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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.
851. |
A particle moves in the X-Y plane according to the equation `vecr=(veci+2vecj)Acosomegat`. The motion of the particle isA. (i), (ii)B. (i), (iii)C. (i), (iii), (iv)D. (iii), (iv) |
Answer» Correct Answer - C `vecr=(hati+2hatj)Acosomegat` `=Acosomegathati+2Acosomegathatj=xhati+yhatj` `x=Acosomegat`, `y=2Acosomegat` `y/x=2impliesy=2x` (straight line), (i) is O.K. `(dvecr)/(dt)=(hati+2hatj)(-Aomegasinomegat)` `veca=(d^2vecr)/(dt^2)=(hati+2j)(-Aomegacosomegat)=-omega^2vecr` `vecaalpha-vecr(SHM)`, (iv) is O.K. sin and cos are periodic functions, (iii) is O.K. |
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852. |
A point mass is subjected to two simultaneous sinusoidal displacements in `x - direction, x_1 (t) = A sin (omega)t and x_2 (t) = A sin ((omega t + (2 pi)/(3))`. Adding a third sinusoidal displacement `x _3 (t) = B sin (omega t + phi)` brings the mas to a complete rest. The values of (B) and (phi) are.A. `sqrt (2 A),(3 pi)/(4)`B. `A, (4 pi)/(3)`C. `sqrt (3 A),(5 pi)/(3)`D. `A, (pi)/(3)` |
Answer» Correct Answer - B Two sinusoidal displacements have amplitude (A) each, with a phase difference of `2 (pi)/(3)`. It is given that sinusoidal displacement `x_3(t)` brings the mass to a complete rest. This is possible when the amplitude of third is (A) and is having a phase difference of `4 (pi)/(3)` with respevt to `x_1(t)` as shown in the figure. |
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853. |
A point mass is subjected to two simultaneous sinusoidal displacements in `x - direction, x_1 (t) = A sin (omega)t and x_2 (t) = A sin ((omega t + (2 pi)/(3))`. Adding a third sinusoidal displacement `x _3 (t) = B sin (omega t + phi)` brings the mas to a complete rest. The values of (B) and (phi) are.A. `sqrt2A,(3pi)/(4)`B. `A,(4pi)/(3)`C. `sqrt3A,(5pi)/(6)`D. `A,(pi)/(3)` |
Answer» Correct Answer - B | |
854. |
What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to vertical height of 10 cm (take `g=9.8m//s^(2)`) A. 2.2 m/sB. 1. 8 m/sC. 1. 4 m/sD. 0.6 m/s |
Answer» Correct Answer - C |
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855. |
A pendulum bob has a speed 3m/s while passing thorugh its lowest position. What is its speed when it makes an angle of `60^0` with the vertical? The length of the pendulum is 0.5m Take `g=10 m/s^2`.A. 3 m/sB. `1/3 m//s`C. `1/2 m//s`D. 2 m/s |
Answer» Correct Answer - D |
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856. |
Due to some force `F_(1)` a body oscillates with period `4//5s` and due to other force `F_(2)` it oscillates with period `3//5s`. If both the forces acts simultaneously in same direction then new period isA. 0.72 secB. 0.64 secC. 0.48 secD. 0.36 sec |
Answer» Correct Answer - C |
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857. |
A horizontal platform with an object placed on it is executing S.H.M. in the vertical direction. The amplitude of oscillation is ` 3. 92 xx 10 ^(-3)` m. what must ve the least period of these oscillations. So that the object is not detached from the platformA. 0.1256 secB. 0.1356 secC. 0.1456 secD. 0.1556 sec |
Answer» Correct Answer - A |
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858. |
Speed `v` of a particle moving along a straight line, when it is a distance X from a fixed point on the line is given by `V^(2)=108-9X^(2)` (all quantities in S.I. unit). ThenA. The motion is uniformly accelereted along the straight lineB. The magnitude of the accelerated at a distance `3 cm` from the fixed point is `0.27 m//s^(2)`.C. The motion is harminic about `x=sqrt(12) m`.D. The maximum displacement from the fixed point is `4 cm`. |
Answer» Correct Answer - B |
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859. |
Which of the following function represent `SHM` ?A. `sin2omegat`B. `sin^(2)omegat`C. ` sin omegat + 2 cos omegat`D. ` sin omegat + cos 2omegat` |
Answer» Correct Answer - A::B::C | |
860. |
The speed v of a particle moving along a straight line. When it is at distance x from a fixed point on the line is `v^(2)=144-9x^(2)`. Select the correct alternativesA. The magnitude of acceleration at a distance 3 units from the fixed point is 27 unitsB. the motion is simple harmonic with `T=(2pi)/(3)` unitsC. The maximum displacement from the fixed point is 4 unitsD. all are correct |
Answer» Correct Answer - D (d) `v=omegasqrt(A^(2)+x^(2))" "`(in SHM) or `v^(2)=omega^(2)A^(2)-omega^(2)x^(2)` Comparing the given equation with this equation we get, `omega^(2)=9` `therefore omega=3=(2pi)/(T)` `therefore T=(2pi)/(3)` units Also , `omega^(2)A^(2)=144` `therefore A=4 "units", |a|=omega^(2)x=(9)(3)=27` units Displacement ` le` distance |
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861. |
A horizontal platform with a mass `m` placed on it is executing SHM slong y-axis. If the amplitude of oscillation is 2.5 cm, the minimum period of the motion for the mass not to be detached from the platform `(g=10(m)/(s^2)`)A. `(10)/(pi)s`B. `(pi)/(10)s`C. `(pi)/(sqrt(10))s`D. `(1)/(sqrt(10))s.` |
Answer» Correct Answer - B::D | |
862. |
The speed v of a particle moving along a straight line, when it is at a distance (x) from a fixed point of the line is given by `v^2=108-9x^2` (all equation are in CGS units):A. the motion is uniformly acclerated along the straight lineB. the magnitude of the acceleration at a distance `3cm` from the fixed point is `27 cm//s`C. The motion is simple harmonic about the given fixed point.D. the maximum displacement displacment from the fixed point is `4 cm`. |
Answer» Correct Answer - B::C | |
863. |
A particle starts oscillating simple harmonically from its equilibrium position then, the ration of kinetic energy and potential energy of the particle at the time `T//12` is: `(T="time period")`A. `2:1`B. `3:1`C. `4:1`D. `1:4` |
Answer» Correct Answer - D |
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864. |
A horizontal plank has a rectangular block placed on it. The plank starts oscillating vertically and simple harmonically with an amplitude of `40 cm`. The block just loses contact with the plank when the latter is at momentary rest. Thin :A. the period of oscillation is `((2pi)/(5))`B. the block weights its true weight `0.5` times its weight on the plank halfway upC. the block weight double weights `1.5` time its weight on the plank halfway downD. The block weight on the plank when the letter moves fastest |
Answer» Correct Answer - A::B::C::D | |
865. |
A horizontal plank has a rectangular block placed on it. The plank starts oscillating vertically and simple harmonically with an amplitude of 40 cm. The block just loses contact with the plank when the latter is at momentary rest Then.A. the period of oscillation is `((2pi)/(5))`B. the block weights its true weight `0.5` times its weight on the plank halfway upC. the block weight double weights `1.5` time its weight on the plank halfway downD. The block weight on the plank when the letter moves fastest |
Answer» Correct Answer - A::B::C::D | |
866. |
The figure shows a graph between velocity and displacement (from mean position) of a particle performing `SHM`: A. the time period of the particle is `1.57s`B. the maximum acceleration will be `40cm//s^(2)`C. the velocity of particle is `2sqrt(21) cm//s` when it is at a distance `1 cm` from the mean position.D. none of these |
Answer» Correct Answer - A::B::C | |
867. |
A particle starts oscillating simple harmonically from its equilibrium position then, the ratio of kinetic energy and potential energy of the particle at the time `T//12` is: `(T="time period")` |
Answer» At,`t=(T)/(12),x A "sin" (2pi)/(T)xx(T)/(12)=A"sin"(pi)/(6)=(A)/(2)` so, KE`=(1)/(2)k(A^(2)-x^(2))=(3)/(4)xx(1)/(2)kA^(2)` and `PE=(1)/(2)kx^(2)=(1)/(4)xx(1)/(2)kA^(2)` `therefore (KE)/(PE)=(3)/(1)` |
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868. |
Acceleration a versus time `t` graph of a body in `SHM` is given by a curve shown below. `T` is the time period Then corresponding graph between kinetic energy `KE` and time `t` is correctly represented by B. C. D. |
Answer» Correct Answer - A `x = A cosomegat` `K.E. = (1)/(2)k(A^(2) - x^(2)) = (1)/(2)kA^(2)sin^(2)omegat` `= (1)/(2) kA^(2)((1-cos2omegat)/(2)) = (kA^(2))/(4) (1 - cos2omegat)` Frequency of `K.E.` is double of acceleration |
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869. |
A particle executes simple harmonic motion along a straight line with an amplitude A . The potential energy is maximum when the displacement isA. `+-A`B. zeroC. `+-A/2`D. `+-A/sqrt2` |
Answer» Correct Answer - A |
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870. |
A particle executes simple harmotic motion about the point `x=0`. At time `t=0`, it has displacement `x=2cm` and zero velocity. If the frequency of motion is `0.25 s^(-1)`, find (a) the period, (b) angular frequency, ( c) the amplitude, (d) maximum speed, (e) the displacement from the mean position at `t=1 s`and (f) the velocity at `t=1 s`. |
Answer» (i) Period , `T=(1)/(f)=(1)/(0.25 s^(-1))=4 s` (ii) Angular frequency, `omega=(2pi)/(T)=(2pi)/(4)=(pi)/(2)"rad"s^(-1)=1.57 "rad"s^(-1)` (iii) Amplitude is the maximum displacement from mean position. Hence, A=2-0=2 cm. (iv) Maximum speed, `v_("max")=Aomega=2.(pi)/(2)=pi cms^(-1)=3.14 cms^(-1)` (v) The displacement is given by `x=A"sin"(omegat+phi)` Initially at t=0, x=2 cm ,then `2=2"sin"phi` or `"sin"phi=1="sin"90^(@)` or `phi=90^(@)` Now at t=3s `x=2"sin"((pi)/(2)xx3+(pi)/(2))=0` (vi) Velocity at x=0 is `v_("max")`. `therefore v_("max")=omegaA=((pi)/(2))(2)=3.14 cms^(-1)` |
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871. |
Maximum acceleration of a particle in SHM is `16 cm//s^(2)` and maximum uelocity is `8 cm//s`.Find time period and amplitude of oscillations. |
Answer» Correct Answer - A::B::C::D `a_(max)=omega^(2) A=16 cm/s^2` …(i) `u_(max)= omega A = 8 cm/s` …(ii) Solving Eqs.(i) and (ii), we get `A = 4cm` and `omega = 2rad//s` Now, `T = (2pi)/(omega) = (2pi)/(2) = (pi) s` or `T~~3.14s` |
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872. |
A particle executes simple harmotic motion about the point `x=0`. At time `t=0`, it has displacement `x=2cm` and zero uelocity. If the frequency of motion is `0.25 s_(-1)`, find (a) the period, (b) angular frequency, ( c) the amplitude, (d) maximum speed, (e) the displacement from the mean position at `t=3 s`and (f) the velocity at `t=3 s`. |
Answer» Correct Answer - A::B::C::D (a) Period `T=(1)/(f)` `=(1)/(0.25 s^(-1))=4 s` (b) Angular frequency `omega=(2pi)/(T)` `=(2pi)/(4)=(pi)/(2) rad//s` `=1.57rad//s` ( c) Amplitude is the maximum displacement from mean position. Hence, `A=2-0=2cm`. (d) Maximum speed `v_(max)=A omega` `=2.(pi)/(2)=pi cm//s` `=3.14 cm//s` (e) At `t=0`, particle starts from extreme position. `:. x=A cos omega t` `=2 cos((pi)/(2)t)` At `t=1sec` `x=0` (f) Velocity at `x=0`, is `v_(max)` i.e. `3.14cm//s`. |
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873. |
Find time period of the function, `y=sin omega t + sin 2omega t + sin 3omega t` |
Answer» Correct Answer - A::B The given function can be written as, `y=y_(1) + y_(2) + y_(3)` Here, `y_(1)=sin omega t` , `T_(1)=(2pi)/(omega)` `y_(2)=sin 2omega t` , `T_(2)=(2pi)/(2omega)=(pi)/(omega)` and `y_(3)=sin 3omegat, T_(3)=(2pi)/(3omega)` `:. T_(1)=2T_(2)` and `T_(1)=3T_(3)` So, the time period of the given function is `T_(1)` or `(2pi)/(omega)`. Because in time `T=(2pi)/(omega)`, first function completes one oscillation, the second function two oscillation and the third three. |
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874. |
Hydrogen atom is exited from ground state to another state with principal quantum number equal to `4` Then the number of spectral line in the emission spectra will beA. 6B. 2C. 3D. 5 |
Answer» Correct Answer - A Number of lines `.^(3)C_(2)=.^(4)C_(2)=((4)(3))/(2)=6` |
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875. |
Energy required for the electron excitation in `Li^(++)` from the first to the third Bohr orbit is:A. `108.8 eV`B. `122.4 eV`C. `12.1 eV`D. `36.3 eV` |
Answer» Correct Answer - A Required energy `=13.6 (Z)^(2) [1/n_(1)^(2)-1/n_(2)^(2)] eV` `=13.6 (3)^(2) [1/1^(2)-1/3^(2)]=108.8 eV` |
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876. |
A particle executes simple harmonic motion. Its instantaneous acceleration is given by `a = - px`, where `p` is a positive constant and `x` is the displacement from the mean position. Find angular frequency of oscillation. |
Answer» Comparing with `a = - omega^(2)x` `omega^(2) = p` `:. omega = sqrt p` |
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877. |
In damped oscillations damping froce is directly proportional to speed of ocillatior .If amplitude becomes half to its maximum value is 1 s, then after 2 s amplitude will be (`A_(0)`- initial amplitude)A. `(1)/(4)A_(0)`B. `(1)/(2)A_(0)`C. `A_(0)`D. `(sqrt(3)A_(0))/(2)` |
Answer» Correct Answer - A (a) Amplitude of damped oscillation is `A=A_(0)e^(-gammat)` `A=(A_(0))/(2)` where t=1s So, `(A_(0))/(2)=A_(0)e^(-gamma1)` or `e^(gamma)=2` After 2 s, `A=A_(0)e^(-gamma^(3))implies A=(A_(0))/(4)` |
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878. |
A pendulum with time period of `1s` is losing energy due to damping. At time its energy is `45 J`. If after completing `15` oscillations, its energy has become `15 J`. Its damping constant (in `s^(-1)`) is :-A. `2`B. `1/15"ln"3`C. `1/2`D. `1/3"ln"3` |
Answer» Correct Answer - D Amplitude of pendulum will very as `A = A_(0)e^(-deltaf), delta =` damping constant `E = 1/(2)kA^(2) = 1/2 kA_(0)e^(-2deltat)` `E = E_(0)e^(-2deltat)` at `t = 0, E_(0) = 45 J` , at `t = 15 s , E = 15 J` so, `15 = 45 e^(-2delta xx 15) rArr delta = 1/3 ln3` |
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879. |
Two particle execute `SHM` of same amplitude of `20 cm` with same period along the same line about the same equilibrium position. The maximum distance between the two is `20 cm`. Their phase difference in radians isA. `2pi//3`B. `pi//2`C. `pi//3`D. `pi//4` |
Answer» Correct Answer - C |
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880. |
A spring having a spring constant k is loaded with a mass m. The spring is cut into two equal parts and one of these is loaded again with the same mass. The new spring constant iA. K/2B. KC. 2KD. `K^(2)` |
Answer» Correct Answer - C |
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881. |
Two particles are in `SHM` in a straight line about same equilibrium position. Amplitude `A` and time period `T` of both the particles are equal. At time `t=0`, one particle is at displacement `y_(1)=+A` and the other at `y_(2)=-A//2`, and they are approaching towards each other. after what time they cross each other?A. `T/6`B. `(5T)/6`C. `T/4`D. `T/3` |
Answer» Correct Answer - A | |
882. |
A block is kept on a rough horizontal plank. The coefficient of friction between the block and the plank is `1/2`. The plank is undergoing SHM of angular frequency 10 rad/s. The maximum amplitude of plank in which the block does not slip over the plank is (g= 10 m/`s^(2)`)A. 4cmB. 5cmC. 10 cmD. 16 cm |
Answer» Correct Answer - B |
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883. |
A plank with a small block on top of it is under going vertical `SHM`. Its period is `2 sec`. The minium amplitude at which the block will separate from plank is :A. `(10)/(pi^(2))`B. `(pi^(2))/(10)`C. `(20)/(pi^(2))`D. `(pi)/(10)` |
Answer» Correct Answer - A |
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884. |
Two particles `P_(1)` and `P_(2)` are performing `SHM` along the same line about the same meabn position , initial they are at their position exterm position. If the time period of each particle is `12` sec and the difference of their amplitude is `12 cm` then find the minimum time after which the seopration between the particle becomes `6 cm` |
Answer» Correct Answer - `t = 2s` The coordinates of the particeles are `x_(1) = A_(1) cos omegat, x_(2) = A_(2) cos omegat` seperation `= x_(1) - x_(2) = (A_(1) - A_(2)) cos omegat = 12 omegat` Now `x_(1) - x_(2) = 6 = 12 cos omegat` `rArr omegat = (pi)/(3)` `rArr (2pi)/(12). t = (pi)/(3)` `rArr t = 2s` |
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885. |
The angular velocities of three bodies in `SHM` are `omega_(1), omega_(2), omega_(3)` with their respective amplitudes as `A_(1),A_(2),A_(3)`. If all three bodies have same mass and maximum velocity thenA. `A_(1)omega_(1)= A_(2)omega_(2)=A_(3)omega_(3)`B. `A_(1)omega_(1)^(2)= A_(2)omega_(2)^(2)=A_(3)omega_(3)^(2)`C. `A_(1)^(2)omega_(1)= A_(2)^(2)omega_(2)=A_(3)^(2)omega_(3)`D. `A_(1)^(2)omega_(1)^(2)= A_(2)^(2)omega_(2)^(2)=A^(2)` |
Answer» Correct Answer - A |
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886. |
Two light spring of force constants `k_(1)` and `k_(2)` and a block of mass m are in one line `AB` on a smooth horizontal table such that one end of each spring is fixed on rigid supports and the other end is free as shown in the figure. The distance `CD` between the spring is `60cm`. If the block moves along `AB` with a velocity `120 cm//s` in between the springs, calculate the period of oscillation of the block. (take `k_(1) = 1.8 N//m`, `k_(2) = 3.2 N//m`, `m = 200 g`) |
Answer» Correct Answer - `2.82 s` Between `C` and `D` block will move with constant speed of `120 cm//s`. Therefore, period of oscillation will be (starting from `C`). `T = t_(CD) + (T_(2))/(2) + t_(DC) + (T_(1))/(2)` Here, `T_(1) = 2pisqrt((m)/(k_(1))` and `T_(2) = 2pisqrt((m)/(k_(2))` and `t_(CD) = T_(DC) = (CD)/(v) = (60)/(120) = 0.5 s` `:. T = 0.5 + (2pi)/(2)sqrt((0.2)/(3.2)) + 0.5 + (2pi)/(2)sqrt((0.2)/(1.8))` `(m = 200 g = 0.2 kg)` `T = 2.82 s` |
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887. |
Two springs, of spring constants `k_(1)` and `K_(2)`, have equal highest velocities, when executing SHM. Then, the ratio of their amplitudes (given their masses are in the ratio `1:2`) will beA. `sqrt(k_(1)//k_(2))`B. `k_(1)//k_(2)`C. `k_(2)//k_(1)`D. `sqrt(k_(2)//k_(1))` |
Answer» Correct Answer - D `(1)/(2)mv^(2) = (1)/(2)K_(1)x_(1)^(2)` `(1)/(2)mv^(2) = (1)/(2)K_(2)x_(2)^(2)` `K_(1)x_(1)^(2) = K_(2)x_(2)^(2)` `(x_(1))/(x_(2)) = sqrt((K_(2))/(K_(1))` |
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888. |
Two particle A and B of equal masses are suspended from two massless springs of spring constants `k_(1)` and `k_(2)`, respectively. If the maximum velocities, during oscillations are equal, the ratio of amplitude of A and B is `(4//3) xx 1000 kg//m^(3)`. What relationship betwen `t` and `t_(0)` is ture?A. `sqrt(k_(1)//k_(2))`B. `k_(1)//k_(2)`C. `sqrt(k_(2)//k_(1))`D. `k_(2)//k_(1)` |
Answer» Correct Answer - C Maximum velocity Given `(v_(max))_(1) = (v_(max))_(2) rArr omega_(1)A = omega_(2)A_(2)` `rArr (A_(1))/(A_(2)) = (omega_(1))/(omega_(2)) = sqrt((k_(2))/(m) xx (m)/(k_(1))) = sqrt((k_(2))/(k_(1)))` |
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889. |
Two light spring of force constants `k_(1)` and `k_(2)` and a block of mass m are in one line `AB` on a smooth horizontal table such that one end of each spring is fixed on rigid supports and the other end is free as shown in the figure. The distance `CD` between the spring is `60cm`. If the block moves along `AB` with a velocity `120 cm//s` in between the springs, calculate the period of oscillation of the block. (take `k_(1) = 1.8 N//m`, `k_(2) = 3.2 N//m`, `m = 200 g`) |
Answer» Correct Answer - B Between `C` and `D` block will move with constant speed of `120 cm//s`.Therefore, period of oscillation will be (starting from C). `T = t_(CD) + T_(2)/2 + t_(DC) + T_(1)/(2)` Here,`T_(1) = 2pi sqrt((m)/(k_(1)))` and `T_(2) = 2pi sqrt((m)/(k_(2))` and `t_(CD) = t_(DC) = (60)/(120) = 0.5s` `:. T = 0.5 + (2pi)/(2)sqrt((0.2)/(3.2)) + 0.5 + (2pi)/(2)sqrt((0.2)/(1.8))` `T = 2.82 s` |
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890. |
Let `K_(1)` be the maximum kinetic energy of photoelectrons emitted by a light of wavelength `lambda_(1)` and `K_(2)` corresponding to `lambda_(`2). If `lambda_(1) = 2lambda_(2)`, thenA. `2K_(1) = K_(2)`B. `K_(1) = 2K_(2)`C. `K_(1) lt (K_(2))/(2)`D. `K_(1) gt 2K_(2)` |
Answer» Correct Answer - C `(K_(2))/(K_(1)) = ((hc)/(lambda_(2)) - phi)/((hc)/(lambda_(1)) - phi) = (2((hc)/(lambda_(2)) - phi))/(((hc)/(lambda_(2)) - 2phi)) rArr K_(1) lt (K_(2))/(2)` |
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891. |
A photon and an electron have equal energy `E . lambda_("photon")//lambda_("electron")` is proportional toA. `E^(0)`B. `E^(1//2)`C. `E^(-1)`D. `E^(-2)` |
Answer» Correct Answer - B `lambda_(1)/lambda_(2)=(h/sqrt(2mE))/((hc)/E)` or `lambda_(1)/lambda_(2) prop E^(1//2)` |
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892. |
Half-lives of two radioactive substances `A` and `B` are respectively `20` minutes and `40` minutes. Initially, he sample of `A` and `B` have equal number of nuclei. After `80` minutes the ratio of the remaining number of `A` and `B` nuclei is :A. `5 : 4`B. `1 : 16`C. `4 : 1`D. `1 : 4` |
Answer» Correct Answer - A `t=80 min=4 T_(A)=2T_(B)` `:.` no. of nuclei of A decayed `=N_(0)-N_(0)/2^(4)=(15 N_(0))/16` `:.` no. of nuclei of B decayed `=N_(0)- N_(0)/2^(2)=(3N_(0))/4` required ratio `=5/4` |
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893. |
A hydrogen atom makes a transition from `n=2` to `n=1` and emits a photon. This photon strikes a doubly ionized lithium atom `(Z=3)` in excited state and completely removes the orbiting electron. The least quantum number for the excited stated of the ion for the process is:A. `2`B. `4`C. `5`D. `3` |
Answer» Correct Answer - B | |
894. |
Statement - 1 : When ultraviolet light is incident on a photocell , its stopping potential is `V_(0)` and the maximum kinetic energy of the photoelectrons is `K_(max)` when the ultraviolet light is replaces by X- rays both `V_(0) and K_(max)`increase Statement - 2 : Photoelectrons are emitted with speeds ranging from zero to a maximum value became of the range of frequencies present in the incident lightA. Statement-1 is ture, Statement-2 is falseB. Statement-1 is ture, Statement- 2 is true, Statement-2 is the correct explanation of Statement-1C. Statement-1 is ture, Statement- 2 is true, Statement-2 is not the correct explanation of Statement-1D. Statement-1 is False, Statement-2 is ture |
Answer» Correct Answer - A | |
895. |
When photon of wavelength `lambda_(1)` are incident on an isolated shere supended by an insulated , the corresponding stopping potential is found to be `V`. When photon of wavelength `lambda_(2)` are used , the orresponding stopping potential was thrice the above value. If light of wavelength `lambda_(3)` is used , carculate the stopping potential for this case.A. `(hc)/(e)[1/(lambda_(3))+(1)/(2lambda_(2))-(1)/(lambda_(1))]`B. `(hc)/(e)[1/(lambda_(3))-(1)/(lambda_(2))-(1)/(lambda_(1))]`C. `(hc)/(e)[1/(lambda_(3))+(1)/(lambda_(2))-(1)/(lambda_(1))]`D. `(hc)/(e)[1/(lambda_(3))+(1)/(2lambda_(2))-(1)/(2lambda_(1))]` |
Answer» Correct Answer - D | |
896. |
When light of wavelength `lambda` is incident on a metal surface, stopping potential is found to be x. When light of wavelength n`lambda` is incident on the same metal surface, stopping potential is found to be `(x)/(n+1)` . Find the threshold wavelength of the metal. |
Answer» Let `lambda_(0)` is the threshold wavelength. The work function is `phi = (hc)/(lambda_(0))` Now, by photoelectric equation `ex = (hc)/(lambda) - (hc)/(lambda_(0))` ……(i) `(ex)/(n+1) = (hc)/(nlambda) - (hc)/(lambda_(0))`……..(ii) From (i) and (ii) `(hc)/(lambda) - (hc)/(lambda_(0)) = (n+1) (hc)/(nlambda) - (n+1) (hc)/(lambda_(0)) rArr (nhc)/(lambda_(0)) = (hc)/(nlambda) rArr lambda_(0) = n^(2)lambda` |
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897. |
If `lambda_(Cu)` is the wavelength of `K_(alpha)` X-ray line of copper (atomic number `29`) and `lambda_(Mo)` is the wavelength of the `K_(alpha)`X-ray line of molybdenum (atomic number `42`),then the ratio `lambda_(Cu)//lambda_(Mo)` is close toA. `1.99`B. `2.14`C. `0.50`D. `0.48` |
Answer» Correct Answer - B `sqrt(C/lambda)=a (Z-b)` `b=1 sqrt(lambda_(Cu)/lambda_(Me))=((Z_(Me)-1)/(Z_(Cu)-1))` `lambda_(Cu)/lambda_(Me)=(41/28)^(2)=2.14` |
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898. |
The electrostatic energy of `Z` protons uniformly distributed throughout a spherical nucleus of radius `R` is given by `E = (3 Z(Z- 1)e^(2))/5( 4 pi e _(0)R)` The measured masses of the neutron `_(1)^(1) H, _(7)^(15) N and , _(8)^(16)O are 1.008665 u, 1.007825 u , 15.000109 u and 15.003065 u, ` respectively Given that the ratio of both the `_(7)^(12) N` and `_(8)^(15) O` nucleus are same , 1 u = = 931.5 Me V`c^(2) ` (c is the speed of light ) and `e^(2)//(4 pi e_(0)) = 1.44 MeV` fm Assuming that the difference between the binding energies of `_7^(15) N and `_(8)^(15) O ` is purely due to the electric energy , The radius of the nucleus of the nuclei isA. `2.85 fm`B. `3.03 fm`C. `3.42 fm`D. `3.80 fm` |
Answer» Correct Answer - C Ealectrostatic energy `= BE_(N) - BE_(0)` `= [[7M_(H) + 8M_(n) - M_(N)] - [8M_(H) + 7M_(n) - M_(o)]] xx C^(2)` `=[-M_(H) + M_(n) + M_(o) - M_(N)]C^(2)` `= [-1.007825 + 1.008665 + 15.003065 - 15.000109] xx 931.5` `+ 3.5359 MeV` `Delta E = (3)/(5) xx (1.44 xx 8 xx 7)/(R) - (3)/(5) xx (1.44 xx 7 xx 6)/(R) = 3.5359` `R = (3 xx 1.44 xx 14)/(5 xx 3.5359) = 3.42 fm` |
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899. |
Assume that the nuclear binding energy per uncleon `(B//A)` versus mass number `(A)` is as shwon in Fig. Use this plot to choose the correct choice `(s)` given below: A. Fusion of two nuclei with mass numbers lying in the range of `1 lt A lt 50` will release energyB. Fusion of two nuclei with mass numbers lying in the range of `51 lt A lt 100` will release energyC. Fission of a nucleus lying in the the mass range of `100 lt A lt 200` will release energy when broken into two equal fragmentsD. Fission of a nucleus lying in the mass range of `200 lt A lt 260` will release energy when broken into two equal fragments |
Answer» Correct Answer - B::D (A) for `1 lt A lt 50`, on fusion mass number for compound nucleus is less than `100` `implies` Binding energy nucleon remains same `implies` No energy is released (B) For `51 lt A lt 100` on fusion mass number for compound nucleus is between `100` & `200` `implies` binding energy per nucleon increases `implies` Energy is released. (C) For `100 lt A lt 200`, on fission, the mass number of product nuclei will be between `50` & `100` `implies` Binding energy per nucleon decreases `implies` No energy is released (D) For `200 lt A lt 260`, on fission, the mass number of product nuclei will be between `100` & `130` `implies` Binding energy per nucleon increases `implies` Energy is released. |
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900. |
As accident in a nuclear laboratory resulting in deposition of a certain amount of radioactive material of half life `18 `days inside the laboratory Tests revealed that the radiation was `64` times more than the permissible level required for save operation of the laboratory what is the minimum number of days after which the laboratory can be considered safe for use?A. 64B. 90C. 108D. 120 |
Answer» Correct Answer - C | |