |
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| 30. |
What are the different modes of communication used for IEC activities? |
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Answer» Project initiation meetings, use of helplines and rural common service centres, wall paintings, door to door contact programmes, schools and colleges, village libraries, engagement of Bharat Nirman Volunteers and Nehru Yuvak Kendras, Engagement of SHGs. |
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| 31. |
What is ‘Green revolution’? Discuss the advantages of green revolution in India. |
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Answer» Increase in Food production through the introduction of high yield crop varieties and the application of modem agricultural techniques is called Green Revolution. 1. Increase in Agricultural production and yield per hectare. 2. Better land use by employing two or three crop pattern. 3. Reduction of imports of food grains. 4. Improves country’s Economic status. |
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| 32. |
Write a note on modes of communication. |
Answer»
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| 33. |
Write the types and distribution of cotton crop in India. |
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Answer» Conditions for Cultivation of cotton: 1. Cotton is a tropical & Sub-tropical crop. It requires high temperature. At the time of growing it requires 21°C to 24°Ç temperature. 2. It requires moderate rainfall of 50 cm to 100 cm. However, it can cultivate in areas of lesser rainfall with the help of irrigation. 3. Deep black soil is well suitable for cotton crop. This soil is commonly known as Black cotton soil. This is capable of retaining moisture. 4. Cotton requires the use of Manures & fertilizers crop rotation helps to maintain fertility of the soil & improve the yield. 5. Cotton cultivation requires large amount of cheap labours for planting, thinning, seeding, picking of cotton. 6. Frosting, Moist weather & heavy rainfall are harmful to the crop. 7. Cotton plant is susceptible to disease & pests. Site requires the use of insecticides & pesticides. 8. The Sunny weather is necessary at the time of harvesting the cotton. Varieties of Cotton: (1) Long Staple Cotton:
(ii) Medium Staple Cotton:
(iii) Short Staple Cotton:
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| 34. |
State mirror formula. Is the same formula applicable to both concave and convex mirrors? Define magnification for a spherical mirror. |
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Answer» Mirror formula: It is a mathematical relation between the object distance is image distance ‘v’ and focal length ‘f of a spherical mirrors. This relation is \(\frac{1}{u}+\frac{1}{v}= \frac{1}{f}\) or \(\frac{1}{v}+\frac{1}{u}= \frac{1}{f}\) This formula is applicable to all concave and convex mirrors Magnification by a spherical mirror: The ratio of the height of the image to the height of the object is called magnification. It is denoted by in m = \(\frac{h'}{h}\) Where h' = image height, h = object height Note: Magnification ‘m’ is also related to the object distance (u) and image distance (v). It can be expressed as m = \(\frac{h'}{h}=\frac{-v}{u}\) A negative sign in the value of magnification indicates that the image is real. A positive sign in the value of magnification indicates that the image is virtual. |
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| 35. |
Describe the land – use pattern of India. |
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Answer» The important types of land use pattern in India are: 1. Forest area 2. Land not available for cultivation 3. Cultivable wasteland 4. Fallow land 5. Net area sown. 1. Forest area : –
2. Land not available for cultivation :
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| 36. |
Define the principal focus of a concave mirror the radius of curvature of a spherical mirror is 20 cm. What is its focal length? Name the mirror that can give an erect and enlarged image of an object. |
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Answer» Principle focus of a concave mirror is a point on the principal axis at which a beam of light incident parallel to the principal axis converges after reflection from the concave mirror. = \(\frac{1}{2}\) x 20 = 10 cm ∴ f = 10 cm Concave mirror gives an erect and enlarged image of an object. |
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| 37. |
What are the types of migration? Analyse types of internal migration of India. |
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Answer» The types of Migration are Internal Migration and International Migration. I. Internal Migration : Movement of people from one region to another within the same country is called internal migration. In India there are four streams of mternal migration. They are 1. Rural to Rural 2. Rural to Urban 3. Urban to Urban 4. Urban to Rural 1. (a) Rural to Rural : This is estimated that about 65.2% of total migration is of this category. Female migrants dominated in this stream, Thus it is an important example for matrimonal migration and it is called women migration. 2. Rural to Urban : Rural to Urban migration (17.6%) is second important type of migration. Rural – Urban migration is caused b both push of the rural areas as well as pull of the urban areas. 3. Urban to Urban : Generally, people like to move from small town with less facility to large cities ith more facilities. 4. Urban to Rural : Urban areas are usually affected by the pollution. The retired and aged people prefer to spend their old age life in nearby villages. Thus people move from ltrban to Rural. 5 International Migration : Movement of people from one country to another across international borders is called International migration. |
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| 38. |
Write a note an the air routes of the world. |
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Answer» 1. The North America – The Greatest air traffic is found in USA with internal and international flights. It has the top airports like Attanta, Chicago, Los Angeles, Dallas. 2. South America – It has far air routes. The Major international airports are: Rio-de-Janeiro, Brasilia, Saopaulo, Santiago and Buenos Aires. 3. Africa – It is served by 2 international Airlines. (a) The East African air route through London, Rome, Cairo, Nairobi and Johannesbrug. (b) Sri Lanka and South East Asia and the Central air route 4. Australia – It is a well developed internal and external air services. Sydney is an important international airport. 5. Asia – China has external links with other countries of the world. Its enroute location is between Europe, Asia, Australia and Africa. 6. The Russian International Air Routes: Russia and other countries of former Soviet Union are well connected by air service. All Countries of the World connects each other through airlines of different comers. |
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| 39. |
Explain the measures to check the growth of population in India. |
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Answer» The government of India has taken several steps to control the growth of population in the country are as :
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| 40. |
Write a note on sex ratio and age structure of the world. |
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Answer» Sex Ratio:
Age structure: Age structure represents the number of people in different age groups. It includes both male & female population.
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| 41. |
Explain the Growth of population in the world. |
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Answer» The growth of population refers to the increases in the number of inhabitants of a country during specific period.
Stages of Population growth: I Three billion – July 1959 II Four billion-April 1974 III Five billion-July 1987 IV Six billion – October 1999 V Seven billion – 12th March 2012. |
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| 42. |
What are the problems caused with rapid growth of population? |
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Answer» Effects of over population:
Measurements to Control the population:
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| 43. |
The fundament particles present in an atom is - (A) Protons and electrons (B) Electrons and neutrons (C) Protons and neutrons (D) Electrons and nucleons |
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Answer» Correct answer is (D) Electrons and nucleons The fundamental particles present in an atom are electrons and nucleons (protons and neutrons). |
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| 44. |
Cathode rays are collection of (A) electrons (B) Protons (C) neutrons (D) atoms |
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Answer» Correct answer is (A) electrons |
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| 45. |
Angular momentum of electron in possible orbit is -(A) L = nh/2π(B) L = nh/π(C) L = 2π/nh(D) L = π/nh |
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Answer» Correct answer is (A) L = nh/2π |
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| 46. |
Which of the following is correct for α –particle ? (A) 1 Protons and 1 neutron (B) 2 Protons and 2 neutron (C) 1 Protons and 3 neutron (D) 2 Protons and 3 neutron |
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Answer» Correct answer is (B) 2 Protons and 2 neutron |
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| 47. |
Explain internet, remote sensing, E-mail and satellite communication. |
Answer»
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| 48. |
The expression for radio active decay is -(A) N = NO e-λT(B) N = NO eλT(C) N = NO e-λ^2T(D) N = NO e-λT^2 |
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Answer» Correct answer is (A) N = NO e-λT |
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| 49. |
In decay of free neutron, name the elementary particle emitted along with proton and electron in nuclear reaction. |
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Answer» nº → p+ + e- + \(\bar{v}\) (β- decay) \(\bar{v}\) = antineutrino (of the electron type) p+ = proton e- = electron nº = free neutron The answer is: Antinutrino |
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| 50. |
In the following nuclear reaction, Identify unknown labelled X.\(^{22}_{11}Na+X\rightarrow ^{22}_{10}Ne+v_e\) |
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Answer» \(^{22}_{11}Na\) + \((^o_{-1}e^-)\) → \(^{22}_{10}Ne\) + \(\bar{v}_e\)(electron antineutrino) \(^o_{-1}e\) (or \(^o_{-1}\beta)\) \(\Rightarrow\) beta particle (electron) The answer is: Electron. |
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