InterviewSolution
Saved Bookmarks
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.
| 51. |
A `2 kg` block moving with `10m//s` strikes a spring of constant `pi^(2)N//m` attached to `2Kg` block at rest kept on a smooth floor. The time for which rear moving block remain in contact with spring will be- A. `sqrt(2)`B. `(1)/(sqrt(2))s`C. `1 s`D. `1/2 s` |
|
Answer» Correct Answer - C Both the blocks remains in contact unitl the spring is in compression. In this time system complete half oscillation. By reduced mass concept time period of system `T = 2pisqrt((mu)/(k)) = 2pisqrt((1)/(pi^(2))) = 2s` `rArr` Required time `= T/2 = 2/2 = 1s` |
|
| 52. |
A `2 kg` block moving with `10m//s` strikes a spring of constant `pi^(2)N//m` attached to `2Kg` block at rest kept on a smooth floor. The time for which rear moving block remain in contact with spring will be- A. `sqrt2 sec`B. `(1)/(sqrt2)sec`C. `1 sec`D. `(1)/(2) sec` |
| Answer» Correct Answer - C | |
| 53. |
The block`A(2kg)` and `B(3kg)` rest up on a smooth horizontal surface are connected by a spring of stiffness `120 N//m`. Identially the spring is underformd. `A` is imparted a velocity of `2m//s` along the line of the spring away from `B`. Find the displacement of `A` at `t` seconds later. |
| Answer» Correct Answer - `0.8t+0.12 sin 10t` | |
| 54. |
The system shown in the figure can move on a smooth surface. The spring is initially compressed by `6 cm` and then released.Find (a) time perod (b) amplitude of `3 kg` block (c ) maximum momentum of `6 kg` block |
| Answer» Correct Answer - (a) `(pi)/(10)sec, (b) `4 cm`, (c ) `2.40 kg m//s` | |
| 55. |
A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15eV. What wil be observed by the detector? (a) 2 photons of energy 10.2 eV (b) 2 photons of energy 1.4 eV (c ) One photon of energy 10.2 eV and an electron of energy 1.4 eV (d) One photon of energy 10.2 eV and another photon of energy 1.4 eVA. `2` photon of energy `10.2 eV`B. `2` photon of energy `1.4 eV`C. One photon of energy `10.2 eV` and an electron of energy `1.4 eV`D. One photon of energy `10.2 eV` and another photon of energy `1.4 eV` |
|
Answer» Correct Answer - C The first photon will excite the hydrogen atom (in ground state) in first excited state (as `E_(2)-E_(1)=10.2 eV`). Hence during de-excitation a photon of `10.2 eV` will be released. The second photon of energy 15 eV can ionise the atom. Hence the balance energy i.e., `(15-13.6)eV=1.4 eV` is retained by the electron. therefore, by the second photon an electron of energy `1.4 eV` willl be released. |
|
| 56. |
An electron collides with a fixed hydrogen atom in its ground state. Hydrogen atom gets excited and the colliding electron loses all its kinetic energy. Consequently the hydrogen atom may emit a photon corresponding to the largest wavelength of the Balmer series. The K.E. of colliding electron will beA. `10.2 eV`B. `1.9 eV`C. `12.1 eV`D. `13.6 eV` |
|
Answer» Correct Answer - C Excitation upto `n=3` is required so that visible length is emitted upon de-excitation. So required energy `=13.6 (1-1/9)=12.1 eV` |
|
| 57. |
For a particle executing (SHM) the displacement (x) is given by `(x = A) cos (omega) t`. 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` |
|
Answer» Correct Answer - A Potential energy is minimum (in this case zero) at mean position `(x = 0)` and maximum at extreme position `(x = +-A)`. At time `t = 0, x= A` Hence, PE should be maximum. Therefore, graph is correct. Further is graph III. PE is minimum at ` x= 0`. Hence this is also correct. |
|
| 58. |
Statement-1 : Mechanical energy of a partical execting SHM is E. Maximum KE of particle may be greater than E. Statement-2 : Potential energy of a system may be negative.A. Statement-1 is True, Statement-2 is True , Statement-2 is a corrrect explanation for Statement-1B. Statement-1 is True, Statement-2 is True , Statement-2 is NOT a corrrect explanation for Statement-1C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - A | |
| 59. |
The graph shows the variation of displacement of a particle execting SHM with time . We infer from this graph that A. the force is zero at time `(3T)/(4)`B. the velocity is maximum at time `(T)/(2)`C. the acceleration is maximum at time TD. th PE is equal to total energy at time `(T)/(2)` |
|
Answer» Correct Answer - D (d) At time `(T)/(2), v=0` `therefore` Total energy =Potential energy |
|
| 60. |
In S.H.M. maximum acceleration is aA. AmplitudeB. EquilibriumC. Acceleration is constantD. None of these |
|
Answer» Correct Answer - A |
|
| 61. |
A particle is oscillating in a stright line about a centre of force `O`, towards which when at a distance `x` the force is `mn^(2)x` where m is the mass, n a constant. The amplitude is `a = 15 cm`. When a distance `(asqrt(3))/(2)` from O, find the new amplitude. |
|
Answer» Correct Answer - A::C As `v = omegasqrt(a^(2) - x^(2))` so `v_(i) = nsqrt(a^(2) - (3a^(2))/(4)) = (na)/(2)` and `v_(2) = (3)/(2) na =- nsqrt(A^(2) - (3a^(2))/(4)` `rArr A = sqrt(3a) = 15sqrt(3) cm` |
|
| 62. |
A particle of mass `m` performs `SHM` along a stright line with frequency `f` and amplitude `A`.A. The average kenetic energy of the particle is zero.B. The average potential energy is `m pi^(2)f^(2)A^(2)`.C. The frequency of ocillation of kinetic energy is `2f`.D. Velocity funciton leads acceleration by `pi//2`. |
|
Answer» Correct Answer - B::C |
|
| 63. |
A particle performs SHM of amplitude A along a straight line. When it is at distance `sqrt(3)/2` A from mean position, its kinetic energy gets increased by an amount `1/2momega^(2)A^(2)` due to an impulsive force. Then its new amplitude becomes.A. `(sqrt(5))/(2)A`B. `(sqrt(3))/(2)A`C. `sqrt(2)A`D. `sqrt(5)A` |
|
Answer» Correct Answer - C at `x = (sqrt(3))/(2)A` `KE = 1/2 momega^(2) (A^(2) - (3)/(4)A^(2)) = 1/8 momega^(2)A^(2)` `KE` is increased by an amount of `1/2momega^(2)A^(2)`. Let now amplitude be `A_(1)` then total `KE` `KE_(1) = 1/8momega^(2)A^(2) + 1/2 momega^(2)A^(2)` `= 5/8momega^(2)A^(2) = 1/2momega^(2)(A_(1)^(2) - (3)/(4)A^(2)) rArr A_(1) = sqrt(2)A` |
|
| 64. |
Phase space deagrams are useful tools . in analyzing all kinds of dynamical problems. They are especially usrful in studying the changes in motion as initial position and momenum are changed. Here we conseder some simple dynamical systems in one dimension. For such systems, phase space is a plane in which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is `x(t) vs. p(t) curve in this plane. The arrow on the curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which positon or momentum upwards (or to right) is poitive and downwards (or to left) is negative. The phace diagram for a ball thrown vertically up from ground is.A. B. C. D. |
| Answer» Correct Answer - D | |
| 65. |
Passage XI) The differential equation of a particle undergoing SHM is given by `a(d^(2)x)/(dt^(2))`+bx = 0. The particle starts from the extreme position. The ratio of the maximum acceleration to the maximum velocity of the particle isA. b/aB. a/bC. `sqrt(a/b)`D. `sqrt(b/a)` |
|
Answer» Correct Answer - D |
|
| 66. |
Passage XI) The differential equation of a particle undergoing SHM is given by `a(d^(2)x)/(dt^(2))`+bx = 0. The particle starts from the extreme position. The time period of osciallation is given byA. `(2pi)/(b)`B. `(2pi)sqrt(b)`C. `(2pi)sqrt(b/a)`D. `(2pi(sqrt(a/b))` |
|
Answer» Correct Answer - D |
|
| 67. |
Write the equation of `SHM` for the sitution shown below : |
|
Answer» General equation of `SHM` can be written as `x = Asin(omegat+phi)` At `t = 0, x = A//2 rArr (A)/(2) = Asinphi rArr phi = 30^(@), 150^(@)` Also at `t = 0, v = -ve, Aomega cosphi = -ve rArr phi = 150^(@)` |
|
| 68. |
A particle executing SHM of amplitude 4 cm and `T=4 s` .The time taken by it to move from positive extreme position to half the amplitude isA. 1 sB. `(1)/(3)` sC. `(2)/(3)` sD. `sqrt((2)/(3))` s |
|
Answer» Correct Answer - C (c)Equation of motion , `y= a"cos"omegat` `implies(a)/(2)=a "cos"omegatimpliesomegat=(pi)/(3)` `implies (2pit)/(T)=(pi)/(3)impliest=((pi)/(3)xxT)/(2pi)=(4)/(3xx2)=(2)/(3) s` |
|
| 69. |
The equation of motion of a particle of mass `1g` is `(d^(2)x)/(dt^(2)) + pi^(2)x = 0`, where `x` is displacement (in m) from mean position. The frequency of oscillation is (in Hz)A. `1//2`B. `2`C. `5sqrt(10)`D. `1//5 sqrt(10)` |
|
Answer» Correct Answer - A `(d^(2)x)/(dt^(2)) = - pi^(2)x` Comparing with `(d^(2)x)/(dt^(2)) = - omega^(2)x` We have, `omega = pi` `:. 2pif = pi` `f = (1)/(2) Hz`. |
|
| 70. |
Wave property of electron implies that they will show diffraction effected . Davisson and Germer demonstrated this by diffracting electron from crystals . The law governing the diffraction from a crystals is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructiely Electron accelerated by potential `V` are diffracted from a crystal if `d = 1 Å and i = 30^(@), V` should be about `(h = 6.6 xx 10^(-34) Js, m_(e) = 9.1 xx 10^(-31) kg , e = 1.6 xx 10^(-19)C)`A. `2000 V`B. `50 V`C. `500 V`D. `1000 V` |
|
Answer» Correct Answer - B `2d cos I =n lambda implies 2d cos i=(h)/sqrt(meV)implies V=50` volt |
|
| 71. |
The left block in figure collides inelastically with the right block and striks to it. Find the amplitude of the resulting simple harmonic motion |
|
Answer» Assuming the collision to last for a small interval only, we can apply the principle of conservation momentum. The common velocity after the collision is `(v)/(2)`. The kinectic energy `= (1)/(2)(2m) ((v)/(2))^(2) = (1)/(4) mv^(2)`. This is also the total energy of viberation as the spring isunstretched at this moment. If the amplitude is `A`, the total energy cen also be written as `(1)/(2)kA^(2)`. Thus, `(1)/(2)kA^(2) = (1)/(4)mv^(2)`, giving `A = sqrt((m)/(2k))v`. |
|
| 72. |
In an experiment, electrons are made to pass through a narrow slit of width `d` comparable to their de Broglie wavelength. They are detected on a screen at a distance `D` from the slit (see figure)`. ` Which of the following graphs can be expected to represent the number of electrons `N` detected as a function of the detector position `y` (y=0 corresponds to the middle of the slit ).A. B. C. D. |
| Answer» Correct Answer - D | |
| 73. |
When a photon of light collides with a metal surface, number of electrons, (If any) coming out isA. only oneB. only twoC. infiniteD. depends upon factors |
|
Answer» Correct Answer - A In photo electric effect only one to one Interaction. |
|
| 74. |
Photon of frequency `v` has a momentum associated with it. If `c` is the velocity of light, the momentum is:A. `v//c`B. `hvc`C. `hv//c^(2)`D. `hv//c` |
|
Answer» Correct Answer - D The relation between energy (E) of a photon and momentum (P) associated with the photon is `E=pc` The corresponding momentum `p=E/c=(hv)/c` |
|
| 75. |
The anode vollage of a photocell is kept fixed . The wavelength `lambda `of the light falling on the cathode varies as followsA. B. C. D. |
|
Answer» Correct Answer - B With the increase in wavelength energy of the photon decreases. Therefore the KE of the electron coming out from the cathode also decreases. Due to which there will be a small decrease in plate current and once `lambda` becomes more than threshold wavelength electron will not come out from the cathode and hence current will become zero. |
|
| 76. |
A body is executing simple harmonic motion with an angular frequency s rad/2 . The velocity of the body at 20 mm displacement, when the amplitude of motion is 60 mm , isA. 40 mm /sB. 60mm/sC. 113 mm/sD. 120 mm/s |
|
Answer» Correct Answer - C |
|
| 77. |
A particle is executing simple harmonic motion with an amplitude A and time period T. The displacement of the particles after 2T period from its initial position isA. AB. 4AC. 8AD. zero |
|
Answer» Correct Answer - D It is the least interval of time after which the periodic motion of a body repreats itself. Therefore, displacement will be zero. |
|
| 78. |
Four simple harmonic vibrations `x_(1) = 8sinepsilont`, `x_(2) = 6sin(epsilont+pi/2)`, `x_(3)=4sin(epsilont_pi)` and `x_(4) = 2sin(epsilon+(3pi)/2)` are superimposed on each other. The resulting amplitude and its phase difference with `x_(1)` are respectively.A. `20, tan^(-)(1/2)`B. `4sqrt(2),(pi/2)`C. `20, tan^(-1)(2)`D. `4sqrt(2),(pi/4) |
|
Answer» Correct Answer - D |
|
| 79. |
Half-life of a radioactive substance A is `4 days`. The probability that a nuclear will decay in two half-lives isA. `1//4`B. `3//4`C. `1//2`D. `1` |
|
Answer» Correct Answer - B After two half life `1//4^(th)` fraction of nuclei will remain undecayed. `3//4^(th)` fraction will decay. Hence, the probability that a nucleus decays in two half lives is `3//4`. |
|
| 80. |
A copper sphere attached to the bottom of a vertical spring is oscillating with time period 10 s. If the copper sphere is immersed in a fluid (assume the viscosity of the fluid is negligilbe) of specfic gravity `(1)/(4)` of that of the copper, then time period of the oscillation isA. 5 sB. 10 sC. 2.5 sD. 20 s |
|
Answer» Correct Answer - B (b) Given, time period, T =10 s `therefore T=2pisqrt((m)/(k))` when viscous the system is immersed in liquid with specfic gravity `(1)/(4)th` of the copper, the liquid is non-viscous , so the time period would not be changed because , there is no change in m and k. |
|
| 81. |
A body of mass `m` attached to a spring which is oscillating with time period `4` seconds. If the mass of the body is increased by `4 kg`, Its timer period increases by `2 sec`. Determine value of initial mass `m`. |
|
Answer» In `I^(st)` case : `T = 2pisqrt((m)/(k)) rArr 4 = 2pisqrt((m)/(k))`….(i) and in `II^(nd)` case: `6 = 2pi sqrt((m+4)/(k))`…..(ii) Divide `(i)` by `(ii) (4)/(6) = sqrt((m)/(m+4)) rArr (16)/(36) = (m)/(m+4) rArr m = 3.2 kg` |
|
| 82. |
A body of mass m attached to a spring which is oscillating with time period 4 s. If the mass of the body is increased by 4 kg, its time period increases by 2 s. Determine value of initial mass m. |
|
Answer» In 1st case , T`=2pisqrt((m)/(k))implies4=2pisqrt((m)/(k))" "…(i)` 2nd case, 6 `=2pisqrt((m+4)/(k))" "(ii)` Divided Eq. (i) by Eq.(ii) `(4)/(6)=sqrt((m)/(m+4))implies (1 6)/(36)=(m)/(m+4)impliesm=3.2 kg` |
|
| 83. |
The length of a second’s pendulum at the surface of earth is 1 m. The length of second’s pendulum at the surface of moon where g is 1/6th that at earth’s surface isA. 1/6 mB. 6 mC. 1/36 mD. 36 m |
|
Answer» Correct Answer - A |
|
| 84. |
Statement-1 If the accelerating potential in an X-ray tube is increased, the wavelength of the characterstic X-rays do not change. because Statement-2 When an electron beam strikes the target in an X-ray tube, part of the kinetic energy is converted into X-ray energy. |
| Answer» Correct Answer - B | |
| 85. |
A student says that he had applied a force `F=-ksqrtx` on a particle and the particle moved in simple harmonic motion. He refuses to tell whether k is a constant or not. Assume that he has worked only with positive x and no other force acted on the particleA. As x increases k increasesB. As x increases k decreasesC. As x increases k remains constantD. The motion cannot be simple harmonic |
|
Answer» Correct Answer - A For SHM `Fprop-x`, for this `kpropsqrtx` |
|
| 86. |
If the earth were a homegeneous sphere and a straight hole was bored in it through its centre, so when a body is dropped in the hole, it will excutes SHM. Determine the time period of its oscillation . Radius of the earth is `6.4 xx 10^(5)` m and `g=9.8 ms^(-2)` |
|
Answer» Given, R`=6.4 xx 10^(5)m ,g=9.8 ms^(-2)` `therefore` Time period of oscillation of body will be `T=2pisqrt((R)/(g))=2pisqrt((6.4xx10^(5))/(9.8))=1605.67 s` |
|
| 87. |
A particle is executing SHM. Then the graph of acceleration as a function of displacement isA. straight lineB. circleC. ellipseD. hyperbola |
|
Answer» Correct Answer - A `a=-omega^2ximpliesxprop-x`, straight line |
|
| 88. |
A particle is executing SHM. Then the graph of velocity as a function of displacement isA. straight lineB. circleC. ellipseD. hyperbola |
|
Answer» Correct Answer - C `v=omegasqrt(A^2-x^2)` `v^2/omega^2+x^2=A^2implies(v^2)/(omega^2A^2)+x^2/A^2=1` v v/s x is ellipse |
|
| 89. |
A mass of `1 kg` attached to the bottom of a spring has a certain frequency of vibration. The following mass has to be added to it in order to reduce the frequency by half :A. `1 kg`B. `2 kg`C. `3 kg`D. `4 kg` |
|
Answer» Correct Answer - C `f_(1) = (1)/(2pi)sqrt((K)/(m_(1)))` `f_(2) = (1)/(2pi)sqrt((K)/(m_(2)))` `f_(2) = (f_(1))/(2)` or `m_(2) = 4m_(1)` or `m_(2) - m_(1) = kg` |
|
| 90. |
A mass m is suspended from a spring of length l and force constant K . The frequency of vibration of the mass is `f_(1)`. The spring is cut into two equal parts and the same mass is suspended from one of the parts. The new frequency of vibration of mass is `f_(2)` . Which of the following relations between the frequencies is correctA. `f_(1) = sqrt2 f_(2)`B. `f_(1) = f_(2)`C. `f_(1) = 2f_(2)`D. `f_(2) = sqrt2 f_(1)` |
|
Answer» Correct Answer - D |
|
| 91. |
The length of a spring is l and its force constant is k . When a weight W is suspended from it, its length increases by x . If the spring is cut into two equal parts and put in parallel and the same weight W is suspended from them, then the extension will beA. 2xB. xC. `x/2`D. `x/4` |
|
Answer» Correct Answer - D |
|
| 92. |
The velocity v and displacement x of a particle executing simple harmonic motion are related as `v (dv)/(dx)= -omega^2 x`. `At x=0, v=v_0.` Find the velocity v when the displacement becomes x.A. `sqrt(v_(0)^(2)+omega^(2)x^(2))`B. `sqrt(v_(0)^(2)-omega^(2)x^(2))`C. `v =root(3)(v_(0)^(3)+omega^(2)x^(3))`D. `v=v_(0)-(omega^(3)x^(3)e^(x^(3)))^(1//3)` |
|
Answer» Correct Answer - B (b) Given, `v(dv)/(dx)=-omega^(2)x` On integrating within the limit` underset(v_(0))overset(v)intvdv=underset(0)overset(x)-omega^(2)xdx` `implies[(v^(2))/(2)]_(v_(0))^(v)=-omega^(2)[(x^(2))/(2)]_(0)^(x)` `implies v^(2)-v_(0)^(2)=-omega^(2)x^(2)impliesv=sqrt(v_(0)^(2)-omega^(2)x^(2))` |
|
| 93. |
A particle of mass 200 g executes a simpel harmonit motion. The resrtoring force is provided by a spring of spring constant `80 N m^-1`. Find the time period.A. 0.31 secB. 0.15 secC. 0.05 secD. 0.02 sec |
|
Answer» Correct Answer - A |
|
| 94. |
A particle of mass 200 g executes a simpel harmonit motion. The resrtoring force is provided by a spring of spring constant `80 N m^-1`. Find the time period.A. 0.93 sB. 0.63 sC. 0.31 sD. None of these |
|
Answer» Correct Answer - C (c) Time period , `T=2pisqrt((m)/(k))` `impliesT=2xx3.14sqrt((200xx10^(-3))/(80))` or, `T=2xx3.14sqrt((1)/(400))` or , `T=(2xx3.14)/(20)` or, T=0.315 |
|
| 95. |
An SHM is give by `y=5["sin"(3pit)+sqrt(3)"cos"(3pit)]`. What is the amplitude of the motion of y in metre ?A. 10B. 20C. 1D. 5 |
|
Answer» Correct Answer - A (a) Given equation is `y=5["sin"3pit+sqrt(3)"cos"3pit]` or `y=5xx2[(1)/(2)"sin"3pit+(sqrt(3))/(2)"cos"3pit]` or `y=10["sin"3pit"cos"(pi)/(3)+"cos"3pit"sin"(pi)/(3)]` or `y=10"sin"(3pit+(pi)/(3))` |
|
| 96. |
Two particles of mass `m_(1) " and " m_(2)` are connected by a spring of natural length L and force constant k. The masses are brought close enough so as to compress the spring completely and a string is used to tie the system. Assume that length of spring in this position is close to zero. The system is projected with a velocity `V_(0)` along the positive x direction. At the instant it reaches origin at time t = 0, the string snaps and the spring starts opening. (a) Show that the mass `m_(1) (or m_(2))` will are perform SHM in the reference frame attached to the centre of mass of the system. Find the time period of oscillation. (b) Write the amplitude of `m_(1) " and " m_(2)` as a function of time. lt( c) Write the X co ordinates of `m_(1) " and " m_(2)` as a function of time |
|
Answer» Correct Answer - (a) T=2pi sqrt((m_(1)m_(2))/((m_(1)+m_(2))k))` ` (b) (m_(2)L)/(m_(1)+m_(2))=A_(1)` `(c ) X_(1)=V_(0^(t))-A_(1)(1-cos omegat);` `X_(2)=V_(0^(t))-A_(2)(1-cos omegat) Where `A_(1)=(m_(2)L)/(m_(1)+m_(2)); A_(2)=(m_(1)L)/(m_(1)m_(2)); omega=sqrt((k(m_(1)+m_(2)))/(m_(1)m_(2)))` |
|
| 97. |
Two masses `m_(1)` and `m_(2)` connected by a light spring of natural length `l_(0)` is compressed completely and tied by a string. This system while moving with a velocity `v_(0)` along `+ve x`-axis pass through the origin at `t=0`. At this position the string snaps. Postion of mass `m_(1)` at time `t` is given by the equation. `x_(1) (t)=v_(0)t-A(1-cosomegat)` calculate : (a) Position of the particle `m_(2)` as a function of time. (b) `l_(0)` in terms of `A`. |
|
Answer» Correct Answer - A::B::C |
|
| 98. |
Two masses `m_(1)` and `m_(2)` concute a high spring of natural length `l_(0)` is compressed completely and tied by a string. This system while conving with a velocity `v_(0)` along `+ve` x-axis pass thorugh the origin at `t = 0`, at this position the string sanps, Position of mass `m_(1)` at time t is given by the equation `x_(1)t = v_(0)(A//1-cosomegat)`. Calculate (i) position of the particle `m_(2)` as a funcation of time, (ii) `l_(0)` in terms of A. |
|
Answer» Correct Answer - (i) `v_(0)t + A(m_(1))/(m_(2))(1-cosomegat)` , (ii) `((m_(1))/(m_(2)) + 1) A` (i) Two massse `m_(1)` and `m_(2)` are connected by a spring of length `l_(0)`. The spring is in compressed position. It is held in this position by a string. When the string snaps, the spring force is brought into operation. The spring force is an internal force w.r.t masses-spring system. No external force is applied on the system. The velocity of centre of mass will not change. Velocity of centre of mass `= v_(0)` `:.` Location/x -coordinate of centre of mass of time `t = v_(0)t` `:. barv = (m_(1)x_(1) + m_(2)x_(2))/(m_(1) + m_(2))` `rArr v_(0)t = (m_(1)[v_(0)t - A(1-cosomegat)]+m_(2)x_(2))/(m_(1) + m_(2))` `rArr (m_(1)+m_(2))v_(0)t=m_(1)[v_(0)t-A(1-cosomegat)]+m_(2)x_(2))` `rArr m_(1)v_(0)t + m_(2)v_(0)t = m_(1)v_(.0)t - m_(1)A(1-cosomegat)] + m_(2)x_(2)` `rArr m_(2)x_(2) = m_(2)v_(0)t + m_(1)A(1-cosomegat)` `rArr x_(2)=v_(0)t + (m_(1)A)/(m_(2))(1-cosomegat)"......"(i)` To express `l_(0)` in terms of A. `:. x_(1) = v_(0)t - A(1-cosomegat) :. (dx_(1))/(dt^(2)) = -Aomega^(2) sinomegat` `:. (d^(2)x^(2))/(dt^(2)) = - Aomega^(2) cosomegat "........"(ii)` `x_(1)` is displacement of `m_(1)` at time t. `:. (d^(2)x_(1))/(dt^(2)) =` acceleration of `m_(1)` at time t. When the spring attains its natural length `l_(0)`, then acceleration is zero and `(x_(2) - x_(1)) = l_(0))` `:. x_(2) x_(1) = l_(0)` , Put `x_(2)` from (i) `rArr [v_(0)t + (m_(1)A)/(m_(2)) (1-cosomegat)] - [v_(0)t - A(1-cosomegat)] = l_(0)` `rArr l_(0) = ((m_(1))/(m_(2)) + 1)A(1-cosomegat)` When `(d^(2)x_(1))/(dt^(2)) = 0, cosomegat = 0` from (ii). `:. l_(0) = ((m_(1))/(m_(2)) + 1)A`. |
|
| 99. |
Two blocks of mass 10 kg and 2 kg are connected by an ideal spring of spring constant `K = 800 N//m` and the system is placed on a horizontal surface as shown. The coefficient of friction between 10 kg block and surface is 0.5 but friction is absent between 2 kg and the surface. Initially blocks are at rest and spring is relaxed. The 2 kg block is displaced to elongate the spring by 1 cm and is then released. (a) Will 10 kg block move subsequently? (b) Draw a graph representing variation of magnitude of frictional force on 10 kg block with time. Time t is measured from that instant when 2 kg block is released to move. |
| Answer» Correct Answer - (a) No | |
| 100. |
A particle of mass m is tied at the end of a light string of length L, whose other end is fixed at point C (fig), and is revolving in a horizontal circle of radius r to form a conical pendulum. A parallel horizontal beam of light forms shadow of the particle on a vertical wall. If the tension in the string is F find - (a) The maximum acceleration of the shadow moving on the wall. (b) The time period of the shadow moving on the wall. |
|
Answer» Correct Answer - `(a) (1)/(m)sqrt(F^(2)-(mg)^(2)), (b) 2pi((m^(2)r^(2))/(F^(2)-(mg)^(2)))^(1//4)` |
|