InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 4201. |
F. Suggested Projects:Mix a few iron fillings, wooden shavings and fiour. Try to separate the ironyou do it?1.Try to separate the iron filings. How would2.Bring a compass to the classroom and try to find the direction of your window, table and chair |
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Answer» Just a move a magnet on the mixture the iron filings will get attracted to it and hence can be separated out. |
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| 4202. |
Weite the adverentBerthielen Soal DRest and motion |
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Answer» Motion:-A body is said be in a state of motion if it changes its position continuously with respect to its surroundings with the passage of time.E.g A bus running on a road, a flying bird etc Rest:- A body is said to be at rest if it does not changes its position continuously with respect to its surroundings with the passage of time.Eg A book lying on a table. Motionandrestare two different states of an object. ... So,motionandrestare relative to a closed frame. Considering the closed frame, an object is said to be atrestif it doesn't change its position at any given time. And,motionis when an object changes its position with respect to time within the closed frame. |
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| 4203. |
what happen th focal length of a mirror when it is broken into small pieces ? |
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Answer» If themirrorisbrokeninto twopieces, itsfocal lengthremains the same .Focal lengthchanges when there is change in the radius of curvature of themirror. ... If themirrorisbrokeninto twopieces, each part of themirrorwill act as an independentmirror. Thus eachbrokenpart will reflect its own image. |
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| 4204. |
11) How does the water kept in an earthen pot remain cool? |
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Answer» In anearthen pot,watergets evaporated quickly through the pores. Cooling is caused by evaporation. Some heat energy is utilised during the process of evaporation. Since this energy is taken from thewateritself, it leads to a lowering of temperature in the remaining amount ofwater. |
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| 4205. |
OURNE11.JOURIEAn express train starts from one station. After 40 min it crosses a small stationwhich is at a distance of 30 km. Find the velocity of train (in m/s) when it crosses |
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Answer» Distance covered by train= 30 km = 30000 m Time taken = 40 min = 40*60= 2400 s Therefore,Velocity of train = Distance /time= 30000/2400= 300/24= 12.5 m/s |
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| 4206. |
5. A train starts from station with an acceleration 1 msA boy is 48 m behind the train with a constant velocity10 ms, the minimum time after which the boy will catchthe train is(a) 4.8 s(b) 8s(c) 10 s(d) 12 s |
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Answer» option B is the correct answer B is the correct option B is the right answer B is the correct option B is the correct answer |
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| 4207. |
Why do we hear the sound produced by humming been while the sound produced by vibration of pendulum is not heard? |
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Answer» It is because thesoundproduced by the pendulum is of much lower frequency than that ofbee. thesoundproduced bybee'swing is continous where as thesoundproducer by the pendulum is just once or twice . thus thesoundfrom the pendulum by the timewe hear, vanishes in the air . |
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| 4208. |
कि... B oo zp,,a.a.p_,i_amua»__mafimm‘.'‘—7%4 = id¥S vexidicadon: |
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Answer» I=V/R v=potential difference or emf r = resistance of the circuit i= current of the circuit |
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| 4209. |
How is sound produced? |
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Answer» The vibration produces sound which is modified into speech by the action of lips,tongue,teeth etc. Sound is produced when something vibrates. The vibrating body causes the medium (water, air, etc.) around it to vibrate. Vibrations in air are called traveling longitudinal waves, which we can hear. the simple answer is sound is made by vibration.when a body vibrate it make a sound. |
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| 4210. |
Define focal plain |
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Answer» Focal plane. : a plane that is perpendicular to the axis of a lens or mirror and passes through the focus. |
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| 4211. |
Ex plain the undnap |
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| 4212. |
1.15Which Organic compound is first prepared from inorganic compounds? |
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Answer» It is widely accepted thatureais the first organic compound to be synthesized from inorganic chemicals. In 1828, the German chemist Friedrich Wöhler obtainedureaartificially by treating silver cyanate withammonium chlorid |
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| 4213. |
TWD Capachtors_이ane pui Seres across a1 5C c |
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| 4214. |
\frac { \operatorname { tan } \frac { \pi } { 9 } - \operatorname { tan } ^ { 3 } \frac { \pi } { 9 } } { 1 - 3 \operatorname { tan } ^ { 2 } \frac { \pi } { 9 } } |
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| 4215. |
Date:elections io M shell islo an atomcual -to -he runke of electors int and stelluchoitis te alomามกา |
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Answer» If the atom has electrons in M shell then the k and L shell must have been fully filled. And as there are same number of electrons in the M shell as that there are in the K shell, so M shell has 2 electrons.Then total number of electrons is2+8+2=12Therefore the following atom is Magnesium. |
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| 4216. |
Check whether the following equation isdimensionally correct or not.I mv2 = mgh |
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| 4217. |
.Thevolumeofa thin brass vessel and that of a solbrass cube are both equal to 1 litre exactly. What willbe their new volumes when heated through 25°C 7αfor brass is 1.9%10-5.c1.Ans. 1001.425 cm in each case. |
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| 4218. |
why is Newton's law of gravitational called as universal law |
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| 4219. |
Why Newton's Law of Gravitation is called Universal Law of Gravitation? |
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Answer» Ans :- It iscalledso because it is applicable on all bodies having mass, and the bodies will be governed by the samelaw, that isnewton's law of gravitation. Thus, as it is applicableuniversally, it iscalledasuniversal law. |
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| 4220. |
icBaUniversal Law of Gravitation |
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| 4221. |
tate the universal law ofgravitation |
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| 4222. |
State the universal law ofgravitation. |
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Answer» Newton’s universal law of gravitation states that every particle in the universe attracts every other particle with a force along a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. |
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| 4223. |
derive the universal law of gravitation. |
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Answer» Statement of Universal Gravitation:The Law of Universal Gravitation states that every object of mass in the Universe attracts every other object of mass with a force which is directly proportional to the product of their masses and inversely proportional to the square of the separation between their centers. The proportionalities expressed by Newton's universal law of gravitation are represented graphically by the following illustration. Observe how the force of gravity is directly proportional to the product of the two masses and inversely proportional to the square of the distance of separation. |
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| 4224. |
ABC is an equilateral triangle with O as its centreF.F and F represent three forces acting alongthe sides AB, BC and AC respectively. If the totaltorque about O is zero then the magnitude of F is[AIPMT (Prelims)2012] 18F+ F2(3) Fi+ Fa(4) F, F |
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| 4225. |
ELECTROMAGNETIC WAVESbr 22. Arrange Infrared, visible, Gamma, X-rays,shiradiowave and microwave in increasingorder of wavelength.Ans |
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Answer» Visible, infrared, gamma, radio waves, x-rays and microwave gammax raysvisibleinfararedmicrowaveradiowave gamma (<1)X-ray (1-10)visible (400-700)Infrared (700-10*5)microwave(10*5-10*8)radio wave (>10*8) gamma X raysvisibleinfraredmicrowaveradiowave |
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| 4226. |
37. (a) Explain why, a cricket player moves his handsbackwards while catching a fast cricket ball.(b) A 150 g ball, travelling at 30 m/s, strikes the palm of aplayer's hand and is stopped in 0.05 second. Find theforce exerted by the ball on the hand. |
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| 4227. |
23. nsmall balls each of mass 'm impinge clastically cach second on a surface with velocity u. The force experiencedby he surface will be:(C) 4 mnu(D) mu2(A) mnu(B) 2 mnu |
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Answer» for 1 ball the impulse provided will be change in momentum = mu -(-mu) = 2mu so F.t = 2mu F = 2mu/t = 2mu now for n ball..the total value is 2mnu. option B |
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| 4228. |
21. If the momentum of an electron is changed by p,then the de Broglie wavelength associated with it changesby 0.5%. The initial momentum of electron will be1) p/200 2) p/100 3) 200p4) 100p |
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Answer» The answer is Option 3. |
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| 4229. |
6. [a] Write three factors on which speed of sound depends[b]A bat can hear sound at frequencies up to 120KHz. Determine the wavelength ofsound in air at this frequency.(Take speed of sound in air as 344m/s.) |
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Answer» Speed of sound depends primarily on thedensityand viscosity of the medium through which it propagates. Sotemperature,humidity(ifnonliquid medium) andpressureare also factors as they impact thedensityand viscosity of the medium, especially if the sound wave propagates through fluids. |
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| 4230. |
A truck of mass 30.000 kg movesThe power of the tosgven g#10ms-1)up an inclined piane of slope1 in 100 at a speed ot 30 хтоの.0㎾4) 2.5 kw |
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| 4231. |
A bat can hear sound at frequencies up to 120KHz. Determine the wavelensound in air at this frequency.(Take speed of sound in air as 344m/s.) |
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Answer» Speed, wavelength, and frequency of a sound wave are related by the following equation: Speed = wavelength * frequencyWavelength = speed / frequencyWavelength = 344/120000 = 28.7 * 10 ^-4 m |
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| 4232. |
A small block of mass 100 g is pressed againstahorizontal spring fixed at one end to compress the springthrough 50 cm (figure S-E11). The spring constant is100 N/m. When released, the block moves horizontallytill it leaves the spring. Where will it hit the ground 2 mbelow the spring?12000000Figure 8-E1l |
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| 4233. |
48. A small block of mass 100 g is pressed against ahorizontal spring fixed at one end to compress the springthrough 50 cm (figure 8-E11). The spring constant is100 N/m. When released, the block moves horizontallytill it leaves the spring. Where will it hit the ground 2 mbelow the spring?Figure 8-E11 |
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| 4234. |
A bullet of mass 100 g is fired with a velocity of 100m/s hit the simple pendulum of mass 500 g to raise itby height H. Find H?(1.00 |
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Answer» 1/2mv^2= mgh1/2*100*(10000)= 500*10*h50= 5000hh= 100m |
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| 4235. |
An object of mass 100 g move around the circumference of circle of radius 2m with constant angular speed 7.5 rad/s. Compute(a) linear speed(b) force directed towards centre.Data: m =100 g= 0.1 kg,r=2 m, w=7.5 rad/s |
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| 4236. |
An object of mass 100 g move around circumference of circle of radius 2m with constant angular speed7.5 rad/s.compute |
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| 4237. |
06. (a) When a ball is thrown vertically upwards, it goes through adistance of 19.6 m. Find the initially velocity of the ball and thetime taken by it to rise to the highest point.( g = 9.8 m/s^2) |
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| 4238. |
0.22 A ball is dropped from a height h on the groundIf the coefficient of restitution is e, the heightto which the ball goes up after it rebounds forthe n" time is.th |
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Answer» thanku |
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| 4239. |
039 A ball falls on the ground from a height of 2.0 m and rebounds upto a height of 1.5 m. Find thecoefficient of restitution. |
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| 4240. |
Name the weakest and strongest force in nature. |
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Answer» Thestrongestforce is the strong nuclearforce,and weakest force is gravitational force. |
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| 4241. |
EQUILIBRIUM OF FORCES |
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Answer» A very basic concept when dealing withforcesis the idea ofequilibriumor balance. ... If the size and direction of theforcesacting on an object are exactly balanced, then there is no netforceacting on the object and the object is said to be inequilibrium thanku sir |
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| 4242. |
4. Is it possible to incréase tie tenpenu5. Why astronout cannot talk to each other on moon as on the earth?ho rinctic rICTRY ?f one mole of an ideal gas is 5/2 RT. What will be its specific heat at constant |
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Answer» The Moon has no tangible atmosphere – at its surface, it is a near vacuum. Sound waves are pressure waves that propagate by vibrating the molecules of a medium, such as air or water. In a vacuum, there is nothing to vibrate, and thus no propagation. When we speak, we are using our mouths to create vibrations in the air in our mouth. That air passes the vibration to the air just outside our mouth, which passes the vibration to the air next to it, and so on, until it reaches the air in our ears, which pass the vibration to our ear drums. Electromagnetic waves can travel through the vacuum of space. They don’t need a medium other than space itself. So, astronauts on the Moon’s surface would talk to each other using radio. |
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| 4243. |
7The sum of the magnitudes of two forces actingat a point is 16 N. The resultant of these forceis perpendicular to the smaller force and has amagnitude of 8 N. If the smaller force is ofmagnitude x, then the value of x is(A)2N(C) 6 N(B) 4 N(D) 7 N |
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Answer» Thank you so much |
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| 4244. |
(d) no Or theyMacimum and minimum values of the resultant of twoforces acting at a point are N and 3 N respectively. Thesmaller force will be equal to :(KCET 1997)(a) 5N(b) 4N(c) 2N(d) IN |
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Answer» hence the maximum value and minimum value of the resultant forces it's 7N and 3N the smallest fleece it's 4N hope this will help you like my answermark as best answer thank you let smaller force is X and large force is Y then acc to sum X+Y=7....... (1) X-Y=3......... (2) there fore(1) +(2) =2X=10 X=5 so, small force is 5N option a is correct answer please like my answer 😀☺😊 The smaller force will be equal to 2solution, :-let the larger force be A and smaller be Bnow,maximum resultant force =A+B=7...(1)And minimum resultant force =A-B=4...(2)adding both equations, we get, 2A=10A=5and smaller force B=7-AB=2hence smaller force =2 B is the smaller force |
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| 4245. |
EXAMPLES BASED ON FORCE AND EQUILIBRIUMTwo forces are such that the sum of their magnitudes is 18 N and their resultant is perpendicular to thesmaller force and magnitude of resultant is 12. Then the magnitudes of the forces are(a) 12 N, 6N4.(b) 13 N, 5N(c) 10 N, 8 N(d)16 N, 2 N |
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| 4246. |
Vector addition is commutative A+B-B Aexplain with example |
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| 4247. |
3. The frequency, which is audible to thehuman ear isa) 50 kHzb) 20 kHzc) 15000 kHzd) 10000 kHz |
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Answer» the frequency which is audible to the human ear is 20 kHZ 20 kHz means 20,000 Hertz 20kHz is answer of this question The human range is commonly given as 20 to 20,000Hz.So option (b) 20 kHz is the right answer. the frequency which is audible to the human ear is 20 KHZ the answer is 20kHz to this question |
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| 4248. |
Infra Sound is the Sound with frequenciesa) Above 20 KHzc) Between 20 KHz and 20 KHz44.b) below 2d) of theseK |
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Answer» lower in frequency than 20 Hzis called infrared sound. option (d) |
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| 4249. |
5.2The slew rate of an op-amp is 0.5 V/usec. Themaximum frequency of a sinusoidal input of2 Vrms that can be handled without excessivedistortion is(a) 3 kHz(b) 30 kHz(c) 200kHz(d) 2 MHz |
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Answer» Option Bas slew rate = 2πfknow2πfk should be less than Slew ratef should be less than Slew k= 2√2 Vassume gain = 1fmax= 0.5/10^-6*2*3.14*2*√2 = 28.13 |
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| 4250. |
Audible range of a human ear is 20 Hz-20 kHz. Express this range in terms of time period. |
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Answer» 1/20 = 0.05second 1/20000- 5 *10^ -5 secondas frequency= 1/time period |
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