

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.
1551. |
Inan electrolyte the current is due to the flow of …….. |
Answer» electrons |
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1552. |
An object of mass 120g is taken upward at the height 5 m, then ………….. Joule work is done. (g=10ms^(-2)) |
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1553. |
The properties of ultrasound that make it useful, are |
Answer» SOLUTION :(i) HIGH energy CONTENT, (II) high directivity. | |
1554. |
It is more difficult to roll a filled tar drum than a empty drum Which has greater inertia ? |
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1555. |
(a) Why can we not associate a potential energy with the frictional force as we did with the gravitational force ? (b) A uniform chain of mass m and length I is lying on a table with (1//4) of its length hanging freely from the edge. Find the amount of work required to be done in dragging the chain on the table completely. |
Answer» SOLUTION :(a) We cannot RECOVER the ENERGY lost due to frictional effects. (b) Mass of `(1//4)` length of the chain =` m//4` the weith `(mgh//4)` of this part of the chain acts an its CG which is at a distance `(i//8)` from the edga. Work done in dragging the chain on the table COMPLETELY = force`xx` DISTACNE = weight of the `(I//4)` length of the chain `xx` distance through its CG is to be moved = `(mg//4) (I/8) = mgl//32`. |
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1556. |
Find out reasons for the situations place a small brick on a plank. When the plank is pulled suddenly the brick remains in the same position as before. |
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1557. |
A 8000 kg engine pulls a trainof 5 wagons, each of 2000 kg along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate (a) The net accelerating force (b) The acceleration of the train (c ) The force ofwagon 1 on wagon 2. |
Answer» Solution :FORCE EXERTED by the engine, F=40000N. Frictional force offered by the track `F_f=5000N`.Net accelerating force ,`Fa=F-F_f` =40000-5000 =-35000N Hence , the net accelerating force is 35000N . Acceleration of the train =a The engine exerts a force of 40000N on all the five wagons. Net acceleration force on the wagons=Fa=35000N Mass of the wagons, m=Mass of a wagon X number of wagons Mass of a wagon =5 `m 2000xx5 =10000 kg` Total mass M=m=10000 kg From Newton.s second LAW of MOTION `F_a=M_a` `a=F_(am)=[35000xx1000=3.5 ms^(2)]` Hence the acceleration of the wagons and the train is `3.5 m//s^2` Mass of all the wagons except wagon 1 is `4 xx 2000 =8000 kg` Acceleration of the wagons `=3.5 m//s^2` Thus , force exerted on all the wagons except wagon 1 `=8000 xx3.5 =28000 N` Therefore , the force exerted by wagon 1 on the remaining four wagons is 28000N,. Hence , the force exerted by wagon 1 on Wagon 2 is 28000 N. |
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1558. |
While doing the experiment on measuring the velocity of a pulse through a stretched string, a student had to choose between a (i) thick silk string and a thick cotton string (ii) stop clock and a table clock. The combination choice that he should prefer is : |
Answer» SILK string and the table clock |
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1559. |
What does the area under velocity-time graph and the time axis represent ? |
Answer» Solution :The AREA under velocity-time GRAPH and the time axis REPRESENTS magnitude of displacement. | |
1561. |
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)) |
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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1563. |
Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 ms^(-1) in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 ms^(-1)in the next 5 s. Calculate the acceleration of the bicycle in both the cases. |
Answer» SOLUTION :In the first case: initial velocity `u = 0,` FINAL velocity `v = 6 MS ^(-1),` time `t = 30 s.` `a = (v - u)/(t)` Subsituting the given values of u, v and t in the above EQUATION, we get `a = (( 6 ms ^(-1) -0 ms ^(-1)))/(30S)= 0.2 ms ^(-2)` In the second case : In the second case: initial velocity `u = 6 ms ^(-1),` final velocity `v = 4 ms ^(-1),` time `t =5s.` Then,` a = (( 4 ms ^(-1) - 6 ms ^(-1)))/(5s) =-0.4 ms ^(-2)` The acceleration of the bicycle in the first case is `0.2 ms ^(-2)` and in the second case, it is `0.4 ms ^(-2).` |
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1564. |
Name three devices constituted o the basis of Pascal's law |
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1565. |
Complete the following sentences: An empty tin container with its mouth closed has an average denstiy equal to that of a liquid. The container is taken 2 m below the surface of that liquid and is left there. Then the container will…...……. |
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1566. |
What is the function of a key (or switch) in an electric circuit ? |
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1567. |
List the safety features while handling with electricity. |
Answer» SOLUTION :Safety features to be FOLLOWED are : Ground CONNECTION TRIP switch Fuse. |
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1568. |
Objects of different masses falling freely from top of a tower, reach surface of the earth simultaneously. |
Answer» Solution :Object falling freely is performing constant accelerated motion, i.e., in time interval of LS, INCREASE in velocity of each object is equal and velocity of the object is ACCORDING to v = GT and HENCE not dependent on their masses. | |
1569. |
By placing tuning forks of different frequencies at the open end of a pipe , it is found that the pipe , has a resonating frequency at 450 Hz and the next harmonic at 750 Hz . Find whether the pipe is closed at one end or open at both ends . Also find the fundamental frequency of the pipe. |
Answer» Solution :(i) Closed pipe , 150 Hz (II) EXPRESSION for pth HARMONIC of open and closedend pipes . (iii) RATIO of harmonics in a pipe closed at one endis `1 : 3 : 5……` |
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1570. |
Inian and Ezhilan argue about the light year. Inian tells that it is 9.46 xx 10^(15)m and Ezhilan argues that it is 0.46 xx 10^(12)km . Who is right ? Justify your anwer . |
Answer» SOLUTION :The magnitude of light YEAR ` = 9 . 46 XX 10^(15)`m . So Inian gave a CORRECT answer . | |
1571. |
Suppose you and your fried are on the moon. Will you be able to hear any sound produced by your friends ? |
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1572. |
The relative density of mercury is ? |
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1573. |
The bob of a simple pendulum of length 1 m has mass 100g and a speed of 1.4 m/ s at the lowest point in its path. Find the tension in the string at this moment. Take g=9.8 m//sec^(2) (AS_(1)) |
Answer» SOLUTION :`1.176 N` | |
1574. |
The diameter of a thin wire can be measured by : |
Answer» Solution :a screw gauge | |
1575. |
A common of mass m_(1) = 12000 kg locatede on a smooth horizontal platform fires a shell of mass m _(2) = 300 kgin horizontal direction with a velocity v _(2)=400 ms//s. Find the velocity of the cannon after it is shot. |
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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1576. |
State Pascal's law |
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1577. |
Pick out the concave and convex mirrors from the following and tabulate them Rear-view mirror, Dentist's mirror, Torch-light mirror, Mirrors in shopping malls, Make-up mirror. |
Answer» SOLUTION :![]() |
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1578. |
Can a single force applied to a body change both its translational and rotational motion? Explain. |
Answer» Solution :(i) Consider the various TYPES of forces actingon a BODY which CAUSE ROTATION. (ii) A body rotates when couple acts on it. (iii) Couple consists of equal and opposite forces acting at two different points. (iv) Couple can be balanced only with equal and opposite couple. |
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1579. |
A wooden plank immerses upto 50% in water. Then ________ % of it is immersed in a liquid of density 0.5 g cm^(-3). |
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1580. |
The brakes applied to a car produce an acceleration of 6 ms^( 2) in the opposite direction to the motion. If the car takes 2s to stop after the application of brakes, calculate the distance it travels during this time. |
Answer» SOLUTION :Here, `a =- 6 ms ^(-2),t =2 s and v =0 ms ^(-1) v =U +at` `0= u + (-6 ms ^(-2)) xx 2s` ` or u = 12 ms ^(-1)` `s = ut + (1)/(2) at ^(2)` `= (12 ms ^(-1)) xx (2s) + (1)/(2) (- 6 ms ^(-2)) xx (2s )^(2)` `= 24 m-12 m= 12 m` Thus, the car will TRAVEL 12 m before it stops after the application of brakes. Can you now appreciate why drivers are cautioned to MAINTAIN some distance between vehicles while travelling on the road ? |
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1581. |
Which of the quantity, velocity or acceleration determines the direction of motion ? |
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1582. |
Can a body rotate even if net force acting on it is zero? Can a single force stop a body from rotation, if the body is rotating under the action of a 'couple'? Explain. |
Answer» SOLUTION :A COUPLE acting on a BODY rotates it, if the body is free to rotate. A coupleis formed by TWO EQUAL forces acting in opposite directions, though their lines of action are not same. The net force acting on a body under the action of a couple is zero, though the body rotates. When a body rotates under the action of a couple, it cannot be stopped from rotating by applying a single force. To stop it from rotating, a couple of equal magnitude but in opposite sense of applied couple is required. | |
1583. |
Name the instrument which can measure accurately the following the diameter of a needle |
Answer» SOLUTION :SCREW GAUGE | |
1584. |
A rocket of mass 3 xx 10^(6) Kg takes off from a launching pad and acquires vertical velocity of 1 km/s and an altitude of 25m calculate its potential energy and kinetic energy |
Answer» SOLUTION :Given m = `3 xx 10^(6)` kg, v = 1 km/s or 1000 m/s and h = 20 m Potential energy = mgh = `3 xx 10^(6) kg xx 10m//s^(2) xx 25m = 7.5 xx 10^(8) J` Kinetic energy = `1/2 mv^(2) = 1/2 xx 3 xx106 xx 1000 m//s xx 1000 m//s = 1.5 xx 10^(12) J` |
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1585. |
Find the pressure at a depth of 10 m in water if tre atmospheric pressure is 100kPa. [1Pa=1N//m] [100kPa=10^(5) Pa=10 ^(5) N//m^(2) =1 atm.] |
Answer» SOLUTION :198 KPA | |
1586. |
A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer. |
Answer» `therefore h=0`. The gravitational force, ACTS vertically downwards. `w=mgh` (`because` Here, W = the WORK DONE by gravitational force) `=mgxx0=0` |
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1587. |
Describe Wilhelm Wien's experiment on canal rays. |
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1588. |
A train was moving at the rate of36 km/h. When brakes are applied to a train running on a straight line, it stops after covering 200 m distance. What would be the retardation produced in the train ? |
Answer» SOLUTION :`4 MS ^(-2)` | |
1589. |
Two people push a car for 3 s with a combined net force of 200 N.(AS_(1)) (a) Calculate the impulse provided to the car. (b) If the car has a mass of 1200 kg, what will be its change in velocity ? |
Answer» SOLUTION :(a) `600 N.s` (B) `0.5 m//s` |
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1590. |
A uniform metre scale of weight 20 gf is supported on a wedge placed at 60 cm mark. If a weight of 30 gf issuspended at 15 cm mark, where should a weight 200 gf be suspended inorder to balance the metre scale? |
Answer» Solution : (I) Apply the law of moments. (ii) Let the weight of 200 gf be suspended at distance 'x'. (iii) moment of force = force `xx` PERPENDICULAR distance of force from the wedge. (iv) Find anti-clockwise and clockwise moments. (v)PRINCIPLE of moments : sumof anti-clockwise moments = sum of clockwise moments. (VI) 67.75 cmmark on scale. |
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1591. |
(MasstoMass calculation) What mass of Al is needed to reduce 10.0 kg of Cr(III) oxide to produce chromium metal? 2Al(l)+Cr_(2)O_(3)(s) overset(Delta)to Al_(2)O_(3)(s)+2Cr(l) |
Answer» Solution :`M(Cr_(2)O_(3))=152.0" g "mol^(-1),1 " mol "Cr_(2)O_(3)=2" mol "Al,M_(Al)=26.98" g "mol^(-1)` Now apply the formulas: `n_(Cr_(2)O_(3))=(10.0xx10^(3)g)/(152" g "mol^(-1))=65.79mol,n_(Al)=2xxn_(Cr_(2)O_(3))=2xx65.78mol=131.58` mol `THEREFORE m_(Al)=n_(Al)*M_(Al)=131.58molxx26.98g" "mol^(-1)=3.550g=3.55`KG |
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1592. |
Explain why, show shoes stop you fromsinking into soft snow. |
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1593. |
The diameter of an iron sphere of mass 10 kg is same as that of aluminium sphere of mass 3.5 kg. Both of them are made to fall freely simultaneously from the top of a tower. When both of them would be at 10 m above the ground, they would have same …………. |
Answer» ACCELERATION |
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1594. |
Kerosene and saline water are taken in two beakers. The same stone is immersed in both liquids. What is the relation between the density of a liquid and buoyancy? |
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1595. |
A certain household has consumed 250 units of energy during a month. How much energy is this in joules. |
Answer» SOLUTION :1 unit of energy is equal to 1 KILOWATT hour 1 unit = 1 kwt(kwh) 1 kwh = `36 xx 10^(6) J` Therefore, 250 units of energy = `250 xx 3. 6 xx 10^(6)` = `9 xx 10^(8) J` |
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1596. |
Why is the land ploughed before the beginning of summer? Does it have any relation to the capillary rise? |
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1597. |
For metal balls A,B,C and D having radius of 2.5 cm each are made of copper, aluminium, gold and iron respectively. The densities of copper, aluminium, gold and iron are 8.9 g//cm^(3),2.7g//cm^(3),19.3g//cm^(3) and 7.8g//cm^(3) respectively. When the balls A,B,C and D are tied to threads, suspended from the hook of a spring balance and immersed completely in strong salty water, one by one, the apparent loss in weight will be: |
Answer» MAXIMUM in BALL B |
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1598. |
An apple falls from a tree because of gravitational between the earth and apple. If F_(1) is the magnitude of force exerted by the earth on the apple and F_(2) is the magnitude of force exerted by apple on earth, then |
Answer» `F_(1)` is very MUCH greater than `F_(2)` |
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1599. |
Immerse the plastic bottle to the bottom of the bucket. Don't you have to exert a force? Why is it so? |
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1600. |
Fill in the blanks. Acceleration due to gravity does not affect the ……… of the object |
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