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. |
Define the homologous series. |
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Answer» Homologous series may be defined as a series of similarly constituted compounds in which the members possess similar chemical characteristics and the two consecutive members differ in their molecular formula by – CH2. |
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| 2. |
A square plate is expanding uniformly, the side is increasing at the rate of 5cm/sec what is the rate at which the area and its perimeter is increasing when the side is 20 cm long? |
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Answer» Area, A = l2 ⇒dA/dt = 2l × dl/dt = 2 × 20 × 5 dA/dt = 200cm2/sec. P = 41 ⇒ dP/dt = 4 dl/dt 4 × 5= 20cm/sec |
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| 3. |
If S = 5t2 +4t – 8.Find the initial velocity and acceleration. |
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Answer» v = 10t + 4 Initial velocity = 0 + 4 = 4 a = 10 |
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| 4. |
Given an account of structure and functions of hind brain. |
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Answer» 1. Structure of hind brain : (1) Hind brain includes cerebellum pons Varolii and medulla oblongata. (2) Cerebellum is 11% of the total brain and is the second largest part of the brain. (3) It has three lobes, median vermis and lateral two cerebellar hemispheres. it hasouter grey and inner white matter. (4) Cerebellar cortex shows sulci and gyri. The inner white matter of cerebellar medullas shows arbor viatae or branching tree-like processes. (5) Pons is the part that connects the two cerebellar hemi-spheres. It has outer white and inner grey matter. Pons is made up of nerve fibres which form bridges between cerebrum and medulla oblongata. (6) Medulla oblongata is the last part of the hind brain which continues further as a spinal cord. It has outer white and inner grey matter. (7) Internally it has posterior choroid plexus. (8) Eight pairs of cranial nerves arise from medulla oblongata. 2. Functions of hind brain. (1) Cerebellum : (1) Cerebellum is a primary centre for the control of equilibrium, posture , balancing and orientation. (2) Neuromuscular activities are regulated by the cerebellum. (3) Coordination of walking, running , speaking , etc. is under the control of hindbrain. (2) Pons : (1) Activities of the two cerebellar hemispheres are coorinated by pons. (2) Nerve fibres cross over in this area and thus the right side of the brain controls the left part of the body and vice versa. (3) Pons controls the consciousness of the brain. (4) Breathing centre is located in pons along with medulla. (3) Medulla oblongata : (1) Medulla oblongata controls all the involuntary activities such as heartbeats, respiration , vasomotor activities (2) Peristalsis and reflex actions such as coughing. sneezing, swallowing , etc. are also under the control of medulla oblongata. (3) Medulla oblongata is essential for all the virtal functions of the body. |
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| 5. |
What is Kerberos? |
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Answer» It is an authentication service developed at the Massachusetts Institute of Technology. Kerberos uses encryption to prevent intruders from discovering passwords and gaining unauthorized access to files. |
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| 6. |
With the help of a neat and labelled diagram, describe anatomy of human eye. Explain the mechanism of vision . |
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Answer» I. Mechanism of Vision : (1) The light rays of visible wavelength pass through the cornea and the lens and are focused on the retina of the eye. (2) The slight is possible due to conjugated proteins present in the rods and the cones. (3) These are photopigments which are composed of opsin (a protein) and retional (Vitamin A derivative). (4) The light induces dissociation of retional from the opsin, which causes a change in the structure of the opsin. (5) This causes the change in the permeability of the retinal cells. (6) It generates action potential which is carried via bipolar cells and ganglion cells and further conducted by the optic nerves to the visual cortex (vision centre) of the brain. (7) The neural impulses are analyzed and the image formed on the retina is thus recognized . |
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| 7. |
What is OSPF? |
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Answer» It is an Internet routing protocol that scales well, can route traffic along multiple paths, and uses knowledge of an Internet's topology to make accurate routing decisions. |
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| 8. |
What is SLIP (Serial Line Interface Protocol)? |
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Answer» It is a very simple protocol used for transmission of IP datagrams across a serial line. |
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| 9. |
What is Proxy ARP? |
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Answer» It is using a router to answer ARP requests. This will be done when the originating host believes that a destination is local, when in fact is lies beyond router. |
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| 10. |
Briefly explain the factors that determine the climate of india. |
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Answer» The average weather condition of place for a long period like 30- 33 years in known in known as climate. India’s climate is said to be “Tropical Monson”. The main factors are monsoon winds. (i) Location: The northern part of India lies in sub-tropical and temperate zone and the part lying to the south of the tropic of cancer come under tropical zone. The tropical zone being nearer to the Equator, experiences high temperature throughout the year, with small daily and annual range. Tropic of Caner 23 1/2° N latitude passes through the centre of the country. So India is situated both in the tropical and temperate region. (ii) Mountain Ranges: The lofty Himalayan Mountains have prevented the cold winds of central Asia, and keep India warm. They are also greatly responsible for the monsoon rains in the country. (iii) Distribution of Land and Water: India is bounded by the Arabian Sea in the west and Bay of Bengal in the east, Indian Ocean in the south. These adjoining seas have influenced the climate of the country considerably. They influence the rainfall of the coastal region. Even the cyclones which originate from these seas regularly affect the weather condition. (iv) The relief features of India also affect the temperature, air pressure, direction and speed of wind, the amount and distribution of rainfall. The windward side of Western Ghats and north east received high rainfall from June to September. (v) Monsoon winds: The climatic conditions of other country are greatly influenced by monsoon winds. The winds blow in a particular direction during one season, but get reversed during the other season. |
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| 11. |
Mention the advantages of ‘LED’ ? |
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Answer» Following are the advantages of LED— (1) Light emitting diods are easily manufactured, (2) LEDs have low cost, (3) LED works at low voltage as compared to the incandesent bulb, (4) No warm up time is taken by them, (5) They can emit monochromatic light as well as bright light. |
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| 12. |
Describe the major types of forests in india. |
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Answer» The peninsular region of India has the largest forest cover with around 57% of the total forest area. According to geo-climatic conditions, forests are classified into: a. Evergreen Forests: These forests are found in the regions of heavy rainfall and high temperature. Tall umbrella shaped trees with dense assemblage is a prominent feature of this forest. The ever green forest always looks green because various species of trees are found here and they shed leaves in different seasons. The hardwood trees, rose wood, white cedar, toon, gurjan, chaplash, ebony, Mahogany, canes, bamboo, shisham etc. These are found in North-east India, Western Ghats, Andaman and Nicobar islands, parts of Assam and some areas of Himalayan foot hills. b. The Deciduous forests: The deciduous forest covers a wide range of rainfall regimes. The trees of these forests seasonally shed their leaves. The Indian deciduous forest is found in a range of landscapes from the plains to the hills. These forests provide shelter to most endangered wild life in the country, such as the Tiger, Asian Elephant, Bison, Gaur etc. The deciduous forest are two types (i) Moist Deciduous forests: The moist deciduous forests are found in wet regions, receiving annual rainfall between 100cm to 200cam and temperature of 25° C to 30° C. The trees of these forests shed their leaves during spring and early summer. They are found on the eastern slopes of the Western Ghats, Chota Nagpur Plateau, the siwaliksetc. (ii) The Dry Deciduous Forests: The dry deciduous forest are found I the areas where annual rainfall is between 50cm to 150 cm and temperature of 25° C to 30° C. Sal is the most significant tree found in this forest. Varieties of acacia and bamboo are also fund here. These forests are found in areas of central Deccan plateau, South-east of Rajasthan, Punjab, Haryana and parts of Uttar Pradesh and Madhya Pradesh. (iii) The mountain forests: As the name indicates these forests are confined to the Himalayan region, where the temperature is less compared to other parts of the country. The trees in this forest are cone shape with needle like leaves. The important trees are oak, fir, pin e spruce, silver fir, deodhar, devdar, juniper, picea chestnut etc. They provide softwood for making country boats, packing materials and sport articles. c. The Desert forests: These forests are found in the areas of very low rainfall. Thorny bushes, shrubs, dry grass, acacia, cacti and babul are the important vegetation found in these forests. The Indian wild date known as ‘Khejurs”, is common in the deserts. They have spine leaves, long roots and thick fleshy stems in which they store water to survive during the long drought. These vegetations are found in Rajasthan, Gujarat, Punjab and Haryana. d. The Mangrove Forests: These forests occur along the river deltas (Ganga, Mahanadi. Godavari and Krishna) of eastern coast and also concentrated in the coastal areas of Katchch, Kathiawar, and Gulf of Khambar. The mangrove forests in the Ganga delta are called Sunder bans because, they have extensive growth of Sundari trees. The trees in these forests are hard, durable and are used in boat making and as fuel. In the recent years mangrove vegetation is being grown I the coastal areas to control effects of tidal waves and coastal erosion. |
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| 13. |
What is the total surface area of a solid hemisphere whose radius is r ? (A) 4πr2 (B) πr2 (C) 2πr2 (D) 3πr2 |
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Answer» The correct option is: (D) 3πr2 |
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| 14. |
Mention applications of Laser technology ? |
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Answer» Following are the applications of laser technology— (1) It is used in industry, (2) It is used in surgery, (3) It is used in communication, (4) It is used in making holograms, (5) It is used in scientific research. |
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| 15. |
The ratio of corresponding sides of similar triangles is 5 : 7, then what is the ratio of their areas ? (A) 25 : 49 (B) 49 : 25 (C) 5 : 7 (D) 7 : 5 |
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Answer» The correct option is:(A) 25 : 49 |
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| 16. |
What are the necessary conditions for obtaining pure spectrum ? |
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Answer» For obtaining pure spectrum following conditions should be satisfied— (1) Slit should be narrow, (2) Rays coming out from convex lens should have parallel incidence, (3) Prism should be in condition of minimum deviation, (4) Emerging light rays should be focussed by achromatic lens. |
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| 17. |
Radius of a sphere is 14 cm. Find the surface area of the sphere. |
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Answer» Surface area of the sphere = 4πr2 = 4 × 22 / 7 × 14 × 14 = 4 × 22 × 2 × 14 = 2464 cm2 |
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| 18. |
Define power of Lens. Write its S.I. unit. |
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Answer» Power of Lens—The power of a lens is defined as the reciprocal of its focal length is meters. P = 1/f(in metre) The S.I. unit of power of lens is ‘dioptre’ (D). |
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| 19. |
Explain the significance of Himalayas. |
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Answer» Significance Of Himalayas 1. Strategic significance. A natural frontier of India with other countries(China, Pakistan, Afghanistan,etc) 2. Climatic significance. Prevent further northward movement of summer monsoon and also prevent cold northern winds from Siberia to enter into India. 3. Agricultural significance. Formation of Himalayas created a trough to its south which is later filled by. the sediments from the Himalayan rivers which is today known as northern plains- Indogangetic plains – Rich agricultural grounds. 4. Economic significance – Himalayan rivers have huge hydroelectric power potential. Moreover, Himalyan timber and medicinal plants have economic significance. 5. Tourist spot – large ecological diversity and hill stations |
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| 20. |
What are eddy currents ? |
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Answer» In 1875, Forcault observed that when a metal piece is kept in variable magnetic field, current are always induced in the plate which is of whirling nature. They are called eddy currents or Forcault's current. Core of transformer is laminated to prevent the induction of these currents. They are used to made moving coil galvanometer dead beat. The direction of eddy currents is given by Lenz's law. |
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| 21. |
Write down the equation of X- axis. |
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Answer» The Correct Answer: y = 0 |
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| 22. |
From the figure find the value of sinθ. |
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Answer» sinθ = AB / BC sinθ = 3 / 5 |
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| 23. |
What is modern Geography? |
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Answer» It deals with ‘Earth and its inhabitants’. |
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| 24. |
What is the shape of the Earth? |
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Answer» The shape of the earth is ‘Oblate spheriod’ or ‘Geoid’. |
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| 25. |
Draw ∠ ARP= 115o and bisect it. |
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Answer» To draw ∠ ARP = 115o . Bisect ∠ ARP. |
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| 26. |
Mention different uses of cyclotron. |
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Answer» Uses of cyclotron are as follows— (1) It is used to produce radioactive material for medical purposes. (2) It is used to bombard the atomic nuclei with highly accelerated particles to study the nuclear reactions. (3) It is used to improve the quality of solids by adding ions |
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| 27. |
Name any two factors of Mechanical weathering. |
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Answer» Temperature, Rainfall, Wind, Ice. |
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| 28. |
Mention two factors on which resisitivity depends. |
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Answer» Two factors on which resisitivity depends are as follows— (1) It depends on the nature of the material of the conductor. (2) It depends on the temp. of the conductor. |
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| 29. |
solve the equation 3x + 2y+ 2 = 0 , 5x - y - 27 = 0 |
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Answer» \(3x+2y+2=0...(1)\) \(5x- y-27=0...(2)\) Multiply equation (2) by (1), we get \(10x-2y-54=0...(3)\) By adding equations (1) and (3), we get \(13x-52=0\) \(\Rightarrow x =\frac{52}{13}=4\) put x = 4 in equation (1) , we get \(12+2y+2=0\) \(\Rightarrow 2y=-14\) \(\Rightarrow y=-7\) Hence, \(x = 4\) & \(y = -7\) is solutions of a given system of equations |
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| 30. |
What is Atmoshpere |
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Answer» The thin layer of gaseous matter encircling the earth in the form of blanket is known as Atmosphere. |
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| 31. |
Observe the adjoining figure and write down one pair of interior angles |
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Answer» (i) ∠d and ∠e or (ii) ∠c and ∠f |
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| 32. |
simply 3^n+3 + 3 ^n / 3^n+3 - 3^n+1 |
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Answer» \(\frac{3^{n+3}+3^n}{3^{n+3}-3^{n+1}}\) \(=\frac{3^n(3^3+1)}{3^{n+1}{(3^2-1)}}\) \(=\frac{1}{3}\times\frac{27+1}{9-1}\) \(=\frac{28}{3\times8}\) \(=\frac{7}{6}\) |
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| 33. |
Which instrument is used to measure depth of the ocean? |
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Answer» Fathamometer. |
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| 34. |
Ponit M is the mid point of seg AB and AB = 14 then AM = ? |
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Answer» Point M is the midpoint of seg AB AM = 1 / 2 × AB ∴ AM = 1 / 2 × 14 ∴ AM = 7 Unit |
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| 35. |
There are three dealers A, B and C in Maharashtra. Suppose, the trade of each of them in september 2018 was as shown in the following table. The rate of GST on each transaction was 5%. Read the table and answer the questions below it.DealerGST collected on the saleGST paid at the time of purchaseITCTax paid to the Govt.Taxbalance with the GovtARs. 5000Rs. 6000Rs. 5000Rs. 0Rs. 1000BRs. 5000Rs. 4000Rs. 4000Rs. 1000Rs. 0CRs. 5000Rs. 5000Rs. 5000Rs. 0Rs. 0(i) How much amount did the dealer A get by sale ? (ii) For how much amount did the dealer B buy the articles ? (iii) How much is the balance of CGST and SGST left with the government that was paid by A ? |
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Answer» (i) The sale of dealer A = 100 / 5 × 5000 = 1,00,000 rupees (ii) The purchase of dealer B = 100 / 5 × 4000 = 80,000 rupees (iii)∴ Balance of CGST paid by A = 1000 / 2 = Rs. 500 and SGST = Rs. 500 |
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| 36. |
15. Which of the following is correct?(1) \( \tan 1>\tan 2 \)(2) \( \tan 2>\tan 1 \)(3) \( \sin 1\cos 2 \) |
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Answer» Answer: (1) tan1>tan2 |
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| 37. |
Find the rank of the following matrix \( A \). \[ A=\left[\begin{array}{cccc} 1 & -1 & 2 & -3 \\ 4 & 1 & 0 & 2 \\ 0 & 3 & 0 & 4 \\ 0 & 1 & 0 & 2 \end{array}\right] \] |
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Answer» \(A=\begin{bmatrix}1&-1&2&-3\\4&1&0&2\\0&3&0&4\\0&1&0&2\end{bmatrix}\) ↔ \(\begin{bmatrix}2&1&-1&-3\\0&4&1&2\\0&0&3&4\\0&0&1&2\end{bmatrix}\) (Applying C1 ↔ C3 Then C2 ↔ C3) ↔\(\begin{bmatrix}2&1&-1&-3\\0&4&1&2\\0&0&3&4\\0&0&0&2\end{bmatrix}\) (Applying R4 → 3R4 - R3) ∴ Rank of matrix A is 4. |
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| 38. |
If α,β,1 are the roots of x3−2x2−5x+6=0 then find α,β |
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Answer» Given equation is x3 – 2x2 – 5x + 6 = 0 since, 1 is root of the equation, therefore (x + 1) is a factor of it. ⇒ (x – 1)(x2 – x – 6) = 0 ⇒ (x – 1)(x2 – 3x + 2x – 6) = 0 ⇒ (x – 1)(x(x – 3) + 2(x – 3)) = 0 ⇒ (x – 1)(x + 2)(x – 3) = 0 ⇒ x = 1, –2 or 3. Hence, 1 –2 & 3 are roots of given equation. Then α & β are –2 & 3. i.e, either α = –2 & β = 3 or α = 3 & β = –2. |
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| 39. |
State superposition principle. Give its one importance. |
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Answer» Superposition principle—It states that—“Total force acting on a given point charge due to a number of point charge around it is the vector sum of the individual forces acting that point charge due to all other point charges.” By using the principle of superposition, we can apply coulomb's law to any collection of point charges. |
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| 40. |
Consider a matrix A = \(\begin{bmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{bmatrix}\). If |2 adj(3 adj(4A-1))| = 2a.3b, then the value of a - 2b is |
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Answer» A = \(\begin{bmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{bmatrix}\) |A| = -1 + 1 + 1 + 1 + 1 + 1 = 4 Now, |2 adj (3adj(4A-1))| = 23|adj(3adj (4A-1))| (\(\because\) |KA| = Kn|A|) = 23|3adj(4A-1)|2 (\(\because\) |adj A| = |A|n-1) = 23(33|adj (4A-1)|)2 (\(\because\) |KA| = Kn|A|) = 23.36 |adj(4A-1)|4 = 23.36(43. (A-1))4 (\(\because\) |KA| = Kn|A|) = 23.36(43. \(\frac14\)) (\(\because\) |A-1| = \(\frac1{|A|}=\frac14\)) = 23.36(42)4 = 23.36.48 = 23.36.216 = 219.36 \(\therefore\) a = 19, b = 6 \(\therefore\) a - 2b = 19 - 12 = 7 |
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| 41. |
Write two basic properties of electric charge. |
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Answer» Two basic properties of charge are as follows— (1) Quantization of electric charge— It is the property of an electric charge by virtue of which any charged body can have charge which is an integral mulitple of the basic charge ‘e’. q = ± ne (2) Conservation of electric charge— Total charge in an isolated system remains conserved. Charges appear in pair of equal and opposite sign. |
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| 42. |
Explain the term noise from the chapter electomagnetic waves and communication system. |
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Answer» 1. Noise is an unwanted signal which interference with original message signal and occur disorder in communication process between a speaker. 2. Communication noise refers to influences on effective communication that influence the interpretation of conversations. |
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| 43. |
A flexible steel cable of total length L and mass per unit length μ hangs vertically from a support at one end.(a) Show that the speed of a transverse wave down the cable is v = √(g(L - x)), where x is measured from the support. (b) How long will it take for a wave to travel down the cable ? |
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Answer» (a) Tension at a point at a distance x from the support is the due to the weight of the cable T = \(\frac{m(L-x)}{L}g\) \(\mu\) = \(\frac mL\) The speed of transverse wave v = \(\sqrt{\frac{T}{\mu}}\) v = \(\cfrac{\frac{m(L-x)g}{L}}{\frac mL}\) v = \(\sqrt{g(L-x)}\) (b) Time taken to transverse a distance dt = \(\frac{dx}{\sqrt{g(L-x)}}\) total time taken t = \(\int\limits_0^L\frac{dx}{\sqrt{g(L-x)}}\) t = 2\(\sqrt{\frac Lg}\) |
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| 44. |
Y1 = a1 sin(wt-2πx/y) and Y2 = a2 sin (wt-2πx/y + €) is |
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Answer» hey there, ● Answer - ∆x = λ(θ+π/2)/2π ● Explanation - Given wave eqns are, y1 = a1sin(ωt-2πx/λ) ...(1) y2 = a2cos(ωt-2πx/λ+θ) y2 = a2sin(ωt-2πx/λ+θ+π/2) ...(2) Comparing (1) & (2), phase difference is ∆θ = θ+π/2 Path difference is given by - ∆x = ∆θ × λ / 2π ∆x = λ(θ+π/2)/2π Hence, path difference is λ(θ+π/2)/2π . Thanks for asking... |
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| 45. |
A quantity of this system which exhibits simple harmonic variation with a period of 1 sec is |
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Answer» The correct answer is potential energy . |
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| 46. |
Which of the adjoining graphs represents ohmic resistance |
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Answer» Correct option (1) Explanation: For ohmic resistance V α I |
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| 47. |
8+16+24+.... +8n=4n(n+1) गणितीय अधिष्ठापन |
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Answer» \(8 + 16 + 24 + \,....+ 8n\) \(= 8(1 + 2 + 3 + ...+ n)\) \(= 8\frac {n(n+ 1)}{2}\) \(\left(\because 1 + 2 + 3 + ....+ n = \frac {n(n + 1)}{2}\right)\) \(= 4n(n+ 1)\) Hence proved. |
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| 48. |
For how many times the time-out can be availed in a volleyball game and what is the time span of a time out? |
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Answer» Two time-out can be availed in one set. The time out is for 30 seconds only. |
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| 49. |
How is a match decided in a volleyball game? |
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Answer» The team that wins two sets out of the three sets is declared the winner. Five sets take place in national and international games. The team that wins three out of five is declared the winner. |
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| 50. |
How does the rotation take place in a volleyball game? |
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Answer» In a volleyball game six players play from each side. Three out of them are in front of the attacking line and three are behind it. The first one takes the service. At the time of service 4, 3, 2 are before the attack line and 5, 6, 1 are behind them. |
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