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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
State the factors on which the atmospheric pressure at a place depends. |
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Answer» SOLUTION :Atmospheric pressure at a place depends on. (i) Height of atmosphere (II) DENSITY of atmosphere and (III) Acceleration due to gravity. |
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| 2. |
If L = 2.06cm +- 0.02cm. |
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Answer» `3.17cm +- 0.05cm`, `:. X +- Delta x = 3.17 +- 0.05` |
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| 3. |
Give anythree differences between progressive waves and stationary waves. A stationary wave is yn - 12 sin 300 t cos 2 x . What is the distance between two nearest nodes ? |
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Answer» Solution :2nd part : Given stationary waveis y = 12 sin 300 t COS 2 X . Comparingthis equation with the stationary wave equation , `y = 2 A sin omega t cos k x `we have , k = 2 . So, distance between two consecutive nodes ` = (lambda)/(2)`, where `lambda`isthe WAVELENGTH . So, ` k = (2pi)/(lambda)"or", (2 pi)/(lambda) = 2 "or", (pi)/((lambda//2)) = 2 ` `:. (lambda)/(2) = (pi)/(2)` so the distance between two nearest nodes is `(pi)/(2) ` . |
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| 4. |
Find the relation between orbital velocity and escape velocity. |
| Answer» SOLUTION :`v_0` = `sqrt2v_0` | |
| 5. |
An open reactangular tank with tank with dimensions 5xx4mxx3m contains water upto a height of 2m is accelerated horizontally along the longer life. (a) Determine the maximum acceleration that can be given without spilling the water. (b) If this acceleration is inscreased by 20%. Calculate the percentage of water spilt over. (c) If initially,the tank is closed at the top and is accelerated horizontally by 9m//s^(2),find the gauge pressure at the bottom of the front and rear walls of the tank. (g=10m//s^(2)) |
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Answer» |
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| 6. |
The modulus of rigidity of a liquid is ……………… . |
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Answer» ZERO |
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| 7. |
At the moment t =0 , a force F = kt ( k is a constant) is applied to a small body of mass m resting on a smooth horizontal plane.The permanent direction of force F , makes an anglebeta with the horizontal . Calculate i) The velocity of the body at the moment of its breaking off the plane . ii) The distance traversed by the body upto this moment |
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Answer» Solution :The free BODY diagram of mass m is shown in figure . For vertical equilibrium ` F sin BETA + N = mg ( or ) As F = kt ` `:. ktsin beta + N = mg `........... (1)  Equation of motion of mass m is `F cos beta = ma` ` or kt cos beta = m (dv)/(dt) ( as " "a=(dv)/(dt)), dv=(kt)/(m) cos beta dt ` INTEGRATING ` int_0^(v) dv = (k)/(m) cos betaint_0^(t) dt ` `[v]_0^(v) = (k)/(m) cos beta [(t^(2))/(2)]_0^(t) implies v= (k cos beta )/( 2 m ) t^(2)` ........... (2) At the moment of breaking off the plane N =0 . If ` tau` is the corresponding time , then ` :.` From (1) `k tau sin beta = mg or tau = (mg)/(k sin beta ) `........... (3) `:.` The velocity at the moment of breaking off the plane is given by putting ` t= tau ` in equation ( 2) , so `(v=v_0)` ` v_0 = (k cos beta )/( 2 m) tau^(2) = ( k cos beta )/(2 m) ((mg)/(k sin beta ))^(2)` `:. v_0=(mg^(2) cos beta)/( 2 k sin^(2) beta)` (b) Equation (2) may be expressed as ` (dx)/(dt) = (k cos beta )/(2m ) t^(2)` Integrating `[x]_0^(x)=(k cos beta)/( 2m) [(t^3)/(3)]_0^(t)` `:. x = (k cos beta )/( 6 m) t^(3)` When the body breaks off the plane , we have ` :. x_0=(k cos beta )/( 6m) ((mg)/( k sin beta ))^(3) = (m^(2)g^(3) cos beta)/( 6k^(2) sin^(3) beta )` |
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| 8. |
The phenomenon used in optical fibres for transmission of light energy is |
| Answer» Answer :A | |
| 9. |
A piece of alloy of gold and silver weighs 2 kg in air and 1.86 kg in water. What is the mass of silver? Density of gold is 19.3 xx 10^(3)kgm^(-3). Density of silver is 10.5 xx 10^(3)kg m^(-3) |
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Answer» Solution :` "" ` Mass of silver = m Volume of silver `= ( " mass")/(" density ") ` ` "" = ( m)/( 10.5 XX 10 ^(3))` Mass of gold ` = 2-m ` Volume of gold = ` (2-m)/( 19.3 xx 10 ^(3)) ` TOTAL volume of alloy` (m)/(10.5 xx 10 ^(-3)) +(2-m)/( 19.3 xx 10 ^(3))` ` "" = [( m)/( 10.5)+(2-m)/(19.3) ]xx (1)/(10 ^(-3))` LOSS of weight of alloy in water = weight of water displaced ` "" = 0.14 = V xx D ` `0.14 = [(m)/( 10.5 ) +( 2-m)/( 19.3) ] xx (1)/( 10 ^(3)) xx 10 ^(3)` ` "" =( 19.3 m +10.5 xx 2 - 10 .5 m )/( 10.5 xx 19.3) ` ` 0 .14 xx 10 . 5 xx 19.3 = 8.8 m + 21` ` m= (0.14 xx 10 . 5 xx 19.3 -21 ) /( 8.8 ) ` ` "" = 0. 8376 kg` |
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| 10. |
The equation of a plane progressive wave is y=10sin2pi(t-0.05x). Where y and x are in an and t in seconds. Calculate the amplitude, frequency, wavelength and velocity of the wave. |
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Answer» Solution :GIVEN: `y=10sin2pi(t-0.005x)""....(1)` The standard eqauation for a harmonic wave is `y=Asin2pi((t)/(T)-(x)/(lamda))""......(2)` Comparing equations (1) & (2), we get, `A=10,(1)/(T)=1,(1)/(lamda)=0.005` (i) Amplitude, A=10 CM [y and A have same UNITS] Frequency, `gamma=(1)/(T)=1Hz` (iii) Wavelength, `gamma=(1)/(0.005)=200cm` [x and `gamma` have same units] (IV) Velocity, `v=gammalamda=1xx200=200cms^(-1)` |
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| 11. |
A motor vehicle travelled the first 1/3 of a distanceat a speed of v_(1)=10kmph, the second 1/3 at a speed of v_(2)=20kmphand the last 1/3 at a speed of v_(3)=10kmph. Determine the mean speed of the vehicle over the entire distance s . |
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Answer» Solution :`v_(1)=10kmph, v_(2)=20,v_(3)=60,v=?` `3/v=1/v_(1)+1/v_(2)+1/v_(3),3/v=1/(10)+1/(20)+1/(60)=0.1+0.05+0.016` `:.v=3/(0.166)=18KMPH` |
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| 12. |
Displacement time equation of a particle executing SHM is x=a sin (wt+pi //6) . Time taken by the particle to go directly from x=-A//2 "to" x=+A//2 is |
| Answer» Answer :A | |
| 13. |
A spring of unstreched length l has a mass m with one end fied to a rigid support. Assuming spring to be made of a uniform wire, the kinetic energy possessed y it if its free end is pulled with uniform velocity v is |
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Answer» `1/2mv^(2)` |
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| 14. |
Three identical rings of masses m each and radius r are kept in contact as shown in Fig. Find the moment of inertia of the system about a common tangent. |
| Answer» Solution :M.I. of RING about a diameter `=MR^(2)//2`, About tangent `=3 XX [(mr^(2)//2) + mr^(2)] = 9mr^(2)//2` | |
| 15. |
0.5 mole of diatomic gas at 27^(@)C is heated atconstant pressure so that its volume is tripled. If R = 8.3 J "mole"^(-1)k^(-1)then work done is |
| Answer» ANSWER :B | |
| 16. |
A body is allowed to fall from a height of 100m . If the time taken for the first 50m ist_(1) andfor the remaining50 m is t_(2) , then : |
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Answer» `t_(1) = t_(2)` |
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| 17. |
A metal rod A of 50 cm length expands by 0.10 cm when its temperature is raised from 0^(@)C to 100^(@)C . Another rod B of a different metal of length60 cm expands by 0.06 cm for the same rise in temperature. A third rod C of 50 cm length is made up of pieces of rods A and B placed end to end expands by0.03 cm on heating from 0^(@)Cto 50^(@)C. Find the length of each portion of composite rod C. |
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Answer» Solution :From the data for rod A, we have `DeltaL=alpha_(A) L DeltaT or alpha_(A)=(DeltaL)/(L DeltaT)=(0.10)/(50xx100)=2xx10^(-5)""^(@)C^(-1)` For ord B, we have `DeltaL = alpha_(B) L DeltaT (or) alpha_(B)=(DeltaL)/(L DeltaT)=(0.06)/(60xx100)=10^(-5)""^(@)C^(-1)` If rod C is made of segments of rod A and B of lengths `l_(1)` and `l_(2)` respectively then we have at `""^(@)C`. `l_(1)+l_(2)=50cm ""...(a)` At `T=50^(@)C, l_(1).+l_(2).=50.03 cm` Thus `alpha_(A)l_(1)DeltaT+alpha_(B) l_(2)DeltaT=0.03cm (or) 2xx10^(-5) XX l_(1)xx50 +10^(-5) xx l_(2)xx50 = 0.03 cm` (or) `2l_(1)+l_(2)=(0.03)/(50)xx 10^(-5) =60cm "" ...(b)` Solving (a) and (b) we get `l_(1)=10 cm and l_(2)=40cm` |
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| 18. |
A ball released from a height of 20m, hits the ground andrebounds to a height of 16 m. The percentage loss of energy during collision is, |
| Answer» ANSWER :A | |
| 19. |
A ball moving translationally collides elastically with another , stationary identical ball. At the moment of impact the angle between the straight line passing the centres of the balls and the direction of the initial motion of the striking ball is equal to phi. Assuming that balls to be smooth find the fraction of the kinetic of the striking ball that is stored as the deformation potential energy at the moment of maximum deformation. |
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Answer» `(1)/(2) COS PHI` |
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| 20. |
Two particles execute SHM of same amplitude and frequency on parallel lines. They cross each another when moving in opposite directions each time their displacement is half their amplitude. What is the phase difference between them? |
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Answer» Solution :If we assume that the particles are initially at the MEAN position, their equationfor displacement. `x=A sin omega t` But `x=A/2"":.A/2=A sin omega`(or `sin omegat=1/2` PHASE `=omegat=30^(@),150^(@)` `( :. sin (180^(@)-theta)=sin theta, sin (180^(@)=30^(@))=sin 30^(@))` ONe of the particles has phase of `30^(@)` and the other has phase of `150^(@)` Phase different between them `=120^(@)=(2pi)/3` radian |
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| 21. |
When does the magnitude of angular momentum will be zero? |
| Answer» SOLUTION :When the reference POINT select on the LINE of action then the magnitude of angular momentum will be zero. | |
| 22. |
In which year didi Hahn and Mietner discover the phenomenon of neutron-induced fission of uranium ? |
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Answer» 1938 |
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| 23. |
When a paper boat is kept on a flowing water surface will there be friction between the paper boat and the water surface? |
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Answer» Solution :YES, if there were no FRICTION, it would not have been CARRIED by the WATER flow. |
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| 24. |
Explain dynamic lift with examples. |
| Answer» Solution :Due to the PRESSURE difference VELOCITY below the BALL will be INCREASES. | |
| 25. |
Calculate the force in excess of its weight required to move away a flat circular plate of radius 5 cm from the surface of water. (Surface tension of water is 70 dyne cm^(-1)) |
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Answer» |
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| 26. |
Explain about amplitude and phase of wave. |
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Answer» Solution :"Magnitude of maximum displacemetn of a PARTICLE taking part in the propagation of wave is called amplitude of wave." It is shown by symbol a or A. ACCORDING to wave equation, `y=a sin (kx - omega t + phi)` Now since the vaue of`sin (omega t - kx + phi)` has exteme values `pm 1` we can write: `y _(max) =a (pm1) =pma` `therefore` Amplitude of wave `= |y_(max)| =a` Amplitude of a wave is always positive. Its SI unit is m and its dimensional formula is `[M^(0) L ^(1) T ^(0)].` Initial phase `(phi): `If we know the initial position of a particle at the origin of wave and direction of its motion on its path a time `t=0,` then we can find out value of initial phase `phi` using, `y (x,t) =a sin (omega t - kx + phi)` Putting `x =0 and t =0, y (0,0) =a in phi` Here knowing `sin phi, ` we can find out initial phase `phi.` In the wave equation , `y=y (x,t) =a sin (omega t - kx + phi)` th ARGUMENT of sine function is `(omega t - kx + phi)` Which is called total phase of a wave at time t at a distance x from its source. It also GIVES total phase of OSCILLATION of a particle at distance x from origin of wave at time t. |
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| 27. |
A glass test tube with a plane base has a diameter of 4 cm and mass 30 g. Centre of gravity of the empty test tube is at a height of 10 cm from the base. Find the amount of water to be taken in the test tube so that when it floats vertially, its centre of gravity shifts to the midpoint of the immersed portionof the tube. |
Answer» Solution :The centre of gravity P of the empty tube is 10 CM above the point O. Let the HEIGHT of water level taken in the tube be h cm. Midpoint of the test tube-water system be at R. Hence, part of the tube immersed in water, AC = `2xxOR`.  Volume of water in the tube = `PI(2)^(2)h=4pihcm^(3)` Mass of the tube + water in it = `(30+4pih)g` From the condition of floatation, `pi(2^(2))xxAC=30+4pih` `THEREFORE" "AC=2xxOR=30/(4pi)+h=d` (say). TAKING moment about the point R, `4pihxxRQ=30xxPR` or, `4pih[OR-OQ]=30xx[OP-OR]` or, `4pih[d/2-h/2]=30[10-d/2]` or, `4pih[1/2(30/(4pi)+h)-h/2]=30[10-1/2(30/(4pi)+h)]` or, h = 8.806 cm Hence the required mass of water = `4pixx8.806=110.66g`. |
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| 28. |
A carrom board [4f t xx 4 ft square] has the queen at the centre. The queen hit by the striker moves to the front edge, rebounds and goes into the hole behind the striking line. Then find the magnitude of displacement of the queen |
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Answer» From the centre of the front edge is `(4)/(3) sqrt(10) ft` |
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| 29. |
DeltaQ = nCdT represents the |
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Answer» change in amount of HEAT contained in a body, as a result temperature CHANGES. |
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| 30. |
The molar heat capacity for an ideal gas a) is zero for an adiabatic process b) is infinite for an isothermal process c) depends only on the nature of the gas for aprocess in which either volume or pressureis constant d) is equal to the product of the molecularweight and specific heat capacity for anyprocess |
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Answer» only a,B are correct |
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| 31. |
A ball A is projected from the ground such that its horizontal range is maximum. Another ball B is dropped from a height equal to the maximum range of A. The ratio of the time of flight of A to the time of fallof B is |
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Answer» `SQRT(2):1` |
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| 32. |
A ray of ligth is incidnet from a denser to rarer medium. The critical angle for total internal reflection is theta_i C. And the Brewster.s angle of incidence is theta_(iB). Such that (sin theta_(iC))/(sin theta_(iB))= eta. The relative refractive index of the two media is. |
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Answer» 0.4 |
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| 33. |
The kinetic energy of freely falling body |
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Answer» is DIRECTLY proportional to the HEIGHT of its FALL |
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| 34. |
The block has to be raised to a height L in the same time t. In which case force required is more ? |
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Answer» Solution :(i) `2F_(1)-mg=ma` `F_(1)=(m)/(2)(g+a)` (ii) `F_(2)-mg=ma(THEREFORE)a_(1)=a_(2))` `F_(2)=mg+ma therefore F_(2)GT F_(1)` 
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| 35. |
A ball falls vertically onto a floor, with momentum p, and then bounces repeatedly. The coefficient of restitution is e. The total momentum imparted by the ball to the floor is |
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Answer» <P>`p(1+E)` |
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| 36. |
Let T_(1) and T_(2) be the time periods of two springs A and B when a mass m is suspended from them separately. Now both the springs are connected in parallel and same mass m is suspended with them. Now let T be the time period in this position. Then |
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Answer» `T = T_(1)+T_(2)` |
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| 37. |
Calculate the efficiency of a petrol engine if its compression ratio is 5. Given y = 1.4. If the power of the engine is 25kW what is its input power ? |
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Answer» Solution :`RHO= 5 , gamma = 1. 4` EFFICIENCY `= eta=1- ((1)/(rho ))^( gamma -1) =1 - ((1)/(5)) ^(1.4 -1) = 47. 5%` Efficiency `= ("Output")/("Input")` Input power `= ( " Output power")/("Efficiency") = ( 25 xx 10 ^(3))/(0.776) = 5.26 xx 10 ^(4) W` |
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| 38. |
Is centre of mass of reality ? |
| Answer» SOLUTION :No. The centre of MASS of a SYSTEM is a hypothetical POINT which acts as a single mass particle of the system for an external force. | |
| 39. |
Figure. gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter14). Give the signs of position, velocity and acceleration variables of the particle at t=0.3s, 1.2s, -1.2s. |
| Answer» Solution :`x LT 0, UPSILON lt 0, a gt 0, x gt 0, upsilon gt 0, a lt 0, x lt 0, upsilon gt 0, a lt 0`. | |
| 40. |
M.I of a solid sphere about its diameter is 64kgm^(2). If that sphere is recast into 8 identical small spheres, then M.I of such small sphere about its diameter is |
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Answer» `8kgm^(2)` |
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| 41. |
Which of the following time measuring devices is most precise ? (A) A wall clock (B) A stop watch (C) A digital watch (D) An atomic clock Give reason for your answer. |
| Answer» Solution :A wall clock MEASURES TIME accurately upto ONE second. A stop watch measures time accurately upto a FRACTION of a second. A digital watch measures time up to a fraction of second. An atomic clock measures time most precisely as its precision is 1s in `10^(13)` s. | |
| 42. |
The ratio of times taken by freely falling body to cover first metre, second metre,... is |
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Answer» `sqrt1 : SQRT2: sqrt3` |
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| 43. |
The relationship between enthalpy and internal energy change is |
| Answer» SOLUTION :ISOTHERMAL EXPANSION | |
| 44. |
The Sun's angular diameter is measured to be 1920". The distance D of the Sun from the Earth is 1.496xx109^(11) m. What is the diameter of the Sun ? |
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Answer» |
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| 45. |
Define one dimensional motion . Give examples . |
| Answer» Solution :ONE dimensional motion is the motion of a PARTICLE moving along a straight line. E.g. Motion of a TRAIN along a straight RAILWAY track. | |
| 46. |
(A) : The total kinetic enery of rolling solid sphere is the sum of translational and rotational kinetic energies.(R ) : For all solid bodies total kinetic energy is always twice translational kinetic energy. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct explanation of (A) |
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| 47. |
Give the expression for the amplitude of a damped oscillation of a particle. Hence discuss the amplitude for driving frequency (a) far from natural frequency and (b) close to natural frequecny. |
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Answer» Solution :Let a time dependent `F(t)=F_(0)cos omega_(d)t`. For the motion of a particle under the combined action of forces such as linear RESTORING FORC, damping force and time dependent driving force. `"i.e."F=F_(t)+F_(d)+F(t)` `"i.e."ma(t)=-kx(t)-bv(t)+F_(0) cos omega_(d)t.` where `x(t)=A cos(omegadt+phi)`. The amplitude A is a function of the forced frequency `omega_(d)` and natual frequency `omega.` `"i.e"A=(F_(0))/({m^(2)(omega^(2)-omega_(0)^(2))^(2)+omega_(d)^(2)b^(2)}^(1//2))` (a) For driving frequency far from natural frequency, the `omega_(d)b lt lt m (omega^(2)-omega_(d)^(2))` `therefore A=(F_(o))/(m(omega^(2)-omega_(d)^(2)))` for `omega=omega_(d), A rarroo` for zero damping (ideal case) (b) For diving frequency, close to natural frequency. `m(omega^(2)-omega_(d)^(2))lt lt omega_(d)b` `A=(F_(0))/(omega_(d)b)` for `omega_(d)` very close to `omega.` Hence maximum possible amplitude DEPENDS on the driving frequecny `A prop (1)/(omega_(d))`. Hence resonance is the condition of OSCILLATOR whose amplitude becomes maximum as the driving frequency approaches the natural frequency. |
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| 48. |
Bob of a simple pendulum executes S.H.M in water with a period 't'_0 while the period of oscillation of the bob is '1' in the air. Neglecting viscous force of water and given that the density of the bob is 4/3 xx 1000kg /m^3, the relationship between 't' and ‘t’_0 is |
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Answer» `t = t_0` |
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| 49. |
Two soap bubbles combine under isothermal conditions to form a single soap bubble. If in this process, the change in volume is V and change in area is S, then |
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Answer» `PV+ TS = 0` |
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| 50. |
A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of contact, B is the centre of the sphere and C is its topmost point Then |
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Answer» `vec(V)_(C)-vec(V)_(A)=2(vec(V)_(B)-vec(V)_(C))` |
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