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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.
| 951. |
A mass M is performing linear simple harmonic motion. Then correct graph for acceleration a and corresponding linear velocity v isA. B. C. D. |
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Answer» Correct Answer - B `v = A omegacos omegat, a = -omega^(2)A sin omegat` `rArr ((v)/(Aomega))^(2) +((a)/(omega^(2) A))^(2) = 1` `rArr` Straight line in `v^(2)` and `a^(2)` |
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| 952. |
A uniform cylinder of length (L) and mass (M) having cross sectional area (A) is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half - submerged in a liquid of density (rho) at equilibrium position. When the cylinder is given a small downward push and released it starts oscillating vertically with small amplitude. If the force constant of the spring is (k), the prequency of oscillation of the cylindcer is.A. `(1)/(2 pi) ((k - A rho g)/(M))^(1//2)`B. `(1)/( 2 pi) ((k + A rho g)/(M))^(1//2)`C. `(1)/(2 pi)((k + rho gL)/(M))^(1//2)`D. `(1)/(2 pi) ((k + A rho g)/(A rho g))^(1//2) |
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Answer» Correct Answer - B If (x) is the displacement then, :. `M omega^2 x= [rho A g + k]x` `rArr omega = [(rho A g + k)/(m)]^(1//2) rArr v = (1)/(2 pi)[(rho A g + k)/(M)]^(1//2)`. |
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| 953. |
A particle of mass (m) is executing oscillations about the origin on the (x) axis. Its potential energy is `V(x) = k|x|^3` where (k) is a positive constant. If the amplitude of oscillation is a, then its time period (T) is.A. proportional to `1//sqrt(a)`B. independent of (a)C. proportional to `sqrt(a)`D. proportional to `a^(3//2)`. |
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Answer» Correct Answer - A `U (x) = k|x|^3` :. `[k] = ([U])/([x^3]) = (M L^2 T^-2)/(L^3) = ML ^-1 T^-2` now time period may depend on ` T prop (mass)^x (amplitude)^y (k)^z` :. `[M^0 L^0T] = [M]^x [L]^y [M L^-1 T^-2]^z` =`[M^x + z L^y -z T^(-2z)]` Equating the powers, we get `-2z = 1 or -1//2` `y - z = 0 or y = z = - 1//2` Hence `T prop (amplitude)^(-1//2) prop a^(1//2)`. |
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| 954. |
A highly rigid cubical block `A` of small mass `M` and side `L` is fixed rigidly on the other cubical block of same dimensions and of modulus of rigidity `eta` such that the lower face of `A` completely covers the upper face of `B`. The lower face of `B` is rigidly held on a horizontal surface . `A` small force `F` is applied perpendicular to one of the side faces of `A`. After the force is withdrawn , block `A` executes faces of `A`. After the force is withdrawn , block `A` exceutes small oscillations , the time period of which is given byA. `sqrt(eta mL)`B. `2pisqrt(m(eta)/(L))`C. `2pisqrt(((mL)/(eta)))`D. `2pisqrt(((m)/(etaL)))` |
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Answer» Correct Answer - D Stress `=(F)/(L^(2)),eta=(F)/(L^(2)phi)` `F=L^(2)eta phi` `F=L^(2)eta((-x)/(L))" "[because "Strain"=phi=-(x)/(L)]` `F=(L eta)(-x)` `rArr" "ma=(Leta)(-x)` `a=((Leta)/(m))(-x)` `therefore" "a prop-omega^(2)x` `(Leta)/(m)(-x)=-omega^(2)x` or `omega=sqrt(((Leta)/(m)))thereforeT=2pisqrt(((m)/(etaL)))` |
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| 955. |
A highly rigid cubical block (A) of small mass (M) and slide (L) is fixed rigidly on to another cubical block (B) of the same dimesions and of low modulus of rigidity (eta) such that the lower face of (A) completely covers the upper face of (B). The lower face of (B) is rigidly held on a horizontal surface. (A) small force (F) is applied perpendicular to one of the sides faces of (A). After the force is withdrawn, block (A) executes small oscillations the time period of which is given by.A. `(2 pi) sqrt(M eta L)`B. (2 pi) sqrt (M eta)/(L)`C. `(2 pi) sqrt (M L)/(eta)`D. `(2 pi)sqrt (M)/(eta L)` |
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Answer» Correct Answer - D When a force is applied on cubical block (A) in the horizontal direction then the lower block (B) will get distorted as shown by dotted lines and (A) will attain a new position (without distrotion as (A) is a rigid body) as shown by dotted lines. `eta = (F//A)/(Delta L//L) = (F)/(A) xx (L)/(Delta L) = (F)/(L^2) xx (L)/(Delta L) = (F) /(L xx Delta L)` rArr `F = eta L Delta L` `eta L` is a constant rArr Force `F prop Delta L` and directed towards the mean position, oscillation will be simple harmonic in nature. Here, `M omega^2 = eta L` `rArr omega = sqrt (eta L)/(M) rArr (2 pi)/(T) = sqrt(eta L)/(M) rArr T = 2 pi sqrt (M)/(eta L)`. |
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| 956. |
Two rigid bodies A and B of masses 1 kg and 2 kg respectively are rigidly connected to a spring of force constant 400 `Nm^(-1)` . The body B rests on a horizontal table. From the rest position , the body A is compressed by 2 cm and then released . Deduce (i) teh frequency of oscillation, (ii) total oscillation energy, (iii) the amplitude of the harmonic vibration of the reaction of the table on body B. |
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Answer» Since it is body A which is oscillation , m=1 kg (i) Frequency of osillation , `v=(1)/(2pi)sqrt((k)/(m))` or `v=(1)/(2pi)sqrt((400)/(1)) Hz=(10)/(pi)Hz=3.182 Hz` (ii) Total oscillation energy, `E=(1)/(2)mA^(2)omega^(2)` or `E=(1)/(2)mA^(2)((k)/(m))=(1)/(2)kA^(2)` `=(1)/(2)xx400xx(0.02)^(2) J=0.08 J (because a=2 cm=0.02 m)` Total force acting on the table `=(1+2 ) kg wt=3 kg wt=3xx9.8 N=29.4 N` This is the mean upward reaction. Now, due to oscillation, the maximum tension developed in the spring is given by `F=kA=400xx 0.02 N=8 N` Thus , the amplitude of the vibration of reaction of the table will vary from (29.4+8) N to (29.4)N. |
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| 957. |
A mass (M), attached to a horizontal spring, executes S.H.M. whith amplitude `A_(1). When the mass (M) passes throgh its mean position then shaller mass (m) is placed over it and both of them move together with amplitude `A_(2). The ration of.A. (a) `(M+m)/M`B. (b) (M/(M+m)^(1/2)`C. (c ) ((M+m)/M)^1/2)`D. (d) M/(M+m)` |
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Answer» Correct Answer - C (c ) The net force becomes zero at the mean point. Therefore, linear momentum must be conserved. :. Mv_(1)=(M+m)v_(2)` `Masqrtk/M=(M+m)A_(2)sqrtk/(m+M):. (V=Asqrtk/M)` `A_(1)sqrtM=A_(2)sqrtM+m :. A_(1)/A_(2)=sqrt(m+M)/M`. |
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| 958. |
Passage XIII) Mr. Anoop having mass 50 ks is standing on a massless platform which oscillates up and down (doing SHM) of frequency 0.5 `sec^(-1)` and amplitude 0.4m. Platforms has a weightng machine fitted in it (Which is also massless). On which Anoop is standing. (Take `(pi^(2))` = 10 and g = 1- `m//s^(2)`. Assume that at t=0 (Platform+Anoop) is at their highest point of oscillation. [More than one option is correct] Which of the following is correct for compression of the spring.A. When reading is 55 kg, then compression will be 1.1mB. When reading is 50 kg, then compression will be 1.0mC. When reading is 50kg, then compression will be 1.1mD. When reading is 55kg, then composition will be 1.0m |
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Answer» Correct Answer - A::B |
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| 959. |
Passage XIII) Mr. Anoop having mass 50 ks is standing on a massless platform which oscillates up and down (doing SHM) of frequency 0.5 `sec^(-1)` and amplitude 0.4m. Platforms has a weightng machine fitted in it (Which is also massless). On which Anoop is standing. (Take `(pi^(2))` = 10 and g = 1- `m//s^(2)`. Assume that at t=0 (Platform+Anoop) is at their highest point of oscillation. [More than one option is correct] Which of the following will be correct for reading of weighting machineA. Maximum reading of weighing machine will be 70 kgB. maximum reading will be at mean positionC. maximum reading will be at lower extremem positionD. Maximum reading will be at upper extreme position. |
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Answer» Correct Answer - A::C |
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| 960. |
A particle starts oscillating simple harmoniclaly from its equilibrium postion. Then the ratio fo kinetic and potential energy of the principle at time `T/12` is (`U_(mean)`=0, T= Time period)A. `2:1`B. `3:1`C. `4:1`D. `1:4` |
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Answer» Correct Answer - B |
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| 961. |
A particle moves such that its acceleration is given by `a = -Beta(x-2)` Here, `Beta` is a positive constnt and x the x-coordinate with respect to the origin. Time period of oscillation isA. `(2pi)sqrt(beta)`B. `2pisqrt(1/beta)`C. `2pisqrt(beta+2)`D. `2pisqrt(1/(beta+2))` |
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Answer» Correct Answer - B |
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| 962. |
Assertion : The graph of total energy of a particle in `SHM` w.r.t. position is a line with zero slope Reason : Total energy of particle in `SHM` remain constant throughout its motionA. 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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| 963. |
A particle moves along X axis such that its acceleration is given by `a = -beta(x -2)`,where `beta` is a positive constant and x is the position co-ordinate. (a) Is the motion simple harmonic? (b) Calculate the time period of oscillations. (c) How far is the origin of co-ordinate system from the equilibrium position? |
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Answer» Correct Answer - (a) Yes , (b) `T=2pi sqrt((1)/(beta)), (c ) 2 unit` |
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| 964. |
STATEMENT-1 : The graph of total energy of a particle in `S.H.M` w.r.t. position is a straight line with zero slope. STATEMENT-2 : Total graph of total energy of a particle in `S.H.M.` remains constant throughout its motion.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. |
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Answer» Correct Answer - A |
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| 965. |
A particle is executing simple harmonic motion with frequency f . The frequency at which its kinetic energy changes into potential energy isA. f/2B. fC. 2 fD. 4 f |
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Answer» Correct Answer - C |
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| 966. |
A test tube of length l and area of cross-section A has some iron fillings of mass M. The test tube floats normally in a liquid of density `rho` with length x dipped in the liquid. A distribuing force makes the tube oscillate in the liquid. The time period of osciallation is given by (neglect the mass of the test tube)A. `(2pisqrt((Mrho)/(Ag))`B. `2pisqrt(x/g)`C. `2pisqrt(l/g)`D. `2pisqrt(M/g)` |
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Answer» Correct Answer - B |
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| 967. |
A particle is performing simple harmonic motion along the x axis about the origin. The amplitude of oscillation is a. A large number of photographs of the particle are shot at regular intervals of time with a high speed camera. It was found that photographs having the particle at `x_(1) + Deltax` were maximum in number and photographs having the particle at `x_(2) + Deltax` were least in number. What are values of `x_(1) " and " x_(2)`? |
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Answer» Correct Answer - `x_(1)=+-a; x_(2)=0` |
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| 968. |
A particle is performing simple harmonic motion with amplitude A and angular velocity `omega` . The ratio of maximum velocity to maximum acceleration isA. `omega`B. `1//omega`C. `omega`D. `Aomeaga` |
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Answer» Correct Answer - B |
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| 969. |
STATEMENT-1 : A particle is performing simple harmonic motion along `X` axis with amplitude `A`. If a graph is plotted between velocity of particle and its `X` coordinate then graph will be elliptical. STATEMENT-2 : Velocity of particle in `SHM` as a fuction of position is given by `v=omegasqrt(A^(2)-X^(2)).`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. |
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Answer» Correct Answer - A |
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| 970. |
A particle is performing a linear simple harmonic motion. If the instantaneous acceleration and velocity of the particle are `a` and `v` respectively, identify the graph which correctly represents the relation between `a` and `v`.A. B. C. D. |
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Answer» Correct Answer - C If `v = v_(0)sin omega t`, then `a = (dv)/(dt) = v_(0) omega cos omega t` `sin omega t = (v)/(v_(0))` …(i) `cos omega t = (a)/(v_(0)omega)` …(ii) Squaring and adding these equations, we get `1 = (v_(2))/(v_(0)^(2) + (a^(2))/(v_(0)omega)` `:. v^(2) = (-(v_(0))/(omega)) a^(2) + v_(0)^(2)` Hence, `v^(2)` versus `a^(2)` equation is a straight line with positive intercept and negative slope. |
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| 971. |
A mass M is performing linear simple harmonic motion. Then correct graph for acceleration a and corresponding linear velocity v isA. B. C. D. |
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Answer» Correct Answer - B |
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| 972. |
Derive the expression for equivalent spring constant of `(i)` parallel combination of `n-`springs `(ii)` series combination of `n-`springs |
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Answer» Correct Answer - (a) (i) Let extension in each spring is `x` then `F_(1) = K_(1)x_(1)` `F_(2) = K_(2)x_(1), F_(3) = K_(3)x = …… = K_(n)x` `F_(1) + F_(2) + F_(3) + ….. + F_(n) = F` `rArr K_(1)x + K_(2)x + K_(3)x + ...... + K_(n)x = K_(eq)x` `rArr K_(eq) = K_(1) + K_(2) + K_(3) + .... + K_(n)` (ii) Let `F` be the force in each spring and `x_(1), x_(2), x_(3)......x_(n)` are the extensions in different springs then `F = K_(1)x_(1) = K_(2)x_(2) = K_(3)x_(3) = ....... = K_(n)x_(n)` `x_(1) + x_(2) + x_(3) .... + x_(n) = x` `rArr (F)/(K_(1)) + (F)/(K_(1)) + (F)/(K_(3)) + ....... + (F)/(K_(n)) = (F)/(K_(eq))` `rArr (F)/(K_(eq)) = (F)/(K_(1)) + (F)/(K_(2)) + (F)/(K_(3)) +.........+ (F)/(K_(n))` |
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| 973. |
If `k_(s)` and `k_(p)` respectively are effective spring constant in series and parallel combination of springs as shown in figure, find `(k_(s))/(k_(p))`. A. `(9)/(2)`B. `(3)/(7)`C. `(2)/(9)`D. `(7)/(3)` |
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Answer» Correct Answer - C The effective spring constant `k_(s)` of this arrangement is `(1)/(k_(s))=(1)/(k)+(1)/(2k)rArr(1)/(2k_(s))=(2+1)/(2k)=(3)/(2k)` `." "k_(s)=(2k)/(3)` The effective spring costant `k_(p)` of this arrangement is `k_(p)=k_(1)+k_(2)=k+2k=3k` `therefore" "(k_(s))/(k_(p))=(2k//3)/(3k)=(2)/(9)` |
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| 974. |
As shown in the figure, two light springs of force constant `k_(1)` and `k_(2)` oscillate a block of mass m. Its effective force constant will be A. `k_(1)k_(2)`B. `k_(1)+k_(2)`C. `(1)/(k_(1)) +(1)/(k_(2))`D. `(k_(1)k_(2))/(k_(1)+k_(2))` |
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Answer» Correct Answer - D (d) The springs are connected in series. So, the effective force constant will be `(1)/(k_(e))=(1)/(k_(1))+(1)/(k_(2))impliesk_(e)=(k_(1)k_(2))/(k_(1)+k_(2))` |
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| 975. |
The spring constants of two springs of same length are `k_(1)` and `k_(2)` as shown in figure. If an object of mass m is suspended and set in vibration , the period will be A. `2pisqrt((mk_(1))/(k_(2)))`B. `2pisqrt((m)/(k_(1)k_(2)))`C. `2pisqrt((m)/(k_(1)-k_(2)))`D. `2pisqrt(m//(k_(1)+k_(2)))` |
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Answer» Correct Answer - D (d) The springs are corrected in parallel. So, the efffective force constant will be `k_(e)=k_(1)+k_(2)` `therefore T=2pisqrt((m)/(k_(e)))=2pisqrt((m)/(k_(1)+k_(2))` |
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| 976. |
Two objects A and B of equal mass are suspended from two springs constants `k_(A)` and `k_(B)` if the objects oscillate vertically in such a manner that their maximum kinetic energies are equal, then the ratio of their amplitudes isA. `(K_(B))/(K_(A))`B. `sqrt((K_(B))/(K_(A)))`C. `(K_(A))/(K_(B))`D. `sqrt((K_(A))/(K_(B)))` |
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Answer» Correct Answer - B Kinetic energy of A =Kinetic energy of B `1/2 K_(A)a^(2)=1/2 K_(B)b^(2)` So ratio of amplitude `=sqrt((K_(B))/(K_(A)))` |
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| 977. |
Two linear SHM of equal amplitudes `A` and frequencies `omega` and `2omega` are impressed on a particle along `x` and `y - axes` respectively. If the initial phase difference between them is `pi//2`. Find the resultant path followed by the particle. |
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Answer» Correct Answer - A::B `x = Asin omega t` …(i) `y = Asin(2omega t + pi//2)` `= Acos 2 omega t` `= A(1 - 2 sin ^(2)omega t)` From Eq. (i), `sin omega t = (x)/(A)` `:. Y = A (1 - (2x^(2))/(A))`. |
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| 978. |
The initial position and velocity of a body moving in SHM with period `T = 0.25s` are `x = 5.0cm` and `v = 218cm//s`. What are the amplitude and phase constant of the motion ? |
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Answer» Correct Answer - A::C::D `omega = (2pi)/(T) = (2pi)/(0.25) = (8pi)rad//s` `x = A sin(omega t + phi)` `U = (dx)/(dt) = omega A cos (omega t + phi)` At `t = 0` `x = Asin phi` `:. 5 = A sin phi` ...(i) `U = omega A cos phi = (8pi) A cos phi` `:. 218 = (8pi A)cos phi` ...(ii) Solving Eqs. (i) and (ii) we get, `A = 10 cm` and `phi = (pi)/(6)` or `30^(@)` |
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| 979. |
Two bolcks `A (5kg)` and `B(2kg)` attached to the ends of a spring contants `1120N//m` are jkkplaced on smooth horizontal plane with the spring undeformed. Simultaneously velocities of `3m//"s ant" 10m//s` along the line of the spring in the same direction are imparted to `A` and `B` then- (a) Find the maximum extension of the spring. (b) When does the first maximum compression occurs after start. A.B.C.D. |
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Answer» Correct Answer - (a) `25 cm`, (b) `3pi//56` seconds |
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| 980. |
Disintegration constant of a radioactive material is `lambda`:A. Its half life equal to `(log_(e)2)/(lambda)`B. It mean life equals to `(1)/(lambda)`C. At time equal to mean life, `63%` of the Initial radioactive material is left undecayedD. After `3-` half 1/3 rd of the initial radioactive material is left undecayed. |
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Answer» Correct Answer - A::B `N = N_(0)e^(-lambdat) rArr (N_(0))/(10) = N_(0)e^(-lambdaT_(1//2))` `rArr T_(1//2) = (log_(e)2)/(lambda) rArr T_(mean) = (1)/(lambda)` |
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| 981. |
The figure shows the results of an experiment involving photoelectric effect. The graphs A, B,C and D relate to a light beam having different wavelengths. Select the correct alternative. A. Beam B has highest frequencyB. Beam C has highest frequencyC. Beam A has highest rate of photoelectric emissionD. Photoelectron emitted by B have highest momentum |
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Answer» Correct Answer - A::B::C::D `"Stopping potential" prop "frequency" prop (1)/("wavelength")` `"Saturation current" prop "rate of photoelectron"` emission. Also, `K.E_(max) = hv - phi, P = sqrt(2mKE)` |
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| 982. |
In a photoelectric experiment, the collector plate is at 2.0V with respect to the emitter plate made of copper `(phi)=4.5eV)`. The emitter is illuminated by a source of monochromatic light of wavelength 2000 `Å`nm.A. The maximum kinetic energy of the photoelctrons reaching the collector is `0`.B. The maximum kinetic energy of the photoelctrons reaching the collector is `3.7` .eV.C. If the polarity of the battery is reversed then answer to part A willl be 0D. If the polarity of the battery is reversed then answer to part B will be `1.7 eV` |
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Answer» Correct Answer - B `K_(max) = (1242)/(200) eV - 4.5 eV` `K_(max) = 1.7 eV` at cathode `K_(max) = (1.7 + 2) eV` at anode If polarity is reversed, no `e^(-)` reach at collector. |
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| 983. |
When photons of energy `4.25 eV` strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy `T_(A)` eV and De-broglie wavelength `lambda_(A)`. The maximum energy of photoelectron liberated from another metal B by photon of energy 4.70 eV is `T_(B) = (T_(A) - 1.50) eV` if the de Brogle wavelength of these photoelectrons is `lambda_(B) = 2 lambda_(A) `, thenA. The work function of A is 2.25 eVB. The work function of B is `4.20 eV`C. `T_(A) = 2.00 eV`D. `T_(B) = 2.75 eV` |
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Answer» Correct Answer - A::B::C (1) `lambda_(A) = (h)/(sqrt(2mT_(A)))` , (2) `lambda_(B) = (h)/(sqrt(2mT_(B)))` (3) `T_(B) = T_(A) - 1.5 eV` , (4) `lambda_(B) = 2lambda_(A)` |
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| 984. |
An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass `M`. The piston and the cylinder have equal cross - section area `A`. When the piston is in equilibrium, the volume of the gas is `(V_0)` and the its pressure is `(P_0)`. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, Show that the piston executes simple harmonic motion and find the frequency of oscillations.A. (a)`1/2_(pi)`B. (b) 1/2_(pi)`C. (c ) 1/2_(pi)sqrt(A^(2)gammaP_(0))/(MV_(0))`D. (d) `1/2_(pi)sqrt(MV_(0)/(AgammaP_(0)` |
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Answer» Correct Answer - C (c ) `(Mg)/A=P_(0)` `F_(0)V_(0)^(gamma)=PV^(gamma)` `Mg=P_(0)A` …(1) `P_(0)Ax_(0)^(gamma) =PA(x_(0)-x)^(gamma)` `P=(_(0)x_(0)^(gamma))/(x_(0)-x)^(gamma)`. |
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| 985. |
An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass `M`. The piston and the cylinder have equal cross - section area `A`. When the piston is in equilibrium, the volume of the gas is `(V_0)` and the its pressure is `(P_0)`. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, Show that the piston executes simple harmonic motion and find the frequency of oscillations. |
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Answer» Correct Answer - `(1)/(2pi)sqrt((gammarho_(0)A^(2))/(MV_(0)))` |
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| 986. |
If the binding energy per nucleon in `._(3)Li^(7)` and `._(2)He^(4)` nuclei are respectively `5.60` MeV and `7.06` MeV, then the ebergy of proton in the reaction `._(3)Li^(7) +p rarr 2 ._(2)He^(4)` isA. `28.24 MeV`B. `17.28 MeV`C. `1.46 MeV`D. `1.46 MeV` |
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Answer» Correct Answer - B In a nuclear reaction the energy remains conservest `P + ._(3)^(7)Li rarr 2(._(2)^(4)He)` energy of proton `+ 7(5.60) = 2(4 xx 7.06)` `rArr` energy of `= 17.28 MeV` |
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| 987. |
A spring has natural length `40 cm` and spring constant `500 N//m`. A block of mass `1 kg` is attached at one end of the spring and other end of the spring is attached to a ceiling. The block is relesed from the position, where the spring has length `45 cm`.A. the block will perform `SHM` of amplitude `5 cm`B. the block will have maximum velocity `30sqrt5 cm//sec`.C. the block will have maximum acceleration `15 m//s^(2)`.D. the mainimum potential energy of the spring will be zero |
| Answer» Correct Answer - B::C::D | |
| 988. |
The distance between the point of suspension and the centre of gravity of a compound pendulum is `l` and the radius of gyration about the horizontal axis through the centre of gravity is `k`, then its time period will beA. `2pisqrt((l+k)/(g))`B. `2pisqrt((l^(2)+k^(2))/(lg))`C. `2pisqrt((l+k^(2))/(g))`D. `2pisqrt((2k)/(lg))` |
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Answer» Correct Answer - B `T = 2pisqrt((l)/(mgl)) = 2pisqrt((l^(2) + k^(2))/(gl))` |
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| 989. |
A `1.00 xx 10^(-20)`kgparticle is vibrating with simple harmonic motion with a period of` 1.00 xx 10^(-5)` sec and a maximum speed of `1.00xx10^(3)` m/s . The maximum displacement of the particle isA. 1.59 mmB. 1.00 mmC. 10 mmD. none of these |
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Answer» Correct Answer - A |
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| 990. |
A particle is vibrating in a simple harmonic motion with an amplitude of 4 cm . At what displacement from the equilibrium position, is its energy half potential and half kineticA. 1 cmB. `sqrt(2)` cmC. 3 cmD. `2sqrt(2)` cm |
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Answer» Correct Answer - D `U=K " or "U=(E)/(2)` `therefore (1)/(2)Kx^(2)=(1)/(2)((1)/(2)KA^(2))` i.e., at `x=+-(A)/(sqrt(2)),U=K` and this situation will occur for four times in one complete period. |
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| 991. |
A particle is vibrating in a simple harmonic motion with an amplitude of 4 cm . At what displacement from the equilibrium position, is its energy half potential and half kineticA. 1 cmB. `sqrt2 cm`C. 3 cmD. `2sqrt2 cm` |
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Answer» Correct Answer - D |
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| 992. |
The total energy of a vibrating particle in `SHM` is `E`. If its amplitude and time period are doubled, its total energy will be :-A. `16E`B. `8E`C. `4E`D. `E` |
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Answer» Correct Answer - D Total energy `e = 1/2momega^(2)a^(2) rArr E prop (a^(2))/(T^(2))` |
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| 993. |
A simple pendulm has a length L and a bob of mass M. The bob is vibrating with amplitude a .What is the maximum tension in the string?A. `mg(1 + A//l)`B. `mg (1 + A//l)^(2)`C. `mg[1 + (A//l)^(2)]`D. `2 mg` |
| Answer» Correct Answer - C | |
| 994. |
The amplitude of a vibrating body situated in a resisting mediumA. decreases linearly with time decreases exponentially with timeB. decreases exponentially with timeC. decreases with time in some other mannerD. remains constant with time |
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Answer» Correct Answer - B `a=a_0e^(-bt)` |
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| 995. |
The length of a second pendulum isA. 99.8 cmB. 99 cmC. 100 cmD. None of these |
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Answer» Correct Answer - C (c) The length of second pendulum is 100 cm. |
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| 996. |
If `N_(0)` is the original mass of the substance of half - life period `t_(1//2) = 5 year` then the amount of substance left after `15` year isA. `(N_(0))/(8)`B. `(N_(0))/(16)`C. `(N_(0))/(2)`D. `(N_(0))/(4)` |
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Answer» Correct Answer - A `N/N_(0)=(1/2)^(n)` where n are is number of half-lives. If `T_(1//2)=5` years then in `15` years, `n=3` `implies N/N_(0)=(1/2)^(3)=1/8 implies N=N_(0)/8` |
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| 997. |
Which of the following radiation has the least wavelength ?A. `gamma`-raysB. `beta`-raysC. `alpha-`raysD. `X-`rays |
| Answer» Correct Answer - A | |
| 998. |
which a `U^(238)` nucleus original at rest , decay by emitting an alpha particle having a speed `u` , the recoil speed of the residual nucleus isA. `(4u)/(238)`B. `-(4u)/(234)`C. `(4u)/(234)`D. `-(4u)/(238)` |
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Answer» Correct Answer - C `U^(238) rarr U^(234)+._(2)He^(4)` Let recoil speed be V then by COLM `234 V=4uimplies V=(4u)/234` |
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| 999. |
A ring of diameter `2m` oscillates as a compound pendulum about a horizontal axis passing through a point at its rim. It oscillates such that its centre move in a plane which is perpendicular to the plane of the ring. The equibvalent length of the simple pendulum isA. `2m`B. `4m`C. `1.5m`D. `3m` |
| Answer» Correct Answer - C | |
| 1000. |
A particle of mass m is suspended with the help of a spring and an inextensible string as shown in the figure. Force constant of the spring is k. The particle is pulled down from its equilibrium position by a distance x and released. (a) Find maximum value of x for which the motion of the particle will remain simple harmonic. (b) Find maximum tension in the string if `x=(mg)/(2k)` |
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Answer» Correct Answer - (a) `(mg)/(k), (b) (3)/(2) mg` |
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