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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.
| 301. |
What do you mean by buoyancy ? |
| Answer» SOLUTION :The UPWARD force EXERTED by a liquid on an object that is immersed in it is known as BUOYANCY | |
| 302. |
An engineer was given a task to measure the rate of increase in pressure at the bottom of an empty cylindrical tank which is filled with water through hose pipe. If the speed of water coming out of the hose pipe is "10 m s"^(-1), diameter and radius of the cylinder and hose pipe are 5 m and 25 cm, respectively, find the result shown by the engineer. (Take g = 10" m s"^(-2)) |
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Answer» Solution :Speed of water coming out of the hose pipe, `"V=10 m s"^(-1)`. `:.` the length of the water column coming out of the hose pipe in one second of time, = 10 m. Radius of the hose = 25 cm = `(1)/(4)m`. `:.` area of cross section of the hosse pipe, `a=pi((1)/(4))^(2)m^(2)`. `:.` VOLUME of water coming out of the hose pipe per second `=v=la=10xxpi((1)/(4))^(2)m^(3)` Hence, the volume of water filled in the tank in one second of time `10xxpi((1)/(4))^(2)m^(3)` Given, diameter of the tank = 5 m `:.` radius of the tank = 2.5 m `:.` area of cross-section of the tank `=pi(2.5)^(2)m^(2)` Let 'H' be the increase in height of the water column, due to water filled in the tank in one second of time. `:.` volume of the water filled in the tank in one second of `=hpi(2.5)^(2)m^(3)` (2) From equation (1) an (2), we get `10xxpi((1)/(4))^(2)=HXX(2.5)^(2)pirArrh=0.1m` `:.` Increase in the height of water column in the tank in one second, h=0.1 m `:.` rate of increase in pressure at the bottom the tank. = hpg Given, `"g=10 m s"^(-2)` and density of water, `p=10^(3)"kg m"^(-3)` `:.` rate of increase in pressure at the bottom of the tank. `=(0.1)xx(10^(3))xx(10)"Pa s"^(-1)=10^(3)"Pa s"^(-1)` |
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| 303. |
A block of iron floats on mercury. Find the fractionof volume which remains immersed in mercury. (Densities of iron and mercury are 7.8gcm^(-3) and 13.6gcm^(-3) respectively). |
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Answer» Solution :Let V be the volume of IRON block and v be its volume immersed in MERCURY. For floatation. Weight of block = Weight of mercury displaced by the immersed portion of block. i.e. `Vxx7.8xxg=vxx13.6xxg` or `v/V=7.8/3.6=0.574` or `v=0.574V` The fraction 0.574 of TOTAL volume will REMAIN immersed in mercury. |
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| 304. |
A block of 2 kg is lifted up through 2mfrom the ground. Calculate the potential energy of the block at that point. |
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Answer» Solution :Mass of theblock, m = 2kg Height raised, h = 2 m Acceleration due to gravity, `g = 9.8m//s^(2)` POTENTIAL energy of the block, P.E. = MG h ` = (2)(9.8)(2)` ` = 39.2` J |
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| 305. |
The pendulum shown in the figure is kept horizontal. It is released from this positions. What is its velocity when it reaches the lowest position? (g=9.8ms^(-2)) |
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Answer» = potential energy at position A `(1)/(2)mv_(max)^(2)=mgh` `therefore v_(max)^(2)=2gh` `therefore v_(max)=sqrt(2gh)` `therefore` Maximum velocity of PENDULUM, `v_(max)=sqrt(2xx9.8xx2.5)` `=sqrt(49)` `=7ms^(-1)` |
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| 306. |
A solid of density 900 kg//m^(3) floats in oil as shown in the given diagram. The oil floats on water of density 1000 kg//m^(3) as shown.The density of oil in kg//m^3 could be: |
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Answer» 850 |
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| 307. |
A seconds' pendulum is taken to a place where acceleration due to gravity falls to one-fourth. How is the time period of the pendulum affected, if at all ? Give reason. What will be its new time period ? |
| Answer» SOLUTION :The time period INCREASES (it is DOUBLED) because `T prop 1/(sqrtg)` . To Its new time period will be 4 s. | |
| 308. |
Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height . |
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Answer» Solution :When a stone is THROWN vertically upwards , it has some initial vleocity (let u). As the stone goes its velocity goes on decreasing (`because` it is moving against the gravity ) and at the highest point i.e., maximum height its velocity become ZERO. Let the stone takes time 't' second to reach at thehighest point. After that stone begins to fall (with zero initial velocity ) and its velocity goes on increasing (since it is moving with the gravity) and it reaches its initial point of projection with the velocity v in the same time (with which it was thrown). So, `{:("Velocity",u,0,-u),("Time",0,t,2t):}` Here, we have TAKEN -u because in the upward motion velocity of stone is in upward direction and in the downward motion, the velocity is in downward direction . The velocity-time GRAPH for the whole journey is show below 
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| 309. |
What do you mean by anomalous expansion of water? |
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| 310. |
What is the importance of universal law of gravitation? |
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Answer» Solution :Universal law of gravitation is important because it tell US about The force that is responsible for binding us to earth The motion of moon AROUND the earth The motion of planets around the sun. The tides formed by rising and falling of WATER level in the ocean are due to the gravitational force EXERTED by both sun and moon on the earth. |
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| 311. |
Explain how the human ear works |
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Answer»
`**` Outer ear : This is also called .pinna.. It collects the sound from the surroundingand DIRECTOR it towards auditory canal. `**` Middle ear : The sound reached the end of the auditory canal where there is a THIN membrane called EARDRUM or tympanic membrane. The sound wave set this membrane to vibrate. These vibrations are amplified by three small bones. hammer, anvil and stirrup. `**` Inner ear :These vibrations reach the COCHLEA in the inner ear and are converted into electrical signalswhich are sent to the brain by the auditorynerve and the brain interprets them as sound. |
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| 312. |
A telescope has an objective of focal length 100 cm and eye piece of focal length6 cm andthe least distanceof distinct vision is 25 cm. The telescope is focused for distinctvision of an object at a distance100 m from the objective. What is thedistance of separationbetween objective and eye piece? |
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Answer» Solution :(i) The image FORMED on the OBJECTIVE is real. Let the image distance be `v_(1)` This image serves as theobject for EYE piece Determine theobject distance for the eye piece `(u_(2))` `:. ` THEDISTANCE between objective and eye piece = `v_(1) + u_(2)` (ii) `106.89` cm |
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| 313. |
The spring will be a maximum potential energy |
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Answer» It is PULLED out |
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| 314. |
A student draws a distance-time graph for a moving scooter and finds that a section of the graph is a horizontal line parallel to the time axis. Which of the following conclusion is correct about this section of the graph ? |
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Answer» the SCOOTER has uniform speed in this section |
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| 315. |
Assertion: The Speedometer of a car or a motor-cycle measures its average speed. Reason: Average velocity is equal to total displacement divided by total time taken. |
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Answer» If both assertion and REASON are TRUE and reason is the CORRECT EXPLANATION of assertion |
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| 316. |
Calculate the amount of charge that would flow in 2 hours through an element of an electric bulb drawing a current of 2.5A. |
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Answer» Solution :GIVEN : Time .t.=2 hours `=2xx60xx60s` `t=7200s` `I= 2.5A` AMOUNT of CHARGE, `Q= Ixxt` `=2.5xx7200` `Q= 18000C` |
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| 317. |
A driver of a car travelling at the speed of 52 km h^(-1)applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 34 kmh ^(-1) in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied ? |
Answer» Solution : For FIRST DRIVER: initial speed `= 52 km h ^(-1)` ` = (52 xx 1000)/(60 xx 60) ms ^(-1) ~~ 14.44 ms ^(-1)` For second driver : initial speed `= 34 km h ^(-1)` `= (34 xx 1000)/(60 xx 60) ms ^(-1) ~~ 9.44 ms ^(-1)` Distance COVERED by the first car before COMING to REST =area of `Delta AOB` `=1/2 xx AO xx OB` `=1/2 xx 14.44 ms ^(-1) xx 5s = 36 .1 m""...(1)` Distance covered by the second driver = area of `Delta COD` `=1/2 xx CO xx OD` `=1/2 xx 9.44 ms ^(-1) xx 10 s ` `= 47.2m ""...(2)` From (1) and (2), the distance travelled by the second car is more. |
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| 319. |
The pitch of a screw ina screw jack A is half that of another screw jack B. for 5 complete rotations of the lever, which one of the jack will lift a car more higher? Which one need more effort to do an equal work if both the jacks have livers of equal length. |
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Answer» <P> Solution :Since the HEIGHT of lifting the car by aJack depends on the PITCH of the screw, and for one rotation of the LEVER, the screw moves up bya distance equal to its pitch.Therefore, greater the pitch, higher the car gets lifted, same is the Jack'2' that it will raise the car high, and same is case of screw jack. `(w)/(E)=(2pi L)/(P)` ` :.E = (W xx P)/(2 pi L)` The work done by The two jacks is equal and the length of their levers is equal, E ` prop` pitch. If pitch increases, effort increases. |
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| 320. |
Why cannot transverse waves be produced in air ? |
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Answer» SOLUTION :(i) Advantages of ULTRA sounds (ii) CHANGES that take place in the MEDIUM when transverse waves propagate . |
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| 321. |
Express the number in exponential notation: (a) 563 (b) 0.0012 (c) 43,900,000 (d) 0.0000007190 |
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Answer» Solution :(a) `5.63xx10^(2)` (B) `1.2xx10^(-3)` (c) `4.39xx10^(7)` (d) `7.19xx10^(-7)`. |
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| 323. |
The frequencies of stationary waves formed in closedend organ pipes are in the ratio 1:3 :5:7….. |
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| 324. |
Measure this time interval using a digital wrist watch or a stopwatch. Calculate the distance of the nearest point of lightning (Speed of sound in air = 346 ms^(-1)) |
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Answer» Solution :SUPPOSE the sound of thunder is heard 2 second after the lightning was seen. `to`The DISTANCE of the NEAREST point of lightning = SPEED `xx` time `= 346 m s^(-1)xx2 s = 692 m` |
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| 325. |
Ankit and Rehman are two friends who live in the same colony. Ankit studies in class IX whereas Rehman is a student of class VII. One day, both Ankit and Rehman were playing cricket with other boys of the colony in the bid ground just outside their colony. At the moment, Ankit and Rehman were standing at two different fielding positions. When batsman hit the ball hard, it went very fast towards Rehman. Rehman stopped and caught the fast moving cricket ball but his hands were hurt badly in stopping the ball. The severe pain in the hands of Rehman made him drop the catch. Next moment, th bastman again hit the ball hard. This time the fast moving ball went straight towards the direction of Ankit. Ankit stopped and caught the fast ball in a particular way without hurting his hands at all. While coming back home after playing cricket, Ankit explained the proper way of catching a fast cricket ball to Rehman without getting the hands hurt. Keeping this advice in mind, Rehman never hurt his hands again while playing cricket. (a) Which physical quantity is vert large in a fast moving cricket ball having high speed? (b) In what way do you think Rehman stopped and caught the fast cricket ball which hurt his hands? (c) Why were Rehman's hands hurt in stopping and catching the fast cricket ball in this way? (d) In what way do you think Ankit stopped and caught the fast cricket ball without hurting his hands? (e) Why were Ankit's hands not hurt in stopping and catching the fast cricket ball in this way? (f) Which law of motion is involved in catching a fast cricket ball? (g) What values are displayed by Ankit in this episode? |
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Answer» SOLUTION :(a) A fast moving cricket ball has a large 'momentum'. (b) Rehman must have stopped (or caught) the fast moving cricket ball suddenly, keeping his hands stationary. Then the large momentum of the fast ball was reduced to zero in a very short time. Due to this, the rate of change of momentum of cricket ball was very large and hence it exerted a large force on Rehman's hands. This large force hurt his hands. (d) Ankit must have moved his hands backwards gradullay (on catching, the fast cricket ball). (e) When Ankit moves back his hands gradually on catching a fast ball, then the time TAKEN to reduce the large momentum of fast ball to zero was increased. Due to more time taken to stop the fast ball, the rate of change of momentum of ball was decreased and hence a small force was exerted on the hands of Ankit. This small force did not hurt Ankit's hands. (f) NEWTON's second law of motion. (G) The values displayed by Ankit in this episode are (i) Awareness (or knowledge) of Newton's second law of motion (ii) Application of knowledge in everyday SITUATIONS, and (iii) Concern for the safety of his friend. |
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| 326. |
Newton's law ofgravitation has been verified experimentally. But there is enough indirect evidence of its truth. For example, gravitational force of attraction of earth is responsible for binding all terrestrial object on earth. The same force is responsible for holding the atmosphere around earth, for rainfall and snowfall on earth. the gravitational force alone is responsible for holding our solar system in place, and so on. Read the above passage and answer the following questions: (i) name any two prediction made on the basis of this law. (ii) what values do you learn from this law ? |
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Answer» Solution :(i)The prediction about solar and junar eclips made on the bassis of this law- always COME out to be true. (ii) From universal law of gravitational, we learn the universal law of LOVE not only for the entire humanity. But also for PLANTS, birds, animals and other living beings. the entire UNIVERSE is a globle village, WHEREIN every creature deserves our love, affection and care. |
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| 327. |
Define : Accelaration |
| Answer» Solution :The CHANGE in VELOCITY of an object PER unit time is CALLED acceleration. | |
| 328. |
A stone of 1 kg is thrown with a velocity of 20 ms^(-1) across the frozen surface of a lake and comes to rest after travelling a distance of 50 m what is the force of friction between the stone and the ice ? |
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Answer» Solution :Initial VELOCITY of the stone , u=20 m/s Final velocity of the stone v=o Distance COVERED by the stone S=50 m Since `V^2-u^2` 2= 2AS `0-20^2=2a xx50` `F=-4ms^(-2)` Force of friction, F=ma=-4N |
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| 329. |
In a residence 4 tubelights each of 40W are operated daily for 5 hour and 3 fans each of 120 W are operated daily for 4 hour. What would be the amount of electricity bill at ₹ 5 per unit for the month of September? |
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Answer» `=4xx40Wxx5h=800Wh` Total electrical energy consumed through three fans in a day `=3xx120Wxx4h` `=1440Wh` `therefore` Total electrical energy consumed in a day = `800Wh+1440Wh` `=2240Wh` `=2.240xx10^(3)Wh` `=2.24kWh` = 2.24 UNIT (`because` 1 kWh = 1 unit) Total units consumed in the month of SEPTEMBER = `30xx2.24` = 67.2 units Now, as CONSUMPTION expense is ₹ 5 per unit, electricity bill for the month of September = `5xx67.2=₹336`. Eletricity bill : ₹ 336 |
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| 330. |
Velocity of a vehicle increases from 5ms to 15 ms^(-1)in 5 s. What is the magnitude of acceleration ? |
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Answer» ` 4m s^(-2)` |
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| 332. |
TheZRP of a physicalbalanceis 10.5whilefindingmass of asubstance.fora weight 34.23 g therestingpointwas found be8.5 when 10 mgwas removed the restingpoint was11.0 . The mostaccuratemass ofthe substanceis "_______" g. |
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Answer» Solution :The relative densityof a bodyis RATIOOF thedensityof thebodyto thedensityof WATER .R.D.Has nounits. Itis a ratioof same physicalquantity. If the UNITSOF masslengthand timeare doubledthere WILLBE thesamechangein thedensitiesof boththebodyand THEWATER . So theratioof thedensityof bodyand density is a ratiowhichis aconstantfor aparticularsubstance . |
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| 333. |
The density of iron is 7.8xx 10^3 "kg m"^(-3) What is itsrelative density? |
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| 334. |
A object of weight 40 N is made to fall freely towards the ground from height 10 m. When it reaches the ground its kinetic energy would be ………….. J. |
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| 335. |
Leave the bottle free. What do you observe? |
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| 336. |
Distinguish between beta -rays and cathode rays. |
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| 337. |
What is constant for an object performing uniform circular motion ? |
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Answer» Acceleration |
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| 338. |
Two bodies fall freely from different heights and reach the ground simultaneously. The time of descent for the first body is t_(1) = 2s and for the second t_(2) = 1s. At what height was the first body situated when the other began to fall? (g=10 m//s^(2)) |
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Answer» SOLUTION :The second body takes 1 second to reach ground. So, we need to find the DISTANCE traveled by the first body in its first second and in two seconds. The distance COVERED by first body in 2s, `h_(1)=1//2 gt^(2) =1//2 xx10xx2^(2)=20 m`. The distance covered in `1s, h_(2)=5 m`. The height of the first body when the other begin to fall h =20-5 =15 m. |
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| 339. |
You are given two identical bars, one of which is magnetised. How will you select the magnetised bar? |
| Answer» Solution :The two bars are SUSPENDED by a silk thread one by one, so that they can swing freely in a HORIZONTAL plane. The bar which comes to REST in the north-south direction will be the magnetised bar . | |
| 340. |
Whatismeantby latentheatof vaporization |
Answer» SOLUTION : Heat energy is absorbed by a solid during melting and an equal amount of heat energy is liberated by the LIQUID during freezing, WITHOUT any temperature CHANGE. It is called latent heat of fusion. In the same manner, heat energy is absorbed by a liquid during vaporization and an equal amount of heat energy is liberated by the vapor during condensation, without any temperature changes. This is called latent heat of vaporization. |
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| 341. |
A football floating on the waves of ocean water is an example of ____ equilibrium. |
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| 342. |
Two children are at opposite ends of an aluminium . One strikes the end of the rod with a stone . Find the ratio of time takes by the sound wave in air and in aluminium to reach the second child. |
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Answer» SOLUTION :`V_(1) = 346m//s ""V_(2) = 6240m//s` Length of aluminium ROD = XM SPEED `= ("distance")/("time ") "" :. " time "=("distance")/("speed")=(x)/(346)` sec Time taken in aluminium `=x/6420` sec Required RATIO `=((x)/(346))/(x/6420)=(x)/(346)xx(6420)/(x) =18.55` |
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| 343. |
A stone is throuwn in a vertically upward direction with a velocity of 5 ms ^(-1). If the acceleration of the stone during its motion is 10 ms ^(-2) in the downword direction, what will be the height attained by the stone and how much time will it take to reach there ? |
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Answer» Solution :Here, `u = 5 ms ^(-1), a =- 10 ms ^(-2)..` (Because the stone is moving in the upward direction). `u =0…` at the highest point. `v + u + at` `THEREFORE 0 = 5 + (1-10)t therefore 10 t =5 therefore t =0.5 sec`  The stone will REACH the height of `1.25 m ` and the time required is `0.5 sec. ` |
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| 344. |
What can you say about the average density of a ship floating on water in relation to the density of water? |
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| 345. |
The displacement of a moving object in a given interval of time is zero. Is the distance travelled by the object zero? Comment |
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Answer» Solution :Suppose an object MOVES from A to B a distance of 1 km on a straight road and then it comes to A from B. Here, the displacement of the object is zero. The distance travelled by the object from A to Bis 1 km and from B to A is 1 km. i.e., the TOTAL distance travelled by the object is 2 km. So, eventhough the displacement is zero, the distance travelled is not zero. |
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| 348. |
A graph of velocity versus time (v - t) for a moving particle on linear path is shown in figure. What would be the displacement of the particle in first 30 s? |
| Answer» Solution :VALUE of DISPLACEMENT = Area enclosed by V -t graph and TIME axis. | |
| 349. |
What conclusion is drawn regarding the magnetic field at a point if a compass needle at that point rests in any direction ? Give reason for your answer. |
| Answer» Solution :Magnetic field is zero. REASON: The earth.s magnetic field at that POINT is NEUTRALISED by the magnetic field of some other magnetised MATERIAL. | |
| 350. |
If an object is placed at the focus of a concave mirror, where is the. image formed? |
| Answer» Solution :IMAGE will be FORMED at INFINITY as real and inverted. | |