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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.
| 51. |
The ratio of the numerical values of the average velocity and average speed of a body is always.A. UnityB. Unity or lassC. unity or moreD. Less than unity |
| Answer» Correct Answer - B | |
| 52. |
The displacement-time graphs of two bodies A and B are OP and OQ respectively. If `anglePOX` is `60^(@)` and `angleQOX` is `45^(@)` the ratio of the velocity of A to that of B isA. `sqrt(3):sqrt(2)`B. `sqrt(3):1`C. `1:sqrt(3)`D. `3:1` |
| Answer» Correct Answer - B | |
| 53. |
A car is moving along a straight (OP). It moves from `O to P` in `18 seconds` amd retuns from `P to Q` in 6 seconds`, where `OP=360 m` and `OQ=240 m What are the car the average velcoty and average speed of the car in going (a) from `Oto P` and back to `Q` ?A. `10ms^(-1),20ms^(-1)`B. `20ms^(1-),10ms^(-1)`C. `10ms^(-1),10ms^(-1)`D. `20ms^(-1),20ms^(-1)` |
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Answer» Correct Answer - A `V_("avg")=("total displacement")/("total time")=(S_(1)+S_(2))/(t_(1)+t_(2))` |
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| 54. |
If a body covers first half of its journey with uniform speed `v_(1)` and the second half of the journey with uniform speed `v_(2)`. Then the average speed isA. `(v_(1)+v_(2))/(2)`B. `(2v_(1)+v_(2))/(v_(1)+v_(2))`C. `(2v_(1)v_(2))/(v_(1)+v_(2))`D. `(L(v_(1)+v_(2)))/(v_(1)v_(2))` |
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Answer» Correct Answer - C Total time `=t_(1)+t_(2)` `=(l)/(2v_(1))+(l)/(2v_(2))=(l)/(2)[(1)/(v_(1))+(1)/(v_(2))]` |
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| 55. |
If a body covers first half of its journey with uniform speed `v_(1)` and the second half of the journey with uniform speed `v_(2)`. Then the average speed isA. `V_(1)+V_(2)`B. `(2V_(1)V_(2))/(V_(1)+V_(2))`C. `(V_(1)V_(2))/(V_(1)+V_(2))`D. `V_(1)V_(2)` |
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Answer» Correct Answer - B Average speed `=(S_(1)+S_(2))/(t_(1)+t_(2))impliesv=(2V_(1)V_(2))/(V_(1)+V_(2))` |
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| 56. |
If a particle travels a linear distance at speed `v_(1)` and comes back along the same track at speed `v_(2)`.A. its average speed is arithemtic mean `(v_(1) + v_(2))//2`B. its avcrage speed is geometric mean `sqrt(v_(1) v_(2))`C. its average speed is harmonic mean `2v_(1)v_(2)//(v_(1) + v_(2))`D. its velocity is zero |
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Answer» Correct Answer - C::D `v_(av) = (2v_(1) v_(2))/(v_(1) + v_(2))` and velocity is zero |
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| 57. |
A person travels along a straight road for the first half length with a constant speed `v_(1)` and the second half length with a constant speed `v_(2)`. The average speed V is:A. `(v_(1) + v_(2))//2`B. `2v_(1) v_(2)//(v_(1) + v_(2))`C. `(v_(1) v_(2))^(1//2)`D. `(v_(2)//v_(1))^(1//2)` |
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Answer» Correct Answer - B Let `t_(1)` and `t_(2)` be time taken to cover first and second half length Then `t_(1) = (x//2)/(v_(1)) = (x)/(2v_(1))` `t_(2) = (x//2)/(v_(2)) = (x)/(2v_(2))` `v_(av) = ((x//2 + x//2))/((x)/(2v_(1)) + (x)/(2v_(2))) = (1)/((1)/(2v_(1)) + (1)/(2v_(2)))` `(1)/(2v_(1)) + (1)/(2v_(2)) = (1)/(v_(av))` `(v_(1) + v_(2))/(2v_(1) v_(2)) = (1)/(v_(av))` `v_(av) = (2v_(1)v_(2))/(v_(1) + v_(2))` |
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| 58. |
An object falling through a fluid is observed to have acceleration given by ` a= g- bv` where ` g= gravitational acceleration and (b) is constant. After a long time of rlease. It is observed to fall with constant speed. What must be the value of constant speed ?A. g/bB. b/gC. bgD. b |
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Answer» Correct Answer - A Here, a = g - bv When an object falls with constant speed `v_c`, its acceleration becomes zero. `therefore g - bv_c = 0` or `v_c = g/b` |
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| 59. |
An object falling through a fluid is observed to have acceleration given by ` a= g- bv` where ` g= gravitational acceleration and (b) is constant. After a long time of rlease. It is observed to fall with constant speed. What must be the value of constant speed ? |
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Answer» When speed becomes constant accleration `a = (dv)/(dt) = 0` Give acceleration `a = g - bv` where, `g =` gravitional accleration Clearly, from above equation as speed incrases accleration will decrease. At a certain speed say `v_(0)`, acceleration will be zero and speed will remain constant. Hence `a = g - bv_(0) = 0` `rArr v_(0) = g//b` |
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| 60. |
If x, v and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T, then, which of the following does not change with time?A. `a T//x`B. `a T + 2 pi v`C. `a T//v`D. `a^(2) T^(2) + 4 pi^(2) v^(2)` |
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Answer» Correct Answer - A `(aT)/(x) = ((omega^(2) x)/(x)) T = omega^(2)T` = constant Hence, this factor `((aT)/(x))` does not change with time. |
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| 61. |
In one dimensional motion, instantaneous speed `v` satisfies `(0 le v lt v_0)`.A. The displacement in time T must always take non-negative valuesB. The displacement x in time T satisfies `-v_(0) T lt x lt v_(0)T`C. The accleration is always a non-negative numberD. The motion has no turning points |
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Answer» Correct Answer - B For maximum and minimum displacement we have to keep in mind the magnitude and direction of maximum velocity. As maximum velocity in positive direction is `v _(0)` maximum velocity in opposite direction is also `v_(0)` . Maximum displacement in one direction `= v_(0)T` Maximum displacement in opposite directions `= -v_(0)T` Hence, `-v_(0)T lt x lt v_(0)T` |
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| 62. |
Fig gives the `x-t` plot of a particle executing one dimensional simple harmonic motion. Give the signs of position, velocity and acceleration variables of the particles at ` t= 0.3 s`, ` 1.2 s, - 1. 2 s`, |
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Answer» Negative, Negative, Positive (at t = 0.3 s) Positive, Positive, Negative (at t = 1.2 s) Negative, Positive, Positive (at t = –1.2 s) For simple harmonic motion (SHM) of a particle, acceleration (a) is given by the relation: `a=-w^2xxwto` angular frequency ... (i) t=0.3 s In this time interval, x is negative. Thus, the slope of the x-t plot will also be negative. Therefore, both position and velocity are negative. However, using equation (i), acceleration of the particle will be positive. t = 1.2 s In this time interval, x is positive. Thus, the slope of the x–t plot will also be positive. Therefore, both position and velocity are positive. However, using equation (i), acceleration of the particle comes to be negative t = – 1.2 s In this time interval, x is negative. Thus, the slope of the x–t plot will also be negative. Since both x and t are negative, the velocity comes to be positive. From equation (i), it can be inferred that the acceleration of the particle will be positive. |
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| 63. |
Fig. shows `x-t` plot of one dimensional motion a particle. Is it correct to say from the graph that the particle moves in a straight line for ` t lt 0` and on a parabolic path form ` t gt 0 `? If not, suggest a suitable physical context for this graph. |
| Answer» The x–t graph of a particle moving in a straight line for t lt 0 and on a parabolic path for t gt 0 cannot be shown as the given graph. This is because, the given particle does not follow the trajectory of path followed by the particle as t = 0, x = 0. A physical situation that resembles the above graph is of a freely falling body held for sometime at a height | |
| 64. |
Look at the graphs (a) to (d) carefully and state, with reasons, with of these cannot possibly represent one dimensional motion of a particle. |
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Answer» (a) The given x-t graph, shown in (a), does not represent one-dimensional motion of the particle. This is because a particle cannot have two positions at the same instant of time. (b) The given v–t graph, shown in (b), does not represent one-dimensional motion of the particle. This is because a particle can never have two values of velocity at the same instant of time. (c) The given v–t graph, shown in (c), does not represent one-dimensional motion of the particle. This is because speed being a scalar quantity cannot be negative. (d) The given v–t graph, shown in (d), does not represent one-dimensional motion of the particle. This is because the total path length travelled by the particle cannot decrease with time. |
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| 65. |
A particle moves along a straight line its velocity dipends on time as `v=4t-t^(2)`. Then for first `5 s`:A. Average velocity is `25//3 ms^(-1)`B. average speed is `10ms^(-1)`C. Average velocity is `5//3 ms^(-1)`D. acceleration is `4ms^(-2)` at `t=0` |
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Answer» Correct Answer - C::D `"Average velocity"=vecv=(int_(0)^(5)vdt)/(int_(0)^(5)dt)=(int_(0)^(5)(4t-t^(2))dt)/(int_(0)^(5)dt)` `=([2t^(2)-(t^(3))/(3)]_(0))/(5)=(50-(125)/(3))/(5)=(25)/(3xx5)=(5)/(3)` For average speed let us put `v=0` which gives `t=0` and `t=4s`. `because"average speed"=(|int_(0)^(4)vdt|int_(4)^(5)vdt|)/(int_(0)^(5)dt)=(|int_(0)^(4)(4t-t^(2))dt|+|int_(4)^(5)vdt|)/(5)` `=([2t^(2)-(t^(3))/(3)]_(0)^(4)+[2t^(2)-(t^(3))/(3)]_(4)^(5))/(5)` `=(|[2t^(2)-(t^(3))/(3)]_(0)^(4)|+|[2t^(2)-(t^(3))/(3)]_(4)^(5)|)/(5)=(13)/(5)ms^(-1)` For acceleration: `a=(dv)/(dt)=(d)/(dt)(4t-t^(2))=4-2t` At `t=0,a=4ms^(-2)` |
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| 66. |
A particle moves with an initial `v_(0)` and retardation alphav, where v is its velocity at any time t. (i) The particle will cover a total distance `v_(0)/alpha`. (ii) The particle will come to rest after time `1/alpha`. (iii) The particle will continue to move for a very long time. (iv) The velocity of the particle will become `v_(0)/2` after time `(1n2)/alpha`A. (i), (ii)B. (ii), (iii)C. (i), (ii), (iv)D. All |
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Answer» Correct Answer - C `(dv)/(dt)=-alpha v` `int_(v_(0))^(v)(dv)/v=-alphaint_(0)^(t) dt` `| log_(e)v|_(0)^(v)=-alpha|t|_(0)^(t)` `log_(e)v-log_(e)v_(0)=-alpha t` `log_(e)(v//v_(0))=-alpha t` `v//v_(0)=e^(-alpha t)` Velocity in terms of t `v=v_(0)e^(-alphat) ...(i)` `(ds)/(dt)=v_(0)e^(-alphat)` `int_(0)^(s) ds=v_(0)int_(0)^(t) e^(-alphat)dt` `s=v_(0)|e^(-alphat)|_(0)^(t)/-alpha=v_(0)/-alpha(e^(-alphat)-e^(0))` `s=v_(0)/alpha(1-e^(-alphat)) ...(ii)` `v=0impliese^(-alphat)=0impliest=oo`, (iii) is O.K. At `t=oo, s=v_(0)/alpha,` (i) is O.K. `v=v_(0)e^(-alphat)impliesv_(0)/2=v_(0) e^(-alphat)` `e^(alphat)=2impliesalphat log_(e)e=log_(e)2` `t=(loge^(2))/alpha=(In 2)/alpha`, (iv) is O.K. |
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| 67. |
Two cars start in a race with velocities `u_(1)` and `u_(2)` and travel in a straight line with acceleration `alpha` and `beta`. If both reach the finish line at the same time, the range of the race isA. `(2(u_(1)-u_(2)))/((beta-alpha)^2)(u_(1)beta-u_(2)alpha)`B. `(2(u_(1)-u_(2)))/(beta+alpha)(u_(1)alpha-u_(2)beta)`C. `((2u_(1)-u_(2))^(2))/((beta-alpha)^(2))`D. `(2u_(1)u_(2))/(betaalpha)` |
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Answer» Correct Answer - A Range`=` dist. Covered before they meet `S=t(u_(1)+(1)/(2)alpha t)=t(u_(2)+(1)/(2)beta t)` Sove for `t` and substitute in the above equation. |
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| 68. |
A particle moves with an initial velocity `v_(0)` and retardation `alpha v`, where v is the velocity at any time t.A. The particle will cover a total distance `(v_(0))/(alpha)`B. the particle will come to rest after time `(1)/(alpha)`C. the particle will continue to move for a along timeD. The velocity of particle will become `(v_(0))/(e)` after time `(1)/(alpha)` |
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Answer» Correct Answer - A::C::D `v.(dv)/(dx) = - alpha v rArr ("or") underset(v_(0))overset(0)int dv = - alpha underset(0)overset(x_(0))int dx` `v_(0) = alpha x_(0) rArr x_(0) = (v_(0))/(alpha)`, `(dv)/(dt) = - alpha v ("or") underset(v_(0))overset(v)int (dv)/(v) = - alpha underset(0)overset(t)int dt` `v = v_(0)e^(-alpha t) ("or") v = 0` for `t = oo` `rArr v = (v_(0))/(e)` when `t = (1)/(alpha)` |
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| 69. |
A particle moves in a straight line as `s =alpha(t-4)+beta(t-4)^(2)`. Find the initial velocity and acceleration. |
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Answer» `s=alpha(t-4)+beta(t-4)^(2)` `=alpha(t-4)+beta(t^(2)-8t+16)` `v=(ds)/(dt)=alpha(1-0)+beta(2t-8)` `a=(dv)/(dt)=beta(2.1)=2beta` At `t=0, v=alpha-8beta` `a=2beta` |
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| 70. |
The position vertor of a particle varies with time as `overline(r)=overline(r_(0)t)(1-alphat)` where `overline(r)_(0)` is a contant vector and `alpha` is a positive constant. The distance travelled by particle in a time interval in which particle returns to its initial position is `(Kr_(0))/(16alpha)`. Determine the value of K? |
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Answer» Correct Answer - 8 `overline(r)=overline(r_(0))t(1-alphat)` `overline(r)=overline(r_(0))-2alpha t overline(r_(0))` `:. overline(v)=0` at `t=(1)/(2alpha)` `:.` distance travelled before coming to original position `:. r=2[overline(r_(0))/(4alpha)]=overline(r_(0))/(2alpha)` `:. k=8` |
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| 71. |
A balloon is ascending at the rate of `9.8 m//s` and is 39.2 m above the ground when a package is dropped. (a) How long does the package take to reach the ground? (b) with what speed does it hit the ground ? `(g = 9.8 m//s^(2))` |
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Answer» Correct Answer - (a) 4 s (b) 29.4 m/s Taking downward direction `+ve , u = - 9.8 m//s, s = 39.2 m " and " a = g = +9.8 m//s^(2)` (a) `39.2 = - 9.8 t + (1)/(2) (9.8) t^(2)` `4.9 t^(2) - 9.8 t - 39.2 = 0` `t^(2) - 2t - 8 = 0` `(t - 4) (t + 2) = 0` `t = 4s` (b) `v = u + g t = - 9.8 xx 4 = 29.4 m//s` |
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| 72. |
Drops of water fall at regular intervals from the roof of a building of height `h=16m`. The first drop striking the ground at the same moment as the fifth drop is ready to leave from the roof. Find the distance betweent he successive drops. |
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Answer» Step I: Time taken by the first drop to touch the ground `=t=sqrt((2h)/(g))` for `h=16m,t=sqrt((16(2))/(g))=4sqrt((2)/(g))` Time interval between two successive drops is `Deltat=((1)/(n-1))t=((1)/(4))t=sqrt((2)/(g))` Where `n=` number of drops Step II: Distance travelled by `1^(st)` drop `S_(1)=(1)/(2)g(4Deltat)^(2)=(1)/(2)g(16)((2)/(g))=16m` Distance travelled by `2^(nd)` drop `S_(2)=(1)/(2)g(3Deltat)^(2)=(1)/(2)g(9)((2)/(g))=9g` Distance travelled by `3^(rd)` drop `S_(3)=(1)/(2)g(2Deltat)^(2)=(1)/(2)g(4)((2)/(g))=4m` Distance travelled by `4^(th)` drop `S_(4)=(1)/(2)g(Deltat)^(2)=(1)/(2)g((2)/(g))=1m` Distance between `1^(st)` and `2^(nd)` drops`=S_(1)-S_(2)=16-9=7`m Distance between `2^(nd)` and `3^(rd)` drops`=S_(2)-S_(3)=9-4=5m` Distance between `3^(rd)` and `4^(th)` drops `=S_(3)-S_(4)=4-1=3m` Distance between `4^(th)` and `5^(th)` drops`=S_(4)-S_(5)=1-0=1m` |
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| 73. |
Which of the following statements are ture?A. A body can have constant speed but varying velocityB. A body can have constant velocity but varying speedC. A body can have acceleration without having velocityD. acceleration is `4 ms^(-2)` at `t = 0` |
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Answer» Correct Answer - A::C::D In circular motion the speed may be constant but velocity is variable. |
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| 74. |
If a body is accelerating:A. it may speed upB. it may speed downC. it may move with same velocityD. it may move with same speed |
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Answer» Correct Answer - A::B::D for an accelerating body speed may increase or decrease |
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| 75. |
A particle executes the motion described by `x (t)=x_(0) (1-e^(-gamma t)) , t gt =0, x_0 gt 0`. (a) Where does the particle start and with what velocity ? (b) Find maximum and minimum values of ` x (t) , a (t)`. Show that ` x (t) and a (t)` increase with time and ` v(t)` decreases with time.A. `x_(0) and 0`B. `x_(0)gamma` and 0C. `0` and `-x_(0)gamma^(2)`D. `x_(0)(1-e^(-gamma))` |
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Answer» Correct Answer - A `x(t)=x_(0)(1-e^(-gammat))` `v(t)=(dx(t))/(dt)=x_(0)gammae^(-gammat)` `a(t)=(dv(t))/(dt)=-x_(0)gamma^(2)e^(-gammat)` |
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| 76. |
Figure shows the displacement time (x-t) graph of a body moving in a straight line which one of the graph shown in figure represents the velocity- time (v-t) graph of the motion of the body.A. B. C. D. |
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Answer» Correct Answer - D 0 to 5 s, velocity is `+ve` and constant 5 to 15 s : slope is zero 15 to 20 s: velocity is `-ve` and constant |
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| 77. |
Two particles A and B start from rest at the origin `x=0` and move along a straight line such that `a_(A)=(6t-3)ms^(-2)` and `a_(B)=(12t^(2)-8)ms^(-2)` where t is in seconds based on the above facts answer the following questions, Total distance travelled by A at `t=4` s isA. 40 mB. 41 mC. 42 mD. 43 m |
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Answer» Correct Answer - B For A `dv_(A)=a_(A)dtimpliesint_(0)^(v_(A))dv_(A)=int_(0)^(t)(6t-3)dtimpliesV_(A)=3t^(2)-3t` `dv_(B)=a_(B)dtimpliesint_(0)^(v_(0))dv_(B)=int_(0)^(t)(12t^(2)-B)dt` `impliesV_(B)=4t^(3)-8t` Let us now calculate the times when A and B are at rest. The particle A is rest `(V_(A)=0)` when `3t^(2)-3t=0impliest=0` and `t=1`s The particle B is at rest `(V_(B)=0)`, when `4t^(3)-8t=0impliest=0s` and `t=sqrt(2)s` The position of particles A and B can be determined using `v=(dx)/(dt)` so `dx_(A)=V_(A)dt` `impliesint_(0)^(X_(A))dx_(A)=int_(0)^(t)(3t^(2)-3t)dt` `impliesX_(A)=t^(3)-(3)/(2)t^(2)` similarly `dX_(B)=V_(B)dt` `impliesint_(0)^(X_(B))dx_(B)=int-(0)^(t)(4t^(3)-8t)dtimpliesX_(B)=t^(4)-4t^(2)` The position of particle A at `t=1` s and 4 s are `X_(A)|t=1s``=1^(3)-(3)/(2)-(3)/(2)(1^(2))=0.5m` `X_(A)|t=4s``=4^(3)-(3)/(2)(4^(2))=40m` Particle A has travelled a total distance given by `d_(A)=2(0.5)+40=41m` The position of particle B at `t=sqrt(2)`s and 4 s are `X_(B)|t=sqrt(2)=(sqrt(2))^(4)-4(sqrt(2))^(2)=-4m` `X_(B)|t=4=(4)^(4)-4(4)^(2)=192m` Particle (B) has travelled a total distance given by `d_(B)=2(4)+192=200m` At `t=4s` s the distance between A and B is `DeltaX_(AB)=192-40=152m` |
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| 78. |
Assertion : In one dimensional motion of an object, path length comes out to be equal to displacement. Reason : Direction of object is changed two times.A. If both assertion and reason are trure and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanations of assertion.C. If assertion is true but reason is false.D. If both assertion and reason is false |
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Answer» Correct Answer - (c ) (c ) : In one dimensional motion, path length is equal to displacement only when direction of motion remains unchanged. |
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| 79. |
A particle moving in a straight in a straight line has velocity and displacement equation as `v = 4 sqrt(1 + s)`, where `v` is in m/s and s is in m. The initial velocity of the particle is:A. 4 m/sB. 16 m/sC. 2 m/sD. zero |
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Answer» Correct Answer - A `v = 4 sqrt(1 + s)` `v^(2) = 16 + 16s` `v^(2) - 16 = 16s` `v^(2) - (4)^(2) = 2 (8) s` `:.` Initial velcoity `= 4 m//s` |
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| 80. |
For motion of an object along the x-axis the velocity v depends on the displacement x as `v=3x^(2)-2x`, then what is the acceleratiobn at `x=2m`.A. `48ms^(-2)`B. `80ms^(-2)`C. `18ms^(-2)`D. `10ms^(-2)` |
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Answer» Correct Answer - B Given `v=3x^(2)-2x`, differentiating v, we get `(dv)/(dt)=(6x-2)(dx)/(dt)=(6x-2)v` `impliesa=(6x-2)(3x^(2)-2x)` now put `x=2` m `impliesa=(6xx2-2)(3(2)^(2)-2xx2)=80ms^(-2)` |
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| 81. |
Two trains, each `50 m` long, are travelling in opposite directions with velocities `10 ms^-1 and 15 ms^-1`. The time of their crossing each other is.A. 8 sB. 4 secC. 2 sD. 6 s |
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Answer» Correct Answer - B `t=(1_(1)+1_(2))/(v_(r))=(1_(1)+1_(2))/(v_(1)+v_(2))` |
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| 82. |
Two trains 121 m and 99 m in length are running in opposite directions with velocities `40 km h^(-1)` and `32 km h^(-1)`. In what time they will completely cross each other?A. 9 sB. 11 sC. 13 sD. 15 s |
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Answer» Correct Answer - (b) Here, `v_A = 40 km h^(-1), v_B = -32 km h^(-1)` Length of train A, `l_A = 121` m, Length of train B, `l_b` = 99 m Relative velocity of two trains is given by `v_(AB) = v_A - v_B = 40 - (-32) = 72 km h^(-1) = 72 xx 5/18 = 20 m s^(-1)` Total distance to be travelled by each train for completely crossing the other train `= 121 + 99 = 220` m Time taken by each train to cross the other train `= 220 / 20 = 11` s Hence the two trains will cross each other in 11 s. |
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| 83. |
Two bodies move in a straight line towards each other at initial velocities `v_(1)` and `v_(2)` and with constant acceleration `a_(1)` and `a_(2)` directed against the corresponding velocities at the initial instant. What must be the maximum initial separation between the bodies for which they meet during the motion ?A. `(v_(1)^(2))/(a_(1)) + (v_(2)^(2))/(a_(2))`B. `((v_(1) + v_(2))^(2))/(2(a_(1) + a_(2)))`C. `(v_(1)v_(2))/(sqrt(a_(1)a_(2)))`D. `(v_(1)^(2) - v_(2)^(2))/((a_(1) - a_(2)))` |
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Answer» Correct Answer - B `v_("rel") = v_(1) + v_(2)` `a_("rel") = (a_(1) + a_(2))`, directed oppposite to motion. Using, `v^(2) - u^(2) = 2 as`, we have `(0)^(2) - (v_(1) + v_(2))^(2) = - 2 (a_(1) + a_(2))s` `s = ((v_(1) + v_(2))^(2))/(2(a_(1) + a_(2)))` |
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| 84. |
The relation `3t=sqrt(3x)+6` describe the displacement of a particle in one direction where x is in metres and t in sec. The displacement, when velocity is zero isA. 24 mB. 12 mC. 5 mD. zero |
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Answer» Correct Answer - D `3t=sqrt(3x)+6` `x=3(t-2)^(2)` `v=(dx)/(dt)=6(t-2)` When `v=0impliest=2 s` Atv `t=2 s, x=0` |
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| 85. |
The displacement of a body along the x-axis depends on time as `sqrt(x)=t+2`, then the velocity of bodyA. increases with timeB. decreases with timeC. independent of timeD. None of these |
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Answer» Correct Answer - A `sqrt(x)=t+2impliesx=(t+2)^(2)` `v=(dx)/(dt)=2(t+2)` As `t` increases, `v` increases. |
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| 86. |
Statement-1 : The relative velocity between any two bodies moving in opposite direction is equal to sum of the velocities of two bodies. Statement-2 : Sometimes relative velocity between two bodies is equal to difference in velocities of the two.A. If both assertion and reason are trure and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanations of assertion.C. If assertion is true but reason is false.D. If both assertion and reason is false |
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Answer» Correct Answer - (b) When two bodies are moving in opposite directions, relative velocity between them is equal to sum of the velocity of bodies. But if the bodies are moving in same direction their relative velocity is equal to difference in velocity of the bodies. |
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| 87. |
The velocity of a particle is `v = v_(0) + g t + ft^(2)`. If its position is ` x=0 ` at ` t= 0 `, then its displacement after unit time `( t = 1 )` isA. `v_(0)+g//2+f`B. `v_(0)+2g+3f`C. `v_(0)+g//2+f//3`D. `v_(0)+g+f` |
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Answer» Correct Answer - C |
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| 88. |
The velocity of a particle is `v = v_(0) + g t + ft^(2)`. If its position is ` x=0 ` at ` t= 0 `, then its displacement after unit time `( t = 1 )` isA. `v_(0) + g//2 + f`B. `v_(0) + 2g + 3f`C. `v_(0) + g//2 + f//3`D. `v_(0) + g + f` |
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Answer» Correct Answer - C `underset(0)overset(x)int dx = underset(0)overset(1)int vt = underset(0)overset(1)int (v_(0) + g t + f t^(2)) dt` `x = [v_(0)t + (g t^(2))/(2) + (f t^(3))/(3)]_(0)^(1)` `x = (v_(0) + (g)/(2) + (f)/(3))` |
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| 89. |
The velocity of a particle is `v = v_(0) + g t + ft^(2)`. If its position is ` x=0 ` at ` t= 0 `, then its displacement after unit time `( t = 1 )` isA. `V_(0)+2g+3f`B. `V_(0)+(g)/(2)+(f)/(3)`C. `V_(0)+g+f`D. `V_(0)+(g)/(2)+f` |
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Answer» Correct Answer - B Given velocity of particle ,`v=v_(0)+g t+ft^(2)` `:.` displacement `s=int vdt` |
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| 90. |
The velocity of a particle is `v = v_0 + g t + ft^2`. If its position is `x = 0` at `t = 0`, then its displacement after unit time `(t = 1)` is.A. `v_(0)+2g+3f`B. `v_(0)+g/2+f/3`C. `v_(0)+g+f`D. `v_(0)+g/2+f` |
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Answer» Correct Answer - B `v=(dx)/(dt)=v_(0)+g t+ft^(2)` `int_(0)^(x) dx=int_(0)^(1)(v_(0)+g t+ft^(2))dt` `x=|v_(0)t+(g t^(2))/2+(f t^(3))/3|_(0)^(1)` `=v_(0)+g/2+f/3` |
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| 91. |
A particle is moving with velocity `v=4t^(3)+3 t^(2)-1 m//s`. The displacement of particle in time `t=1 s` to `t=2 s` will beA. 21 mB. 17 mC. 13 mD. 9 m |
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Answer» Correct Answer - A `v=(ds)/(dt)=4t^(3)+3t^(2)-1` `int_(0)^(s) ds=int_(1)^(2)(4t^(3)+3t^(2)-1)dt` `s=|t^(4)+t^(3)-t|_(1)^(2)` =`{(2)^(4)+(2)^(3)-(2)}-{(1)^(4)+(1)^(3)-(1)}` `=22-1=21 m` |
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| 92. |
Assertion : Two bodies of unequal masses `m_1 and m_2` are dropped from the same height. If the resistance offered by air to the motion of both bodies is the same, the bodies will reach the earth at the same time. Reason : For equal air resistance, acceleration of fall of masses `m_1 and m_2` will be different.A. If both A and B are true and R is the correct explanation of AB. If both A and R are true but R is not correct explantion of AC. If A is true but R is falseD. If A is false but R is true |
| Answer» Correct Answer - D | |
| 93. |
A: A body is released from a height. As it is falling vertically downwards, at some position, it explodes into fragments under purely internal force. Centre of mass of the system of fragments will keep moving along the original vertical line and also accelerate downwards with an acceleration g. R: Whenever linear momentum of a system is conserved, its centre of mass always remains at rest.A. If both A and B are true and R is the correct explanation of AB. If both A and R are true but R is not correct explantion of AC. If A is true but R is falseD. If A is false but R is true |
| Answer» Correct Answer - C | |
| 94. |
Which of the following statements `is//are` correct ?A. If the velocity of body changes it must have some acceleration.B. If the speed of a body changes it must have some acceleration.C. If the body has acceleration its speed must changeD. If the body has acceleration its speed may change. |
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Answer» Correct Answer - A::B::D `a=(dv)/(dt)`, if velocity changes, definetly there will be acceleration if speed changes then velocity also changes so definietly there will be acceleration. Acceleration may be due to change in the direction of velocity only and not magnitude. If body has acceleration its speed may changes if acceleration is due to change in magnitude of velocity. |
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| 95. |
A tennis ball is released so that it falls vertically to the floor and bounces again. Taking velocity upwards as positive, which of the following graphs best represents the variation of its velocity `v` with time `t`?A. B. C. D. |
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Answer» Correct Answer - C Intitally velocity increase along `-ve` axis then it becomes `+ve` after collision. |
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| 96. |
A rocket is fired and ascends with constant vertical acceleration of `10 m//s^(2)` for 1 minute. Its fuel is exhausted and it continues as a free particle. The maximum altitude reached is `(g=10 m//s^(2))`A. 18 kmB. 36 kmC. 72 kmD. 108 km |
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Answer» Correct Answer - B `h=(1)/(2)at^(2),v=at` max.height`=H=(v^(2))/(2g)` Total distance from the ground`=(H+h)` `=(1)/(2)at^(2)(1+(a)/(g))` |
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| 97. |
A car is moving wth a velocity of `40 m//s` the driver sees a stationary truck a head at a distance of 200 m. After some reaction time `Delta t` the breaks are applied producing a (reaction) retardation of `8 m//s^(2)`. What is the maximum reaction time to avoid collision ? |
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Answer» The car before coming to rest (v=0) `v_(2)=u^(2) +2 "as" to 0(40)^(2)-2 xx 8 xx s` `impliess=100m` The distance travelled by the car is `100m` To avoid the clash, the remaining distance `200-100=100m` must be covered by the car with uniform velocity `40 m//s` during the reaction time `Deltat`. `(100)/(Deltat)=40impliesDeltat=2.5s` The maximum reaction time `Deltat=2.5s` |
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| 98. |
A body moves from one corner of an equilateral Delta of side 10 cm to the same corner along the sides. Then the distance and displacement are respectively.A. 30 cm & 10 cmB. 30 cm & 0 cmC. 0 cm & 30 cmD. 30 cm & 30 cm |
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Answer» Correct Answer - B Displacement`=`shortest distance between initial point and final point. |
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| 99. |
If the distance between the sun and the earth is `1.5xx10^(11) m` and velocity of light is `3xx10^(8) m//s`, then the time taken by a light ray to reach the earth from the sun isA. 500 sB. 500 minutesC. 50 sD. `5xx10^(3)s` |
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Answer» Correct Answer - A `s=vt,t+(s)/(v)` |
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| 100. |
A body X is thrown vertically upwards with an initial speed 45 m/s. Another body Y is also thrown vertically upwards with an initial speed 27 m/s. During the last `(1)/(2)` sec of motion of each body, speed of each reduces by the same value. R : Both bodies are moving with same valueA. If both A and B are true and R is the correct explanation of AB. If both A and R are true but R is not correct explantion of AC. If A is true but R is falseD. If A is false but R is true |
| Answer» Correct Answer - A | |