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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.
| 51. |
A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms^(-2),with what velocity will it strike the ground ? After what time will it strike the ground? |
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Answer» Solution :Here, `u =0,a = 10 ms ^(-2), s =20 m.` `2as = v ^(2) - u^(2)` `therefore 2 xx 10 ms ^(-2) xx 20 m = v ^(2) -0` `therefore v ^(2) = 400 m ^(2) s ^(-2) therefroe v = 20 ms ^(-1)` `v = u + at` `therefore 20 ms ^(-1) = 0 + 1 0 ms ^(-2) xx t` `therefore 20 ms ^(-1) = 10 ms ^(-2) xx t` `therefore t = (20 ms ^(-1))/(10 ms ^(-2)) =2s ` The ball will strike the GROUND with a velocity of `20 ms ^(-1).` It will strike the ground after 2s. |
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| 52. |
Find the frequency of a wave whose time period is 0.002 second |
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| 53. |
Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minute 30 second and then turns around and jogs 100 m back to point in another 1 minute. What are Joseph's average speeds and velocities in jogging (a) from A to B and (b) from A to C? |
Answer» Solution : (a) From A to B, distance COVERED = 300 m = displacement, time taken = 2 min 30 SEC `= 2 xx 60 + 30 = 150 sec ` Average speed `= ("TOTAL distance covered")/("Total time taken")` `= (300 m )/( 150 s ) = 2 ms ^(-1)` Average velocity `= ("Displacement ")/("Time taken")` `= (300 m)/(150)s ) = 2 ms ^(-1)` (b) From A to B to C, distance covered `= (300 + 100) m = 400 m` displacemenet `= AB -BC = (300 -100) = 200 m` time taken `= 2 min 30 sec + 1 min = (150 + 60) s = 210 s` Average speed `= ("Total distance covered")/("Total time taken")` `= (400)/(210) ~~ 1.90 ms ^(-1)` Average velocity `= ("Displacement")/("Time taken") = (200 )/(210) ~~ 0.95 ms ^(-1)` (a) Average speed `=1.90 ms ^(-1)` Average velocity `= 0.95 m s ^(-1)` |
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| 54. |
Suppose if the body is moved from the surface of the earth to the centre. What happens? |
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| 55. |
When brakes are applied to a bus, the retardation produced is 25 cm s^(-2) and the bus takes 20 s to stop. Calculate : (i) the initial velocity of bus, and (ii) the distance travelled by bus during this time. |
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| 56. |
The global climate has always fluctuated. Millions of years ago, some parts of the world that are now quite warm, were converted with ice, and over recent centries, average temperatures have risen and fallen in cycles. What is new, however, is that current and future climatechange will be caused not just by natural events but also by activities of human beings. Suggest three simple ways to help save the climate of out planet. |
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Answer» Solution :1. Drive LESS. Walk Carpoole or take Public Transport. You will save emission of 1.5 KG of `CO_2` for every 5 km you do not drive. 2. Use renewable energy. Use solar water heater instead of an ELECTRIC geyser. A 100 litre solar water heater can sace around 1500 units of electricity every year. For lighting, use batteries that can be charged by sunlight. A solar cooker cooks food without losingits essential natrients. 3. Check your tyres. Keeping your tyres properly inflated improves the FUEL efficiency of your vehical. Every litre of petrol saved, keeps 2.5 kg `CO_2` out of atmosphere. Using radial tyres will help you save 3% to 7% of fuel. |
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| 57. |
A mug full of water appears light as long as it is under water in the bucket than whenit is outside water . ? Why |
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| 58. |
A car travels first 30 km with a uniform speed of 60 km h^(-1) and then next 30 km with a uniform speed of 40 km h^(-1). Calculate : (i) the total time of journey, (ii) the average speed of the car. |
| Answer» SOLUTION :(i) 175 min (ii) 48 km `H^(-1)` | |
| 59. |
Find the water equivalent of paraffin oil if 100 kg of paraffin oil absorbs 4180xx10^(3)J to raise its temperature from 300 K to 320 K. (Take specific heat of water as 4.18 J g^(1)""^(@)C^(-1)) |
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Answer» Solution :(i) Recall the definition of water equivalent. Use the formula, `Q= C DELTA theta` where 'C' is the heat capacity and `Delta theta` is the rise in TEMPERATURE. Calculate the rise in temperature from the GIVEN data. Is the rise in temperature equal on Celsius scale and Kelvin scale ? (ii) 50 KG |
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| 60. |
What is the time period of satellite near the earth's surface? (neglect height of the orbit of satellite from the surface of the earth)? |
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Answer» Solution :The force on the satellite due to earth is given by `F=(GmM)/(R^(2))` M-Mass of earth `=6xx10^(24) kg` m-mass of satellite, R-radius of earth `=6.4xx10^(6) m` Let v be the speed of the satellite `v=(2pi R)/(T) RARR T=(2 pi R)/(v)` Required CENTRIPETAL force is provided to satellite by the gravitational force hence `F_(C )=(MV^(2))/(R)` But `F_(C)=(GMm)/(R^(2))` ACCORDING to Newton.s law of gravitation. i.e., `(GMm)/(R^(2))=(m(2pi R)^(2))/(T^(2)R)` `rArr T^(2) =(4 pi^(2) R^(3))/(GM)`, as mass of the earth (M) and G are constants the value of T depends only on the radius of the earth. `rArr T^(2) alpha R^(3)` Substituting the values of M, R and G in above equation we get, T = 84.75 minutes. Thus the satellite revolving around the earth in a circular path near to the earth.s surface takes 1Hour and 24.7 minutes approximately to complete one revolution around earth. |
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| 61. |
n' different liquids, which do not react chemically, are mixed to from a homogeneous mixture. If the densities of the liquids are p_(1),p_(2), . . . . . . . . . . . . P_(n), respectively, then the density of the homogeneousmixture when (a) the masses of the liquids forming the mixture are equal (b) the volumes of the liquids forming the mixture are equal |
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Answer» Solution :(1) Find the FORMULA for density. Then for GIVEN homogeneos mixture of 'n' liquids, `d=(m_(1)+m_(2)+m_(3)+ . . . . . +m_(n))/(v_(1)+v_(2)+. . . ..+v_(n))` (1) Then, for density of each liquid will be, `d_(1)=(m_(1))/(v_(1))` `d_(2)=(m_(2))/(v_(2))` `d_(n)=(m_(n))/(v_(n))` (2) (ii) The MASSES of the liquids forming the mixture are equal then, `d_(m)=(m+m+m+. . . . . +m)/(v_(1)+v_(2)+ . . . . .+v_(n))` `(nm)/(v_(1)+v_(2)+ . . . . .+v_(n))`(3) Convert the volumes of (3) in terms of DENSITIES. (iii) If volumes of liquids forming the mixture are equal. `"i.e.,"V_(1)=V_(2)=V_(3). . . . . .=V_(n)=V`(4) Substitute values of (4) in (1). Converts the masses and of liquids in terms of their densities. Substitute these and find density of the homogeneous mixture. |
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| 62. |
A bullet initially moving with a velocity 20 m s^(-1) strikes a target and comes to rest after penetrating a distance 10 cm in the target. Calculate the retardation caused by the target. |
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| 63. |
Describe with the help of a diagram , how compressions and rarefactions are produced m air near a source of sound. |
Answer» SOLUTION :
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| 64. |
Give an example of uniform acceleration. |
| Answer» Solution :The MOTION of a freely falling body is an EXAMPLE of uniformly ACCELERATED motion. | |
| 65. |
When brakes are applied to a car running on a straight road, retardation of 4 m s^(-2) is produced. It stops after 3 s. Calculate the distance travelled by car after brakes are applied. |
| Answer» SOLUTION :Use `V = u + at and s = ut + (1)/(2) at ^(2)` | |
| 66. |
A car accelerates at a rate of 5 "m s"^(-2). Find the increase in its velocity in 2 s. |
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| 68. |
If a stone and a pencil are dropped simultaneously in vacuum from the top of a tower, which of the two will reach the ground first ? Give reason. |
| Answer» Solution :Both will REACH the GROUND SIMULTANEOUSLY, since acceleration DUE to gravity is same (= g) on both. Also, the opposing VISCOUS force of air is absent in vacuum. | |
| 69. |
A train travels with a speed of 60 km h^(-1)' from station A to station B and then comes back with a speed 80 km h^(-1)' from station B to station A. Find : (i) the average speed, and (ii) the average velocity of train. |
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| 70. |
A cannon of mass m_(1) = 12000 kg located on a smooth horizontal platform fires a shell of mass m _(2) = 300 kgin horizontal direction with a velocity v _(2)=400 m//s. Find the velocity of the cannon after it is shot. |
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Answer» Solution :Since the pressure of the powder gases in the bore of the cannon is an iternal force the net external force ACTING on cannon during the firing is ZERO. Let `v _(1)` be the velocity of the cannon after shot. The initial momentum of system is zero. The final momentum of the system `=m _(1) v_(1) + m _(2) v_(2)` From the CONSERVATION of linear momentum, We GET, `m _(1) v_(1) +m_(2) v_(2) =0` `m_(1)v_(1) =-m_(2)v_(2)` `v _(1) =-m_(2) v_(2) //m_(1)` Substituting the given values in the above equation, we get `v _(1) =- ((300kg) x (400m//s))/(112 000 kg)` `=-10 m//s.` Thus the velocity of cannon is 10 m/s after the slot, Here .-. sign indicates that the canon moves in a direction opposite to the motion of the bullet. |
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| 71. |
An object of mass 40kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is halfway down. (Take g=10ms^(-2).) |
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Answer» Potential ENERGY `E_(p)=mgh` `=40kgxx10ms^(-2)xx5m` `=2000J` When it is allowed to fall, its potential energy gets converted into KINETIC energy. When the OBJECT is halfwary down `s=h=2.5m` Let its velocity be `vms^(-1)`. `v^(2)-u^(2)=2gh` `therefore v^(2)-(0)^(2)=2xx10xx2.5` `therefore v^(2)=50` Kinetic energy `E_(k)=(1)/(2)mv^(2)` `=(1)/(2)xx40xx50=1000J` |
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| 72. |
In which case does the hydrometer sink more? In liquids of greater density or those with a lower density? |
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| 73. |
What are the force experienced by an object inside a fluid? |
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| 74. |
What is the basis of Pascal's law? |
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| 75. |
Select the scalars and vectors from the following: Velocity, distance, acceleration, work, mass, retardation. |
| Answer» Solution :Scalars : DISTANCE, work, MASS. VECTORS : VELOCITY, acceleration, retardation. | |
| 76. |
A uniform metallic rod PQ of length 2 m is acted upon by two forces A and B along the directionsas shown in the figure. Find the magnitude and position of the resultant normal force that actson the rod. |
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Answer» Solution :The forces that act normally to the rod are ` A cos 60^(@) = 90 xx (1)/(2)=45N` ` B cos 60^(@) = 30 xx (1)/(2) = 15 N`. As the two forces are unlike parallel forces, the resultant of these two forces is outside the rod at a point 'O' and in the direction of greater force, i.e. , A.  Magnitude of R = 45 - 15 = 30 N POSITION of R: Let 'R' is at 'O' andata distance of X m from'P'. Then ` 45 xxx=15xx(2+x)` `3x=2+x` `2x=2` ` x = 1 m ` |
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| 77. |
Two identical pieces, one of ice (density =900kgm^(-3)) and other of wood (density =300kgm^(-3)) float on water. a. Which of the two will have more volume summerged inside water? b. Which of the two will experience more upthrust due to water? |
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| 78. |
State the laws of flotation. |
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Answer» Solution :LAWS of FLOTATION are, 1. The weight of a floating body in a fluid is EQUAL to the weight of the fluid DISPLACED by the body. 2. The centre of gravity of the floating body and the centre of buoyancy are in the same vertical line. |
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| 79. |
In a pressure cooker, the food is cooked faster because .......... . |
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Answer» increased pressure LOWERS the BOILING point |
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| 80. |
Refraction of light at air - water interface Put a straight pencil into a tank of water or breaker of water at an angle of 45^@ and look at it from one side and above. How does the pencil look now? |
| Answer» Solution :The PENCIL APPEARS to be BENT at the surface of WATER. | |
| 81. |
A manometer is connected to gas container. Then the mercury level rises by 2 cm in the arm of the manometer which is not connected to the container. If the atmospheric pressure is 76 cm of mercury, then the pressure of the gas is ________ cm of mercury. |
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Answer» 80 `:.` the pressure of the gas atmospheric pressure + pressure due to 4 cm of Hg = 76 cm of Hg+4 cm of Hg=80 cm of Hg |
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| 82. |
A road roller of 200 kg wt slides on ground when pushed by a lever AB of length 1m, as shown in the figure. The force required to slide the roller acts at a distance of 5cmfrom the fulcrum. If the coefficient of friction between the roller and the ground is sqrt(2) find the effort required to move the roller. (Takeg= 10 m s^(-2)) |
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Answer» Solution :The VARIOUS forces acting on the road roller are shown below in figure. Let the force required to slide the roller be F. The component of the force F, to overcomethe force of friction `f_(s)` is `F cos theta = F cos theta 45^(@)=(F)/(sqrt(2)) " " (1)`  The force of friction `f_(s)= mu N ` where` N = mg - F sin theta` substituting mg`=200xx10=2000 N` ` sin theta = sin = 45^(@)` `N= 2000 -(F)/(sqrt(2))` The frictionalforce `f_(s)` `f_(s)= mu N = sqrt(2)xx(200-(F)/(sqrt(2)))` `f_(s)=2000 sqrt(2)-F "" (2)` Equationg (1) and (2) `(F)/(sqrt(2))=2000 sqrt(2)-F` `F=4000-sqrt(2)F` `F=(4000)/(sqrt(2)+1)=(4000)/(sqrt(2)+1)xx(sqrt(2)-1)/(sqrt(2)-1)` `=4xx414=1656N` Applyingthe lawof MOMENTS to the lever, E `xx` EFFORT arm = load `xx` load arm `Exx1= 1656 xx 0.05` `E=82.8 N ~= 83 N` (approximately) |
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| 83. |
The leaning tower of PISA does not collapse inspite of being in a slanting position. Explain. |
| Answer» Solution :The LINE joining the centre of GRAVITY of the BUILDING and the centre of the Earth FALLS within the BASE of the building. So, the building is in stable equilibrium. | |
| 84. |
How is the current flowing in a conductor changed if the resistance of conductor is doubled keeping the potential difference across it the same ? |
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| 86. |
. In which of the following cases of motions, the distance moved and the magnitude of the displacement are equal?i. If the car is moving on a straight roadii. If the car is moving in circular pathiii. The pendulum is moving to and froiv. The earth is moving around the sun |
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Answer» only 1 |
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| 87. |
The unit of potential difference is : |
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Answer» ampere |
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| 88. |
Give 2 uses of ultrasound. |
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Answer» It is used in echo-cardiogaraphy, the ultrasonic waves are madeto reflect from various parts of the heart and FORM the IMAGE of the heart. |
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| 89. |
How much momentum will a dumbbell of mass 10 kg transfer to the floor if it falls from a height of 80 cm ? Take its downward acceleration to be 10 ms^(-2). |
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Answer» SOLUTION :Mass of the dumbbell m =10 kg Distance covered by the dumbbell , s=80 cm =0.8 m Acceleration in the downward direction `a=10 m//s^2` Initial velocity of the dumbbell,u=o FINAL velocity of the dumbbell (when it was about to hit the floor) =v According to the third equatons of motion `v^2=u^2+2as` `v^2=0=2(10)0.8` v=4 m/s HENCE themomentum with which the dumbbell hits the floor is `=mv =10xx4 =40 kg ms^(-1)` |
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| 90. |
The driver of a car approaching a cliff with a uniform velocity of 15 m s^(-1) sounds the horn and the echo is heard by the driver after 3 seconds . If the velocity of sound is 330 m s^(-1) , calculate the distance between the cliff and the point where the horn was sounded ? Also calculate the distance between the cliff and the point where the echo is heard ? |
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Answer» Solution :(i) 517.5 m , 472.5 m (ii) Equation for echo and velocity of SOUND (iii) Draw a rough linear figure DEPICTING the positions of the car approaching the cliff when it sounded the horn , and when the echo is heard by its DRIVER and the position of thecliff . From the figure find the distance travelled by the sound in the given time interval . For this consider the distance travelled by the car in the given time interval . Using the formula , velocity of sound = `("Distance travelled by the sound")/("time to heat the echo")` estimate the distance of the car from the cliff when it sounded the horn . |
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| 91. |
Which of the following is not true about simple machines? They |
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Answer» save ENERGY. |
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| 92. |
Take a glass of water and put some ice cubes inte it. Observe it for some time. What happens? The ice cubes melt and disappear. Why did it happen? |
| Answer» SOLUTION : It is because HEAT energy in the water is TRANSFERRED to the ice. | |
| 93. |
An object is placed between two identical convex mirrors X and Y of focal length 15 cm a themidpoint on their common principal axis. If the two mirrors are separated by a distance of 20 cm, determine thedistance of thefirst two images formed in the mirror Y . |
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Answer» Solution :(i) Find THEIMAGE distance by using `1/f=1/u+1/v` Now the FIRST image formed by THEMIRROR x acts as thevirtual OBJECT for themirror 'y'. `:.` The object distance for theformation of second image `= V_(1) = d` Where `V_(1)`is the first image distance and 'd' is thedistance between twomirrors. (ii) 6cm, 7.5cm |
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| 94. |
An object 2 cm high is placed at a distance of 16 cm from a concave mirror which produces a real image 3 cm high. Find the position of the image. |
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Answer» Solution :Height of object `h_1 =2` CM Height of real image `h_2 = -3` cm Magnification `m = (h_2)/(h_1) = (-3)/(2) = -1.5` We know that, `m = (-v)/(u )` Here, object distance `u= -16 cm` Substituting the value, we get `-1.5 = -(v)/((16)) rArr 1.5 = (v)/( 16) rArr = 16 XX (-1.5) = -24 c` The position of image is 24 cm in front of the mirror (negative sign indicates that the image is. on the LEFT side of the mirror). |
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| 95. |
Kinetic energy of a car, having mass 1000 kg is 1,12,500 J. Driver applies brakes when an obstacle is sighted and car comes to halt after travelling 100 m distance (without meeting with an accident). Calculate frictional force. |
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| 96. |
Which instrument is used to test the purity of milk? |
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| 97. |
State which of the following situations are possible and give an example for each of these: an object moving in a certain direction with an acceleration in the perpendicular direction. |
| Answer» SOLUTION :OBJECT has CIRCULAR PATH in MOTION. | |
| 98. |
What is the magnitude of the gravitational force between the earth and a 1 kg object emits surface? (Mass of the earth is 6xx10^(24)kg and radius of the earth is 6.4 xx 10^(6) m). |
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Answer» Solution :GIVEN that MASS of the body, m=1kg Mass of the Earth, `M=6xx10^(24)kg`. Radius of the earth, `R=6.4xx10^(6)` Now and of the gravitational FORCE between the earth and the body can be the `F=G(Mxm)/(r^(2))=(6.67xx10xx6xx10xx1)/((6.4xx10^(6))^(2))` `=(6.67xx6xx10)/(6.4xx64.4)=9.8N`. |
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| 99. |
A liquid flows into a vessel initially empty at a steady rate of 70 cm^(3) s^(-1) . Thepictorial repressentation (graph) of theincrease in themass ofthe vessel with time isgiven below. (i) What doesthe point Arepresent ? (ii) What doesthe horizontalline beyond B indicate ? (iii) find thecapacity of thevessel. (iv)Find thedensityof theliquid . |
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Answer» Solution : (i) calculateL.C. `=("pitch ")/("NUMBEROF DIVISION on circular scale ")` (ii) P.S.R = nthdivision`xx`pitch (iii)Observed reading=P.S.R +(H.S.R`xx`L.C.) (IV) Zeroerror =[H.S.R. + (N -n) `xx`L.C.] (v) Correctionsis positiveif error is negative (vi)Correctmeasurement= observed reading+ correction. (vii) Correct measurement =7.25 mm |
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| 100. |
The speed of rotation of an electric motor can be increased by decreasing the area of the coil |
| Answer» Solution :Correct Statement: The SPEED of rotation of COIL can be INCREASED by INCREASING the area of the coil. | |