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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.
| 37801. |
Select the correct pair from the following parameters. |
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Answer» WORK and temperature |
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| 37802. |
Explain in detail the kinetic interpretation of temperature . |
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Answer» Solution :We KNOW that , `P=1/3N/V mbar(v^2)` `PV=1/3Nmbar(v^2)` Comparing the equation (1) with ideal gas equation PV = NkT, `NkT=1/3Nmbar(v^2)` `kT =1/3mbar(v^2)""...(2)` MULTIPLY the above equation by 3/2 on both sides, `3/2kT = 1/2 mbar(v^2)""....(3)` R.H.S of the equation (3) is called average kinetic energy of a single molecule `bar(KE)` . The average kinetic energy per molecule `bar(KE)= in =3/2kT ""...(4)` It is implied from equation (3) that the temperature of a gas isa measure of the average translational kinetic energy per molecule of the gas. |
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| 37803. |
(A) : The speed of revolution of an artificial satellite revolving very near the earth is 8 kms^(-1) (R) : Orbital velocity of a satellite, become independent of height of satellite. |
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Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
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| 37804. |
In moving bicycle frictional force on wheel will be in …... Direction. |
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Answer» |
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| 37805. |
A particle is vibrating in SHM with an amplitude of 4cm. At what displacement from the equilibrium position it has half potential and half kinetic |
| Answer» Answer :D | |
| 37806. |
Solve the following with regard to significant figures. (a) sqrt(4.5 - 3.31)(b) 5.9 xx 10^(5) - 2.3 xx 10^(4) (c) 7.18 + 4.3(d) 6.5 + 0.0136 |
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Answer» Solution :(a) Among the two, the least NUMBER of significant after decimal is one `sqrt(4.5 - 3.31) = sqrt(1.19) = 1.09` (B) The number of minimum significant figures is 2 `5.9 xx 10^(5) - 2.3 xx 10^(4) = 5.9 xx 10^(5) - 0.23 xx 10^(5) = 5.67 xx 10^(10) = 5.7 xx 10^(10)` (C) The lowest least number of significant digit after decimal is one 7.18 + 4.3 = 11.48 Rounding off we GET 11.5 (d) The lowest least number of significant digit after decimal is one 6.5 +0.0136 = 6.5136 = 6.5 |
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| 37807. |
The difference in the value of 'g' at pole and at a lattitude is (3)/(4) R omega^(2) then latitude angle is |
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Answer» `60^(@)` |
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| 37808. |
A body is throw up with a velocity 'u'. It reaches maximum height 'h'. If its velocity of projection is doubled the maximum height it reaches is _________ |
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Answer» 4H |
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| 37809. |
State which of the following statements are false. A scalar quantity is one that |
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Answer» is CONSERVED in a process |
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| 37810. |
The magnitude of the resultant of the two orthogonal vectors of 3 units and 4 units is : |
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Answer» 7 `= sqrt(9 +16) = sqrt(25) = 5` |
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| 37811. |
A ball of mass m falls vertically from a height h and collides with a block of equal mass m moving horizontally with a velecity upsilon on a surface. The coefficient of kinetic friction between the block and the surface is 0.2, while the coefficient of restitution e between the ball and the block is 0.5. There is no friction acting between the ball and the block. The velocity of the blockdecreases by : |
| Answer» Answer :D | |
| 37812. |
On fogy day two drivers spot in front of each other when 80 m apart, they were travelling 70 kmph and 80 kmph. Both apply brakes simultaneously which retards the car at the rate of 5 m//s^2 . Which of the following statement is correct? |
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Answer» The COLLISION will be AVERTED |
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| 37813. |
Scent sprayer is based on |
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Answer» Charle's LAW |
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| 37814. |
n moles of an ideal gas is made to undergo the cycle 1-2-3-4-1 as shown in figure. Process 3-4 is a straight line. The gas temperatures in states 1,2 and 3 are T_1,T_2 and T_3 respectively. Temperature at 3 and 4 are equal. Determine the work done by the gas during the cycle. |
| Answer» SOLUTION :`1/2nR[(T_3)/(T_2)+(T_2)/(T_1)-2](T_2 -T_1)` | |
| 37815. |
A body of imparted motion from rest to move in a straight line if it is then obstructed by an opposite force then |
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Answer» the body will necessarilly CHANGE direction |
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| 37816. |
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2m ling is set into oscillation. The speed of the bob at its mean position is 1ms^(-1). Then the trajectory of the bob if the string is cut when the bob is (a) at one its extreme positions, (b) at its mean position. |
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Answer» Solution :(a) We KNOW that at each EXTREME position, velocity of the bob is zero. If the string is cut at the extreme position, it is only under the action of .g.. Hence the bob will fall vertically downwards. (b) At the mean position,velocity of the bob is 1 m/s. along the tangent to the are, which is in the horizontal direction. If he string is CUR at mean position, the bob will behave as a horizontal projectile. |
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| 37817. |
Four charges, each of value q, are placed at the four corners of a square of side a. What is the potential at the centre of the square |
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Answer» Solution : Consider four masses each of MASS m at the corners of a square of side l,We have four mass pairs at DISTANCE l and two diagonal pairs at distance`sqrt(2)`l hence, W(R)= - 4 `(Gm^(2))/(l) - 2 (Gm^(2))/(sqrt(2)l)` = - `(2.Gm^(2))/(l ) (2 + (1)/(sqrt(2))) = - 5.41 (Gm^(2))/(l)` The gravtational potential at the centre of the square `(r = sqrt(2)l//2)` is U(r)= - `4 sqrt(2) (Gm)/(l)` . 
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| 37818. |
A rain drop of mass m_(0) starts falling from rest and it collects water vapour and grows . If it gains lambda kg//s,find its velocity at any instant. |
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Answer» Solution :`d/(dt) (MV) =MG,(DM)/(dt)=lambda "or" m=lambdat+K` where K is a constant when `t=0, m=m_(0) :. M=m_(0)+lambdat` so,`d/(dt){(m_(0)+lambdat)v}=(m_(0)+lambdat)G` Integrating, `(m_(0)+lambdat)v=int(m_(0)+lambdat)gdt =(m_(0)t+(lambdat^(2))/2)g+C ` where c is a constant when `t=0, v=0``:. C=0` `:. (m_(0)+lambdat) v=(m_(0)t+(lambdat^(2))/2)g , v =(g(m_(0)t+(lambdat^(2))/2)/(m_(0)+lambdat)=(g(t+(lambdat^(2))/(2m_(0))/(1+(lambdat)/m_(0)))` Thismean velocitychanges w.r.t.time |
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| 37819. |
A bodyis displacedby 2min Z - direction by a force (-4,2,6)N . Find the work done ………. |
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Answer» Solution :`W = vec(F) . Vec(d) = ( - 4hat(i) +2 hat(j) +6hat(K)).(2hat(k))` ` = - 4xx0 +2 xx 0 + 6 xx2 ` = 12 J |
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| 37820. |
State and prove perpendicular axis theorem. |
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Answer» Solution :Perpendicularaxistheorem : theperpendicularaxistheoremholdsgoodforplanelaminarstartsthattheis equalto thesumof momentsof inertia about twoperpendicularaxeslyingin theplaneof thebodysuch thatall thethreeaxesmutuallyperpendicularand havea commonpoint Let theX and Y- axesliein the planeandZ-axisperpendicularto theplaneof thelaminarobject . ifthe momentsof inertiaof thebodyaboutX and Y-axesare ` I_(X) andI_Y`respendicularand havea common POINT. Let theX ANDY axesliein theplaneand Z - axis,thenthe perpendicularaxistheoremcouldbe expressedas , `I_(Z)= I_(X )+I_(Y)` to provethistheorem. let usconsidera planelaminer objectof negligiblethicknesson whichliesthe origin(O ).The Xand Y- axes lieon theplaneand Z-AXISIS perpendicularto itas shownin figure.Thelamina isconsidered tothemadeup ofa largenumberofparticlesof massm . Letuschooseone suchpartivle at apointP whichhascoordinates(x, y)at adistancer from O .  Themomentof inertiaof theparticleaboutZ- axisis `mr^2 ` thesummationof hteaboveexpressiongivesthe momentofinertiaof thelaminaaboutZ- axisas , `L_(Z)= summr^2` Here,`r^2=x^2+y^2` then ` I_(z)= summ(x^2+y^2)` ` I_(Z)= summx^2+ summy^2` in theaboveexpression , theterm` sum mx^2` is the moment ofinertiaof thebodyabouttheY -axisand similary the term` summy^2` is themomentof inertiaaboutX - axisthus ` I_(X) = summy^2and I_(Y)= sum mx^2` Substitutingin theequationof `I_Z`gives ` I_(Z )= I_(X ) +I_(Y)` Thusthe perpendicularaxistheoremis proved . |
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| 37821. |
The damping force on an oscillator is directly proportional to the velocity . The units of the constant of proportionality are |
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Answer» `kg ms^(-1)` `K= F/V = (Kg ms^(-2))/(ms^(-1)) = kg s^(-1)` |
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| 37822. |
Two bodies are thrown from the same point with the same velocity of 50 ms^(-1). If their angles of projection are complimentary angles and the difference of maximum heights is 30 m, their maximum heights (g = 10 ms^(-2)) |
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Answer» 50 m and 80 m |
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| 37823. |
A drop of water of volume 0.05cm^(3) is pressed between two glass plates, as a consequence of which it spreads and occupies an area of 40cm^(2). If the surface tension of water is 70 dyne/cm, then the normal force required to separate out the two glass plates will be in Newton |
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Answer» 90 |
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| 37824. |
A coin placed on a rotating turn just slips if it is placed at a distance of 4cm from the centre. If the angular velocity of turn table is doubled, it will just slip at a distance of |
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Answer» 1cm |
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| 37825. |
A block regulated by a seconds pendulum keeps correct time. During summer the length of the pendulum increases to 1.01 m How much will the clock gain or lose in one day? |
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Answer» Solution :The time PERIOD of a simple pendulum is `T=2sqrt(l/G)=((2pi)/g)(l^(1//2))` differentiating T w.r.t l and dividing by T on both slides we get `(gT)/T=1/2(dl)/l` In case of a SECONDS pendulum T=2 s and `l=g/(pi^(2)0=0.9927m` Also `dl=1.01-0.9927=0.0173m` `:.(dT)/2=1/2(0.0173/0.9927)` i.e. The change in time period or change in time for 1 oscillation `dT=0.0173/0.9927s` `dT=1/2 (dl)/lT=1/2(dl)/l 86,400`PER day `=1/20.0173/0.9927xx86400=752.9s` When the length INCREASES the time period of the pendulum increases. So the clock loses time or mass slow. |
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| 37826. |
A force vecF = 3 hati - 2 hatj + 4hatkdisplaces a body from a point A(8, -2, -3) to the point B(-2, 0,6). The work done is |
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Answer» 1 UNIT `= - 30 - 4 = 36 = 2 ` units |
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| 37827. |
What happens when a hot needle is placed between two match sticks placed on the surface of water? |
| Answer» SOLUTION :The MATCH sticks move AWAY from each other as the surface TENSION decreases due to increase in temperature. | |
| 37829. |
A particle of mass m is attached to a spring of spring constant 'k' and has a natural frequency omega_(0). An external force F(t) proportional to cosomegat (omega != omega_(0)) is applied to the oscillator. The maximum displacement of the oscillator will be proportional to |
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Answer» `m/(omega_(0)^(2)+omega^(2))` |
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| 37830. |
A solid sphere is moving and it makes an elastic collision with another stationary sphere of half ot its own radius. After collision it comes to rest. The ratio of the densities of the meterials of second sphere and first sphere is |
| Answer» Answer :C | |
| 37831. |
When an object is said to be in motion ? |
| Answer» Solution :When any SUBSTANCE makes DISPLACEMENT with RESPECT to surrounding substance, it is called in MOTION. | |
| 37832. |
One side of a square is measured as 16.7 cm to an accuracy of 0.1 cm. What is the percentage error in area? |
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Answer» |
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| 37833. |
mu_(0) and epsilon_(0) denote the permeability and permittivity of free space, the dimensions of mu_(0) epsilon_(0)are |
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Answer» `[LT^(-1)]` `C= (1)/(SQRT(mu_(0)epsilon_(0))` or `mu_(0)epsilon_(0)= (1)/(c^(2)) :. [mu_(0)epsilon_(0)] = (1)/([LT^(-1)]^(2))= [L^(-2)T^(2)]` |
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| 37834. |
A rat and horse are to be projected from the earth into space. State whether the velocity is the same or different projecting each animal. Justify. |
| Answer» SOLUTION :VELOCITY is the same because escape velocity is independent of mass of body that is PROJECTED into space | |
| 37835. |
Water is filled into a container with hexagonal cross section of side 6 cm. If the surface tension of water is 0.075"Nm"^(-1) then the surface energy of water will be |
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Answer» `8.5 XX 10^(-4)` J |
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| 37836. |
Three metal rods of same lengths and same area of cross-section having conductivities 1, 2, 3 units are connected in series. Then their effective conductivity will be |
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Answer» 2 UNITS |
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| 37837. |
The time period of a simple harmonic motion is 8 s At t=0. it is at the mean position.The ratio of the distance travelled bi it in the first and second secondsis, |
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Answer» `(1)/(2)` |
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| 37838. |
A particle of mass m moving with certain velocity collides elastically head on with a particle of mass 4m at rest. The percentage of K.E. transferred is |
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Answer» 0.75 |
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| 37839. |
Sand drops from a stationary hopper at the rat of 5 kgs^(-1) onto a conveyor belt moving with a constant speed of 2 ms^(-1). What is the force required to keep the belt moving and what is the power delivered by the motor moving the belt ? |
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Answer» Solution :Impulse J = Ft = mv, `F=(mv)/(t)` `v=2 ms^(-1), (m)/(t)=5 kgs^(-1) RARR F=10 N` POWE `P=(W)/(t)=(Fs)/(t)=FV = 10xx2=20` WATT. |
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| 37840. |
The ladder shown in the figure is light and stands on a frictionless horizontal surface. Arms AB and BC are of equal length and M and N are their mid points. Length of MN is half that of AB. A man of mass M is standing at the midpoint of BM. Find the tension in the mass less rod MN. Consider the man to be a point object. |
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Answer» |
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| 37841. |
A boat takes 4 hrs to travel certain distance in a river in down stream and it takes 6 hrs to travel the same distance in upstream. Then the time taken by the boat to travel the same distance in still water is |
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Answer» 4.8 hrs |
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| 37842. |
What will be the percentage differene between the frequency of the sound heard by an oberver and the actual frequency of the sound heard by an an observer and the actual frequency of the whistle of a train approaching at a speed of 100 ms^(-1) towards the observer? Velocity of sound in air=330 ms^(-1) |
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Answer» |
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| 37843. |
A spherical hole is made in a lead sphere of radius R, such that its surface touches the outside surface of the lead sphere and passes through its centre. The mass of the sphere before hollowing was M. With what gravitational force will the hollowed-out lead sphere attract a small sphere of mass m, which lies at a distance d from the centre of the lead sphere on the straight line connecting the centres of the sphere and of the hollow? |
Answer» Solution : Volume of the sphere removed `v = 4/3 PI (R/2)^3` MASS of the sphere removed `M. = (M)/(4/3 pi R^3) xx 4/3 pi (R/2)^3 = M/8` The FORCE on the small sphere of mass m = Force due to sphere of RADIUS R with the distance d -Force due to sphere of radius R/2 with the distance (d-R/2) `i.e., F= (GMM)/(d^2) - (GM.m)/((d -R//2)^2)` ` = (GMm)/(d^2)- (G(M/8)m)/((d - R//2)^2)` ` = GMm [(1)/(d^2) - (1)/(8(d-R//2)^2) ] = (GMm)/(d^2) [1- (1)/(8(1-R//2d)^2) ]` |
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| 37844. |
When wave propagating in one medium, enters another medium, which of the characteristic do remain constnat ? |
| Answer» SOLUTION :FREQUENCY ANGULAR frequency and PERIODIC TIME. | |
| 37845. |
A capillary tube of 0.5 mm bore stands vertically in a wide vessel containing a liquid surface tension 30 dynes/cm. The liquid throughly wets the tube and has a specific gravity 0.8 calculate the rise of liquid in the tube [g=10mS^(-2)] |
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| 37846. |
What is capillarity? Obtain an expression for the surface tension of a liquid by capillary rise method. |
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Answer» Solution :The rise or fall of a liquid in a narrow tube is called capillarity. Let us consider a CAPILLARY tube which is HELD vertically in a beaker containing water, the water rises in the capillary tube to a height h due to surface tension.  The surface tension force `F_(T)` acts along the tangent at the point of contact downwards and its reaction force upwards. Surface tension T, is resolved into two components. (i) Horizontal COMPONENT T `sintheta` and (ii) Vertical component T `costheta` acting upwards, all along the whole circumference of the meniscus. Tota upward force `=(Tcostheta)(2pir)=2pirTcostheta` Where `theta` is the angle of contact, r is the radius of the tube. Let `RHO` be the density of water and h be the height to which the liquid rises inside the tube. Then, `{:(("the volume of"),("liquid column"),("in the tube,V")):}={:(("volume of the"),("liquid column of"),("radius r height h")):}+{:(("volume of the liquid of radius"),("r and height h-volume of"),("the hemisphere of radius of r")):}` `V=pir^(2)h+(pir^(2)xxr-(2)/(3)pir^(3))` `rArrV=pir^(2)h+(1)/(3)pir^(3)` The upward force supports the weight of the liquid column above the free surface, therefore, `2pirTcostheta=pir^(2)(h+(1)/(3)r)rhog` `rArr""T=((h+(1)/(3)r)rhog)/(2costheta)` If the capillary is a very fine tube ofradius (i.e., radius is very small) then `(r)/(3)` can be neglected when it is compared to the height h. Therefore, `T=(rrhogh)/(2costheta)` |
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| 37847. |
Calculate the specific heat capacities at constant volume and at constant pressure for helium. Given gamma = (5)/(3) , R = 8.3joule mol ^(-1) K ^(-1). |
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Answer» <P> Solution :`C _(P) - C _(V) = R, C _(P) = gamma C _(gamma) C _(V) = R//gamma -1 = 83 [(5//3)-1]=12.45 J mol ^(-1) K ^(-1) .C _(P) = C _(V)+ R =20.75 J mol ^(-1) K ^(-1).` |
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| 37848. |
In a process on a system, the initial pressureand volume are equal to the final pressure and volume a) The initial temperature must be equal to thefinal temperature b) The initial internal energy must be equal tothe final internal energy c) The net heat given to the system in theprocess must be zero d) The net work done by the system in theprocess must be zero |
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Answer» only a,B are CORRECT |
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| 37849. |
Assume that liquid drop evaporates by decreasing in its surface energy so that temperature remains unchanged. The surface tension of liquid drop is T, density of liquid is rho and L is latent heat of vapouration of liquid. What should be the minimum radius of the drop for this to be possible |
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Answer» `(T)/(rhoL)` |
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| 37850. |
What are interial frames ? |
| Answer» SOLUTION :Inertialframesare thereferenceframeswherean OBJECT isfreefrom allforcesthen itmoveswithconstantvelocity(or)in uniformmotionalonga STRAIGHTLINE. EG: EARTH | |