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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.
| 901. |
Monochromatic radiation of wavelength `lambda_(1) = 3000 Å` falls on a photocell operating in saturating mode. The correspondin spectral sensitivity of photocell is `J = 4.8 xx 10^(-3) A//W`. When another monochromatic radiation of wavelength `lambda_(2) = 1650 Å` and power `P = 5 xx 10^(-3) W` is incident, it is found that maximum velocity of photoelectrons increases `n = 2` times. Assuming efficiency of photoelectron generation per incident photon to be same for both the cases, calculate (i) threshold wavelength for the cell. (ii) saturation current in second case. |
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Answer» Correct Answer - (i) `4125Å` (ii) `13.3 muA` Energy of photo with `lambda_(1) = 3000 Å = (hc)/(lambda_(1)) = 4.14 eV` Energy of photo with `lambda_(1) = 1650 Å = (hc)/(lambda_(2)) = 7.53 eV` Power of source `= 5 xx 10^(-3) W` `overset(.)(N)_(2) = (5 xx 10^(-3))/(7.53 xx 10^(-19) xx 1.6) = 4.15 xx 10^(15)` `overset(.)(N)_(1)` for `1W = (1)/(4.14 xx 1.6 xx 10^(-19)) = 1.5 xx 10^(18)` Current `= 4.8 xx 10^(-3) A` `rArr overset(.)(eta) = (4.8 xx 10^(-3))/(1.6 xx 10^(-19)) = 3 xx 10^(16)` `eta = (3 xx 10^(16))/(1.5 xx 10^(16)) = 2 xx 10^(-2) = 2%` `overset(.)(eta)_(2) = 4.15 xx 10^(16) xx 0.02 = 8.3 xx 10^(13)` Current `= overset(.)(n)e = 13.3 muA` Also `v_("max" 1650) = 2V_("max" 5000)` `rArr KE_("max" 1650) = 4KE_("max" 3000)` `rArr 4(4.14 - phi) = (7.53 - phi)` `rArr 163.56 - 7.53 = 3phi` `rArr 16.56 - 7.53 = 3phi` `rArr phi = (9.03)/(3) = 3.01 eV, lambda_(th) = (he)/(E) = 4126 Å` |
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| 902. |
Light coming from a discharge tube filled with hydrogen falls on the cathode fo t he photoelectric cell. The work function of the surface of cathode is `4eV`. Which one of the following values of the anode voltage (in volts) with respect to the cathode will likely to make the photo current zero?A. `-4`B. `-6`C. `-8`D. `-10` |
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Answer» Correct Answer - D Maximum photon energy `= 13.6 eV ("emitted")` So `K_(max) = 13.6 - 4 = 9.6 eV` Hence stopping potential is `- 9.6 V` So `- 10 V` can stop |
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| 903. |
The electron in a hydrogen atom makes a transition `n_(1) rarr n_(2)`, where `n_(1)` and `n_(2)` are the principle quantum numbers of the two states. Assume the Bohr model to be valid. The time period of the electron in the initial state is eight times that in the final state. the possible values of `n_(1)` and `n_(2)` areA. `n_(1) = 4, n_(2) = 2`B. `n_(1) = 8, n_(2) = 2`C. `n_(1) = 8, n_(2) = 1`D. `n_(1) = 6, n_(2) = 3` |
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Answer» Correct Answer - A::D Time period, `T_(n)=(2pi r_(n))/v_(n)` (in `n^(th)` state) i.e. `T_(n) prop r_(n)/v_(n)` But `r_(n) prop n^(2)` and `v_(n) prop 1/n` Therefore, `T_(n) prop n^(3)` given `T_(n1)=8T_(n2)` Hence, `n_(1)=2n_(2)` |
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| 904. |
Light of wavelength `lambda_(ph)`falls on a cathode plate inside a vacuum tube as shown in the figure .The work function of the cathode surface is `phi` and the anode is a wire mesh of conducting material kept at distance d from the cathode. A potential different V is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is `lambda_(e) ` which of the following statement (s) is (are) true? A. For large potential difference `(V gtgt phi//e), lambda_(e)` is approximately halved if V is made four timesB. `lambda_(e)` increases at the same rate as `lambda_(ph)` for `lambda_(ph) lt hc//phi`C. `lambda_(e)` is approximately halved, if d is doubledD. `lambda_(e)` decreases with increase in `phi` and `lambda_(ph)` |
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Answer» Correct Answer - A `K_("max")= (hc)/lambda_(Ph) - phi` kinetic energy of `e^(-)` reaching the anode will be `K=(hc)/lambda_(Ph)- phi+eV` Now `lambda_(e)=(h)/sqrt(2mK)=(h)/sqrt(2m((hc)/lambda_(Ph)-phi+eV))` If `eV gt gt phi` `lambda_(e)=(h)/sqrt(2m((hc)/lambda_(Ph)+eV))` If `V_(f)=4V_(i)` `(lambda_(e))_(f)=((lambda_(e))_(i))/(2)` |
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| 905. |
Let `m_(p)` be the mass of a poton , `M_(1)` the mass of a `_(10)^(20) Ne` nucleus and `M_(2)` the mass of a `_(20)^(40) Ca` nucleus . ThenA. `M_(2) = 2 M_(1)`B. `M_(2) gt 2M_(1)`C. `M_(2) lt 2M_(1)`D. `M_(1) lt 10(m_(n) + m_(p))` |
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Answer» Correct Answer - C::D Due to mass defect (which is finally responsible for the binding energy of the nucleus), mass of a nucleus is always less than the sum of masses of its constituent particles. `._(1.0)^(20)Ne` is made up of `10` protons plus `10` neutrons. Therefore, mass of `._(10)^(20)Ne` nucleus, `M_(1) lt 10 (m_(p)+m_(n))` Also, heavier the nucleus, more is the mass defect. Thus, `20 (m_(n)+m_(p))-M_(2) gt 10 (m_(p)+m_(n))-M_(1)` `implies 10 (m_(p)+m_(n)) gt M_(2)-M_(1) implies M_(2) lt M_(1)+10(m_(p)+m_(n))` Now since `M_(1) lt 10 (m_(p)+m_(n)) :. M_(2) lt 2M_(1)` |
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| 906. |
A particle performing `SHM` takes time equal to `T` (time period of `SHM`) in consecutive appearances at a perticular point. This point is :A. An extreme positionB. The mean positionC. Between positive extreme and mean positionD. Between negative extreme and mean position |
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Answer» Correct Answer - A Position where we see the particle once in a time period that is only extreme position. Twice through every other position |
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| 907. |
For a particle performing `SHM`A. The kinetic energy is never equal to the potential energyB. the kinetic energy is always equal to the potential energyC. The average kinetic in one time period is equal to the average potential in this periodD. The avarage kinetic energy in any time interval is equal to average potential energy in that interval. |
| Answer» Correct Answer - C | |
| 908. |
A point particle if mass `0.1 kg` is executing SHM of amplitude `0.1 m`. When the particle passes through the mean position, its kinetic energy is `8 xx 10^(-3)J`. Write down the equation of motion of this particle when the initial phase of oscillation is `45^(@)`.A. `0.1 cos(4t + (pi)/(4))`B. `0.1 sin(4t + (pi)/(4))`C. `0.4 sin(t + (pi)/(4))`D. `0.2 sin ((pi)/(2) + 2t)` |
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Answer» Correct Answer - B From equilibrium `(1)/(2)momega^(2)A^(2) = 8 xx 10^(-3) rArr (1)/(2) xx 0.1 xx omega^(2) xx (0.1)^(2) = 8 xx 10^(-3) rArr omega = 4` So, equation of `SHM` is `x = A sin(omegat + phi) = 0.1 sin (4t + (pi)/(4))`. |
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| 909. |
The `KE` and `PE` , at is a particle executing `SHM` with amplitude `A` will be equal when its displacement isA. `sqrt(2)A`B. `(A)/(2)`C. `(A)/(sqrt(2))`D. `sqrt((2)/(3))A` |
| Answer» Correct Answer - C | |
| 910. |
A particle executes simple harmonic motion with a frequency. (f). The frequency with which its kinetic energy oscillates is.A. f// 2B. fC. 2 fD. `4 f |
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Answer» Correct Answer - C During one complete oscillation, the kinetic energy will become maximum twice. Therefore the frequency of kinetic energy will be (2 f). |
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| 911. |
A particle executes simple harmonic motion with a frequency. (f). The frequency with which its kinetic energy oscillates is.A. f/2B. fC. 2fD. 4f |
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Answer» Correct Answer - C |
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| 912. |
The potential difference applied to an X-ray tube is 5k V and the current through it is 3.2 mA. Then, the number of electrons striking the target per second is. (a) `2xx10^16` (b) `5xx10^6` (c ) `1xx10^17` (d) `4xx10^15`.A. `2 xx 10^(14)`B. `5 xx 10^(6)`C. `1 xx 10^(12)`D. `4 xx 10^(15)` |
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Answer» Correct Answer - A `i=q/t=(n e)/t implies n= (I t)/e` Substituting `=3.2xx10^(-3) A` `e=1.6xx10^(-19) C` and `t=1 s` We get, `n=2xx10^(16)` |
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| 913. |
In figue `k=100Nm^-1 M=1kg and F=10N`. a. Find the compression of the spring in the equilibrium position. b. A sharp blow by some xternal agent imparts a speed of `2ms^-1` to the block towards left. Find the sum of the potential energy of the spring and the kinetic energy of the block at ths instant. c. find the time period of the resulting simple harmonic motion. d. Find the ampitude. e. Write the potential energy of the spring when the block is at the left extreme. f. Write teh potential energy of the spring when teh block is at the right extreme. The anwer of b, e, and f are different. Explain why this does not violate the principle of conservation of energy. |
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Answer» Correct Answer - A::B::C::D Given `k=100BN/m, m=1kg and F=10 N a. In the equilibrium position `Compression `=delta=F/k=10/100` `=0.1m=10cm` b. The below imparts a speed of 2 m/s to the block towards left. `:.+P.E.+K.E.=1/2kdelta^2+1/2Mv^2` `=1/2x100xx(0.1)^2+1/2x1xx4` `=0.5+2=2.5J` `c. Time period `2pisqrt(M/k)` `=2pi sqrt(M/100)=pi/5 sec` d. Let the amplitude be x whigh means the distasnce betwen the mean position and the extreme position. So in the extreme position, compression of the spring is `(x-K)` Since in SHM the totl energy remains constant, `1/2k(x+delta)^2=1/2kdelta^2+1/2mv^2+Fx` `=2.5+10x [because 1/2kdelta^2+1/2mv^2=2.5]` so, `50(x+0.1)^2=2.5+10x` `:. 50x^2+0.5+10x=2.5+10x` `=:. 50x^2=2` `rarr x^2=2/50=4/100` `rarr x=2/50x=20m` e. potential energy at the left extreme is given by `P.E. =1/2k(x+delta)^2` `=1/2x100xx(0.1+0.2)^2` `=50xx(0.09)=4.5J` f. Potential energy at the right extreme is given by `P.E. 1/2k(x+delta)^2-F(2x)` `[2x`=distance between two extreme)` `=4.5-10(0.4)=0.5J` The different values is b, e, and f do not violate law of conservastion of energy as the work is done by the exterN/Al force 10 N. |
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| 914. |
Two blocks of masses `m` and `2m` rest on a frictionless horizontal surface. They are connected by an ideal spring of relaxed length `l` and stiffness constant `k`. By means of a massless thread connecting the blocks the spring is held compressed to a length `l//2`. the whole system is moving with speed `v` in a direction perpendicular to the length of the spring. the thread is then burnt. Answer the following questions in terms of `l,km,v` The time period of oscillation of the system is :A. `2pisqrt((2m)/(3k))`B. `2pisqrt((3m)/(2k))`C. `2pisqrt((3m)/(k))`D. `pisqrt((2m)/(3k))` |
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Answer» Correct Answer - A |
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| 915. |
A simple pendulum of length L has a bob of mass m. The bob is connected to light horizontal spring of force constant k. The spring is relaxed when the pendulum is vertical (see fig (i)). (a) The bob is pulled slightly and released. Find the time period of small oscillations. Assume that the spring remains horizontal. (b) The spring is replaced with an elastic cord of force constant k. The cord is relaxed when the pendulum is vertical (see fig (ii)). The bob is pulled slightly and released. Find the time period of oscillations. |
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Answer» Correct Answer - `(a) 2 pi ((g)/(l)+(k)/(m))^(-(1)/(2)), (b) pi ][((g)/(l)+(k)/(m))^(-(1)/(2))+((g)/(l))^(-(1)/(2))]` |
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| 916. |
A simple pendulum suspended from the ceiling of a trans has a time period `T` when the train is at rest. If the train is accelerating uniformly at `a` then its time periodA. increaseB. decreaseC. remain unaffectedD. become infinite |
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Answer» Correct Answer - B `T=2pisqrt((L)/((g^2+a^2)^(1//2))` |
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| 917. |
An assembly of identicl spring mass system is placed on a smooth horizontal surface as shown. Initially the springs are relaxed. The left mass is displaced to the left while the right mass is displaced to the right and released. The resulting collision is elastic. The time period of the oscillatins of the system is. A. `2pisqrt((2M)/k))`B. `2pisqrt(M/(2k))`C. `2pisqrt(M/k)`D. `pisqrt(M/k)` |
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Answer» Correct Answer - D |
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| 918. |
(i) In the shown arrangement, both springs are relaxed. The coefficient of friction between `m_(2) " and " m_(1)` is `mu`. There is no friction between `m_(1)` and the horizontal surface. The blocks are displaced slightly and released. They move together without slipping on each other. (a) If the small displacement of blocks is x then find the magnitude of acceleration of `m_(2)`. What is time period of oscillations ? (b) Find the ratio `(m_(1))/(m_(2))` so that the frictional force on `m_(2)` acts in the direction of its displacement from the mean position. (ii) Two small blocks of same mass m are connected to two identical springs as shown in fig. Both springs have stiffness K and they are in their natural length when the blocks are at point O. Both the blocks are pushed so that one of the springs get compressed by a distance a and the other by `a//2`. Both the blocks are released from this position simultaneously. Find the time period of oscillations of the blocks if - (neglect the dimensions of the blocks) (a) Collisions between them are elastic. (b) Collisions between them are perfectly inelastic. |
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Answer» Correct Answer - (a) ` a=((k_(1)+k_(2))/(m_(1)+m_(2)))x ; T=2pisqrt((m_(1)+m_(2))/(k_(1)+k_(2)))` (b) `(m_(1))/(m_(2)) gt (k_(1))/(k_(2)); T=2pisqrt((m)/(k))` " for both the blocks in both cases". |
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| 919. |
A uniform rod of length 1.00 m is suspended through an end and is set into oscillation with small amplitude under gravity. Find the time period of oscillation. |
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Answer» For small amplitude the angular motion is nearly simple harmonic and the time period is given by `T=2pisqrt((1)/(mg(l//2)))=2pisqrt(((ml^(2)//3))/(mg(l//2)))=2pisqrt((2l)/(3g))=pisqrt((2xx1.00m)/(3xx10m//s^(2)))=(2pi)/(sqrt(15))s`. |
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| 920. |
A body executes simple harmonic motion. The potential energy (P.E), the kinetic energy (K.E) and energy (T.E) are measured as a function of displacement `x`. Which of the following staements is true?A. (a) K.E. is maximum when `x=0`.B. (b) T.E is zero when `x=0`C. (c ) K.E is maximum when x is maximumD. (d) P.E is maximum when `x=0` |
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Answer» Correct Answer - A (a) `K.E.=1/21/2 momega^(2)(a^(2)-x^(2)`. Where `x=0`, K.E is maximum and is equal to `1/2momega^(2)a^(2)`. |
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| 921. |
Statement-1 : The periodic time of a hard spring is less as compared to that of a soft spring. Statement-2 : The periodic time depends upon the spring constant, and spring constant is large for hard springA. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false |
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Answer» Correct Answer - A |
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| 922. |
A body executes simple harmonic motion. The potential energy (P.E), the kinetic energy (K.E) and energy (T.E) are measured as a function of displacement `x`. Which of the following staements is true?A. PE is maximum when x=0B. KE is maximum when x=0C. TE is zero when x=0D. KE is maximum when x is maximum |
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Answer» Correct Answer - B KE is maximum at mean position , i.e. at x=0 |
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| 923. |
Assertion : Damped vibrations indicate loss of energy Reason : The loss may be due to friction , air resistance ectA. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false |
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Answer» Correct Answer - B |
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| 924. |
STATEMENT-1 : Frequency of spring mass system in two cases shown in the figure is same. STATEMENT-2 : Frequency of a spring mass system `f=(1)/(2pi)sqrt((K)/(M))`, which does not depend upon acceleration due to gravity. A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - A | |
| 925. |
A fusion reaction of the type given below `._(1)^(2)D+._(1)^(2)D rarr ._(1)^(3)T+._(1)^(1)p+DeltaE` is most promissing for the production of power. Here D and T stand for deuterium and tritium, respectively. Calculate the mass of deuterium required per day for a power output of `10^(9) W`. Assume the efficiency of the process to be `50%`. Given : `" "m(._(1)^(2)D)=2.01458 am u," "m(._(1)^(3)T)=3.01605 am u` `m(._(1)^(1) p)=1.00728 am u` and `1 am u=930 MeV`. |
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Answer» Correct Answer - (a) `4MeV, 17.6 MeV` (b) `7.2 MeV` (c) `0.384 %` `2(._(1)^(2)D) rarr ._(1)^(3)T+._(1)^(1)P` Mass defect `DeltaM=M_("Product")-M_("Reactant")` `={(3.016049)+1.00785}-2[2.014102]` `=4.023899-4.028204` `implies Deltam=4.3xx10^(-3)` amu and `1` amu `rarr 931.5` `E=Deltamc^(2)=4.01 MeV` `.^(3)T_(1)+._(1)^(2)D rarr ._(2)^(4)He+._(0)^(1)n` `Deltam`(mass defect)`=DeltaM_("Product")-DeltaM_("Reactant")` `=[4.002603+1.008665]-[3.016049+2.014102]` `=[5.011268-5.030151]` `Deltam=0.018883` `E=Deltam(931.5) implies 17.58 MeV` `E_("deutron")=(DeltaE_("total"))/3=7.2 MeV` `("Total Energy")/("Total Mass")implies n=(Deltam_(1)+Deltam_(2))/(3_(1)^(2) D)` `implies n=((0.004305+0.018883)/(3(2.014102)))xx100=n=0.384%` |
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| 926. |
A small bottle contains powered beryllium Be & gaseous radon which is used as a source of `alpha`-particles. Netrons are produced when `alpha-`particles of the radon react with beryllium. The yield of this reaction is `(1//4000)` i.e. only one `alpha`-particle out of induced the reaction. Find the amount of radon `(Rn^(222))` originally introduced into the source, if it prouduces `1.2 xx 10^(6)` neutrons per second after `7.6` days. [`T_(1//2)` of `R_(n) = 3.8` days] |
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Answer» Correct Answer - `3.3 xx 10^(-6) g` Let the amount of `R_(x)^(222)` be `N_(0), N_(7.6)=N_(0)/2^(2)=N_(0)/4` Nuclei left after `7.6` day `=N_(0)/4` Decay constant `=(ln 2)/t_(1//2)=0.693/(3.8xx24xx3600)//sec` `lambda=2.11xx10^(-6) sec` Activity `=N_(0)/4 lambda=` No. of `alpha` particles No. of neitrons produced `=(N_(0)xx2.11xx10^(-6))/(4xx4000)=1.2 xx10^(6)` `N_(0)=9.095xx10^(15)` nuclei `=3.354xx10^(-4) g` |
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| 927. |
A hollow cylindrical shell of radius R has mass M. It is completely filled with ice having mass m. It is placed on a horizontal floor connected to a spring (force constant k) as shown. When it is disturbed it performs oscillations without slipping on the floor. (a) Find time period of oscillation assuming that the ice is tightly pressed against the inner surface of the cylinder. (b) If the ice melts into non viscous water, find the time period of oscillations. (Neglect any volume change due to melting of ice) |
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Answer» Correct Answer - (a) 2pi sqrt((4M+3m)/(2K));` (b) ` 2pi sqrt((2M +m)/(K))` |
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| 928. |
The total energy of a particle, executing simple harmonic motion is. where x is the displacement from the mean position, hence total energy is independent of x.A. `prop X`B. `prop X^(2)`C. independent of `X`D. `prop X^(1//2)` |
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Answer» Correct Answer - C |
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| 929. |
The total energy of a particle, executing simple harmonic motion is. where x is the displacement from the mean position, hence total energy is independent of x.A. (a) independent of x.B. (b) `propx^(2)`C. (c ) prop x`D. (d) propx^(1//2)` |
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Answer» Correct Answer - A (a) At any instant the total energy is `1/2kA^(2)=constant, where `A=amplitude hence total energy is independent ofx. |
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| 930. |
The total energy of a particle, executing simple harmonic motion is. where x is the displacement from the mean position, hence total energy is independent of x.A. `propx`B. `propx^(2)`C. lndependent of xD. `propx^(1//2)` |
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Answer» Correct Answer - C |
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| 931. |
A body is executing simple harmonic motion. At a displacement x (from its mean position) its potential energy is `E_(1)` and at a displacement y its potential energy is `E_(2)`. The potential energy is E at displacement (x+y) . Then:A. `sqrt(E ) = sqrt(E_(1))-sqrt(E_(2))`B. `sqrt(E ) = sqrt(E_(1))+sqrt(E_(2))`C. `E = E_(1)-E_(2)`D. `E=E_(1)+E_(2)` |
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Answer» Correct Answer - B |
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| 932. |
A body is executing simple harmonic motion. At a displacement `x` its potential energy is `E_(1)` and at a displacement `y` its potential energy is `E_(2)` The potential energy `E` at displacement `(x+y)` isA. `sqrtE=sqrtE_(1)- sqrtE_(2)`B. `sqrtE=sqrtE_(1)+ sqrtE_(2)`C. `E=E_(1)+ E_(2)`D. `E=E_(1)- E_(2)` |
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Answer» Correct Answer - B |
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| 933. |
A point describes simle harmonic motion in a line 4 cm long. The velocity of the point while passing through the cente of the line is 12cm per second. Find the period. [Hint: use formula v = `omegasqrt(a^(2)-a^(1))`] |
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Answer» Correct Answer - 1.047s |
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| 934. |
A diatomic molecule has atoms of masses `m_(1)` and `m_(2)`. The potential energy of the molecule for the interatomic separation `r` is given by `V(r )=-A+B(r-r_(0))^(2)`, where `r_(0)` is the equilibrium separation, and `A` and `B` are positive constants. the atoms are compressed towards each other from their equilibrium positons and released. what is the vibrational frequency of the molecule? |
| Answer» Correct Answer - A::B | |
| 935. |
The displacement - time graph of a particle executing SHM is shown in figure. Which of the following statements is//are true ? A. The velocity is maximum at `t = T//2`B. The acceleration is maximum at `t = T`C. The force is zero at `t = 3T//4`D. The kinetic energy equals the total oscillation energy at `t = T//2` |
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Answer» Correct Answer - B::C `v or KE = 0` at `y = +- A` `v` or `KE =` maximum at `y = +- 0` `F` or a is maximum at `y = +- A` `F` or a is zero at `y = 0` |
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| 936. |
In previous problem, if the body is suspended from the two springs connected in parallel, the time period will beA. `sqrt(T_1T_2)`B. `(T_1T_2)/(sqrt(T_1^2+T_2^2))`C. `sqrt((T_1^2T_2^2)/(2))`D. `(2T_1T_2)/(sqrt(T_1^2+T_2^2))` |
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Answer» Correct Answer - B `k_(eq)=k_1+k_2` `(1)/(T^2)=(1)/(T_1^2)+(1)/(T_2^2)=(T_2^2+T_1^2)/(T_1^2T_2^2)` `T=sqrt((T_1^2T_2^2)/(T_1^2+T_2^2))=(T_1T_2)/(sqrt(T_1^2T_2^2)` |
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| 937. |
The maximum velocity a particle, executing simple harmonic motion with an amplitude 7 mm, 4.4 m//s. The period of oscillation is.A. `0.01s`B. `10 s`C. `0.1 s`D. `100 s` |
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Answer» Correct Answer - A |
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| 938. |
The maximum velocity a particle, executing simple harmonic motion with an amplitude 7 mm, 4.4 m//s. The period of oscillation is.A. (a) `0.01 s`B. (b) `10 s`C. (c ) `0.1s`D. (d) `100s` |
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Answer» Correct Answer - A (a) Maximum velocity, `v_(max)=aomega, v_(max)=axx2_(pi)/T` rArr `T=(2pia)/v_(max)=(2xx3.14xx7xx10^(-3))/(4.4)~~0.01 s`, |
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| 939. |
The maximum velocity a particle, executing simple harmonic motion with an amplitude 7 mm, 4.4 m//s. The period of oscillation is.A. `100 s`B. `0.01 s`C. `10 s`D. `0.1 s` |
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Answer» Correct Answer - B `Aomega = v_(max)` `T = (2pi)/(omega) = (2piA)/(v_(max)) = 0.01` sec. |
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| 940. |
Two particles (A) and (B) of equal masses are suspended from two massless spring of spring of spring constant `k_(1)` and `k_(2)`, respectively, the ratio of amplitude of (A) and (B) is.A. `sqrt(k_(1)//K_(2))`B. `K_(1)//k_(2)`C. `sqrt(k_(2)//K_(1))`D. `k_(1)//k_(2)` |
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Answer» Correct Answer - C |
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| 941. |
Two masses 8 kg 4 kg are suspended together by a massless spring of spring constant `1000 Nm^(-1)` . When the masses are in equilibrium 8 kg is removed without disturbing the system . The amplitude of oscillation isA. 0.5 mB. 0.08 mC. 0.4 mD. 0.04 m |
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Answer» Correct Answer - B (b) `x_(12)=(12xx10)/(1000)=0.12 m" "(x=(mg)/(k))` `x_(4)=(4xx10)/(1000)=0.04 m` `therefore A=x_(12)-x_(4)=0.08 m` |
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| 942. |
A harmonic oscillation of force constant `4xx10^(6) Nm^(-1)` and amplitude 0.01 m has total energy 240 J. What is maximum kinetic energy and minimum potential energy? |
| Answer» `k=4 xx 10^(6)N//m , A=0.01m`, Total energy =240 J minimum potential energy =total energy -maximum kinetic energy=40J | |
| 943. |
Two particles (A) and (B) of equal masses are suspended from two massless spring of spring of spring constant `k_(1)` and `k_(2)`, respectively, the ratio of amplitude of (A) and (B) is.A. (a) `sqrtk_(1_/k_(2_`B. (b) `k_(2)/k_(1)`C. (c ) `sqrtk_(2_/k_(1)`D. (d) k_(1)/k_(2)` |
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Answer» Correct Answer - C (c ) Maximum velocity is SHM=Aomega=Asqrtk/m` `[:. Omega=sqrtk/m` Here the maximum velocity is same and m is also same :. `A_(1) sqrtk_(1)=A_(2)sqrtk_(2)` `A_(1)/A_(1)=sqrtk_(2)/k_(1)`. |
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| 944. |
The potential energy of a particle executing S.H.M. is 2.5 J, when its displacement is half of amplitude. The total energy of the particle will be |
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Answer» PE`=(1)/(2)kx^(2)implies(1)/(2)k(A^(2))/(4)=2.5 " "(because x=(A)/(2))` where , x is the displacement of the particle and A is the amplitude. `implies` Total energy `=(1)/(2)kA^(2)=2.5xx4=10 J` |
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| 945. |
In a simple harmonic oscillator, at the mean positionA. (a) kinetic energy is minimum, potential energy is maximum.B. (b) both kinetic and potential energies are maximum.C. (c ) kinetic energy is maximum, potential energy is minimum.D. (d) both kinetic and potenteal energies are minimum. |
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Answer» Correct Answer - C (c ) `K.E=1/2k(A^(2)-x^(2)), `U=1/2kx^(2)` At the mean position `x=0` :. `K.E.=1/2kA^(2)=Maximum and`U=0`. |
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| 946. |
In a simple harmonic oscillator, at the mean positionA. Kinetic energy is minimum, potential energy is maximumB. Both kenetic and potential energies are maximumC. kinetic energy is maxium, potentail energy minimumD. Both kenetic and potential energies are minimum |
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Answer» Correct Answer - C |
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| 947. |
A samll mass executes linear `SHM` about `O` with amplitude a and period `T`. Its displacement from `O` at time `T//8` after passing through `O` is:A. `a//8`B. `a//2sqrt2`C. `a//2`D. `a//sqrt2` |
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Answer» Correct Answer - D |
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| 948. |
The total energy of a particle executing S.H.M. is 80 J . What is the potential energy when the particle is at a distance of 3/4 of amplitude from the mean positionA. 60 jB. 10 jC. 40 jD. 45 j |
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Answer» Correct Answer - D |
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| 949. |
In a simple harmonic oscillator, at the mean positionA. Kinetic energy is minimum, potential energy is maximumB. Both kinetic and potential energies are maximumC. Kinetic energy is maximum, potential energy is minimumD. Kinetic energy is maximum, potential energy is minimum |
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Answer» Correct Answer - C |
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| 950. |
A block has a L shaped stand fixed to it. Mass of the block with the stand is M. At the free end of the stand there is a spring which carries a ball of mass m. With the spring in its natural length, the ball is released. It begins to oscillate and the stand is tall enough so that the ball does not hit the block. (a) Find maximum value of mass (m) of the ball for which the block will not lose contact with the ground? (b) If the stand is not tall enough and the ball makes elastic impact with the block, will your answer to part (a) change? |
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Answer» Correct Answer - For both (a) and (b) the block will not lose contact with the ground for any value of m. |
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