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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.
| 501. |
If you put one ice cube in a glass of water and another in a glass of alcohol, what would you observe? |
| Answer» SOLUTION :Explain your observations. Ice FLOUTS in water and not in alcohol. This is because the density of ice is `0.917` g/ccwhich is lower than that of water which is 1. Whereas, the density of ethanol (alcohol) is only `0.789`g/ccwhich is lesser than ice, hence it FLOATS in alcohol. | |
| 502. |
Sound waves travel very fast in _____. |
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Answer» in LIQUIDS |
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| 503. |
What is a horse power? |
| Answer» Solution :A horse POWER (HP) is ANOTHER unit of power, 1 hp = 746 watts | |
| 504. |
A book of mss 1 kg is raised through a height 'h'. If the potential energy increased by 49 J, find the height raised. |
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Answer» SOLUTION :The INCREASE in POTENTIAL ENERGY = mgh That is , mgh = 49 J (1) (9.8)h = 49 J The HEIGHT raised, `h = (49) //(1xx9.8)` = 5m |
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| 505. |
An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms^(-1) to 8 ms^(-1) in 6 s . Calculate the initial and final momentum of the object . Also find the the magnitude of the force exerted on the object. |
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Answer» Solution :Initial velocity of the obeject u=5 m/s FINAL velocity of the OBJECT v=8 m/s Mass of the object m=100 KG TIME taken by the object to acceleration ,t=6s Initial momentum =mu=`100 xx 5 =500 kg ms^(-1)` Final momentum=mv=`100 xx 8 =800 kg ms^(-1)` Force exerted on the object ,F=mv-mu/t =m(v-u)/t=800-500 =300 /6 =50 N initial momentum of the object is `500 kg ms^(-1)` Final momentum of the object is `800 kg ms^(-1)` Force exerted on the object is 50 N |
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| 506. |
A solid body weighs 2.10 N in air. Its relative density is 8.4. How much will the body weigh if placed (i) in water (ii) in a liquid of relative density 1.2? |
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Answer» Solution :(i) Given: Weight of the BODY in air `W_(1)=2.10N`. R.D. of body `=8.4`, weight of body in WATER `W_(2)=?` `R.D.=(W_(1))/(W_(1)-W_(2))"":.""8.4=2.1/(2.1-W_(2))` or `8.4(2.1-W_(2))=2.1` or `W_(2)=(2.1xx7.4)/8.4=1.85N` Thus weight of body in water `=1.85N` (ii) Upthrust due to water `=W_(1)-W_(2)=2.10-1.85` `0.25N` Upthrust due to liquid `=` Upthrust due to water `xx` R.D. of liquid `=0.25xx1.2=0.30N` `:.` Weight of body in liquid `=` Weight of body in air - Upthrust due to liquid `=2.10-0.30=1.8N` ALTERNATIVE method: Let weight of body in liquid be x N. Then R.D. `=("Weight of body in air")/("Weight of body in air - Weight of body in liquid")xx` R.D. of liquid or `8.4=2.1/(2.1-x)xx1.2` or `4(2.1-x)=1.2` or `x=7.2/4=1.8N` |
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| 507. |
The momentam of a light and heavy objects are same. Find the ratio of their kinetic energy. Whose kinetic energy is more? |
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Answer» `=(1)/(2)mv^(2)xx(m)/(m)` `=(1)/(2)(m^(2)v^(2))/(m)` `=(p^(2))/(2M)(because p=mv)` Here, the momentum of two OBJECTS is same. So `E_(k)prop(1)/(m)` `therefore` For lighter object `E_(k_(1))prop(1)/(m_(1))` `therefore` For heavier object `E_(k_(2))prop(1)/(m_(2))` `therefore (E_(k_(1)))/(E_(k_(2)))=(m_(2))/(m_(1))` But, `m_(2)gtm_(1)` `therefore (m_(2))/(m_(1))gt1` ``therefore (E_(k_(1)))/(E_(k_(2)))gt1` `therefore E_(k_(1))gtE_(k_(2))` |
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| 508. |
Akhtar, kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when as insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar saidthat since the motorcar was moving with a larger velocity, it exerted, a large force on the insect. And as a result the insect died. Rahul while putting an exnirely new explanation said that both the motorcar and the insect experiennced the same force and a change in their momentum. Comment on these suggestions. |
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Answer» SOLUTION :Akhtar STATEMENT is CORRECT. `m_(1)u_(1) + m_(2)u_(2) = m_(1)v_(1) + m_(2)v_(2)`. |
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| 509. |
Dischargeis stopped in a discharge tube at a pressure nearly equal to______. |
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| 510. |
When a body falls freely towards the Earth, then its total energy : |
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Answer» INCREASES |
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| 511. |
Is the acceleration due to gravity of earth g a constant ? Discuss. Calculate the acceleration due to gravity on the surface of a satellite having a mass of 7.4 xx 10 ^(22) and a radius of 7.4 xx 10 ^(22) kg and a radius of 1.74 xx 10^(6)m G(=6.7 xx 10 ^(11) Nm^(2)//kg ^(2)).Which satellite do you think it could be ? |
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| 512. |
A train covers 's' distance at the constant speed 30 km h^(-1)Then it covers the same distance in opposite direction at constant speed 45 km h^(-1), then its average speed would be ......... km h ^(-1). (37.5, 36, 25) |
| Answer» Solution :AVERAGE speed `= ("Total distance")("Total TIME") = (s +s)/(( s )/(30 ) + (s)/(45)) = (2s )/((s (45 + 30))/(30 xx 45 ))= (2s (30 xx 45))/( s (75)) = (2700)/(75) = 36 km H ^(-1)` | |
| 513. |
A standing passenger falls backwards when the bus starts suddenly. Which Newtons law gives the above concept.State the law. |
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| 515. |
Distinguish between Distance and Displacement. |
Answer» SOLUTION :
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| 516. |
A train moving with a velocity of 20 m s^(-1)' is brought to rest by applying brakes in 5 s. Calculate the retardation. |
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| 517. |
Name and define the SI unit of current. |
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Answer» ampere |
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| 518. |
The least count of a vernier callipers is 0.01 cm and its zero error is + 0.02 cm. While measuring the length of a rod, the main scale reading is 4.8 cm and sixth division on vernier scale is in line with a marking on the main scale. Calculate the length of the rod. |
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Answer» Solution :Given, L.C. = 0.01cm, zero ERROR = + 0.02 cm, main SCALE READING = 4.8 cm, and number of vernier division coinciding with the main scale division is 6, so, p = 6 Length of the ROD = observed reading - zero error = [main scale reading + (coinciding vernier division `XX` L.C.)] -` zero error. `= [(4.8 cm ) + (6xx 0.01 cm)] + (-0.02 cm)` `=(4.8 cm + 0.06 cm) - (+ 0.02 cm)` `= 4.86 cm - 0.02 cm = 4.84 cm`. |
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| 519. |
A student lists the following precautions for the experiment on determining the velocity of a pulse propagated through a stretched string: (A) The string should not be stretched too tight. (B) The counting of the pilse journeys must start from zero and not from one. (C) The string should be stretched straight in contact with the table. (D) The amplitude of the pulse should be kept appreciably high. The incorrect entry, in this list of precautions, is the precaution listed as: |
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Answer» A |
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| 520. |
Name some equipments that use electromagnetism for functioning. |
| Answer» SOLUTION :MANY of the medical equipments such as scanners, x-ray equipments and other equipmetns ALSO use PRINCIPLE of electromagnetism for their functioning. | |
| 521. |
Why are the magnetic poles of a bar magnet not situated at the ends, but located slightly inside ? (or) why is the effective length of a bar magnet different from its actual length |
| Answer» SOLUTION :The magnetic pole is a point inside a magnetic where it ATTRACTIVE power is maximum. Where do the magnetic lines of FORCE ENTERING or leaving at the ends of a magnet appear to come from ? | |
| 522. |
The velocity of light is maximum in |
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Answer» vacuum |
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| 523. |
Give reason We slip on a mossy surface. |
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| 524. |
State whether the following statement is true or false: Sound produced by a vibrting body travels to our ears by the actual movement of air. |
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| 525. |
Rajan went on a World Tour with his family during the summer holidays. Actually, it was a group tour consisting of 40 persons arranged by a leisure travel company. Thomas Cook'. Today the whole group was taken to a sea that lies between Is real and Jordan, and has become a famous tourist spot. On reaching the seashore, everyone was surprised to see a person floating in this sea water in the sitting position and even reading a newspaper in this position. The Guide, Mr. Jose, who accompanied the group, asked if anyone could explain this strange observation. Rajan is a science student of class IX. Rajan took a little of sea water in his palm and put it into his mouth for a moment (and then spit it out). After thinking for a while, he could answer Mr, Jose's question. Rajan then explained everything very clearly to all the persons in the group. Everyone appreciated his knowledge of science. (a) What is the name of the sea which lies between Isreal and Jordan? (b) Why is this sea called by this name? (c) What did Rajan find when he put a little of water from this sea into his mouth for a moment? (d) Explain why, a person can float in this sea water in the sitting position. (e) Why is it not possible to float in sitting position in the water of Indian sea? (f) What values are displayed by Rajan in this episode? |
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Answer» Solution :(a) Dead sea. (b) It is called 'Dead sea' because DUE to its very high salt content, no living things (plants and animals) can exist in it. (c) When Rajan put a little of this sea water in this mouth, he found it to be extermely salty (having a lot of SALTS dissolved in it). (d) The water of Dead sea has an extremely large amount of salts dissolved in it. Due to large amounts of dissolved salts, the density of this sea water is very,very high. Because of extremely high density, the water of Dead sea exerts a very,VERT large upward 'buoyant force' (or upthrust) that MAKES a person float in it sitting up. (e) Indian sea water contains much less dissolved salts than the water of Dead sea. So, the salty water of Indian sea exerts much less upward 'buoyant force' (or upthrust) which is not able to support the weight of a person in sitting up position. (f) The various values displayed by Rajan in this episode are (i) Awareness of buoyant force EXERTED by liquids (ii) Knowledge of factors affecting the magnitude of buoyant force (such as density of liquid) (iii) Application of knowledge in solving problems, and (iv) Analytical mind (in checking sea water). |
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| 526. |
For two cars A and B in motion distance time graph is shown: (a) State initial positions of car A and car B. (b) When and at what distance from the origin, these two cars would meet? (c) Find final speed of car A and car B. |
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Answer» Solution :(i) The initial position of CAR A = ORIGIN .O. The initial positon of car B is 500 m from origin .O. (ii) At 40 s and 600 m away from .O. (ii)` 15 MS ^(-1), 2.5 ms ^(-1)` |
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| 527. |
Which of the statements is true in case of an astronomical telescope? |
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Answer» The image of the object is ERECT when VIEWED through it. |
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| 528. |
Acceleration can get negative value also |
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| 529. |
As an air bubble rises from the bottom of a large water storage tank to free surface of water, the radius of the air bubble increases from 6 mm to 10 mm. The temperature of the water at the surface is 42^(@)C and its bottom is 27^(@)C. Find the depth of the water tank. (Take density of water =1g cm^(-3), g=10 ms^(-2), 1 atmospheric pressure = 760 mm of Hg, density of mercury =13.6 g cm^(3) |
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Answer» Solution :(i) Consider the pressure `(P_(1))," volume "(V_(1))" and temperature "(T_(1))` of the air in the bubble in TERMS of S.I. units. `P_(1)=1 atm = 10.336` m of water `V_(1)=(4)/(3)xxpi r_(1)^(3)=(4)/(3)xxpi((10)/(1000)m)^(3)` `T_(1)=(42+273)K` Find the volume `(V_(2))" and temperature "(T_(2))` of the air bubble at the bottom of the tank. `V_(2)=(4)/(3)xxpi r_(2)^(3)=(4)/(3)xxpi((6)/(1000)m)^(3)` `T_(2)=(27+273)K` Apply the value of `P_(2)` (in terms of pressure exerted by water COLUMNS) from (1) Then, the height of the water colomn, `H=(P_(2)-P_(1))` m (ii) `35.273` m |
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| 530. |
When spring is compressed its potential energy ……….. |
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Answer» REMAINS constant |
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| 531. |
On which law, hydraulic brak works? Name other three devices that works on the basis of this law. State the law. |
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| 532. |
Which of the following graphs represents uniform motion of an object ? |
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| 534. |
A rubber ball floats on water with its 1/3 rd volume outside water. What is the density of rubber? |
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| 535. |
{:(,"Column - I", ,"Column - II"),((a),"Tuning fork",(i),"The point where density of air is maximum"),((b), "Sound",(ii),"Maximum displacement from the equilibrium position"),((c ), "Compressions", (iii), "The sound whose frequency is greater than 20,000 Hz"),((d),"Amplitude",(iv),"Longitudinal wave"),((e ), "Ultrasonics",(v),"Production of sound"):} |
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| 536. |
Arrange the following speeds in increasing order: 10 m s^(-1), 1 km min^(-1) 18 km h^(-1) |
| Answer» SOLUTION :18 KM `h^(-1), 10 m s^(-1)`, 1 km `MIN^(-1)` | |
| 537. |
The smoke from the gun barrel is seen 2 second before the explosion is heard. If the speed of sound in air is 340 ms^(-1), calculate the distance of observer from gun. State the approximation used. |
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Answer» Solution :Given : Speed of SOUND `=340 ms^(-1) ,` time = 2s Speed `=("distance travelled")/("time period")` `THEREFORE` Distance = speed `xx` time `=340xx2 = 680 m` Approximation: The speed of light `(= 3 xx 10^8 m s^(-1))` is much larger in comparison to the speed of sound `(= 340 m s^(-1))`, therefore we can assume that the light takes NEGLIGIBLE time and sound takes 2 s to REACH the observer |
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| 538. |
An object of mass , m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest. |
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Answer» Solution :KINETIC energy of an object of mass, moving with a velocity, v is given by the EXPRESSION Ek = `1/2 mv^(2)` To bring the object to rest, `1/2 mv^(2)` amount of WORK is requied to be DONE on the object. |
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| 539. |
What is SONAR ? |
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| 540. |
A conductor carries a current of 0.2 A. (a) Find the amount of charge that will pass through the cross section of conductor in 30s. (b) How many electrons will flow in this time interval if charge on one electron is 1.6xx 10^(-19)C? |
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Answer» Solution :Given : `I = 0.2 A, t = 30 s` (a) `"CHARGE" = "Current" xx "time"` or `Q = Ixx t = 0.2 xx 30 = 6 C` (b) If `n` electrons flow and `e` is the charge on one electron, then total charge passed `Q = ne` `:. n = (Q)/(e) = (6)/(1.6 xx 10^(-19)) = 3.75 xx 10^(19)` |
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| 541. |
Ilustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventua- lly come to rest? What happens to its energy eventually ? Is it a viloation of the law of conservation of energy. |
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Answer» Solution :The law of conservation of energy STATES that energy can be neither created nor DESTROYED. It can only be conserved from one from to another. Consider the case of an oscillating pendulum. When a pendulum moves from its mean position P to either of its extreme positions A or B, it rises through a height h above the mean level P. At this point, the kinetic energy of the bob changes completely into potential energy. The kinetic energy becomes zero and the bob possesses only potential energy. As it moves TOWARDS point P, its potential energy decreases PROGRESSIVELY. Accordingly, the kinetic energy increases. As the bob prossesses only kinetic energy. This process is repeated as long as the pendulum oscillates. The bob does not oscillate forever. It comes to rest because air resistance resists its motion. The pendulum loses its kinetic energy to overcome this friction and stops after same time. The law of conservation of energy is not violated because the energy lost by the pendulum to overcome friction is gained by its surrounding. Hence, the TOTAL energy of the pendulum and the surrounding system remain conserved. |
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| 542. |
A body weighs 82.1 gf in air, 75.5 gf in water and 73.8 gf in a liquid. A. Find the relative density of the liquid. B. How much will it weigh if immersed in a liquid of relative density 0.87 ? |
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Answer» Solution :a. Given WEIGHT of the BODY in air `W_(1)=82.1gf` Weight of the body in liquid `=W_(2)=73.8gf` Weight of the body in water `W_(3)-75.5gf` R.D. of liquid `=(W_(1)-W_(2))/(W_(1)-W_(3))` `=(82.1-73.8)/(82.-75.5)=8.3/6.6=1.26` b. Given R.D. of liquid `=0.87, W_(1)=82.1gf, W_(2)=?,W_(3)=75.5gf` From RELATION R.D. `=(W_(1)-W_(2))/(W_(1)-W_(3))` `0.87=(82.1-W_(2))/(82.1-75.5)` or `82.1-W_(2)=0.87xx6.6=5.742` `:.W_(2)=82.1-5.742=76.358` `=76.4gf` |
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| 543. |
For verifying the laws of reflection of sound, a student has to choose from: (i) a black polished metal sheet or a white thermocole sheet. (ii) a 0.5 m long tube of diameter 3 cm or a 1.5m long tube of diamater 20 cm He should prefer to choose the: |
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Answer» METAL SHEET and the 0.5m LONG tube |
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| 544. |
Area of the region under graph of velocity-time for an object performing uniform motion on a linear path, suggests distance covered by the object during given time interval |
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Answer» Solution :The SPEED v of the OBJECT performing uniform motion on a linear path REMAINS constant. $ The GRAPH of velocity-time for it is obtained straight line parallel to t-axis. Area of region COVERED under this graph = Area of a rectangle or square = speed x time = distance |
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| 545. |
When a fresh eggs is put into a beaker filled with water, it sinks. When the same egg is put in a strong salty water, then if floats. Which of the following is the incorrect statement in this context? |
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Answer» salt water ENTERS into EGG by osmosis and makes it lighter |
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| 546. |
Given below are two statements. Which of the state meants is/are true ? Statement-A : when the south pole of a bar magnet points towards the geographic north pole, the neutral points are along the equatorial line Statement B: At the neutral point, the Earth's magnetic field is zero . |
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Answer» A is TRUE, B is false |
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| 547. |
A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road, then the work done against the gravitational force would be …………. J (Take g=10ms^(-2)) |
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| 548. |
How is the frequency of a wave related to its time period ? |
| Answer» SOLUTION : FREQUENCY = 1/time PERIOD | |
| 549. |
A coil of insulated copper wire is wound around a piece of soft iron and current is passed in the coil from a battery. What name is given to the device so obtained ? Give one use of the device mentioned by you. |
| Answer» SOLUTION :ELECTROMAGNET, ELECTRIC RELAY | |
| 550. |
Define pascal's law. Explain the application of pascal's law in our daily life. |
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Answer» <P> Solution :Pascal.s law states that the external pressure applied on an incompressible liquid is transmitted uniformly throughout the liquid. Pascal.s law became the basis for one of the important machines ever developed, the hydraulic press. It consists of two cylinders of different cross-sectional areas. They are fitted with pistons of cross-sectional areas "a" and "A". The OBJECT to be compressed is placed over the PISTON of large cross-sectional area A. The force `F_(1)`is applied on the piston of small cross-sectional area a .The pressure produced by small piston is transmitted equally to large piston and a force `F_(2)`actson A which is much larger than `F_(1)`.Pressure on piston of small area .a. is given by, `P = F_(1)//A_(1)` ..................... (1) Applying I.ascal.s law, the pressure on large piston of area A will be the same as that on small piston. Therefore, `P = F_(2)//A_(2)` ........................(2) Comparing equations (1) and (2).we get `F_(1)//A_(1) = F_(2)//A_(2) " or " F_(2) = F_(1)x A_(2)//A_(1)` Since, the ratio `A_(2) //A_(1)` is GREATER than 1, the force `F_(2)`that acts on the larger piston is greater than the force `F_(2)` acting on the smaller piston. Hydraulic systems working in this way are known as force multipliers. 
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