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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.
| 12351. |
A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance d of 1.2 m from the person. In the following, state of the lift's motion is given in List I and the distance where the water jet hits the floor of the lift is given in List II. Match the statements from List I with those in List II and select the correct answer using the code given below the lists. |
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Answer» `{:(P,Q,R,S),(2,3,2,4):}` `d=V sqrt((2h)/(g))= sqrt(gh) sqrt((2H)/(g))= 2 sqrt(h(H))` We can see that above relation is true for all effective values of g with RESPECT to elevator except zero. Because when lift falls freely then effective acceleration due to gravity becomes zero and water doesn.t come out. |
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| 12352. |
If an external force and the frictional force acting on a body cancel each other and keep the body at rest, the frictional force is |
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Answer» Rolling friction |
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| 12353. |
A ring consisting of two parts ADB and ACB of same conductivity k carries an amount of heat H. The ADB part is now replaced with another metal keeping the temperatures T_(1) and T_(2) constant. The heat carried increases to 2H. What should be the conductivity of the new ADB part? Given (ACB)/(ADB)=3 : |
| Answer» Answer :A | |
| 12354. |
Two capillary tubes of same diameter are put vertically one each in two limbs whose relative densities are 0.8 and 0.6 and surface tension are 60 and 50 dune/cm respectivley. Ratio of heights of liquids in the two tubes is |
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Answer» 44478 |
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| 12355. |
Two discs of moments of inertia I-1 and I_2 about their respective axes , rotaiting with angular frequencies omega_1 and omega_2 respectively, are brought into contact face to face with their axes of rotation coincident. The angular frequency of the composite disc will be |
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Answer» `I_(1)omega_1 + (I_(1)omega_2)/I_1 + I_2` `L_(i) = I_(1)omega_(1)+I_(2)omega_(2)` When two discs are brought into contact face to face (one on top of the other) and their axes of rotation coincide, the moment of inertia, i.e., `I=I_(1)+I_(2)` Let `OMEGA` be the final angular speed of the system. The final angualr momentum of the system is `L_(f) = Iomega = (I_(1)+I_(2))omega` As no EXTERNAL torque acts on the system, therefore according to the law of conservation of angular momentum, we get `L_(i)= L_(f)` `I_(1)omega_(1)+ I_(2)omega_(2) = (I_(1)+I_(2))omega` `therefore omega = (I_(1)omega_(1) + I_(2)omega_(2))/(I_(1)+I_(2))` |
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| 12356. |
Monica was watching the night sky. She saw a star , moving towards her, with increase in brightness. After some few minutes when she watched closely, she found it was the light from a flight in the sky. Whe was surprised, but initially the flight looked stationary, after soem time it was glowing brightly miving towards her. So she went and asked her father. Why this effect occured? |
| Answer» Solution :Any MOVING object, which is PERPENDICULAR to our eyes sight will look LIKE a stationary thing for a while. But as the moving object chagnes its angle of vision, its movement will be known. | |
| 12357. |
Mean free path of a gas molecule is : |
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Answer» INVERSELY proportional to diameter of the molecule where n = Number of molecules per unit volume d = Diameter of the molecules As `PV=k_BNT:.n=N/V` `n = P/(k_BT)` `:.lamda=(k_BT)/(sqrt2pid^2P)` where, `k_B` = BOLTZMANN constant P = Pressure of gas T = Temperature of the gas |
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| 12358. |
The root mean square velocity of a gas molecule of mass m at a given temperature is proportional to |
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Answer» `m^@` |
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| 12359. |
The correct order in which the ratio of SI unit to CGS unit increases is a) Power b) Surface tension c) Pressure d) Force |
| Answer» Answer :A | |
| 12360. |
Water at 4^(@) C is filled to the brim of two beakers A and B of glass of negligible coefficient of expansion. A boy performed experiments by beating the beaker A and cooling the beaker B. He observes |
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Answer» LEVEL of water in 'B' WENT down and water in 'A' flows out |
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| 12361. |
Figure shows the graph of velocity versus displacement of a particle executing simple harmonic motion. Find the period of oscillation of the particle |
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Answer» SOLUTION :`x_("max")= A= 10cm` and `v_("max")= omegaA= 0.6 MS^(-1)` `:. Omega= (v_("max"))/(v_("max"))= (6 xx 10^(-1))/(10xx10^(-2))= 6rads^(-1)` `:. T= (2PI)/(omega)= (2pi)/(6)= (PI)/(3) "sec"` |
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| 12362. |
The oxygen molecule has a mass of 5.30xx10^(-26)kg and a moment of inertia of 1.94xx10^(-46 )kgm^(2) about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is (2)/(3) of its kinetic energy of translation. Find the average angular velocity of the molecules. |
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Answer» Solution :Rotational kinetic energy = `(2)/(3)` TRANSLATIONAL kinetic energy `(1)/(2)IOMEGA^(2)=(2)/(3)xx(1)/(2)MV^(2)` `omega^(2)=(2)/(3)xx(mv^(2))/(I)` `therefore omega=sqrt((2)/(3)(mv^(2))/(I))` `=sqrt((2)/(3)xx(5.3xx10^(-26)xx(500)^(2))/(1.94xx10^(-46)))` `=sqrt((265xx10^(-22))/(3xx1.94xx10^(-46)))` `therefore omega=sqrt(45.53xx10^(24))` `therefore omega=6.7477xx10^(12)" rad s"^(-1)~~6.75" rad s"^(-1)` |
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| 12363. |
As the temperating is increases, the time period of pendulum |
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Answer» INCREASES as its effective length increases even THOUGH its centre of man (CM) still reaming at the centre of the bob |
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| 12364. |
A ladders is more likely to slip when a person is near the top than when he is near the bottom. The friction between the ladder and floor decreases as he climbs up. |
| Answer» Solution :The ladder can be ROTATED about the point of contact of the ladder with the ground. When the LABOURER is atthe top of the ladder, the lever ARM of the FORCE is large. Hence the turning effect on the ladder will be large. | |
| 12365. |
Determine the mean velocityand the mean acceleration of a point in 5 and 10 seconds if it moves as shown in the Fig. |
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| 12366. |
A particle is moving in a circle of radius R in such a way that at any instant the normal and tangential components of the acceleration are equal. If its speed at t=0 is u_(0) the time taken to complete the first revolution is : |
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Answer» `R//u_(0)` |
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| 12367. |
A cube of coefficient of linear expansion alpha_(s) is floating in a bath containing a liquid of coefficient of volume expansion gamma_(1). When the temperature is raised by DeltaT, the depth upto which the cube is submerged in the liquid remains the same, Find the relation between alpha_(s)andgamma_(1) showing all the steps. 1) gamma=alpha_(s) 2) gamma=3alpha_(s) 3) 2gamma=alpha_(s) 4) gamma=2alpha_(s) |
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Answer» Solution :When the TEMPERATURE is increased, volume of the cube will increase while density of liquid will decrease. The depth upto which the cube is SUBMERGED in the liquid remains the same, hence the upthrust will not CHANGE. F = F. `thereforeV_(1)rho_(L)g=V_(1).rho_(L).g" "(V_(1)="volume immersed")` `therefore(Ah_(1))(rho_(L))(g)=A(1+2alpha_(s)DeltaT)(h_(1))xx(rho_(L)/(1+gammaDeltaT))g` Solving this equation, we GET `gamma_(1)=2alpha_(s)` |
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| 12368. |
A particle is acted simultaneously by mutually perpendicular simple harmonic motion x = a cos omegat and y = a sin omegat The trajectory of motion of the particle will be |
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Answer» an ellipse |
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| 12369. |
A block of mass m_(1)=4kg lying on a plane inclined at an angle of 30^(@), is connected to another freely suspended block of mass m_(2)=6kg with the help of a string passing over a smooth pulley as shown in the figure. The acceleration of the each block is (g=10 m//s^(2)) |
| Answer» Answer :C | |
| 12370. |
A carnot engine, where efficiency is 40% takes heat from a source maintained at a temperature of 500K. It is have an engine of efficiency60%. Then, the intake temperature for the same exhaust(sink) temperature must be |
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Answer» Solution :`eta=1-T_(2)/T_(1),0.4=1-T_(2)/500` `rArr T_(2)=300K` `0.6=1-T_(2)/T_(1)=1-300/T_(1)^(1)` `rArrT_(1)^(1)=300/0.4=750K` |
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| 12371. |
A block of mass 'm' is attached to the light spring of force constant k and released when it is in its natural length. Find amplitude of subsequent oscillations, |
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Answer» Solution :The MAXIMUM DISPLACEMENT of the spring in subsequent MOTION will be `(2mg)/(k)`The equilibrium POSITION of the system will occur at the extension of `(mg)/(k)` `:. "amplitude"= (2mg)/(k)-(mg)/(k)= (mg)/(k)` |
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| 12372. |
In the problem what force should the man exert on the rope to get his correct weight on the machine ? |
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Answer» Solution :`T-cancel(M)G+cancel(M)g=Ma` `(because R=MG)a=(T)/(M)` `T-Mg-mg=ma` `T-m(T)/(M)=(m+M)g` `T=((m+M)Mg)/(M-m)=((30+60)60XX10)/(60-30)` `= (""^(3)cancel(90)XX600)/(cancel(30))=1800N` 
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| 12373. |
The difference in the value of .g. at poles and at a place lattitude 45° is: |
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Answer» `ROMEGA^(2)` |
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| 12374. |
The time period of a geostationary satellite is 24 h, at a height 6R_(E) (R_(E) is radius of earth) from surface of earth. The time period of another satellite whose height is 2.5 R_(E) from surface will be |
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Answer» `(12)/(2.5)h` `T^(2) prop R^(3)` `T^(2) prop (R_(E) + h)^(3)` `(T_(1)^(2))/(T_(2)^(2)) = ((R_(E) + 6R_(E))^(3))/((R_(E) + 2.5 R_(E))^(3))` `(T_(1)^(2))/(T_(2)^(2)) = (7^(3))/((7)/(2))^(3)` `(T_(1)^(2))/(T_(2)^(2)) = 8` `T_(2) = (T_(1))/(2sqrt(2))` `T_(2) = (24)/(2sqrt(2))` `T_(2) = 6sqrt(2) h` Hence option (b) is correct. |
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| 12375. |
A source of sound of frequency 2000 Hz moves towards a wall velocity u = 6 cm//s and the wall moves towards the source at v = 3 cm//s. What is the beat frequency registered by a receiver moving towards the wall at w = 2 cm//s, it being in between the source and the wall ? Velocity of sound = 340 m//s |
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| 12376. |
Calculate the difference in temperature between the water at the top and bottom of a water fall 200 m high. Specific heat capacity of water 42000 J kg ^(-1) K ^(-1). |
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Answer» <P> SOLUTION :P.E. of water at a height `h = mgh``= m XX 9.8 xx 200 = 19 60 MJ` Using this P.. LET the temperature of water rise by `Delta T` `Delta Q = mc Delta T` `1960 m = m xx 4200 xx Delta T` `Delta T = (19 60)/(4200) = 0.4667 K` |
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| 12377. |
Given reason if he objectis pressed hard on the surface where it is placed as result it is more difficult to move the object |
| Answer» Solution :IFTHE OBJECT is pressedhardon thesurfacethen thenormalforceactingon theobjectwillincrease. Asa result , it ismoredifficultto MOVETHE object . | |
| 12378. |
A ball is moving with uniform translatory motion. This means that |
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Answer» it is at rest In a uniformtranslatorymotionall partsof theball MOVESIN samedirectionwith samevelocityand thisvelocityis constant. Thisisrepresentedin adjacentdiagram 
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| 12379. |
A transverse wave is represented by y = Asin(omegat - kx). For what value of the wavelength is the wave velocity equal to the maximum particle velocity ? |
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Answer» `(PIA)/(2)` Moreover `V _(max) =v ` (GIVEN) `therefore A omega = (omega )/(k)` `therefore A = (1)/( k ) = (lamda )/( 2pi // lamda) = (lamda)/( 2pi )` `therefore lamda = 2pi A` |
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| 12380. |
Two small spheres A and B of equal radius but different masses of 3m and 2m are moving towards each other and impinge directly. The speeds of A and B before collision are, respectively, 4u and u. The collision is such that B experiences an impulse of magnitude 6mcu, where c is a constant. Determine a. the coefficient of restitution, b. the limits for the value of c for which such collision is possible. |
Answer»  `m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)v_(2)` Putting the values we get `(3m)(4u)-(2m)u=3mv_(1)+2mv_(2)` `implies 3v_(1)+2v_(2)=10u`……..i Using equation of restitution `v_(2)-v_(1)=5ue` ..........ii From i and ii `v_(2)=2u+3eu` Now impulse on `m_(2)`: `J=2mv_(2)-(-2MU)` Put the value of `J` and `v_(2)` to get `e=c-1` As `0leele1implies1lecle2` |
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| 12381. |
The width of a river is 3sqrt3 km. A boat has a velocity of sqrt3km//s and starts to cross the riverin a direction perpendicular to the bank of th erver. If the river is flowing with a velocity 2km//sec, find the distance drifted by the boat. |
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Answer» |
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| 12382. |
A cricket ball sometimes rebounds from the cricket pitch with a velocity greater than which it was bowled with by a bowler. How can it be possible? |
| Answer» Solution :If the cricket ball spins just before it hits theground then this spinning kinetic energy of translation of the ball. As a result, the ball rebounds from the cricket PITCH with a GREATER VELOCITY by VIRTUE of this spinning or ROTATIONAL kinetic energy. | |
| 12383. |
A block of mass m be placed on a frictionless surface of a table. Springconstants of a spring joint on both side are k_(1)" and "k_(2) if block. Leave the block by slight displaced it, then its angular frequency will be………. |
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Answer» `((k_(1)+k_(2))/(m))^(1/2)` and `omega = sqrt((k_(eq))/(m))= ((k_(1)+k_(2))/(m))^(1/2)`. |
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| 12384. |
A Particle of mass M is moving in a straight line with uniform speed v parallel to x-axis in X-Y plane at a height .h. from the X-axis. Its angular momentum about the origin is |
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Answer» zero |
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| 12385. |
A pieces of ice is floating in water, taken in a beaker. What will happen to thelevel of water when all the ice melts? What will happen if the beaker is filled with water but with liquid(a) denser than water (b) lighter than water? |
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Answer» Solution :If M gram of ice is floating in a liquid of desity `rho_(L)` then for its EQUALIBRIUM Weight of ice = thrust i.e. `Mg=V_(D)rho_(L)g` So the volume of liquid DISPLACED by the floating ice `V_(D)=(M//rho_(L))` Now if M gram ice melts completely, water formed will have mass M GRAMS(as mass is CONSERVED). Now if `rho_(w)` is the density of water, the volume of water formed will be `V_(F)=(M//rho_(w))` Here the liqid is water, i.e. `rho_(L)=rho_(w)`, so water displaced by floating ice is equal to water formed by melting of whole ice and hence the level fo water wil remain unchanged. Further more: a. If `rho_(L)gtrho_(w),(M//rho_(L))LT(M//rho_(w))`, i.e. `V_(D)ltV_(F)` i.e. water displaced by floating ice will be lesser than water formed and so thelevel fo lilquid in the beaker will rise. b.If `rho_(L)ltrho_(w),(M//rho_(L))gt(M//rho_(w))` i.e. `V_(D)gtV_(F)` i.e. water displaced by floating ice will be more than water formed and so the level of liquid in the breaker will fall. |
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| 12386. |
A particle executes shm of amplitude 'a'. (i) At what distance from the mean position is its kinetic energy equal to its potential energy? (ii) At what points is its speed half the maximum speed ? |
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Answer» |
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| 12387. |
The adjacent sides of a parallelogram is represented by vectors 2 hat(i) + 3 hat(j) and hat(i) + 4hat(j). The area of the parallelogram is |
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Answer» 5 units |
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| 12388. |
Starting with the same initial conditions anideal gas expands from volume V_(1)to V_(2)in three different ways. The work done by the gas is w_(1)if the process is purely isothermal, W_(2)if purely isobaric and W_(3)if purely adiabatic. Then |
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Answer» `W_(2) GT W_(1) gt W_(3)` |
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| 12389. |
An object of mass m is tied to a string of length l and a variable horizontal force is applied on it which is initially is zero and gradually increases until the string makes ann angle theta with the vertical. Workdone by the force F is |
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Answer» `MGL(1-sin THETA)` |
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| 12390. |
A and B are two satellites revolving round the earth in circular orbits have time periods Shr and 1hr respectively. The ratio of their radius of orbits |
| Answer» ANSWER :C | |
| 12391. |
A wire mesh consisting of very small squares is viewed at a distance of 8cm through a magnifying converging lens of focal length 10cm, kept close to the eye. The magnification produced by the lens is |
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Answer» 5 |
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| 12392. |
A long spring when stretched by 2cm its potential energy is U. If the same spring is stretched by 8cm the potential energy of the spring is ---- |
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Answer» 8U |
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| 12393. |
Two identical drops of water are falling through air with a steady speed of 'V' each. If the drops coalesce to form a single drop, what is the new terminal velocity? |
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Answer» SOLUTION :From CONSERVATION of mass, `(4)/(3)piR^(3)xxrho=(4)/(3)pir^(3)xxrho+(4)/(3)pir^(3)xxrho` or `R=(2^(1//3))r. and V_(T)propr^(2)` (stokes law) `(V^(1))/(V)=(R^(2))/(r^(2))=2^(2//3) therefore V^(1)=2^(2//3)V`. |
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| 12394. |
An object is attached to a vertical unstretched light spring suspended from a ceiling and slowly lowered to its equilibrium position. This stretches the spring by x. If the same object is attaced to the same vertical spring but permitted to fall suddenly, the spring stretched by |
| Answer» ANSWER :C | |
| 12395. |
A closed cubical box is completely filled with water and is accelerated horizontally towards right with an acceleration a. The resultant normal force by the water on the top of the box. |
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Answer» PASSES through the CENTRE of the top |
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| 12396. |
A body is projected up with velocity u. It reaches a point in its path at times t_(1) and t_(2) seconds from the time of projection. Then (t_(1)+t_(2)) is |
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Answer» `(2U)/(G)` |
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| 12397. |
Coefficient of. cubical expansion of a solid is (0.000027//^(@)C ). If the temperature is measured on Fahreheit scale, numerical value of coefficient of linear expansion of solid is |
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Answer» 0.000009`//^(@)`F |
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| 12398. |
In a mercury rhermometer the ice point (lower fixed point) is marked as 10^(@) and the steam point (upper fixed point ) is marked as 130^(@). At 40^(@)C tmperature, what will this thermometer read? |
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Answer» `78^(@)` `(40-100)/(100-0)=(T-10)/(130-10)` `""(40)/(100)=(T-10)/(120)or 2/5=(T-10)/(120)` `""5T-50=240rArr5T=290` `rArr""T=(290)/5=58^(@)` |
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| 12399. |
One mole of a diatomic ideal gas ( gamma = 1.4) is taken through a cyclic process starting from point 'A'. This process A to B is an adiabatic compression. B to C is isobaric CtoD an adiabatic expansion and D to A is isochoric. The volume ratios are (V_A)/(V_B) =16 and (V_C)/(V_B) = 2 and the temperature at 'A' is T_A = 300^@ K. Calculate the temperature of the gas at the points 'B' and 'D' and find the efficiency of the cylcle. |
| Answer» SOLUTION :`300 XX (16)^(0.4 ) ,791.3 K ,61.4%` | |
| 12400. |
Rubone of your palms on a metal surfacefor soe time, say 30-40s. Place the otherpalm on an unrubbed proption of the surfaceand thenthe rubbedportion of the metalsurface .The rubbed protion will fellwarmer . Nowrepeat the sameona wooden surfaceis largerthan thatof themetal surface .Why ? |
| Answer» SOLUTION :Metalsare goodconductorsof heat . Heatgenerateddueto rubbing of palm spreads rapidlyfrom rubbedregions tootherregionsof themetal SURFACE . Woodon the otherhand is a bad CONDUCTOR . Theheatgeneratedat the rubbedregion can notmoverapidlyto otherareasof the WOODEN surface . Hence rubbed andunrubbed regions exhibita distinct temperaturediffernece . | |