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. |
Find the power gain of an amplifier whose power gain in 50 dB. |
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Answer» Gain in dB = 10 log10 A |
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| 2. |
What are advantages of negative feedback. |
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Answer» (i) Stabilisation of the gain. (ii) Increase in input impedance. (iii) Decrease in output impedance. (iv) Increase in bandwidth. (v) Reduction in noise and distortion. |
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| 3. |
An amplifier has a bandwidth of 300 KHz and voltage gain of 100. Calculate its bandwidth if 10% negative feedback is introduced. |
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Answer» BW = (1 + Aβ)BW |
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| 4. |
Find voltage of an amplifier in dB scale if its voltage gain is 10s. |
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Answer» Gain in dB = 20 log Av |
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| 5. |
If the bandwidth of an amplifier is 20 kHe and open-loop gain is 200 and β = 0.1, find the gain of the amplifier with negative feedback. |
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Answer» A= 200, |
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| 6. |
How is the gain bandwidth product of an amplifier? |
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Answer» Gain bandwidth of an amplifier is always a constant. |
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| 7. |
Rain water, which falls on a flat rectangular surface of length 6 m and breadth 4 m is transferred into a cylindrical vessel of internal radius 20 cm. What will be the height of water in the cylindrical vessel if a rainfall of 1 cm has fallen?[ use π = 22/7] |
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Answer» Volume of rainfall = lbh = 6 × 4 × 0.01 = 0.24 m3 ∵ Volume of water in cylindrical vessel = volume of rainfall ∴ πr2h = 0.24 ⇒ h = \(\frac{0.24\times7}{22\times0.2^2}\) ⇒ h ≈ 1.91m = 191 cm |
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| 8. |
A small error in the measurement of the quantity having the highest power (in a given formula), will contribute maximum percentage error in the value of the physical quantity to whom it is related. Explain why? |
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Answer» The quantity which has maximum power (n) should be measured most accurately because any error in the measurement of this quantity is multiplied n times in the final result. |
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| 9. |
If earth contracts to half of its present radius what would be length of the day at equator? |
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Answer» \(I_1= \frac{2}{5}MR^2,I_2=\frac{2}{5}M(\frac{R}{2})^2\) or \(I_2=\frac{I_1}{4}\) \(I_1w_1=I_2w_2\) \(I(\frac{2π}{T_1})=\frac{1}{4}I(\frac{2π}{T_2})\) \(T_2= \frac{T_1}{4}=\frac{24}{4}=6\, hours\) |
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| 10. |
There is a stick half of which is wooden and half is of steel, (i) it is pivoted at the wooden end and a force is applied at the steel end at right angle to its length (ii) it is pivoted at the steel end and the same force is applied at the wooden end. In which case is the angular acceleration more and why? |
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Answer» I (first case) > 1 (Second case) ∴ \(\tau\) r = l α ⇒ α (first case) < α (second case) |
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| 11. |
Define center of gravity. |
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Answer» The center of gravity of a body is the point at which the entire weight of the body acts, irrespective of the position and orientation of the body. |
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| 12. |
Explain the types of equilibrium with suitable examples. |
Answer»
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| 13. |
Three blocks of masses m1 = 2 kg, m2 = 3 kg and m3 = 5 kg are suspended with string, which is passed over a pulley as shown in the figure. Calculate the tension T1, T2 and T3, when the whole system (a) is moving upward with an acceleration of 2 ms-2, and (b) is held stationary. |
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Answer» Here, m1 = 2 kg, m2 = 3 kg and m3 = 5 kg (a) When the system is moving upward with an acceleration 'a' where a = 2 ms-2. As per the fig., consider the free body diagram of the system of three blocks. Then, T1 - (m1 + m2 + m2) g = (m1 + m2 + m3) a or, T1 = (m1 + m2 + m3) (a + g) ...(i) or, T1 = (2 + 3 + 5) x (2 x 9.8) = 118 N Consider the free body diagram of blocks of masses m2 and m3. Then T2 - (m2 + m3)g = (m2 + m3) x a ....(ii) or, T2 = (m2 + m3)(a + g) ⇒ T2 = (3 + 5) x (2 x 9.8) = 94.4 N Consider the free body diagram of block of mass m3. Then, T3 - m3g = m3a ....(iii) or, T3 = m3(g + a) = 5 x (2 + 9.8) = 59 N (b) When the system is held stationary: Obviously, a = 0, setting a = 0, in the equations (i), (ii) and (iii), we have T1 = (m1 + m2 + m3) g = (2 + 3 + 5) x 9.8 = 98 N T2 = (m2 + m3) g = (3 + 5) x 9.8 = 78.4 N and T3 = m3g = 5 x 9.8 = 49 N |
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| 14. |
The speed of the center of a wheel rolling on a horizontal surface is vQ. A point on the rim in level with the center will be moving at a speed of speed of,(a) zero (b) v0 (c) \(\sqrt2\) v0 (d) 2 v0 |
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Answer» Correct answer is (c) \(\sqrt2 \) v0 |
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| 15. |
Define torque and mention its unit. |
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Answer» Torque is defined as the moment of the external applied force about a point or axis of rotation. The expression for torque is, \(\vec \tau\)= \(\vec r\) x \(\vec F\) |
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| 16. |
While rolling, the path of center of mass of an object is – (a) straight line(b) parabola(c) hyperbola (d) circle |
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Answer» (a) straight line |
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| 17. |
अर्द्धचालकों की चालकता (i) ताप पर निर्भर नहीं करती (ii) ताप बढ़ने पर घटती है (iii) ताप बढ़ने पर बढ़ती है (iv) ताप घटने पर घटती है |
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Answer» (iii) ताप बढ़ने पर बढ़ती है। |
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| 18. |
किसी n-प्रकार के अर्द्धचालक में आवेश वाहक होते हैं(i) केवल इलेक्ट्रॉन (ii) केवल कोटर (होल) (iii) दोनों, अल्प संख्या में इलेक्ट्रॉन तथा अधिक संख्या में कोटर (होल) (iv) दोनों, अधिक संख्या में इलेक्ट्रॉन तथा अल्प संख्या में कोटर (होल) |
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Answer» (iv) दोनों, अधिक संख्या में इलेक्ट्रॉन तथा अल्प संख्या में कोटर (होल) |
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| 19. |
जर्मेनियम को किस प्रकार से p-टाइप का अर्द्धचालक बनाया जाता है? |
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Answer» इसमें त्रिसंयोजी अपद्रव्य (ऐलुमिनियम) मिलाकर। |
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| 20. |
शुद्ध अर्द्धचालक में जब कोई अपद्रव्य मिलाया जाता है, तो क्या होता है? |
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Answer» चालकता बढ़ जाती है। |
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| 21. |
कार्बन, सिलिकॉन और जर्मोनियम, प्रत्येक में चार संयोजक इलेक्ट्रॉन हैं। इनकी विशेषता ऊर्जा बैड अन्तराल द्वारा पृथक्कृत संयोजकता और चालन बैंड द्वारा दी गई हैं, जो क्रमशः (Eg)c, (Eg)s; तथां (Eg) Ge के बराबर हैं। निम्नलिखित में से कौन-सा प्रकथन सत्य है? (a) (Eg)si <(Eg) Ge <(Eg)c (b) (Eg)c <(Eg) Ge > (Eg)st (c) (Eg)c > (Eg)s >(Eg) Ge (d) (Eg)c = (Eg)si = (Eg)Ge |
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Answer» चालन बैंड तथा संयोजकता बैंड के बीच ऊर्जा अन्तराल कार्बन के लिए सबसे अधिक, सिलिकॉन के लिए उससे कम तथा जर्मेनियम के लिए सबसे कम होता है; अतः (c) प्रकथन सत्य है। |
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| 22. |
एक के पश्चात् एक श्रेणीक्रम सोपानित में दो प्रवर्धक संयोजित किए गए हैं। प्रथम प्रवर्धक की वोल्टता लब्धि 10 और द्वितीय की वोल्टता लब्धि 20 है। यदि निवेश संकेत 0.01 वोल्ट | है तो निर्गम प्रत्यावर्ती संकेत का परिकलन कीजिए। |
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Answer» यहाँ A1 = 10 तथा A2 = 20 Vi = 0.01 वोल्ट अतः कुल वोल्टता लाभ A = A1 x A2 = 10 x 20 = 200 परन्तु A = vo/vi ⇒ निर्गत वोल्टता V0 = A x Vi V0 = (200 x 0.01) वोल्टे = 2 वोल्ट |
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| 23. |
What is a cyclotron? State its principle of working. |
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Answer» A cyclotron is a cyclic magnetic resonance accelerator in which an alternating potential difference of a few kV is used to accelerate light positive ions such as protons, deuterons, α-particles, etc., but not electrons, to very high energies of the order of a few MeV. It was developed by E. O. Lawrence and M. S. Livingston in 1932. Principle: The cyclotron employs the principle of synchronous acceleration to accelerate charged particles which describe a spiral path at right angles to a constant magnetic field and make multiple passes through the same alternating p.d., whose frequency is the same as the frequency of revolution of the particles. |
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| 24. |
Which one among the following shows particle nature of light?(A) Photoelectric effect(B) Interference(C) Refraction(D) Polarisation |
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Answer» The answer is (A) Photoelectric effect |
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| 25. |
The mass of a photon at rest is(A) zero(B) 1.67 x 10-35kg(C) 1 amu(D) 9 x 1031 kg |
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Answer» The answer is (A) zero |
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| 26. |
क्या निकट दृष्टिदोष अथवा दीर्घ दृष्टिदोष आवश्यक रूप से यह ध्वनित होता है कि नेत्र ने अपनी समंजन क्षमता आंशिक रूप से खो दी है? यदि नहीं, तो इन दृष्टिदोषों का क्या कारण हो सकता है? |
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Answer» यह आवश्यक नहीं है कि निकट दृष्टिदोष अथवा दूर दृष्टिदोष केवल नेत्र के आंशिक रूप से अपनी समंजन क्षमता खो देने के कारण ही उत्पन्न होता है। यह नेत्र गोलक के सामान्य आकार से बड़ा अथवा छोटा होने के कारण भी उत्पन्न हो सकता है। |
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| 27. |
Define the following terms 1. celestial objects 2. celestial sphere 3. horizon |
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Answer» 1. celestial objects : The objects in the sky are called celestial objects. 2. celestial sphere : The imaginary sphere centered around us is called celestial sphere. 3. horizon : The imaginary circle along which the sky appears to meet the ground is called the horizon. |
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| 28. |
State whether the following statement are true or False :1. Moon has no atmosphere.2. Titan is the satellite of Jupiter. |
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Answer» 1. True 2. False |
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| 29. |
State whether the following statement are true or False :1.Aries (Mesha) is zodiacal constellations.2. Mars contains oxygen in its atmosphere. |
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Answer» 1. True 2. True |
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| 30. |
State whether the following statement are true or False :1. Saturn has highest density.2. Venus can be seen over head in the night sky. |
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Answer» 1.False 2.False |
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| 31. |
The constellation that helps to locate north star is. (a) Cassiopeia (b) Orion (c) Taurus (d) Leo |
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Answer» (a) Cassiopeia |
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| 32. |
The poles of this planet faces the sun for longer periods of time than any other planet. (a) Mars (b) Uranus (c) Jupiter (d) Saturn |
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Answer» Correct answer is (a) Mars |
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| 33. |
Polar ice caps of the planet mars are due to. (a) Solidified water (b) solidified carbon dioxide (c) volcanoes erupted near the poles (d) cover of dust particles that reflect |
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Answer» (b) solidified carbon dioxide |
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| 34. |
The second biggest planet of the solar system has. (a) lowest density (b) density equal to water (c) highest density (d) density equal to earth. |
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Answer» (a) lowest density |
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| 35. |
Give reason for the followings.1. A person in England can recognize the pole star easily than a person in Bangalore 2. Mercury in difficult to observe 3. A day on Venus is longer than its year 4. Green house effect is prominent on Venus |
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Answer» 1. A person in England can recognize the pole star easily than a person in Bangalore : Pole star appears to be stationary above the Earth’s North Pole, since England lies in the Northern side Northern hemisphere a person in England can recognize pole star than a person in Bangalore. 2. Mercury in difficult to observe : It is very difficult to observe mercury because it is very close to the sun and most the time it is hidden by the glare of the sun. However it can be seen just before sunrise and just after sunset. 3. A day on Venus is longer than its year : The period of rotation of Venus is 243 days and the period of revolution is 224 days. Since period of rotation is longer than period of revolution, a day on Venus is longer than its year. 4. Green house effect is prominent on Venus : Venus is composed mostly of carbon dioxide. There are several layers of clouds many kilo meters thick composed of sulfuric acid. This dense atmosphere of carbon-dioxide blocks the long wavelength infrared radiation and heat is trapped inside, hence the green house effect is prominent on Venus. |
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| 36. |
Explain our solar system? |
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Answer» Our solar system comprises of the following: 1. The sun which is the center of the solar system. 2. Eight planets are mercury, Venus, Earth, mars, Jupiter, saturnuranus and Neptune, 3. The Asteroid belt between the orbits of mars and Jupiter 4. Satellites of the planets, which are also Called the moons of the planets. 5. Comets such as Halley. 6. Artificial Satellites and space crafts, |
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| 37. |
Express the following in km. 1. 10 light minutes 2. 10 light days 3. 10 Light year |
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Answer» 1. light minutes 10 × 60 × 300000 = 180000000 km = 180 million km 2. 10 light days 10 × 24 × 60 × 60 × 300000 = 259300000000 km = 259200 million km 3. 10 light years 10 × 365 × 24 × 60 × 60 × 300000 = 94608000000000 = 94608000 million km. |
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| 38. |
State the Newton’s laws of motion. Write any two illustrations. |
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Answer» Newton’s first law of motion states that, everybody continues to be in its state of rest or uniform motion along a straight line unless compelled to change its state by an external force. Newton’s second law of motion states that the rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force. Newton’s third law of motion states that for every action there is. an equal and opposite reaction. Illustrations for Newton’s third law of motion are
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| 39. |
Which satellite has active volcanoes? |
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Answer» In a satellite of Jupiter has active volcanoes. |
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| 40. |
How are satellites different from planets? |
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Answer» Planets: 1. planets are natural celestial objects 2. planets revolve round the sun 3. planets are big in size Satellites: 4. Satellites may be natural or artificial objects. 5. Satellites revolve round the planets. 6. Satellites are smaller in size than planets. |
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| 41. |
Name the following: 1. Most fascinating planet with set of ring system as viewed from a telescope. 2. Nearest planet to earth. 3. Nearest planet to the sun 4. Biggest planet of the solar system |
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Answer» 1. Saturn 2. Venus 3. Mercury 4. Jupiter |
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| 42. |
Which planet is called morning star or evening star? |
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Answer» Venus planet is called morning star or evening star. |
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| 43. |
Give an example of a system with variable mass. |
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Answer» Rocket propulsion. |
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| 44. |
How many types of friction are there? |
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Answer» There are two types of friction : (i) Static friction : This type of friction acts till the body does not move, therefore this type of friction is known as static friction. The maximum static friction is known as “Limiting friction”. (ii) Dynamic or kinetic friction : When the body starts moving,then the acting friction is known as dynamic or kinetic friction. Its value is some less than the limiting friction. |
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| 45. |
Explain how will you find out the direction of static friction? |
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Answer» Static Frictional Force : Static friction is the friction that exists between a stationary object and the surface on which it resting. Static friction keeps an object at rest. It must is overcome to start moving the object. Imagine trying to push an object across the floor. You push on it with a small force, but it does not move, this is because it is not accelerating. However, according to Newton’s second law, the object must move with an acceleration. Now, as the body remains at rest, it implies that an opposing force equal to the applied force must have come into play resulting in zero net force on the object. This force is called static friction. It is denoted by Fs. Thus, static friction is the opposing force that comes into play when one body tends to move over the surface of another, but the actual motion has not started.
Limiting Friction : If the applied force is increased, the force of static friction also increases. If the applied force exceeds a certain (maximum) value, the body starts moving. This maximum value of static friction up to which the body does not move, is called limiting friction. 1. The magnitude of limiting: friction between any two bodies in contact is directly proportional to the normal reaction between them. Where the constant of proportionality p s is called the coefficient of static friction. Its value depends upon the nature of surfaces of the two bodies in contact that means whether dry or wet; rough or smooth; polished or non polished. For example, when two polished metal surfaces are in contact, ps = 0.2, when these surface are lubricated, μs = 0.1 2. Direction of the force of limiting friction is always opposite to the direction in which one body is at the verge of moving over the other. 3. Coefficient of Static Friction : |
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| 46. |
Define the coefficients of static and kinetic friction. How will you find out their value? |
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Answer» 1. Static Frictional Force : Static friction is the friction that exists between a stationary object and the surface on which it resting. Static friction keeps an object at rest. It must is overcome to start moving the object. Imagine trying to push an object across the floor. You push on it with a small force, but it does not move, this is because it is not accelerating. However, according to Newton’s second law, the object must move with an acceleration. Thus, static friction is the opposing force that comes into play when one body tends to move over the surface of another, but the actual motion has not started.
Limiting Friction : If the applied force is increased, the force of static friction also increases. If the applied force exceeds a certain (maximum) value, the body starts moving. This maximum value of static friction up to which the body does not move, is called limiting friction. 1. The magnitude of limiting- friction between any two bodies in contact is directly proportional to the normal reaction between them. μs = 0.1 2. Direction of the force of limiting friction is always opposite to the direction in which one body is at the verge of moving over the other. 3. Coefficient of Static Friction : 2. Kinetic Friction : We know that when the applied force on a body is small, it may not move but as the applied force becomes greater than the force of limiting friction, the body is set into motion. The force of friction acting between the two surfaces in contact which are moving relatively, so as to oppose their motion, is known as kinetic frictional force. or fk ∝ μkN ………….. (1) i. e., Coefficient of kinetic friction is always less than coefficient of static friction. Thus we require more force to start a motion than to maintain it against friction. This is because once the motion starts actually; inertia of rest has been over come. Also when motion has actually started, irregularities of one surface have little time to get locked again into the irregularities of the other surface. |
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| 47. |
Whose value is more μs or μk? |
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Answer» μs > μk because fs > fk. |
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| 48. |
Of two contact surfaces, whose value is more : Static or kinetic friction? |
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Answer» Static because fs > fk i. e., static friction is more than dynamic friction. |
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| 49. |
Fill in the blanks with suitable words:Radio signals are emitted by the planet ………… |
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Answer» Radio signals are emitted by the planet Jupiter. |
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| 50. |
Fill in the blanks with suitable words:Winter season constellation is ………… |
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Answer» Winter season constellation is Orion. |
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