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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.
| 751. |
A particle is subjected to two simple hasrmonic motions in the same direction having equal amplitudes and equal frequency. If the resultant amplitude is equal to the amplitude of the individual motions, find the phase difference between the individual motions. |
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Answer» Let the ampitudes of the individual motions be A each. The resultant asmplitude is also A. If the phase difference between the two motion is delta `A=sqrt(A^2+A^2+2A.A.cosdelta)` or `=Asqrt(2(1+cosdelta))=Acosdelta/2` or `cosdelta/2=1/2` `or delta=2pi/3` |
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| 752. |
A particle doing simple harmonic motion amplitude `= 4 cm` time period `= 12sec` The ratio between time taken by it in going from its mean position to `2cm ` and from `2cm` to extreme position isA. 1B. `1/3`C. `1/4`D. `1/2` |
Answer» Correct Answer - D
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| 753. |
A horizontal disc is oscillating in its own plane harmonically with amplitude a and period T. A body placed on the disc is about to slip on it. What is the coefficient of friction between the disc and the body? |
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Answer» Correct Answer - `4pi^(2)a//T^(2)g` |
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| 754. |
A plank rests symmetrically on two cylinders which are separated by a distance of 2l and rotate with uniform speed in opposite direction. If the plank is displaced a little and then released, show that the motion of the plank is simple harmonic. Calculate the period of oscillations of the peak. The co-efficient of friction between the plank and either cylinder is `mu`. |
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Answer» Correct Answer - `2pisqrt(l//mug)` |
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| 755. |
A uniform table of mas M stays horizontally and symmetrically on two wheels rotatig in opposite directions figure. The separsation between the wheels is L. The friction coefficeint between each whee, and the plate is `mu`. Find the time period of oscilationof the pate if it is slightly displaced along its length and released. |
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Answer» Correct Answer - B Let x be the displacement of the plank towards left. Now the centre of gravity is also displaced through x. `R_1+R_2=mg` Taking moment about g we get `R_1(l/2-x)=R_2(1/2-x)` `=(mg-R_1)(l/2+x)…………1 `ltbr.gt `S, R_1(l/2-x)=(mg-R_1)(l/2+x)` `rarr R_1 1/2-R_1x=` `mgl/2-R_1x+mgx-R_1l/2` L `rarr R-1l/2+R_1l/2=mg(x+1/2)` `rarr R_1(l/2+l/2)=mg((2x+1)/2)` `rarr R_1l=mg(2x+1) R_1=mg((1+2x)/(2l)` .........i Now, `F_1=muR_1=mumg((1+2x))/(2l)` similarly `F_2=muR=mumg((1+2x))/(2l)` similarlty `F_2=muR_2=muMg((1+2x))/(2l)` sin `F_1gtF_2` `rarrF_1-F_2=ma-(2mu(mg)/l)x` `a/x=2mu=(mg)/l=omega^2` `rarr =omegamusqrt(g/l)` `: Time period `=2pisqrt(l/(2mug))` |
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| 756. |
If lt E gt and lt U gt denote the average kinetic and the average potential energies respectively of mass describing a simple harmonic motion, over one period, then the correct relation isA. `ltE gt = ltUgt`B. `ltEgt = 2ltUgt`C. `ltE gt = -2ltUgt`D. `ltEgt = -ltUgt` |
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Answer» Correct Answer - A Option ( A) correct answer |
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| 757. |
The potential energy of a particle executing SHM varies sinusoidally with frequency `f`. The frequency of oscillation of the particle will beA. `(f)/(2)`B. `(f)/sqrt(2)`C. `f`D. `2f` |
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Answer» Correct Answer - A Kinetic energy and potential energy in SHM oscillate with double frequency. |
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| 758. |
Mark the wrong statementA. All S.H.M.’s have fixed time periodB. All motion having same time period are S.H.MC. In S.H.M. total energy is proportional to square of amplitudeD. Phase constant of S.H.M. depends upon initial conditions |
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Answer» Correct Answer - B |
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| 759. |
Einstein in `1905` proppunded the special theory of relativity and in `1915` proposed the general theory of relativity. The special theory deals with inertial frames of reference. The general theory of relativity deals with problems in which one frame of reference. He assumed that fixed frame is accelerated w.r.t. another frame of reference of reference cannot be located. Postulated of special theory of realtivity ● The laws of physics have the same form in all inertial systems. ● The velocity light in empty space is a unicersal constant the same for all observers. ● Einstein proved the following facts based on his theory of special relativity. Let v be the velocity of the speceship w.r.t a given frame of reference. The obserations are made by an observer in that reference frame. ● All clocks on the spaceship wil go slow by a factor `sqrt(1-v^(2)//c^(2))` ● All objects on the spaceship will have contracted in length by a factor `sqrt(1-v^(2)//c^(2))` ● The mass of the spaceship increases by a factor `sqrt(1-v^(2)//c^(2))` ● Mass and energy are interconvertable `E = mc^(2)` The speed of a meterial object can never exceed the velocity of light. ● If two objects A and B are moving with velocity u and v w.r.t each other along the `x`-axis, the relative velocity of A w.r.t. `B = (u-v)/(1-uv//v^(2))` One cosmic ray particle appraches the earth along its axis with a velocity of `0.9c` towards the north the and another one with a velocity of `0.5c` towards the south pole. The relative speed of approcach of one particle w.r.t. another is-A. `1.4 c`B. `0.9655 c`C. `0.8888c`D. `c` |
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Answer» Correct Answer - B Ralative speed `= ((u+v)/(1+(uv)/(c^(2))))= (0.9c + 0.5c)/(1+((0.9c)(0.5c))/(c^(2))) = (1.4c)/(1.45) = 0.9655c` |
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| 760. |
Einstein in `1905` proppunded the special theory of relativity and in `1915` proposed the general theory of relativity. The special theory deals with inertial frames of reference. The general theory of relativity deals with problems in which one frame of reference. He assumed that fixed frame is accelerated w.r.t. another frame of reference of reference cannot be located. Postulated of special theory of realtivity ● The laws of physics have the same form in all inertial systems. ● The velocity light in empty space is a unicersal constant the same for all observers. ● Einstein proved the following facts based on his theory of special relativity. Let v be the velocity of the speceship w.r.t a given frame of reference. The obserations are made by an observer in that reference frame. ● All clocks on the spaceship wil go slow by a factor `sqrt(1-v^(2)//c^(2))` ● All objects on the spaceship will have contracted in length by a factor `sqrt(1-v^(2)//c^(2))` ● The mass of the spaceship increases by a factor `sqrt(1-v^(2)//c^(2))` ● Mass and energy are interconvertable `E = mc^(2)` The speed of a meterial object can never exceed the velocity of light. ● If two objects A and B are moving with velocity u and v w.r.t each other along the `x`-axis, the relative velocity of A w.r.t. `B = (u-v)/(1-uv//v^(2))` A stationary body explodes into two fragments each of rest mass `1 kg` that move apart at speed of `0.6c` relative to the original body. The rest mass of the original body is -A. `2 lg`B. `2.5 kg`C. `1.6 kg`D. `2.25 kg` |
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Answer» Correct Answer - B `m_(0)c^(2) = (m_(0.1)c^(2))/(sqrt(1-(v/c)^(2))) + (m_(02)c^(2))/(sqrt(1-(v/c)^(2))) = 1/0.8 + 1/0.8 = 2.5 kg` |
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| 761. |
A meacury arec lamp provides `0.1` watt of ultra-violet radiation at a wavelength of `lambda=2537 Å` only. The photo tube (cathode of photo electric device) consists of potessium and has an effective area of `4 cm^(2)`. The cathode is located at a distance of `1 m` from the radiation source. The work funcation for potassium is `phi_(0) = 2.22 eV`. According to classical theory, the radiation from arc lamp spreads out uniformaly in space as spherical wave. What time of expource to the radiation should be required for a potassium atom (radius `2Å`) ub tge cathode to accumulate sufficient enerfy to eject a photo-electron?A. `352` secondB. `176` secondC. `704` secondsD. None time lag |
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Answer» Correct Answer - A UV energy flux at a distance of `1m = (0.1)/(4pi xx 1^(2))` cross section (effective area) of atom `= pi xx (2 xx 10^-10)^(2) = 4pi xx 10^(20) m^(2)` Energy required to eject a photoelectron from potassium `= 2.2 eV -= 2.2 xx 1.6 xx 10^(-19) J`. `rArr "Exposure time" = (2.2 xx 1.6 xx 10^(-19))/(((0.1)/(4pi xx 1^(2))) (4pi xx 10^(-20))) = 352` seconds |
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| 762. |
A particle moves in `Xy` plane according to the law `X=a sinomegat and y=a (1-cosomegat)` where a and `omega` are constants. The particle traces-A. a parabolaB. a traight line equally inclined to `X` and `y` axesC. a circleD. a distance proportional to time |
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Answer» Correct Answer - C::D |
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| 763. |
Vertical displacement of a plank with a body of mass m on it is varying according to the law `y=sinomegat+sqrt(3)cosomegat`. The minimum value of `omega` for which the mass just breaks off the plank and the moment it occurs first time after t=0, are given by (y is positive towards vertically upwards).A. `sqrt(g/2),sqrt(2/6),(pi)/(sqrt(g))`B. `g/sqrt(2),2/3sqrt(pi)/g)`C. `sqrt(g/2),(pi)/3)sqrt(2/g)`D. `sqrt(2g),sqrt((2pi)/(3g))` |
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Answer» Correct Answer - A |
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| 764. |
Eqautions `y=2A cos^(2) omegat` and `y=A (sinomegat+sqrt(3) cosomegat)` represent the motion of two particles.A. Only one of these is `S.H.M`B. Ratio of maximum speed is `2:1`C. Ratio of maximum speed is `1:1`D. Ratio of maximum speeds is `1:1` |
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Answer» Correct Answer - C |
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| 765. |
The motion of a particle varies with time according to the relation `y=a (sinomegat + cos omegat)` ,thenA. the motion is oscillatory but not SHMB. the motion is SHM with amplitude aC. the motion is SHM with amplitude `sqrt2a`D. None of these |
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Answer» Correct Answer - C `y=a sin omegat+a cos omegat+a sin(omegat+(pi)/(2))` The resultant motion is SHM but resultant amplitude is `A=sqrt(a^(2)+a^(2)+2a^(2)cos .(pi)/(2))=sqrt(2a)` |
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| 766. |
A particle executes SHM along a straight line so that its period is 12 s. The time it takes in traversing a distance equal to half its amplitude from its equilibrium position isA. 6 sB. 4 sC. 2 sD. 1 s |
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Answer» Since, particle starts from mean position. `therefore" x=A sin omegat` `therefore" "(A)/(2)=A sin omegat_(0)" or " (1)/(2)=sin omegat_(0)` or `sin. (2pi)/(T)t_(0)=sin. (pi)/(6)` or `(2pi)/(T)t_(0)=(pi)/(6)` `therefore" "t_(0)=(T)/(12)=(12)/(12)=1s` |
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| 767. |
The scale of a spring balance reading from 0 to 10 kg is 0.25 m long. A body suspended from the balance oscillates vertically with a period of `pi//10` second. The mass suspended is (neglect the mass of the spring)A. 10 kgB. 0.98 kgC. 5 kgD. 20 kg |
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Answer» Correct Answer - B |
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| 768. |
Two bodies `P` and `Q` of equal masses are suspended from two separate massless springs of force constants `k_(1)` and `k_(2)` respectively. If the two bodies oscillate vertically such that their maximum velocities are equal. The ratio of the amplitude of `P` to that of `Q` isA. `sqrt((k_(1))/(k_(2))`B. `(k_(1))/(k_(2))`C. `sqrt(k_(2))/(k_(1))`D. `(k_(2))/(k_(1))` |
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Answer» Correct Answer - C `omega_(1) A_(1) = omega_(2) A_(2)` `:. (A_(1))/(A_(2)) = (omega_(2))/(omega_(1)) = sqrt(k_(2)//m)/(sqrt(k_(1)//m)) = sqrt(k^(2)/(k_(1)` |
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| 769. |
A rectangular plate of sides a and b is suspended fropm a ceilling by two parallel strings of length L each figure. The separation between the strings is d. The plate is displaced slightly in its plane keeping the strings tight. Show that it will execute harmonic motion. Find the time period. |
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Answer» Correct Answer - B Here we have to consider oscillation of centre of mass. Driving force, F=mg sin theta` Acceleration =a=s/m=gsintheta` `:. A=gtheta=g(m/L)` [wehre g and L and const]. `:. A alpha x ` So the motion is simple Harmonic. `Time period `t=2pisqrt(displacement/Acceleration)` `=2pi sqrt(x/(gx/L))=2pi sqrt(L/g)` |
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| 770. |
A vehicle is moving on a circular path of radius R with constant speed `sqrt(gR)`. A simple pendulum of length l hangs from the ceilling of the vehicle. The time period of oscillations of the pendulum isA. `2pisqrt(l/g)`B. `2pisqrt(l/(2g))`C. `2pisqrt(sqrt(2l)/g)`D. None of these |
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Answer» Correct Answer - B |
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| 771. |
A simple pendulum of length l is suspended from the ceilling of a car moving with a speed v on a circular horizontal road of radius r. a. Find the tension in the string when it is at rest with respect to the car. b.Find the time period of small oscillation. |
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Answer» Correct Answer - `2pisqrt(l/((g^2+v^4)/r^2)^(1/2))` a. From the free body diagram `T=sqrt((mg)^2+((mv^2)/r)^2)` `=m sqrt((g^2+v^4)/r^2)` `=msqrt((g^2+v^4)/r^2)=ma` where a=acceleration `a=((g^2+v^4)/r^2)^(1/2)` The time period of small acceleration is given by `T=2pisqrt(l/a)` `=2pisqrt(l/((g^2+v^4)/r^2)^(1/2))` |
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| 772. |
How many head-on elastic collisions must a neutron have with deuterium nuclei to reduce it energy from `6.561 MeV` to `1 keV`?A. 4B. 5C. 8D. 3 |
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Answer» Correct Answer - A `("Energy loss")/("Initial KE") = (4m_(1)m_(2))/((m_(1) + m_(2))^(2)) = (4(1)(2))/((1+2)^(2)) = 8/9` After `1^(st)` collision `DeltaE_(1) = 8/9 E_(0)` After `2^(nd)` collision `DeltaE_(2) = (8)/(9)E_(1)`, After `n^(th)` collision `DeltaE_(n) = (8)/(9)E_(n1)` Adding all the losses `DeltaE = DeltaE_(1) + E_(2) + "......." DeltaE_(n) = 8/9 (E_(0) + E_(1) + "......"E_(n-1))`: here `E_(1) = E_(0) - DeltaE_(1) = E_(0) - 8/9E_(0) = 1/9 E_(0)` `E_(2) = E_(1) - DeltaE_(2) = E_(1) (8)/(9) E_(1) = (1)/(9)E_(1) = (1/9)^(2)E_(0)` and so on `rArr DeltaE = 8/9[E_(0) + (1/9)^(2)E_(0)+(1/9)^(2)E_(0)+"......"+(1/9)^(n-1)E_(0)] = 8/9 [(1-1/(9^(n)))/(1-1/9)]E_(0) = (1-(1)/(9^(n)))E_(0)` `E_(0) = 6.561 MeV, DeltaE = (6.561 - 0.001)MeV rArr (6.561 - 0.001)/(6.561) = 1 - (1)/(9^(n)) rArr (1)/(6561) = (1)/(9^(n)) rArr n = 4` |
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| 773. |
The angular frequency of a spring block system is `omega_(0)`. This system is suspeded from the ceilling of an elevator moving downwards with a constant speed `v_(0)`. The block is at reast realtive to the elevator. Lift is suddenly stopped. Assuming the downward as a positive direction, choose the wrong statement :A. the amplitude of the block is `(v_(0))/(omega_(0))`B. the initial phase of the block is `pi`C. the equation of motion for the block is `(v_(0))/(omega_(0)) sin omega_(0)t`D. the maximum speed of the block is `v_(0)` |
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Answer» Correct Answer - B Maximum speed `v_(0) rArr A = (v_(0))/(omega^(0))` So equation of motion `x = (v_(0))/(omega_(0)) sin(omega_(0)t)` |
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| 774. |
A body is moving in a room with a velocity of `20 m//s` perpendicular to the two walls separated by `5` meters These is no friction and the collisions with the walls are elastic .The motion of the body isA. Not periodicB. Periodic but not simple harmonicC. Periodic and simple harmonicD. Periodic with variable time perio |
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Answer» Correct Answer - B |
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| 775. |
A particle of mass `m` is attached with three springs `A,B` and `C` of equal force constancts `k` as shown in figure. The particle is pushed slightly against the spring `C` and released. Find the time period of oscillation. . |
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Answer» Correct Answer - A `OP = x` `/_ POM = /_ PON ~~ 45^(@)` `y = xcos 45^(@) = (x)/(sqrt2)` Net restoring force, `F_(net) = - [kx + 2ky cos 45^(@)]` ` = - (2k)x (as y = (x)/(sqrt2))` `:. k_(e) = 2k` Now, `T = 2pi sqrt((m)/(k_(e)) = 2pi sqrt((m)/(2k))`. |
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| 776. |
A force of 6.4 N stretches a vertical spring by 0.1 m. The mass (in kg) that must be suspended from the spring so that it oscillates with a time period of `(pi)/(4)` second.A. `(pi/4)` kgB. 1 kgC. `(1/ pi)`D. 10b kg |
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Answer» Correct Answer - B |
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| 777. |
A stone is swinging in a horizontal circle of diameter `0.8m` at `30 rev//min`. A distant light causes a shadow of the stone on a nearly wall. The amplitude and period of the SHM for the shadow of the stone areA. `0.4m,4s`B. `02 m. 2 s`C. `0.4 m, 2s`D. `0.8 m, 2 s` |
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Answer» Correct Answer - C |
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| 778. |
A stone is swinging in a horizontal circle of diameter `0.8m` at `30 rev//min`. A distant light causes a shadow of the stone on a nearly wall. The amplitude and period of the SHM for the shadow of the stone areA. `0.4m, 4s`B. `0.2m, 2s`C. `0.4m, 2s`D. `0.8m, 2s` |
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Answer» Correct Answer - C `A = radius = ("Diameter")/(2) = 0.4m` `f = (30)/(60) = (1)/(2)rps` or `(1)/(2)Hz ` `:. T = (1)/(f) = 2s`. |
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| 779. |
A uniform circular disc of radius R oscillates in a vertical plane about a horizontal axis. Find the distance of the axis of rotation from the centre of which the period is minimum. What is the value of this period? |
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Answer» The time period of a compound pendulum is the minimum when its length is equal to the radius of gyration about its centre of gravity , i.e., l=k. Since, the moment of inertia of a disc about an axis perpendicular to its plane and passing through its centre is equal to, `I=MK^(2)=(1)/(2)MR^(2)implies K=(R)/(sqrt(2))` Thus, the disc oscillate with the minimum time period when the distance of the axis of rotation from the centre is `(R)/(sqrt(2))` . And the value of this minimum time period will be `T_("min")=2pisqrt((2R//sqrt(2))/(g)) =2pisqrt((sqrt(2)R)/(g)` or `T_("min")~~2pisqrt((1.414R)/(g))` |
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| 780. |
A thin uniform rod of length l is pivoted at its upper end. It is free to swing in a vertical plane. Its time period for oscillation of small amplitude isA. `2pisqrt((l)/(g))`B. `2pisqrt((2l)/(3g))`C. `2pisqrt((3l)/(2g))`D. `2pisqrt((l)/(3g))` |
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Answer» Correct Answer - B (b) `T=2pisqrt((I)/(mgl//2))=2pisqrt(((ml^(2)//3))/((mgl//2)))( because I_("rod")=(ml^(2))/(3))=2pisqrt((2l)/(3g))` |
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| 781. |
A disc of radius `R` is pivoted at its rim. The period for small oscillations about an axis perpendicular to the plane of disc isA. `2pisqrt(R )/(g)`B. `2pi sqrt ((2R)/(g))`C. `2pi sqrt ((2pi)/(3g))`D. `2pi sqrt((3R)/(2g))` |
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Answer» Correct Answer - D `I = (1)/(2)mR^(2) + mR^(2) = (3)/(2)mR^(2)` `T = 2pi sqrt((1)/(mgl))` Here `l = R` |
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| 782. |
A disc of radius `R` and mass `M` is pivoted at the rim and it set for small oscillations. If simple pendulum has to have the same period as that of the disc, the length of the simple pendulum should beA. `(5)/(4)` RB. `(2)/(3)` RC. `(3)/(4)` RD. `(3)/(2)` R |
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Answer» Correct Answer - D (d)`T=2pisqrt((I)/(MgL))=2pisqrt((l)/(g))` or `l=(I)/(ML)=((3//2MR^(2)))/(MR)=(3)/(2)R` |
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| 783. |
A rod of length `l` and mass `m`, pivoted at one end, is held by a spring at its mid - point and a spring at far end. The spring have spring constant `k`. Find the frequency of small oscillations about the equilibrium position. |
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Answer» Correct Answer - A::B::D Restoring torque `tau = - (kl theta)l - (k(l)/(2)theta)(l)/(2) = - (5)/(4)kl^(2)theta` Now, `((ml^(2))/(3))alpha = - ((5)/(4)kl^(2))theta` `f = (1)/(2pi)sqrt|(alpha)/(theta)| = (1)/(2pi)sqrt((15k)/(4m))` |
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| 784. |
Order of magnitude of density of uranium nucleus is , [m = 1.67 xx 10^(-27 kg]`A. `10^(20) kg//m^(3)`B. `10^(17) kg//m^(3)`C. `10^(14) kg//m^(3)`D. `10^(11) kg//m^(3)` |
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Answer» Correct Answer - B Radius of a nucleus is given by `R=R_(0)A^(1//3)` (where `R_(0)=1.25xx10^(-15) m)` `=1.25 A^(1//3)xx10^(-15) m` Here, A is the mass number and mass of the uranium nucleus will be `m~~Am_(p)` `m_(p)=` mass of proton `=A(1.67xx10^(-27) kg)` `:.` Density `rho=("mass")/("Volume")=m/(4/3 pi R^(3))=(A(1.67xx10^(-27)kg))/(A(1.25xx10^(-15) m)^(3))` `implies rho~~2.0xx10^(17) kg//m^(3)` |
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| 785. |
Imagine an atom made of a proton and a hypothetical particle of double the mass of the electron but having the same change as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle of the first excited level. the longest wavelength photon that will be emitted has wavelength [given in terms of the Rydberg constant `R` for the hydrogen atom] equal toA. `9//5R`B. `36//5R`C. `18//5R`D. `4//R` |
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Answer» Correct Answer - C In hydrogen atom `E_(n)= - (Rhc)/n^(2)`. Also `E_(n) prop m` where m is the mass of the electron. Here, the electron has been replaced by a particle whose mass is double of an electron. Therefore, for this hypothetical atom energy in `n^(th)` orbit will be given by `E_(n)= -(2Rhc)/n^(2)` The longest wavelength `lambda_("max")` (or minimum energy) photon will correspond to the transition of particle from `n=3` to `n=2`. `:. (hc)/lambda_("max")=E_(3)-E_(2)=Rhc(1/2^(2)-1/3^(2))` `implies lambda_("max")=18/(5R)` |
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| 786. |
Binding energy per nucleons vs mass curve for nucleus is shown in the figure `W,X,Y`and `Z` are four nuclei indicated on the curve . The process that would release energy is A. `Y rarr 2Z`B. `W rarr X + Z`C. `W rarr 2Y`D. `X rarr Y + Z` |
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Answer» Correct Answer - C Energy is released in process when total binding energy of the nucleus (=binding energy per nucleon `xx` number of nucleons) is increased or we can say, when total binding energy of products is more than the reactants. By calculation we can see that only in case of option (c), this happens. Given : `W rarr 2Y` Binding energy of reactants `=120xx7.5` `=900 MeV` and binding energy of products `=2(60xx8.5)=1020 MeV gt 900 MeV`. |
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| 787. |
A nuclear of mass `M +deltam ` is at rest and decay into two daughter nuclei of equal mass `(M)/(2)` each speed is `c` The binding energy per nucleon for the nucleus is `E_(1)` and that for the daugther nuclei is `E_(2)` ThenA. `E_(1) = 2E_(2)`B. `E_(2) = 2E_(1)`C. `E_(1) gt E_(2)`D. `E_(2) gt E_(1)` |
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Answer» Correct Answer - D Because energy is releasing `implies` Binding energy per nucleon of product `gt` that of parent `implies E_(2) gt E_(1)`. |
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| 788. |
For a simple pendulum, a graph is plotted its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one of the following represents these correctly?A. (a).B. (b).C. (c ).D. (d). |
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Answer» Correct Answer - D (d) `K.E=1/2k(A^(2)-d^(2)` and P.E.=1/2kd^(2)` At mean position `d=0`. At extremes positions `d=A`. |
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| 789. |
For a particle executing SHM, the displacement `x` is given by `x = A cos omegat`. Identify the graph which represents the variation of potential energy `(PE)` as a function of time `t` and displacement `x`. (a) `I, III` (b) `II, IV` (c ) `II, III` (d) `I, IV` |
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Answer» Correct Answer - A Potential energy is minimum (in this case zero) at mean position `(x = 0)` and maximum at extreme positions `(x = + A)`. At time `t = o`, `x = A`. Hence, `PE` should be maximum. Therefore, graph `I` is correct. Furhter in graph III, `PE` is minimum at `x = 0`. Hence, this is also correct. |
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| 790. |
The displacement of a particle varies with time as `x = 12 sin omega t - 16 sin^(2) omega t` (in cm) it is motion is `S.H.M.` then its maximum acceleration isA. `12 omega^(2)`B. `36 omega^(2)`C. `144 omega ^(2)`D. `sqrt(192 omega^(2)` |
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Answer» Correct Answer - B |
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| 791. |
The S.H.M. of a particle is given by the equation `y=3 sin omegat + 4 cosomega t` . The amplitude isA. 7B. 1C. 5D. 12 |
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Answer» Correct Answer - C |
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| 792. |
For a particle executing simple harmonic motion, which of the following statements is not correctA. vertical oscillations of a springB. motion of simple pendulumC. oscillation of liquid column in a U-tubeD. vertical oscillation of a wooden plank floating in a liquid |
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Answer» Correct Answer - C (c) The motion of planets around the sun is periodic but not simple harmonic motion. |
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| 793. |
The amplitude of damped oscillator becomes half in one minute. The amplitude after 3 minutes will be `1//x` times the original, where x isA. 8B. 2C. 3D. 4 |
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Answer» Correct Answer - A The variation in amplitude of damped harmonic oscilator with time is given by `A=A_(0)e^(-bt)` `A_(0)`=Initial amplitude, b=damping factor It is given that after 1 minute, `A_(1)=A_(0)//2=A_(0)e^(-bt)` `:. 2=e^(b)` `:.` after 3 minutes, `A_(3)=A_(0)//x=A_(0)e^(-3b)` `:. x=2^(3)` |
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| 794. |
The amplitude of a damped oscillator becomes half in one minutes. The amplitude after 3 minutes will be 1/x times of the original . Determine the value of x. |
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Answer» Amplitude of damped oscillations is `A=A_(0)e^(-gammat)` `" "`(from `x=x_(m)e^(-gammat)`) As `A=A_(0)//2 " at " t=1` minute, So `(A_(0))/(2)=A_(0)e^(-gamma) " or "e^(gamma)=2` After 3minutes the amplitude will be `A_(0)//x` So `(A_(0))/(x)=A_(0)e^(-gamma3)` or `x=e^(3y)=(e^(gamma))^(3)=2^(3)=8` |
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| 795. |
A block is executing SHM on a rough on a horizontal surface under the action of an external variable force. The force is plotted against the position x of the particle from the mean position. |
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Answer» Correct Answer - A::B::C::D x is positive in II & IV , is positive if `f_(ext) gt 0` |
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| 796. |
Passage XII)_ A particle of mass m is constrained to move along x-axis. A force F acts on the particle. F always points toward the position labelled E. For example, when the particle is to the left of E,Fpoints to the right. The magnitude of F is consant except at point E where it is zero. The systm is horizontal. F is the net force acting on the particle. the particle is displaced a distance A towards left from the equilibrium position E and released from rest at t=0. Velocity-time graph of the particle is A. B. C. D. |
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Answer» Correct Answer - A |
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| 797. |
The left block in figure moves at a speed v towards the right block placed in equilibrium. All collilsions to take place are elastic and the surfaces are frictionless. Show that motion of the two blocks are periodic. Find the timepoerid of these periodic motions. Neglect the widths of the blocks. |
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Answer» Correct Answer - B When the block A moves with velocity V and collides with the block B, it transfers all energy to the block B. (because it is an elastic collision). The block A will move a distance x against the spring, again the spring, again the block b will return to the origiN/Al point and completes half of the oscillation. So, the time period of b is `(2pisqrt(m/k))/2=m/k` The block B collides with the block A and comes o rest at that point. The block AS again moves a further distance L to return to its origiN/Al position. The time taken is `L/V+L/V=2(L/V)` So the period of the periodic motion is `2L/V+pi sqrt((m/k))` |
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| 798. |
If a spring has time period T, and is cut into (n) equal parts, then the time period of each part will be.A. `Tsqrtn`B. `T//sqrtn`C. `nT`D. `T` |
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Answer» Correct Answer - B |
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| 799. |
A particle moves along the `X-`axis according to the equation `x = 10 sin^(3)(pit)`. The amplitudes and frequency of component `SHMs` are.A. amplitude `30//4, 10//4 :` frequencies `3//2, 1//2`B. amplitude `30//4, 10//4 :` frequencies `1//2, 3//2`C. amplitude `10, 10 ,` frequencies `1//2, 1//2`D. amplitude `30//4, 10 :` frequencies `3//2, 2` |
| Answer» Correct Answer - B | |
| 800. |
For a particle acceleration is defined as `veca=(-5xhati)/(|xx|)` for `X ne 0 "and" veca =0 "for" X=0`. If the particle is initially at rest `(a,0)` what is period of of motion of the particle.A. `4sqrt(2a//5)sec.`B. `8sqrt(2(a//5))sec.`C. `2sqrt(2a//5) sec.`D. cannot be determined |
| Answer» Correct Answer - A | |