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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.
| 251. |
A uniform thin cylindrical disk of mass M and radius R is attaached to two identical massless springs of spring constatn k which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance d from its centre. The axle is massless and both the springs and the axle are in horizontal plane. the unstretched length of each spring is L. The disk is initially at its equilibrium position with its centre of mass (CM) at a distance L from the wall. The disk rolls without slipping with velocity `vecV_0 = vacV_0hati.` The coefficinet of friction is `mu.` The centre of mass of the disk undergoes simple harmonic motion with angular frequency `omega` equal to -A. `sqrt((k)/(M))`B. `sqrt((2k)/(M)`C. `sqrt((2k)/(3M))`D. `sqrt((4k)/(3M))` |
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Answer» Correct Answer - D `F_("net") = -(4kx)/(3) = -M(omega^(2)x)` `omega = sqrt((4k)/(3M))` |
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| 252. |
A uniform thin cylindrical disk of mass M and radius R is attaached to two identical massless springs of spring constatn k which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance d from its centre. The axle is massless and both the springs and the axle are in horizontal plane. the unstretched length of each spring is L. The disk is initially at its equilibrium position with its centre of mass (CM) at a distance L from the wall. The disk rolls without slipping with velocity `vecV_0 = vacV_0hati.` The coefficinet of friction is `mu.` The centre of mass of the disk undergoes simple harmonic motion with angular frequency `omega` equal to -A. `sqrt((k)/(M))`B. `sqrt((2k)/(M)`C. `sqrt((2k)/(3M))`D. `sqrt((4k)/(3M))` |
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Answer» Correct Answer - D `F_("net") = -(4kx)/(3) = -M(omega^(2)x)` `omega = sqrt((4k)/(3M))` |
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| 253. |
A brass cube of side a and density `sigma`is floating in mercury of density `rho`. If the cube is displaced a bit vertically, it executes S.H.M. Its time period will beA. `2pi sqrt((sigmaa)/(rho g))`B. `2pi sqrt((rhoa)/(sigmag))`C. `2pi sqrt((rhog)/(sigmaa))`D. `2pi sqrt((sigmag)/(rhoa))` |
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Answer» Correct Answer - A |
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| 254. |
A uniform rod of length `L` and mass `M` is pivotedat the centre. Its two ends are attached to two springs of equal spring constants. `k`. The springs as shown in the figure, and the rod is free to oscillate in hte horizontal plane. the rod is gently pushed through a small angle `theta` in one direction and released. the frequency of oscilllation is- A. `(1)/(2pi)sqrt((2k)/(M))`B. `(1)/(2pi)sqrt((k)/(M))`C. `(1)/(2pi)sqrt((6k)/(M))`D. `(1)/(2pi)sqrt((24k)/(M))` |
| Answer» Correct Answer - C | |
| 255. |
Two bodies (M) and (N) of equal masses are suspended from two separate massless springs of spring constants (k_1) and (k_2) respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of vibration of (M) to the of (N) is.A. (k_1)/k_2)B. sqrt(k_1)//(k_2)C. (k_2)/(k_1)D. sqrt(k_2)//k_1) |
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Answer» Correct Answer - D Both the bodies oscillate in simple harmonic motion, for the maximum velocities will be Given that `v_1 = v_2 rArr a_1 omega_1 = a_2 omega_2` :. `a_1 xx (2 pi)/(T_1) = a_2 xx (2 pi)/(T_2)` `rArr (a_1)/(a_2) = (T_1)/(T_2) = (2 pi sqrt(m)/(k_1))/(2 pi sqrt(m)/(k_2)) = sqrt(k_2/(k_1))`. |
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| 256. |
Two bodies (M) and (N) of equal masses are suspended from two separate massless springs of spring constants (k_1) and (k_2) respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of vibration of (M) to the of (N) is.A. `k_1/k_2`B. `k_2/k_1`C. `sqrt(k_1/k_2)`D. `sqrt(k_2/k_1)` |
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Answer» Correct Answer - D `v_(max)=A_Momega_M=A_Nomega_N` `A_M/A_N=omega_N/omega_M=sqrt((k_2lm)/(k_1lm))=sqrt(k_2/k_1)` |
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| 257. |
Two light springs of force constants (k_1 and k_2) and a block of mass (m) are in one line (AB) on a smooth horizontal table such that one end of each spring is fixed on rigid supports and the other end is free as shown in the figure. The distance (CD) between the free ends of the springs is (60 cms). If the block moves along (AB) with a velocity (120 cm//sec) in between the springs, calculate the period of oscillation of the block `k_1 = 1.8 N//m, k_2 = 3.2 N//m, m = 200 gm)`. . |
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Answer» Correct Answer - B::C The mass will strike the right spring, compress it. The K.E. of the mass will convert into P.E. of the spring. Again the spring will return to its natural size thereby verting its P.E. to K.E. of the block.The time taken for this process will `beT/2`, where T=2pi sqrtm/k. :. t_(l) =T/2=pisqrtm/k_(2) =pisqrt(0/2)/(3.2)=0.785 sec` The block will move from A to B without any acceeleration. The time taken will be `t_(2) =(60)/(120) =0.5` Now the block will compress the left spring and then the spring again attains its natural length. ength. The time taken will be oscillation. The time taken for doing so `t_(3)=pi sqrtm/k_(l) =pisqrt(0.2)/(1.8) =1.05 sec`. Again the block moves from B to A, completin one oscillation. The time taken for doing so `t_(4) =(60)/(120) =0.5` :. The complete time of oscillation will be `=t_(l)+t_(2)+t_(3)+t_(4)` `=0.785+0.51.05+05` `=2.83(app.). |
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| 258. |
On a smooth inclined plane, a body of mass M is attached between two springs. The other ends of the springs are fixed to firm supports. If each spring has force constant K , the period of oscillation of the body (assuming the springs as massless) is A. `2pi(m/(2K))^(1//2)`B. `2pi((2M)/(K))^(1//2)`C. `2pi (Mgsintheta)/(2K)`D. `2pi((2Mg)/(K))^(1//2)` |
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Answer» Correct Answer - A |
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| 259. |
An equilateral prism of mass m is kept on a smooth table between two identical springs each having a force constant of k. The two springs have their lengths perpendicular to the inclined faces of the prism and are constrained to remain straight. The ends of the springs have light pads aligned parallel to the faces of the prism, and distance between pads and the incline faces is d. The prism is imparted a velocity v to the right. Find time period of its oscillation. |
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Answer» Correct Answer - `(8d)/(sqrt(3v))+4pi sqrt((m)/(3k))` |
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| 260. |
Two identical blocks 1 and 2, each of mass m, are kept on a smooth horizontal surface, connected to three springs as shown in the figure. Each spring has a force constant k. Under suitable initial conditions, the two blocks oscillate in phase and their respective displacement from the mean position is given by `x_(1)=A sin omega t " and " x_(2)=A sin omega t` (i) Suggest one such initial condition that will result in such oscillation. (ii) Find `omega` |
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Answer» Correct Answer - (ii) ` omega=sqrt((k)/(m))` |
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| 261. |
A massless spring, having force constant k, oscillates with frequency n when a mass m is suspended from it. The spring is cut into two equal halves and a mass `2m` is suspended from one half. The frequency of oscillation will now beA. nB. 2nC. `n//sqrt2`D. `n^((2)^(1//2)` |
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Answer» Correct Answer - A |
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| 262. |
A spring has a certain mass suspended from it and its period for vertical oscillations is `T_(1)`. The spring is now cut into two equal halves and the same mass is suspended from one of the half. The period of vertical oscillation is now `T_(2)`. The ratio of `T_(2)//T_(1)` isA. `T/2`B. `T/sqrt2`C. `sqrt(2T)`D. 2T |
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Answer» Correct Answer - B |
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| 263. |
An object suspended from a spring exhibits oscillations of period T. Now the spring is cut in two halves and the same object is suspended with two halves as shown in figure. The new time period of oscillation will become A. `T/sqrt(2)`B. 2TC. T/2D. `T/(2sqrt(2))` |
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Answer» Correct Answer - C |
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| 264. |
A mass `M`, attached to a horizontal spring, excutes `SHM` with a amplitude `A_(1)`. When the mass `M` passes through its mean position then a smaller mass `m` is placed over it and both of them move together with amplitude `A_(2)`, the ratio of `((A_(1))/(A_(2)))` is :A. `(M)/(M + m)`B. `(M + m)/(M)`C. `((M)/(M + m))^(1//2)`D. `((M + m)/(M))^(1//2)` |
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Answer» Correct Answer - D `C.O.L.M.` `MV_(max) = (m + M)V_("new") , V_(max) = A_(1)omega_(1)` `V_("new") = (MV_(max))/((m + M))` Now, `V_("new") = A_(2).omega_(2)` `(MA_(1))/((m + M))sqrt((K)/(M)) = A_(2)sqrt((K)/((m + M)))` `A_(2) = A_(1)sqrt((M)/((m + M))), (A_(1))/(A_(2)) = ((m + M)/(M))^(1//2)` |
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| 265. |
A mass `M` attached to a horizontal spring executes `SHM` with an amplitude `A_(1)`. When mass `M` passes through its mean position a smaller mass `m` is placed over it and both of them move togther with amplitude `A_(2)`. Ratio of `((A_(1))/(A_(2)))` is:A. `(M)/(M + m)`B. `(M + m)/(M)`C. `((M)/(M + m))^(1//2)`D. `((M + m)/(M))^(1//2)` |
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Answer» Correct Answer - D `C.O.L.M.` `MV_(max) = (m + M)V_("new") , V_(max) = A_(1)omega_(1)` `V_("new") = (MV_(max))/((m + M))` Now, `V_("new") = A_(2).omega_(2)` `(MA_(1))/((m + M))sqrt((K)/(M)) = A_(2)sqrt((K)/((m + M)))` `A_(2) = A_(1)sqrt((M)/((m + M))), (A_(1))/(A_(2)) = ((m + M)/(M))^(1//2)` |
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| 266. |
Hydrogen `(_(1)H^(1))` Deuterium `(_(1)H^(2))` singly omised helium `(_(1)He^(1))` and doubly ionised lithium `(_(1)Li^(6))^(++)` all have one electron around the nucleus Consider an electron transition from `n = 2 to n = 1` if the wavelength of emitted radiartion are `lambda_(1),lambda_(2), lambda_(3),and lambda_(4)`, repectivelly then approximetely which one of the following is correct ?A. `lambda_(1) = lambda_(2) = 4lambda_(3) = 9lambda_(4)`B. `lambda_(1) = 2lambda_(2) = 3lambda_(3) = 4lambda_(4)`C. `4lambda_(1) = 2lambda_(2) = 2lambda_(3) = lambda_(4)`D. `lambda_(1) = 2lambda_(2) = 2lambda_(3) = lambda_(4)` |
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Answer» Correct Answer - A In Bohr model `1/lambda=Rz^(2)[1/n_(1)^(2)-1/n_(2)^(2)]implies z prop 1/z^(2)` `lambda_(1) : lambda_(2) : lambda_(3) :lambda_(4) :: 1/1:1/1:1/4:1/9` `implies lambda_(1)=lambda_(2)=4lambda_(3)=9lambda_(4)` |
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| 267. |
A particle is executing a simple harmonic motion its maximum acceleration is a and maximum velocity is `beta` .Then its time of vibration will beA. `(2pialpha)/(beta)`B. `(2pi beta)/(alpha)`C. `2pialphabeta`D. `(pibeta)/(alpha)` |
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Answer» Correct Answer - A `alpha=A omega` `beta=omega^(2) A ` `T=-(2pi alpha)/(beta)` |
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| 268. |
An electron of mass `m` and charge `e` initially at rest gets accelerated by a constant electric field `E`. The rate of change of de-Broglie wavelength of this electron at time `t` ignoring relativistic effects is |
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Answer» Correct Answer - `(-h)/(inEt^(2))` `lambda_("de-brogile")=(h)/sqrt(2mK)=h/(mv)` `v=u+at=0+((e E)/m)timplies lambda=h/(e Et)` `implies (d lambda)/(dt)=-(h)/(e Et^(2))` |
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| 269. |
A body of mass `5g` is executing SHM with amplitude `10cm`, its velocity is `100cm//s`. Its velocity will be `50cm//s` at a displacement from the mean position equal toA. `5cm`B. `5sqrt3`C. `10sqrt3cm`D. `15sqrt3cm` |
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Answer» Correct Answer - B `A=10cm`, `v_(max)=100cm//s`, `v=50cm//s`, `x=?` `v_(max)=Aomegaimplies100=10omegaimpliesomega=10rad//s` `v=omegasqrt(A^2-x^2)` `50=10sqrt((10)^2-x^2)` `x=5sqrt3cm` |
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| 270. |
Pulse rate of a noumal person is 75 per minute. The time period of heart isA. `0.8`sB. `0.75`sC. `1.25s`D. `1.75s` |
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Answer» Correct Answer - A The heat frequency of heart is `v=(75)/(1 "min")=(75)/(60s)=1.25=1.25Hz` The time period of heart is `T=(1)/(v)=(1)/(1.25s^(-1))=0.8s` |
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| 271. |
A particle of mass m= 5g is executing simple harmonic motion with an amplitude `0.3`m and time period `pi//5` second. The maximum value of force acting on the particle isA. 5 NB. 4 NC. `0.5` ND. `0.15`N |
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Answer» Correct Answer - D We know maximum acceleratiom, `a_("max")=omega^(2)A=(4pi^(2))/(T^(2))A` `=(4pi^(2))/(((pi)/(5))^(2))xx0.3=30m//s^(2)` Maximum force, `F_("max")=ma_("max")=(.5)/(1000)xx30=0.15N` |
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| 272. |
Phase space deagrams are useful tools . in analyzing all kinds of dynamical problems. They are especially usrful in studying the changes in motion as initial position and momenum are changed. Here we conseder some simple dynamical systems in one dimension. For such systems, phase space is a plane in which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is `x(t) vs. p(t) curve in this plane. The arrow on the curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which positon or momentum upwards (or to right) is poitive and downwards (or to left) is negative. The phace diagram for a ball thrown vertically up from ground is.A. (a).B. (b).C. (c ).D. (d). |
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Answer» Correct Answer - D (d) When the ball is throw upwards, at the point of throw (O) the linear momentum is in upwards derection (and has a maximum value ) and the position is zero. At the time passes, the ball moves upwards and its momentum goes on decreasing and the position becomes zero at the topmost point (A). As the time increases, the ball starts moving down with an increasing linear momentum in the downward derection. These characteristics are represented by graph)^(d). |
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| 273. |
When a particle of mass m moves on the x-axis in a potential of the form `V(x) =kx^(2)` it performs simple harmonic motion. The correspondubing time period is proprtional to `sqrtm/h`, as can be seen easily using dimensional analusis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of `x=0` in a way different from `kx^(2)` and its total energy is such that the particle does not escape toin finity. Consider a particle of mass m moving on the x-axis. Its potential energy is `V(x)=ax^(4)(agt0)` for |x| neat the origin and becomes a constant equal to `V_(0)` for |x|impliesX_(0)` (see figure). . The acceleration of this partile for `|x|gtX_(0)` is (a) proprtional to `V_(0)` (b) proportional to.A. (a) proprtional to `V_(0)`B. (b) proportional to `V_(0)/(mX_(0))C. (c ) proportional to `sqrtV_(0)/(mX_(0)`D. (d) zero`. |
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Answer» Correct Answer - D (d) `F=(-dV(x))/(dx)` As `V(x)=constamt for `xgtX_(0)` Since `F=0,a=0`. |
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| 274. |
When a particle of mass m moves on the x-axis in a potential of the form `V(x) =kx^(2)` it performs simple harmonic motion. The correspondubing time period is proprtional to `sqrtm/h`, as can be seen easily using dimensional analusis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of `x=0` in a way different from `kx^(2)` and its total energy is such that the particle does not escape toin finity. Consider a particle of mass m moving on the x-axis. Its potential energy is `V(x)=ax^(4)(agt0)` for |x| neat the origin and becomes a constant equal to `V_(0)` for |x|impliesX_(0)` (see figure). The acceleration of this partile for `|x|gtX_(0)` is (a) proprtional to `V_(0)` (b) proportional to.A. propotional to `V_(0)`B. propotional to `V_(0)/(mX_(0))`C. propotional to `sqrt((V_(0))/(mX_(0)))`D. zero |
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Answer» Correct Answer - D `F = -(dU)/(dx)` as for `|x| gt x_(0) V = V_(0) =` constant `rArr (dU)/(dx) = 0 rArr F = 0`. |
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| 275. |
When a particle of mass m moves on the x-axis in a potential of the form `V(x) =kx^(2)` it performs simple harmonic motion. The correspondubing time period is proprtional to `sqrtm/h`, as can be seen easily using dimensional analusis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of `x=0` in a way different from `kx^(2)` and its total energy is such that the particle does not escape toin finity. Consider a particle of mass m moving on the x-axis. Its potential energy is `V(x)=ax^(4)(agt0)` for |x| neat the origin and becomes a constant equal to `V_(0)` for |x|impliesX_(0)` (see figure). For periodic motion of small amplitude A,the time period (T) of thes particle is proportional to.A. `Asqrt((m)/(alpha))`B. `(1)/(A)sqrt((m)/(alpha))`C. `Asqrt((alpha)/(m))`D. `(1)/(A)sqrt((alpha)/(m))` |
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Answer» Correct Answer - B `V = alphaX^(4)` `T.E. = (1)/(2) momega^(2)A^(2) = alphaA^(4)` (not stricltly applicable just for dimension matching it is used) `omega^(2) = (2alphaA^(2))/(m) rArr T prop (1)/(A)sqrt((m)/(alpha))` |
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| 276. |
Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each other moving in opposite directions each time their displacement is half their amplitude. Their phase difference is |
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Answer» Let equations of motion of two particles be `x=Asinomegat` `x=Asin(omegat+phi)` (ii) When `x=A//2`, from (i), `sin omegat=1//2=sinpi//6` `omegat=pi//6` When `x=A/2`, from (ii), `sin (omegat+phi)=1/2=sinpi//6` or `sin ((5pi)/(6))` `omegat+phi=pi//6` or `(5pi)/(6)` (a) If `omegat+phi=pi//6impliespi//6+phi=pi//6impliesphi=0` (not possible) (b) If `omegat+phi=5pi//6impliespi//6+phi=(5pi)/(6)impliesphi=(2pi)/(3)=120^@` |
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| 277. |
Two particles A and B execute simple harmonic motions of period T and `5T//4`. They start from mean position. The phase difference between them when the particle A complete an oscillation will beA. `pi//2`B. `0`C. `2pi//5`D. `pi//4` |
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Answer» Correct Answer - C `x_A=Asin((2pi)/(T_1)t)`, `x_B=Asin((2pi)/(T_2))t` Phase difference `|phi|=((2pi)/(T_1)-(2pi)/(T_2))t` At `t=T`, `|phi|=((2pi)/(T)-(2pi)/(5T//4))T` `=2pi(1-4/5)=(2pi)/(5)` |
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| 278. |
Two particles P and Q start from origin and execute simple harmonic motion along X-axis with same amplitude but with periods `3s` and `6s` respectively. The ratio of the velocities of P and Q when they meet isA. `1:2`B. `2:1`C. `2:3`D. `3:2` |
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Answer» Correct Answer - B |
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| 279. |
A particle executes simple harmonic motion with an amplitude of 4 cm . At the mean position the velocity of the particle is 10 cm/ s . The distance of the particle from the mean position when its speed becomes 5 cm/s isA. `sqrt3 cm`B. `sqrt5 cm`C. `2sqrt3 cm`D. `2sqrt5 cm` |
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Answer» Correct Answer - C |
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| 280. |
The time taken by a particle performing `SHM` on a straight line to pass from point `A` to `B` where its velocities are same is `2` seconds .After another `2` seconds it return to `B` The time period of oscillation is (in seconds)A. `2`B. `8`C. `6`D. `4` |
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Answer» Correct Answer - B |
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| 281. |
The time taken by a particle performing SHM to pass from point A and B where it is velocities are same is `2:3`. After another 2 s it returns to B. The time period oscillation is (in seconds)A. 2B. 8C. 6D. 4 |
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Answer» Correct Answer - B |
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| 282. |
An X-ray tube, operated at a potential difference of 40 kV, produce heat at the rate of 720 W.Assuming 0.5% of the energy of incident electrons is converted into X-rays , calculate (i)The number of electron per second striking the target (ii)The velocity of the incident electrons . |
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Answer» Correct Answer - (i) `1.1 xx 10^(17)` (ii) `1.2 xx 10^(8) m//s` (i) Operating voltage `=40 kV, 0.5 %` energy for x ray `:. 99.5/100xxnxxexxV=720` `n=(720)/(1.6xx10^(-14)xx0.995xx40xx10^(3))=1.1xx10^(17)` (ii) Velocity of incident `e^(-)implies 1/2 mv_(e)^(2)=eV` `v_(e)=sqrt((2eV)/m)=sqrt((2xx1.6xx10^(19)xx40xx10^(3))/(9.1xx10^(-31)))` `=1.2xx10^(8) m//s` |
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| 283. |
In a mixture of `H-He^(+)` gas (`He^(+)` is singly ionized `He` atom), `H` atoms and `He^(+)` ions are excited to their respective first excited states. Subsequently, `H` atoms transfer their total excitation energy to `He^(+)` ions (by collisions). Assume that the Bohr model of atom is exctly valid. The ratio of the kinetic energy of the `n=2` electron for the `H` atom to that of `He^(+)` ion is:A. `1//4`B. `1//2`C. `1`D. `2` |
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Answer» Correct Answer - A Ratio of kinetic energy `K_(1)/K_(2)=((Z_(1)//n_(1))^(2))/((Z_(2)//n_(2))^(2))` Since `n_(1)=n_(2)=2` & `Z_(1)=1` for `H, Z_(2)=2` for `He^(+) implies K_(1)/K_(2)=1/4` |
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| 284. |
Scientists are working hard to develop nuclear fusion reactor Nuclei of heavy hydrogen, `_(1)^(2)H` , known as deuteron and denoted by `D`, can be thought of as a candidate for fusion rector . The `D-D` reaction is `_(1)^(2) H + _(1)^(2) H rarr _(2)^(1) He + n+` energy. In the core of fusion reactor, a gas of heavy hydrogen of `_(1)^(2) H` is fully ionized into deuteron nuclei and electrons. This collection of `_1^2H` nuclei and electrons is known as plasma . The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually , the temperature in the reactor core are too high and no material will can be used to confine the to plasma for a time `t_(0)` before the particles fly away from the core. If `n` is the density (number volume ) of deuterons , the product` nt_(0) `is called Lawson number. In one of the criteria , a reactor is termed successful if Lawson number is greater then `5 xx 10^(14) s//cm^(2)` it may be helpfull to use the following boltzmann constant `lambda = 8.6 xx 10^(-5)eV//k, (e^(2))/(4 pi s_(0)) = 1.44 xx 10^(-9) eVm` In the core of nucleus fusion reactor , the gas become plasma because ofA. strone nuclear force acting between the deuteronsB. Coulomb force acting between the deutronsC. Coulomb force acting between deutron electron pairsD. the high temperature maintained inside the reactor core |
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Answer» Correct Answer - D Due to the high temperature developed as a result of collision & fusion causes the core of fusion reactor to plasma. |
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| 285. |
Scientists are working hard to develop nuclear fusion reactor Nuclei of heavy hydrogen, `_(1)^(2)H` , known as deuteron and denoted by `D`, can be thought of as a candidate for fusion rector . The `D-D` reaction is `_(1)^(2) H + _(1)^(2) H rarr _(2)^(1) He + n+` energy. In the core of fusion reactor, a gas of heavy hydrogen of `_(1)^(2) H` is fully ionized into deuteron nuclei and electrons. This collection of `_1^2H` nuclei and electrons is known as plasma . The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually , the temperature in the reactor core are too high and no material will can be used to confine the to plasma for a time `t_(0)` before the particles fly away from the core. If `n` is the density (number volume ) of deuterons , the product` nt_(0) `is called Lawson number. In one of the criteria , a reactor is termed successful if Lawson number is greater then `5 xx 10^(14) s//cm^(2)` it may be helpfull to use the following boltzmann constant `lambda = 8.6 xx 10^(-5)eV//k, (e^(2))/(4 pi s_(0)) = 1.44 xx 10^(-9) eVm` Assume that two deuteron nuclei in the core of fusion reactor at temperature energy `T` are moving toward each other, each with kinectic energy `1.5 kT` , when the seperation between them is large enough to neglect coulomb potential energy . Also neglate any interaction from other particle in the core . The minimum temperature `T` required for them to reach a separation of `4 xx 10^(-15) m ` is in the rangeA. deuteron density `= 2.0 xx 10^(12) cm^(-3)`, confinement time `= 5.0 xx 10^(-3) s`B. deuteron density `= 8.0 xx 10^(14) cm^(-3)`, confinement time `= 9.0 xx 10^(-1) s`C. deuteron density `= 4.0 xx 10^(23) cm^(-3)`, confinement time `= 1.0 xx 10^(11) s`D. deuteron density `= 1.0 xx 10^(24) cm^(-3)`, confinement time `= 4.0 xx 10^(12) s` |
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Answer» Correct Answer - B,C,D `nt_(0)=8xx10^(14)xx9xx10^(-1)=7.2xx10^(14) gt 5xx10^(14)` `nt_(0)=4xx10^(23)xx1xx10^(11)=4xx10^(34) gt 5xx10^(14)` `nt_(0)=1xx10^(24)xx4xx10^(12)=4xx10^(36) gt 5xx10^(14)` |
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| 286. |
Find the Q - value for the nuclear reaction: `._(2)He^(4)+._(7)N^(14) rarr ._(8)O^(17)+._(1)H^(1)` `m(._(2)He^(4))=4.0039 "amu" m(._(7) N^(14))=14.0075 "amu" m(._(8)O^(17))=17.0045` amu `m(._(1)H^(1))=1.0082` amuA. `2.54` MeVB. `1.54` MeVC. `5` MeVD. `8` MeV |
| Answer» Correct Answer - B | |
| 287. |
The damging force on an oscillator is directly proportional to the velocity. The units of the constant of propor-tionality areA. `kgms^(-1)`B. `kgms^(-2)`C. `kgs^(-1)`D. `kgs` |
| Answer» Correct Answer - C | |
| 288. |
The maximum energy of the electrons released in photocell is independent of -A. Frequency of incident light.B. Intensity of incident light.C. Naure of cathode surface.D. None of these. |
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Answer» Correct Answer - B `K.E._(max) = underset("frequency of light")(hv) - underset("depend on properies of cathode")(phi)` |
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| 289. |
Photoelectric effect takes place in element A. Its work funcation is `2.5 eV` and threshold wavelength is `lambda`. An other element B is having work function of `5 eV`. Then find out the wavelength that can produce photoelectric effect in B.A. `lambda//2`B. `2lambda`C. `lambda`D. `3lambda` |
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Answer» Correct Answer - A `(phi_(1))/(phi_(2)) = (lambda_(2))/(lambda_(1))` |
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| 290. |
the funcation ` sin^(2)(omegat)` repesentsA. A simple harmonic motion with a period `2 pi//omega`B. A simple harmonic motion with a period `pi //omega`C. A periodic buy not simple harmoic motion with a period `2pi// omega`D. A periodic but not simple harmonic, motion with a period `pi//omega` |
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Answer» Correct Answer - D |
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| 291. |
A large horizontal surface moves up and down in SHM with an amplitude of 1 cm . If a mass of 10 kg (which is placed on the surface) is to remain continually in contact with it, the maximum frequency of S.H.M. will beA. 0.5 HzB. 1.5 HzC. 5 HzD. 10 Hz |
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Answer» Correct Answer - C |
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| 292. |
Assertion : The spring constant of a spring is k. When it is divided into n equal parts, then springconstant of one piece is k/n. Reason : The spring constant is independent of material used for the spring.A. 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 assertion is false but reason is true |
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Answer» Correct Answer - D |
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| 293. |
A simple harmonic oscillator has amplitude A, angular velocity `omega`, and mass m . Then, average energy in one time period will beA. `(1)/(4)m omega^(2)A^(2)`B. `(1)/(2)m omega^(2)A^(2)`C. `m omega^(2)A^(2)`D. zero |
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Answer» Correct Answer - A Average energy `=(int_(0)^(T)"u dt")/(int_(0)^(T)dt)=(1)/(T)int_(0)^(T)"u dt"` `=(1)/(2T)int_(0)^(T)momega^(2)A^(2)cos^(2)(omegat+phi)dt=(1)/(4)momega^(2)A^(2)` |
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| 294. |
Two tunnels - `T_(1) " and " T_(2)` are dug across the earth as shown in figure. One end of the two tunnels have a common meeting point on the surface of the earth. Two particles `P_(1) " and " P_(2)` are oscillating from one end to the other end of the tunnels. At some instant particles are at mid point of their tunnels as shown in figure. Then - (a) Write phase difference between the particle `P_(1) " and " P_(2)`. Can the two particles ever meet? (b) Write the ratio of maximum velocity of particle `P_(1) " and " P_(2)`. |
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Answer» Correct Answer - (a) `180^(@),No " " (b) 2:1` |
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| 295. |
Determine the natural frequency of vibration of the `100N` disk. Assume the disk does not slip on the inclined surface. |
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Answer» In equilibrium, `mg sin theta = kx_(0)`…(i) When displaced by `x`, `E = (1)/(2) mv^(2) + (1)/(2) I omega^(2) + (1)/(2)k(x + x_(0))^(2) - mgx sin theta` Since, E = constant `(dE)/(dt) = 0` `0 = mv((dv)/(dt)) + I omega ((d omega)/(dt)) + k (x + x_(0))(dx)/(dt) - mg sin theta (dx)/(dt)` Substituting, `(dv)/(dt) = a, omega = (v)/(R), I = (1)/(2)mR^(2)` `(d omega)/(dt) = alpha = (a)/(R),(dx)/(dt) = v` and `kx_(0) = mg sin theta` We get, `3ma = - 2kx` `:. f = (1)/(2x)sqrt|(a)/(x)| = (1)/(2pi) sqrt((2k)/(3m))` Substituting the values, `f = (1)/(2pi)sqrt(((2 xx 200)/(3 xx 100))/(9.8))` `= 0.56 Hz`. |
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| 296. |
The amplitude of damped oscillator becomes half in one minute. The amplitude after 3 minutes will be `1//x` times the original, where x isA. 2x 3B. `2^(3)`C. `3^(2)`D. `3xx 2^(2)` |
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Answer» Correct Answer - B |
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| 297. |
A nucleus with Z =92 emits the following in a sequence: `alpha,beta^(-),beta^(-),alpha,alpha,alpha,alpha,alpha,beta^(-),beta^(-),alpha,beta^(+),beta^(+),alpha`. The Z of the resulting nucleus isA. `76`B. `78`C. `82`D. `74` |
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Answer» Correct Answer - B `8 alpha` have been emitted `4 beta^(-)` have been emitted `2 beta^(+)` have emitted `alpha` reduces atomic number by 2 `beta^(-)` increases atomic number by 1 `beta^(+)` increases by 1 So, `Z_(eff) = 92 - (8 xx 2) + (4 xx 1) - (2 xx 1) = 96 - 18 = 78` |
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| 298. |
Which of the following cannot be emitted by radioactive substances during their decay ?A. ProtonsB. NeutrinosC. Helium nucleiD. Electrons |
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Answer» Correct Answer - A A nucles during decay emits `alpha (He^(2+)), beta` (electrons). `gamma` or neutrino, it does not emit protons during decay. |
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| 299. |
The potential energy of a particle execuring S.H.M. is 2.5 J, when its displacement is half of amplitude. The total energy of the particle beA. 18 JB. 10 JC. 12JD. 2.5J |
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Answer» Correct Answer - B Total energy =`1/2 m omega^(2)A^(2)` Potential energy=`1/2 m omega^(2) x^(2)` `2.5 =1/2 m omega^(2)(A/2)^(2)` so total energy =10 J |
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| 300. |
A particle of mass m is dropped from a great height h above the hole in the earth dug along its diameter.A. The motion of the particle is simple harmonicB. The motion of the particle is periodicC. The speed of the particle at the centre of earth equals `sqrt((2GM)/((R+h)))` , where R and M are the radius and mass of the earth respectivelyD. The speed of the particle at the centre of earth equals `sqrt((GM(R+3h))/(R(R+h)))`, where R and M are the radius and mass of the earth respectively |
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Answer» Correct Answer - D (d) The motion is simple harmonic only inside earth. Further `(1)/(2) mv^(2)=-(GMm)/((R+h))+(3)/(2)(GMm)/(R)` `implies v=sqrt((GM(R-h))/(R(R+h)))` |
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