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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.
| 101. |
Displacement-time equation of a particle execution SHM is x=A sin(`omegat+pi/6)` Time taken by the particle to go directly from `x = -A/2 to x = + A/2 is `A. `pi/(2omega)`B. `pi/(2omega)`C. `(2pi)/(omega)`D. `(pi)/(omega)` |
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Answer» Correct Answer - A |
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| 102. |
A particle performs SHM in a straight line. In the first second, starting from rest, it travels a distance a and in the next second it travels a distance b in the same direction. The amplitude of the SHM isA. a-bB. `(2a-b)/3`C. `(2a^(2))/(3a-b)`D. None of these |
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Answer» Correct Answer - C |
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| 103. |
An electron in hydrogen atom first jumps from second excited state to first excited state and then from first excited state to ground state. Let the ratio of wavelength, momentum and energy of photons emitted in these two cases be a, b and c respectively, ThenA. `z = 1//x`B. `x = 9//4`C. `y = 5//27`D. `z = 5//27` |
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Answer» Correct Answer - B `E_(1) = (hc)/(lambda) [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` For second excited state to first excited state `E_(1) = (hc)/(lambda) [1/4 - 1/9] rArr (hc)/(lambda) ((5)/(36))` For first excited state to ground excited state `E_(2) = (hc)/(lambda) [1-(1)/(4)] rArr (hc)/(lambda) [(3)/(4)]` (A) `(E_(1))/(E_(2)) = (5)/(27)` , (B) `(E_(1))/(E_(2)) = ((hc)/(lambda_(1)))((lambda_(2))/(hc)) = (lambda_(2))/(lambda_(1)) = 27/5` (C) `P prop 1/(lambda) rArr (P_(1))/(P_(2)) = 5/27` |
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| 104. |
Which of the following is/are correct for nuclear reactor?a)A typical fission is represented by `""_(92)""^(235)U=""_(0)""^(1)nto""_(54)""^(140)Ba+""_(36)""^(93)Kr+` Energyb)Heavy water (`D""_(2)O`) is used as moderator in preference to orduinary water (`H""_(2)O`) because hydrogen may capture neutrons, while D would not do thatc)Cadmium rods increse the reactor power when they go in and decrease when they go outwardsd)Slower neutrons are more effective in causing fission than faster neutrons in the case of `""_(235)U`A. A typical fission represented by `._(92)U^(235)+._(0)n^(1) rarr._(56)Ba^(143)+._(36)Kr^(93)+"energy"`B. Heavy water is used as moderator in preference to ordinary water because H may capture neultrons. While D would notC. Cadmium rod increase the reactor power when they go in, decrease when they go outwardD. Slower neutron are more effective in causing fission than faster neutrons in case of `U^(235)` |
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Answer» Correct Answer - A::B::D A is balanced both in mass number & atomic no. |
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| 105. |
In a horizontal spring-block system force constant of spring is k = 16N/m, mass of the block is 1 kg. Maximum kinetic energy of the block is 8J. ThenA. amplitude of osciallation is 1mB. at half the amplitude, potential energy stored in the spring is 2J.C. at half the amplitude kinetic energy is 6JD. angular frequency of oscillation is 16 rad/s |
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Answer» Correct Answer - A::B::C |
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| 106. |
In case of a simple pendulum, time period versus length is depicted byA. B. C. D. |
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Answer» Correct Answer - B |
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| 107. |
In case of a simple pendulum, time period versus length is depicted byA. B. C. D. |
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Answer» Correct Answer - B (b)`T=2pisqrt((l)/(g))` or `T^(2)prop l` i.e., T-l graph is a parabola |
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| 108. |
A tunnel has been dug through the centre of the earth and a ball is released in it. It will reach the other end of the tunnel afterA. 84.6 minB. 42.3 minC. 1 dayD. will not reach the other end |
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Answer» Correct Answer - B (b)Ball will execute SHM inside the tunnel with time period `T=2pisqrt(R//g)=84.63` min Hence, time do reach the ball from one end to the other end of the tunnel `t=(84.63)/(2)=43.3` min. |
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| 109. |
Passage I) In simple harmonic motion force acting on a particle is given as `F=-4x`, total mechanical energy of the particle is 10 J and amplitude of oscillations is 2m, At time t=0 acceleration of the particle is `-16m/s^(2)`. Mass of the particle is 0.5 kg. At x=+1m, potential energy and kinetic energy of the particle areA. 2 J and 8 JB. 8 J and 2 JC. 6 J and 4 JD. 4 J and 6 J |
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Answer» Correct Answer - D |
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| 110. |
Passage I) In simple harmonic motion force acting on a particle is given as `F=-4x`, total mechanical energy of the particle is 10 J and amplitude of oscillations is 2m, At time t=0 acceleration of the particle is `-16m/s^(2)`. Mass of the particle is 0.5 kg. Displacement time equation equation of the particle isA. `x=2sin2t`B. `x= 2sin4t`C. `x=2cos2t`D. None of these |
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Answer» Correct Answer - D |
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| 111. |
A cart of mass `2.00kg` is attached to the end of a horizontal spring with force constant `k = 150N//m`. The cart is displaced `15.0cm` from its equilibrium position and released. What are (a) the amplitude (b) the period ( c) the mechanical energy (e) the maximum velocity of the cart ? Neglect friction. |
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Answer» Correct Answer - A::B::C::D (a) A = initial displacement from mean position `= 15cm` (b) `T = 2pi sqrt((m)/(k)) = 2pi sqrt ((2.0)/(150)) = 0.726 s` ( c) `F = (1)/(T) = 1.38Hz` (d) `E = (1)/(2)kA^(2)` `= (1)/(2) xx 150 xx (0.15)^(2) = 1.69 J` (e) `v_(max) = omega A` `= ((2pi)/(T)) A = ((2pi)/(0.726))(0.15)` ` = 1.30 m//s` |
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| 112. |
A block of mass m is connected with two ideal pullies and a massless spring of spring constant K as shown in figure. The block is slightly displaced from its equilibrium position. If the time period of oscillation is `mupisqrt(m/K)`. Then find the value of `mu`. |
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Answer» Correct Answer - 8 |
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| 113. |
The composition of two simple harmonic motions of equal periods at right angle to each other and with a phase difference of `pi` result in the displacement of the particle alongA. straight lineB. circleC. ellipseD. figure of eight |
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Answer» Correct Answer - A `x=A_1sinomegat`, `y=A_2sin(omegat+pi)=-A_2sinomegat` `y/x=-A_2/A_1` (straight line) |
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| 114. |
Three arrangements are shown in figure. (a) A spring of mass `m` and stiffness `k` (b) A block of mass `m` attached to massless spring of stiffness `k` ( c) A block of mass `(m)/(2)` attached to a spring of mass `(m)/(2)` and stiffness `k` If `T_(1)`, `T_(2)` and `T_(3)` represent the period of oscillation in the three cases respectively, then identify the correct relation.A. `T_(1) lt T_(2) lt T_(3)`B. `T_(1) lt T_(3) lt T_(2)`C. `T_(1) gt T_(3) gt T_(2)`D. `T_(3) lt T_(1) lt T_(2)` |
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Answer» Correct Answer - B `T = 2pi sqrt ((m + m_(s)//3)/(k))`(in general) Here `m_(s)` = mass of spring and `m` = mass of block. |
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| 115. |
A body of mass `0.10kg` is attached to vertical massless spring with force constant `4.0 xx 10^(3)N//m`. The body is displaced `10.0cm` from its equilibrium position and released. How much time elapses as the body moves from a point `8.0 cm` on one side of the equilibrium position to a point `6.0cm` on the same side of the equilibrium position ? |
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Answer» Correct Answer - A::C::D `omega = sqrt ((k)/(m)) = sqrt ((4 xx 10^(3))/(0.1)) = 200 rad//s` `X = 10 sin (200 t)` `6 = 10 sin 200 t_(1)` `:. 200 t_(1) = sin^(-1)((3)/(5)) = 0.646 rad` `:. t_(1) = 3.23 ms` `8 = 10 sin 200t_(2)` or `200 t_(2) = sin^(-1)((4)/(5)) = 0.925 rad` `:. t_(2) = 4.62 ms` `:. Delta t = 1.4 ms = 1.4 xx 10^(-3)s`. |
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| 116. |
Two identical balls (A) and (B) each of mass (0.1 kg), are attached to two identical massless springs. The spring - mass system is constrained to move inside a riged smooth pipe bant in the form of a circle as shown in Fig. The pipe is fixed in a horizontal plane. The centres of the balls can move in a circle of radius (0.06 pi) meter. Each spring has a natural length `zof0.06 pi` meter and spring spring constant5 `0.1 N//m`. Initially, both the balls are displaced by an angle `theta = pi// 6` radian with respect to the diameter (pQ) of the circle (as shown in Fig.) and released from, rest. . (i) Calculate the frequency of oscillation of ball (B). (ii) Find the speed of ball (A) when (A) and (B) are at the two ends of the diameter (PQ). (iii) What is the total energy of the system. |
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Answer» Correct Answer - A::B::C::D (i) As both the balls displaced by `thetapi//6` radian with espect to the diameter `PQ` of the circle and released from rest. It results into compression of spring in upper segment and an equal elongation of spring in lower segment. Let it x. Pband QA denote x in the figure. Compression =Rtheta= elongation =x` :. Force exerted by each spring on each ball=2 kx` :. Total force on each ball due to two spring `4kx` :. Restoring torque about origin `=O=-(4kx)R` :. tau=-4k (Rtheta) R, where theta =Anfular displacement or `tau=-4kR^(2)theta` Since torque (tau) is proprtional to theta, each ball executes angular SHM about the centreO. Again, `tau=-4kR^(2)theta` or `l alpha=-4kR^(2)theta` where alpha=angular acceleration` or `(mR^(2))alpha=-4kR^(2)theta or alpha=-((4k)/m)theta` :. Frequency `f=1/2_(pi)sqrtalpha/theta` :. `Frequency of each ball=1/2_(pi)sqrt(4k)/m` `=1/2_(pi)sqrt(4xx0. 1)/(0.1)=1/(pi)sec^(-1)` (ii) Let velocity at the mean position be v_(max)` Loss in elastic potential energy=Gain in kinetic energy` `[1/2K(2R(pi)/2)]` `2[1/1xx0.1(0.02pi)^(2)] `=3.95xx10^(-4) J`. |
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| 117. |
A simple pendulum has time period (T_1). The point of suspension is now moved upward according to the relation `y = K t^2, (K = 1 m//s^2)` where (y) is the vertical displacement. The time period now becomes (T_2). The ratio of `(T_1^2)/(T_2^2)` is `(g = 10 m//s^2)`.A. `(5)/(6)`B. `(6)/(5)`C. `1`D. `(4)/(5)` |
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Answer» Correct Answer - B `T_(1) = 2pi sqrt((l)/(g)) = 2pi sqrt((l)/(10))` upward accelertion `(d^(2)y)/(dt^(2)) = 2k = 2 xx 1 = 2 m//s^(2)` `:.` Acceleration `w.r.t.` point of suspension `= 12 m//s^(2)` `T_(2) = 2pisqrt((l)/(12)) :. (T_(1))/(T_(2)) = sqrt((12)/(10)) :. ((T_(1))/(T_(2)))^(2) = (6)/(5)` `T_(1) = 2pi sqrt((l)/(g)) = 2pi sqrt((l)/(10))` |
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| 118. |
A solid uniform cylinder of mass `M` attached to a massless spring of force constant k is placed on a horizontal surface in such a way that cylinder can roll without slipping. If system is released from the stretched position of the spring, then the period will be- A. `2pisqrt((M)/(k)`B. `2pisqrt((3m)/(2k))`C. `2pisqrt((M)/(2k))`D. `2pisqrt((2M)/(3k))` |
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Answer» Correct Answer - B Let small angular displacement of cylinder be `theta` then restoring torque `Ialpha = -k(Rtheta) R` where `I = 3/2 MR^(2)` `rArr (d^(2)theta)/(dt^(2)) + (kR^(2))/((3)/(2)MR^(2)) theta = 0 rArr (d^(2)theta)/(dt^(2)) + (2k)/(3M)theta = 0 rArr omega^(2) = (2k)/(3m)` |
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| 119. |
A cage of mass M hangs from a light spring of force constant `k`. A body of mass m falls from height h inside the cage and stricks to its floor. The amplitude of oscillations of the cage will be- A. `((2mgh)/(k))^(1//2)`B. `((k)/(2mgh))^(1//2)`C. `(mg)/(k)`D. `((mg)/(k))^(1//2)` |
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Answer» Correct Answer - A `mgh = 1/2(M + m) v^(2) = 1/2 kx^(2)` `mgh = 1/2 kx^(2) rArr x = [(2mgh)/(k)]^(1//2)` |
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| 120. |
A ball is suspended by a thread of length l at the point O on an incline wall as shown. The inclination of the wall with the vertical is α.The thread is displaced through a small angle β away from the vertical and the ball is released. Find the period of oscillation of pendulum. Consider both cases a. `alpha gt beta` b. `alpha lt beta`Assuming that any impact between the wall and the ball is elastic.A. `2pisqrt((L)/(g))`B. `2pisqrt((L)/(g))[pi + 2 sin^(-1)(alpha/beta)]`C. `2pisqrt((L)/(g))[pi/2 + sin^(-1)(alpha/beta)]`D. None of the above |
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Answer» Correct Answer - A As `beta lt alpha` so `T = 2pisqrt((L)/(g))` |
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| 121. |
Passage X) A 2kg block hangs without vibrating at the bottom end of a spring with a force constant of 400 N/m. The top end of the spring is attached to the ceiling of an elevator car. The car is rising with an upward acceleration of `5m/`s^(2)` when the accleration suddenly ceases at t=0 and the car moves upward with constant speed. (g=10m/`s^(2)`) What is the angular frequency of oscillation of the block after the acceleration ceases?A. `pi//4 rad`B. `pi//2 rad`C. `pi rad`D. `3pi//2 rad` |
| Answer» Correct Answer - D | |
| 122. |
A particle free to move along the `x-`axis has potential energy given by `U(x) = k[1 - e^(-x^(2))]` for `-oo le x le + oo`, where `k` is a positive constant of appropriate dimensions. Then select the incorrect optionA. at points away from the origin, the particle is in unstable equilibrium.B. for any finite non-zero value of `x`, there is a force directed away from the originC. If its total mechanical enerfy is `k//2`, it has its minimum kinetic energy at the origin.D. for small displacements from `x = 0`, the motion is simple harmonic. |
| Answer» Correct Answer - A::B::C | |
| 123. |
A `2kg` block hangs without vibrating at the bottom end of a spring with a force constant of `400 N//m`. The top end of the spring is attached to the ceiling of an elevator car. The car is rising with an upward acceleration of `5 m//s^(2)` when the acceleration suddenly ceases at time `t = 0` and the car moves upward with constant speed `(g = 10 m//s^(2)` The amplitude of the oscillation isA. `7.5 cm`B. `5 cm`C. `2.5 cm`D. `1 cm` |
| Answer» Correct Answer - C | |
| 124. |
A ball is suspended by a thread of length l at the point O on an incline wall as shown. The inclination of the wall with the vertical is α.The thread is displaced through a small angle β away from the vertical and the ball is released. Find the period of oscillation of pendulum. Consider both cases a. `alpha gt beta` b. `alpha lt beta`Assuming that any impact between the wall and the ball is elastic.A. `2sqrt((t)/(g))[sin^(-1)((alpha)/(beta))]`B. `2sqrt((l)/(g))[(pi)/(2)+sin^(-1)(alpha/(beta))]`C. `2sqrt((l)/(g))[cos^(-1)(alpha/(beta))]`D. `2sqrt((l)/(g))[cos^(-1)(-alpha/(beta))]` |
| Answer» Correct Answer - B::D | |
| 125. |
A ball is suspended by a thread of length l at the point O on an incline wall as shown. The inclination of the wall with the vertical is α.The thread is displaced through a small angle β away from the vertical and the ball is released. Find the period of oscillation of pendulum. Consider both cases a. `alpha gt beta` b. `alpha lt beta`Assuming that any impact between the wall and the ball is elastic.A. `2sqrt((t)/(g))[sin^(-1)((alpha)/(beta))]`B. `2sqrt((l)/(g))[(pi)/(2)+sin^(-1)(alpha/(beta))]`C. `2sqrt((l)/(g))[cos^(-1)(alpha/(beta))]`D. `2sqrt((l)/(g))[cos^(-1)(-alpha/(beta))]` |
| Answer» Correct Answer - B::D | |
| 126. |
A `2kg` block hangs without vibrating at the bottom end of a spring with a force constant of `400 N//m`. The top end of the spring is attached to the ceiling of an elevator car. The car is rising with an upward acceleration of `5 m//s^(2)` when the acceleration suddenly ceases at time `t = 0` and the car moves upward with constant speed `(g = 10 m//s^(2))` What is the angular frequencyof the block after the acceleration ceases ?A. `10sqrt(2) rad//s`B. `20 rad//s`C. `20sqrt(2) rad//s`D. `32 rad//s` |
| Answer» Correct Answer - A | |
| 127. |
A particle is executing SHM according to the equation `x = A cos omega t`. Average speed of the particle during the interval `0 le t le (pi)/(6omega)` isA. `(sqrt(3)Aomega)/(2)`B. `(sqrt(3) A omega)/(4)`C. `(3Aomega)/(pi)`D. `(3 A omega)/(pi)(2 - sqrt(3))` |
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Answer» Correct Answer - D `t = (pi)/(6omega)` `:. omega t = (pi//6)` According to the equation, `X = A cos omega t` `X = A` at `t = 0` and `X = (sqrt (3) A)/(2)` at `omega t = pi//6` `:.` Distance travelled ` = A - (sqrt 3)/(2) A` `= (2 - sqrt3)(A)/(2)` Average speed `= ("distance")/("time")` `= ((2 - sqrt3)A//2)/((pi//6omega)) = (3omega A)/(pi) (2 - sqrt3)`. |
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| 128. |
Three simple harmonic motions in the same direction having the same amplitude and same period are superposed. If each differ in phase from the next by `45^(@)`, thenA. the resultant amplitude is `(1+sqrt(2))`aB. the phase of the resultant motion relative to the first is `tan^(-1)(1/2)`C. the energy associated with the resulting motion is `(3+2sqrt(2))` times the energy associated with any single motion.D. the resulting motion is not simple harmonic |
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Answer» Correct Answer - A::C |
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| 129. |
If a `SHM` is given by `y = (sinomegat + cos omegat)m`, which of the following statement are tureA. The amplitude is `1m`B. The amplitude is `(sqrt(2))m`C. Time is considered from `y = 1 m`D. Time is considered from `y = 0` |
| Answer» Correct Answer - B::C | |
| 130. |
Three simple harmonic motions in the same direction having the same amplitude and same period are superposed. If each differ in phase from the next by `45^(@)`, thenA. the resultant amplitude is `(1 + sqrt(2))a`B. the phase of the resultant motion motion relative to the first is `90^(@)`C. the energy associated with the resulting motion is `(3+2sqrt(2))` time the energy associated with any single motion.D. the resulting motion is not simple harmonic. |
| Answer» Correct Answer - A::C | |
| 131. |
A particle executing linear `SHM`. Its time period is equal to the smallest time interval in which paricle acquires a particular velcity `overset(vec)(v)`, the magnitude of `overset(vec)(v)` may be :A. ZeroB. `V_(max)`C. `(V_(max))/(2)`D. `(V_(max))/(sqrt(2))` |
| Answer» Correct Answer - B | |
| 132. |
Position vector of a particle as a function of time is given by ` vecR= (a sin omegat hati + (a cos omegat) hat j + (b sin omega_(0)t) hat k ` The particle appears to be performing simple harmonic motion along z direction, to an observer moving in xy plane. (a) Describe the path of the observer. (b) Write the distance travelled by the observer himself in the time interval he sees the particle completing one oscillation. |
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Answer» Correct Answer - (a) circle of radius a. , (b) `2 pi a ((omega)/(omega_(0)))` |
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| 133. |
The position of a particle at time moving in `x-y` plane is given by `vec(r) = hat(i) + 2 hat(j) cos omegat`. Then, the motikon of the paricle is :A. open a straight lineB. on an ellipseC. periodicD. SHM |
| Answer» Correct Answer - A::C::D | |
| 134. |
The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of photon with the most enegy ? A. IIIB. IVC. ID. II |
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Answer» Correct Answer - A The enrgy of emitted photon is directly proportional to the difference of the two energy levels. This differences is maximum between level (2) and level (1) hence photon for maximum energy will be librated for this transition only. |
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| 135. |
Find the elastic potential energy stored in each spring shown in figure, when the block is in equilibrium. Also find the time period of vertical oscillation of the block. |
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Answer» Correct Answer - `=2pi sqrt(M(1/k_1+1/k_2+1/k_3))` `PE_1=(m^2g^2)/(2k_1)` `PE_2=(m^2g^2)/(2k_2)` `PE_3=(m^2g^2)/(2k_3)` `k_1, k_2, k_3` are in series `=1/k=1/k_1+1/k_2+1/k_3` `rarr K=(k_1k_2k_3)/(k_1k_2+k_2k_3+k_3k_1)` time period `T=2pisqrt(m/k)` `=2pi sqrt(M(1/k_1+1/k_2+1/k_3))` Now force =weight =mg :. At `k_1` spring `x_1=(mg)/k_1` similarly `x_2=(mg)/k_2` and `x_3=(mg)/k_3` `:. PE_1=1/2k_1x_1^2` `=1/2k_1((Mg)/K_1)^2` `=1/2k_1(m^2g^2)/k_1^2` `=1/2 (m^2g^2)/k_1=(m^2g^2)/(2k_1)` similarly `PE_2=(m^2g^2)/(2k_2)` and `PE_3=(m^2g^2)/(2k_3)` |
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| 136. |
Three spring mass systems are shown in figure. Assuming gravity free spece, find the time period of oscillations in each case. What should be the answer if space is not gravity free ? |
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Answer» Correct Answer - (a) `2pisqrt((m)/(k_(1) + k_(2))), k_(eq.) = k_(1) + k_(2) ;` `2pisqrt((m)/(k_(1) + k_(2)))` `k_(eq.) = k_(1) + k_(2) ;` (c) `2pisqrt((m(k_(1) + k_(2)))/(k_(1)k_(2)))` `k_(eq.) = (k_(1)k_(2))/(k_(1) + k_(2))` (a) When spring is stretched by `x` then restoring force. `F = K_(1) x + K_(2) x` `F = K_(eq) x` `K_(eq) x = K_(1) + K_(2) x` `K_(eq) = K_(1) + K_(2)` `T = 2pisqrt((m)/(K_(eq))) = 2pisqrt((m)/(K_(1) + K_(2)))` (b) When bock is displaced by `x` from mean position then restoring force. `F = K_(1) x + K_(2) x` `K_(eq) x = K_(1) x + K_(2) x` `K_(eq) = K_(1) + K_(2)` `T = 2pisqrt((m)/(K_(eq))) = 2pisqrt((m)/(K_(1) + K_(2)))` (c) When block is displaced `y x` and extension is upper spring is `x_(1)`, extension in lower spring is `x_(2)` then `F = K_(1)x_(1) rArr x_(1) = (F)/(K_(4))` `F = K_(2)x_(2) rArr x_(2) = (F)/(K_(2))` `F = K_(eq)x rArr x = (F)/(K_(eq))` `x = x_(1) + x_(2) rArr (F)/(K_(eq)) = (F)/(K_(1)) + (F)/(K_(2))` `K_(eq) = (K_(1)K_(2))/(K_(1) + K_(2))` `T = 2pisqrt((m)/(K_(eq))) = 2pisqrt((m(K_(1) + K_(2)))/(K_(1)K_(2)))` When space in not gravity free than answer do not changes as time period of spring mass system is independent of gravity. |
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| 137. |
A vertical spring-mass system with lower end of spring is fixed, made to undergo small oscillations. If the spring is stretched by `25cm`, energy stored in the spring is `5J`. Find the mass of the block if it makes `5` oscillations each second. |
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Answer» Correct Answer - `(16)/(10pi^(2)) = 0.16 Kg` `(1)/(2)Kx^(2) = 5` `(1)/(2) k xx ((25)/(100))^(2) = 5` `k = 160` `omega = 10pi = sqrt((K)/(m))` `m = (K)/(100pi^(2))` `m = (160)/(100pi^(2)) = (16)/(10pi^(2)) = 0.16 kg` |
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| 138. |
Three spring mass systems are shown in figure. Assuming gravity free spece, find the time period of oscillations in each case. What should be the answer if space is not gravity free ? |
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Answer» Correct Answer - (a) `2pisqrt((m)/(k_(1) + k_(2))), k_(eq.) = k_(1) + k_(2) ;` `2pisqrt((m)/(k_(1) + k_(2)))` `k_(eq.) = k_(1) + k_(2) ;` (c) `2pisqrt((m(k_(1) + k_(2)))/(k_(1)k_(2)))` `k_(eq.) = (k_(1)k_(2))/(k_(1) + k_(2))` (a) When spring is stretched by `x` then restoring force. `F = K_(1) x + K_(2) x` `F = K_(eq) x` `K_(eq) x = K_(1) + K_(2) x` `K_(eq) = K_(1) + K_(2)` `T = 2pisqrt((m)/(K_(eq))) = 2pisqrt((m)/(K_(1) + K_(2)))` (b) When bock is displaced by `x` from mean position then restoring force. `F = K_(1) x + K_(2) x` `K_(eq) x = K_(1) x + K_(2) x` `K_(eq) = K_(1) + K_(2)` `T = 2pisqrt((m)/(K_(eq))) = 2pisqrt((m)/(K_(1) + K_(2)))` (c) When block is displaced `y x` and extension is upper spring is `x_(1)`, extension in lower spring is `x_(2)` then `F = K_(1)x_(1) rArr x_(1) = (F)/(K_(4))` `F = K_(2)x_(2) rArr x_(2) = (F)/(K_(2))` `F = K_(eq)x rArr x = (F)/(K_(eq))` `x = x_(1) + x_(2) rArr (F)/(K_(eq)) = (F)/(K_(1)) + (F)/(K_(2))` `K_(eq) = (K_(1)K_(2))/(K_(1) + K_(2))` `T = 2pisqrt((m)/(K_(eq))) = 2pisqrt((m(K_(1) + K_(2)))/(K_(1)K_(2)))` When space in not gravity free than answer do not changes as time period of spring mass system is independent of gravity. |
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| 139. |
A block suspended from a spring is released when the spring is unstretched. Then choose the correct options. A. Block starts oscillating simple harmonicallyB. Throughout the motion block is acceleratedC. Maximum acceleration of block is gD. In the upward motion of block, spring is detached from the block at its mean position, then block will rise upto more height from where it was released. |
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Answer» Correct Answer - A::C::D |
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| 140. |
The spring shown in figure is unstretched when a man starts pulling on the cord. The mass of the block is `M`. If the man exerts a constant force `F`, find (a) the amplitude and the time period of the motion of the block, (b) the energy stored in the spring when the block passes through the equilibrium position and (c) the kinetic energy of the block at this position. |
| Answer» Correct Answer - (a) `(F)/(k), 2pisqrt((M)/(k))`, (b) `(F^(2))/(2k)`, (c) `(F^(2))/(2k)` | |
| 141. |
A mass m is suspended by means of two coiled spring which have the same length in unstretched condition as in figure. Their force constant are k 1 and k 2 respectively. When set into vertical vibrations, the period will be A. `2pi sqrt(((m)/(k_(1) k_(2)))`B. `2pi sqrt(m(k_(1) /k_(2))`C. `2pi sqrt(m/(k_(1) -k_(2))`D. `2pi sqrt(m/(k_(1)+ k_(2))` |
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Answer» Correct Answer - D |
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| 142. |
Two springs have force constants `k_(1) " and " k_(2) (k_(1) gt k_(2))`. On which spring is more work done, if (i) they are stretched by the same force and (ii) they are stretched by the same amount ? |
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Answer» (i) Let each spring be stretched through a distance y. Work done on second spring `=(1)/(2) k_(2)y^(2)` Work done on first spring `(1)/(2)k_(1)y^(2)`. But `k_(1) gt k_(2)` (given) `therefore (1)/(2)k_(1)y^(2) gt (1)/(2)k_(2)y^(2)` So, more work is done on the first spring. (ii) Let each spring be stretched by the same force F. Let `y_(1)` and `y_(2)` be the extension in the first and second springs respectively. Then, `y_(1)=(F)/(k_(1))` and `y_(2)=(F)/(k_(2))` Work done on first spring `=(1)/(2)k_(1)y_(1)^(2)=(1)/(2)k_(1)((F)/(k_(1)))^(2)=(1)/(2)(F^(2))/(k_(1))` Similarly, the work done on the second spring `(1)/(2)(F^(2))/(k_(2))` But `k_(1) gt k_(2) " "`(given) `therefore (1)/(2)(F^(2))/(k_(1)) lt (1)/(2)(F^(2))/(k_(2))` So, more work is done on the second spring. |
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| 143. |
The force constants of two springs are`K_(1) and K_(2)`. Both are stretched till their elastic energies are equal. If the stretching forces are ` F_(1) and F_(2) then F_(1) : F_(2)` isA. `K_(1) : K_(2)`B. ` K_(2) : K_(1)`C. `sqrt(k_(1)) : sqrt(K_(2))`D. ` K_(1)^(2) : K_(2)^(2)` |
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Answer» Correct Answer - C |
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| 144. |
A particle moving on x - axis has potential energy `U = 2 - 20x + 5x^(2)` joule along x - axis. The particle is relesed at `x = -3`. The maximum value of `x` will be (`x` is in metre)A. `5m`B. `3 m`C. `7 m`D. `8 m` |
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Answer» Correct Answer - C `F = - (dU)/(dx) = 20 - 10x` `F = 0` at `x = 0` So, from `x = - 3` to `x = 2`, amplitude in `5`. Hence, other extreme position will be `x = 2 + 5 = 7` |
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| 145. |
Two springs have their force constants `K_(1)` and `K_(2)` and they are stretched to the same extension. If `K_(2) gt K_(1)` work done isA. More in spring AB. more in sping BC. Equal in bothD. Noting can be said |
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Answer» Correct Answer - A |
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| 146. |
The spring as shown in figure is kept in a stretched position with extension `x` when the system is relesed. Assuming the horizontal surface to be frictionless, the frequency of oscillation is A. `(1)/(2pi) sqrt([(k(M + m))/(Mm)])`B. `(1)/(2pi) sqrt([(mM)/(k(M + m))])`C. `(1)/(2pi) sqrt([(kM)/(m + M)])`D. `(1)/(2pi) sqrt ([(km)/(M + m)])` |
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Answer» Correct Answer - A `f = (1)/(2pi) sqrt((k)/(mu))` Where, `mu` = reduced mass ` = (M m)/(M + m)`. |
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| 147. |
The displacement of two identical particles executing SHM are represented by equations. `x_(1) = 4sin(10t+pi/6)` and `x_(2)=5cosepsilont` For what value of epsilon of both the particles is same?A. 16 unitB. 6 unitC. 4 unitD. 8 unit |
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Answer» Correct Answer - D |
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| 148. |
A block A of mass m connected with a spring of force constant k is executing SHM.The displacement time equation of the block is x=`x_(0) = a sinomegat`. An indentical block B moving towards negative x -axis with velocity `v_(0)` collides elastically with block A at time t=0. Then A. displacement time equation of A after collision will be `x=x_(0) - v_(0)sqrt(m/k)sinomegat`B. displacement time equation of A after collision will be `x=x_(0)+v_(0)sqrt(m/k)sinomegat`C. Velocity of b just after collision will be a `omega` towards positive x-direction.D. velocity of B just after collision will be `v_(0)` towards positive x-direction. |
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Answer» Correct Answer - A::C::D |
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| 149. |
The total mechanical energy of a particle of mass `m` executing `SHM` with the help of a spring is `E = (1//2)momega^(2)A^(2)`. If the particle is replaced by another particle of mass `m//2` while the amplitude `A` remains same. New mechanical energy will be :A. `sqrt(2)E`B. `2E`C. `E//2`D. `E` |
| Answer» Correct Answer - D | |
| 150. |
A man with a wrist watch on his hands fall from the top of a tower. Does the watch give correct time? |
| Answer» Correct Answer - Yes, the motion in the wristwatch depends on spring action and has nothing to do with acceleration due to gravity. | |