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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. |
From the following displacement-time graph find out the velocity of a moving body. A. `(1)/(sqrt(3)) m//s`B. `3 m//s`C. `sqrt(3) m//s`D. `(1)/(3) m//s` |
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Answer» Correct Answer - C In first instant you will apply `v = tan theta` and say, `v = tan 30^@ = (1)/(sqrt(3)) m//s`. But it is wrong because formula `v = tan theta` is valid when angle is measured with time axis. Here angle is taken from displacment axis. So angle from time axis `90^@ - 30^@ = 60^@` Now `v = tan 60^@ = sqrt(3)`. |
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| 52. |
The position vector of a aprticle is given as `vecr=(t^2-4t+6)hati+(t^2)hatj`. The time after which the velocity vector and acceleration vector becomes perpendicular to each other is equal toA. 2 secB. 1 secC. 1.5 secD. 5 sec |
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Answer» Correct Answer - B `vec r =(t^2 - 4t + 6) hat i + t^2 hatj , vec v = (d vec r)/(dt)` =`(2t - 4)hat i + 2t hat j, vec a = (d vec v)/(dt) = 2 hat i + 2 hat j` If `vec a` and `vec v` are perpendicular, `vec a. vec v = 0` `(2 hat i + 2 hat j).((2 t - 4) hat i + 2t hat j) = 0` `8t - 8 = 0, t = 1 sec`. |
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| 53. |
Figure shows a velocity-time graph. This shows that. A. the body is at restB. the body starts from rest and moves with uniform velocityC. the body has some initial velocity and moves with uniform accelerationD. the motion is retarded |
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Answer» Correct Answer - C Body as clear from the graph has some initial velocity and then it moves with uniform acceleration as the slope of straight line is constant. |
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| 54. |
A stone is thrown upwards with a velocity `50 ms^-1`. Another stone is simultaneously thrown downwards from the same location with a velocity `50 ms^-1`. When the first stone is at the highest point, the velocity of the second stone is `("Take" g = 10^-2)` :A. ZeroB. `50 ms^-1`C. `100 ms^-1`D. `150 ms^-1` |
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Answer» Correct Answer - C `t = (50)/(g) = (50)/(10) = 5 s` `v = -50 - 10 xx 5 = -100 m//s`. |
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| 55. |
A man in a balloon throws a stone downwards with a speed of `5 m//s` with respect to balloon. The balloon is moving upwards with a constant acceleration of `5 m//s^2`. Then velocity of the stone relative to the man after `2` seconds is. A. `10 m//s`B. `30 m//s`C. `15 m//s`D. `35 m//s` |
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Answer» Correct Answer - D Relative velocity of stone `= 5 m//s` Relative acceleration of stone =` 10 + 5 = 15 m//s^2` `v = u+ at = 5 + 15 xx 2 = 35 m//s` relative velocity after `t = 2` seconds is `35 m//s`. |
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| 56. |
A plank is moving on ground with a velocity `v` and a block is moving on the plank with respect to it with a velocity `u` as shown in figure. What is the velocity of block with respect to ground ? A. `v - u` towards rightB. `v - u` towards leftC. `u` towards rightD. none of these |
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Answer» Correct Answer - A `vec v_b = vec v_(b,p) + vec v_p` =`(-u) + v`(toward sign) `= v - u`. |
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| 57. |
A juggler keeps on moving four balls in the air throwing the balls after regular intervals. When one ball leaves his hand `("speed" = 20 ms^-1)` the positions of other balls (height in m) `("Take" g = 10 ms^-2)`.A. `10,20,10`B. `15, 20,15`C. `5, 15,20`D. `5,10,20` |
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Answer» Correct Answer - B Time taken by same ball to return to the hands of juggler `= (2 u)/(g) = (2 xx 20)/(10) = 4 s`. So he is throwing the balls after each `1 s`. Let at some instant he is throwing ball number `4`. Before `1 s` of it he throws ball `3`. So height of ball `3` : `h_3 = 20 xx 1 - (1)/(2) 10(1)^2 = 15 m` Before `2 s`, he throws ball `2`. So height of ball `2` : `h_2 = 20 xx 2 - (1)/(2) 10(2)^2 = 20 m` Before `3 s`, he throws ball `1`. So height of ball `1` : `h_1 = 20 xx 3 - (1)/(2) 10(2)^2 = 20 m`. |
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| 58. |
A juggler maintains four balls in motion, making each to them to rise a height of `20 m` from his hand. What time interval should he maintain, for the alphaer distance between them.A. 3 sB. `(3)/(2) s`C. 1 sD. 2 s |
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Answer» Correct Answer - C From `v^2 = u^2 - 2 as` we have `0 = u^2 - 2(10)(20)` or `u = 20 m//s` Also `v = u - at` or `0 = 20 - 10 t` or `t = 20 s` So, the ball returns to the hand of the juggler after `4 s`. To maintain alphaer distance, the balls must be thrown up at an interval of `(4)/(4)` or `1 s`. |
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| 59. |
A body is dropped from a height of `40 m`. After it crosses half distance, the acceleration due to gravity ceases to act. The body will hit ground with velocity `("Take" g = 10 m//s^2)` :A. `20 m//s`B. `10 m//s`C. `2 m//s`D. `15 m//s` |
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Answer» Correct Answer - A Suppose `v` be the velocity of the body after falling through half the distance. Then `s =(40)/(2) = 20 m, u = 0 and g = 10 m//s^2` `v^2 = u^2 + 2gh = 0^2 + 2 xx 10 xx 20` `rArr v = 20 m//s`. When the acceleration due to gravity ceases to act, the body travels with the uniform velocity of `20 m//s`. So it hits the ground with velocity `20 m//s`. |
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| 60. |
Assertion: The slope of displacement-time graph of a body moving with high velocity is steeper than the slope of displacement-time graph of a body with low velocity. Reason: Slope of displacement-time graph = Velocity of the body.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - A Since slope of displacement-time graph measures velocity of an object. |
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| 61. |
A particle starting from rest has a constant acceleration of `4 m//s^2` for 4 s. It then retards uniformly for next 8 s and comes to rest. Find during the motion of particle (a) average acceleration (b) average speed and (c) average velocity.A. `4 m//s^2`B. zeroC. `8 m//s^2`D. `- 4 m//s^2` |
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Answer» Correct Answer - B Average acceleration is zero, because change in velocity is zero. |
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| 62. |
Assertion: In a uniformly accelerated motion, acceleration the graph is straight line with positive slope. Reason: Acceleration is rate of change of velocity.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assersion is false but reason is true. |
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Answer» Correct Answer - D Positive slope indicates that acceleration increases uniformly with time. It is not uniform. |
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| 63. |
In a car race, car `A` takes `t_0` time less to finish than car `B` and passes the finishing point with a velocity `v_0` more than car `B`. The cars start from rest and travel with constant accelerations `a_1 and a_2`. Then the ratio `(v_0)/(t_0)` is equal to.A. `(a_1^2)/(a_2)`B. `(a_1 + a_2)/(2)`C. `sqrt(a_1 a_2)`D. `(a_2^2)/(a_1)` |
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Answer» Correct Answer - C Let `s` be the distance traveled by each car. Then `sqrt(2 a_1 s) -sqrt(2 a_2 s) = v_0 and , sqrt((2 s)/(a^2)) - sqrt((2s)/(a_1)) = t_0` `:. (v_0)/(t_0) = (sqrt(a_1) - sqrt(a_2))/(1/(sqrt(a_2))-(1)/(sqrt(a_1)))= sqrt(a_1 a_2)`. |
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| 64. |
A car moving at `160 km//h` when passes the mark-A, driver applies brake and reduces its speed uniformly to `40 km//h` at mark-C. The mark are spaced at equal distances along the road as shown below. At which part of the track the car has instantaneous speed of `100 km//h` ? Neglect the size of the car. A. A mark-BB. Between mark-A and mark-BC. between mark-B and mark-CD. insufficient information to decide. |
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Answer» Correct Answer - C From `A` to `C` : `(40)^2 - (160)^2 = 2a (AC)` From `A` to `D` : `(100)^2 - (160)^2 = 2a (AD)` Here `D` is the point where speed is `100 km//h`. From above `AD = (13)/(20) (AC) and AB = (AC)/(2)` We see that AD gt AB`. |
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| 65. |
From the top of a tower, a particle is thrown vertically downwards with a velocity of `10 m//s`. The ratio of the distances, covered by it in the `3rd` and `2nd` seconds of the motion is `("Take" g = 10 m//s^2)`.A. `5 : 7`B. `7 : 5`C. `3 : 6`D. `6 : 3` |
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Answer» Correct Answer - B `S_(3rd) = 10 + (10)/(2)(2 xx 3 -1) = 35 m` `S_(2nd) = 10 + (10)/(2) (2 xx 2 - 1) = 25 m rArr (S_(3rd))/(S_(2nd)) = (7)/(5)`. |
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| 66. |
In which of the graphs both net displacement and velocity are negative ? A. Graph (iii) and (iv)B. Graph (i) and (ii)C. Graph (i) onlyD. Graph (iv) only |
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Answer» Correct Answer - C For graph (i) net displacement and velocity are both negative. |
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| 67. |
Which of the following is a one-dimensional motion ?A. Landing of an aircraftB. Earth revolving around the sunC. Motion of wheels of a moving trainsD. Train running on a straight track |
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Answer» Correct Answer - D All other motions are not along the straight line except (d). |
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| 68. |
Figure shows the position the graph for a particle in one dimensional motion. Which of the graphs in fugure represent the variation in the velocity of the particle with time ? A. .B. .C. .D. . |
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Answer» Correct Answer - C Initially slope is positive, so velocity is positive. Then slope is zero, so velocity is zero. The slope is negative so velocity is negative. |
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| 69. |
A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to `v(x) = beta x^(-2 n)` where `beta` and `n` are constant and `x` is the position of the particle. The acceleration of the particle as a function of `x` is given by.A. `- 2 n beta^2 x^(-2 n - 1)`B. `- 2n beta^2 x ^(- 4n - 1)`C. `-2 n beta^2 x^(-2 n + 1)`D. `-2 b beta^2 e^(-4 n + 1)` |
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Answer» Correct Answer - B We are given velocity of the particle `v(x) = beta x^(- 2n)` We know acceleration `a = v (dv)/(dx)` `a = beta x^(- 2n) (d)/(dx) (beta x^(- 2n))` =`beta^2 x^(- 2 n) (- 2 n)x^(- 2 n - 1) = -2 n beta^2 x^(-2 n - 1 - 2n)` `a = - 2n beta^2 x^(-4 n - 1)`. |
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| 70. |
It takes one minute for a passenger standing on an escalator to reach the top. If the escalator does not move it takes him `3` minute to walk up. How long will it take for the passenger to arrive at the top if he walks up the moving escalator ?A. 30 secB. 45 secC. 40 secD. 35 sec |
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Answer» Correct Answer - B Let speed of elevator be `v_e`. `t_1 = (L)/(v_e) rArr 1 min = (L)/(v_e)` Let speed of person relative to elevator be `v_P`, then `t_2 = (l)/(v_p) rArr 3 min = (L)/(v_p)` when the escalator is moving : `t_3 = (L)/(v_e + v_p) = (L)/((L)/(1 min) + (L)/(3 min)) = (3)/(4) min = 45 s`. |
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| 71. |
An elevator is accelerating upwards with a constant acceleration a `ms^-2`. If a coin is dropped in it by a passenger, then.A. the coin starts to fall downwards instantaneously as observed by the passenger.B. the coin starts to fall downwards instantaneously as observed by a ground observer.C. first, coin continues to move up with same acceleration of a `ms^-2` for some time and then starts falling with gravitational acceleration as observed by a ground observer.D. None of these |
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Answer» Correct Answer - A After releasing, the coin will come under the effect of gravity only, so it will move with acceleration due to gravity `g` downwards. When the coin in dropped it has velocity in upwards direction same as that of elevator w.r.t. a person on ground, the coin first will go up and then come down. But w.r.t. a passenger in elevator, it will seem to be falling downwards always because passenger himself has velocity upwards. |
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| 72. |
In `Q.36`, average velocity during the motion of the particle is.A. zeroB. `8 m//s`C. `2 m//s`D. `4 m//s` |
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Answer» Correct Answer - B Average velocity and average speed are same because particle travels in the same direction throughout. |
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| 73. |
In previous problem, average speed during the motion of the particle is.A. `8 m//s`B. zeroC. `4 m//s`D. `16 m//s` |
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Answer» Correct Answer - A Distance travelled `s = (1)/(2) a_1 t_1^2 + (1)/(2) a_2 t_2^2 = (1)/(2) 4(4)^2 + (1)/(2)(2)(8)^2` [Acceleration in second part will be half of first part because time taken is double] `rArr s = 96 mv_(av) = (96)/(4 + 8) = 8 m//s`. |
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| 74. |
When a ball is thrown up vertically with velocity `v_0`, it reaches a maximum height of h. If one wishes to triple the maximum height then the ball should be thrown with velocityA. `sqrt(3 V_o)`B. `3 V_o`C. `9 V_o`D. `3//2 V_o` |
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Answer» Correct Answer - A `H_(max) prop u^2` :. `u prop sqrt(H_(max))` i.e., to triple the maximum height, ball should be thrown with velocity `sqrt(3) u`. |
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| 75. |
A ball is thrown vertically upwards. Which of the following plots represents the speed-time graph of the ball during its height if the air resistance is not ingnored ?A. .B. .C. .D. . |
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Answer» Correct Answer - C As ball is thrown upwards velocity decreases as height increases. At the highest point velocity is zero after that body starts downward motion, its speed increases. |
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| 76. |
Two stones are thrown up simultaneously from the edge of a cliff ` 240 m ` high with initial speed of ` 10 m//s and 40 m//s` respectively . Which of the following graph best represents the time variation of relative position of the speed stone with respect to the first ? ( Assume stones do not rebound after hitting the groumd and neglect air resistance , take ` g = 10 m//s^(2))` ( The figure are schematic and not drawn to scale )A. .B. .C. .D. . |
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Answer» Correct Answer - C For the second stone time required to reach the ground is given by `y = ut - (1)/(2) g t^2` `- 240 = 40 t - (1)/(2) xx 10 xx t^2` :. `5t^2 - 40 t - 240 = 0` `(t - 12)(t + 8) = 0` :. `t = 12 s` For the first stone : `- 240 = 10 t - (1)/(2) xx 10 xx t^2` :. `- 240 = 10t - 5t^2` `5 t^2 - 10 t - 240 = 0` `(t - 8)(t + 6) = 0` `T = 8 s ` During first `8` seconds both stones are in are : :. `y_2 - y_1 = (u_2 - u_1) t = 30 t` So, graph of `(y_2 - y_1)` against `t` is a straight line. After `8` seconds. `y_2 = u_2 t - (1)/(2) g t^2 - 240` Stones two has acceleration with respect to stone one. Hence graph (c ) is the correct description. |
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| 77. |
A body is dropped from a height `h`. If `t_1 and t_2` be the times in covering first half and the next distances respectively, then the relation between `t_1 and t_2` is.A. `t_1 = t_2`B. `t_1 = 2t_2`C. `t_1 = 3t_2`D. `t_1 = (t_2)/((sqrt(2) - 1))` |
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Answer» Correct Answer - D `t_1 = sqrt((2(h//2))/(g)) = sqrt((h)/(g)) rArr t_1 + t_2 = sqrt((2h)/(g))` `rArr t_2 = sqrt((2h)/(g)) - t_1 = (sqrt(2 ) - 1) sqrt((h)/(g))` `rArr (t_1)/(t_2) = (1)/(sqrt(2) - 1) rArr t_1 = (t_2)/(sqrt(2) -1)`. |
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| 78. |
A stone falls freely under gravity. It covered distances `h_1, h_2` and `h_3` in the first `5` seconds. The next `5` seconds and the next `5` seconds respectively. The relation between `h_1, h_2` and `h_3` is :A. `h_1 = 2h_2 = 3h_3`B. `h_1 = (h_2)/(3) = (h_3)/(5)`C. `h_2 = 3h_1` and `h_3 = 3h_2`D. `h_1= h_2 = h_3` |
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Answer» Correct Answer - B For a particle released from a certain height the distance covered by the particle in relation with time is given by, `h = (1)/(2) g t^2` For first `5 sec, h_1 = (1)/(2) g(5)^2 = 125` Further next `5 sec, h_1 + h_2 = (1)/(2) g (10)^2 = 500` `rArr h_2 = 375` `h_1 + h_2 + h_3 = (1)/(2) g(15)^2 = 1125` `rArr h_3 = 625` `h_1 = 3h_1, h_3 = 5h_1` or `h_1 = (h_2)/(3) = (h_3)/(5)`. |
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| 79. |
A body starts from rest and moves with constant acceleration. The ratio of distance covered by the body in `nth` second to that covered in `n` second is.A. `(1)/(n)`B. `(2 n - 1)/(n^2)`C. `(n^2)/(2n - 1)`D. `(2n - 1)/(2 n^2)` |
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Answer» Correct Answer - B `S_n = u + (a)/(2)(2n - 1) = (a)/(2)(2n - 1)" "(because u = 0)` Total distance covered in `n` seconds `S = un + (1)/(2) an^2 = (1)/(2) an^2 , So (S_n)/(S) = (2n - 1)/(n^2)`. |
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| 80. |
Refer `Q.95`. If the man takes `t_1 and t_2` to move to and fro journey respectively, the time taken by him to go downstream (while the man does not swim) is :A. `(2 t_1 t_2)/(t_(1)+t_(2))`B. `sqrt(t_1^2 + t_2^2)`C. `sqrt(t_1 t_2)`D. `(2 t_1 t_2)/(t_1 - t_2)` |
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Answer» Correct Answer - D velocity of man relative to water = v Velocity of water flow = u Velocity of man relative to ground in upstream `= v - u` Velocity of man relative to ground in downstream `= v + u` Now for to and fro motion, time taken to cover distance `d` in upstream `t_1 = (d)/(v - u)` `1/t_(1)=v/d-u/d` ...(i) And time taken to cover distance `d` in downstream `t_2 = (d)/( v+ u)` `(1)/(t_2) = (v)/(d) + (u)/(d)` ...(ii) Now time taken to cover distance `d` in downstream When ` v= 0` `t_3 = (d)/(u)` Subtracting equation (i) from equation (ii) we get `(1)/(t_2) - (1)/(t_1) = (2u)/(d)` Now put value of `u//d` from equation (iii) `(t_1 - t_2)/(t_1 t_2) = (2)/(t_3) rArr t_3 = (2 t_1 t_2)/(t_1 - t_2)`. |
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| 81. |
A pebble is dropped from rest from the top of a till cliff and falls `4.9 m` after `1.0 s` has elapsed. How much farther does it drop in the next `2.0 s` ? `("Take" g - 9.8 m//s^2)`.A. 9.8B. 19.6 mC. 39 mD. 44 m |
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Answer» Correct Answer - C We take downward as the positive direction with `y = 0 and t = 0` at the top of the cliff. The freely falling pebble then has `v_0 = 0 and a = g = +9.8 m//s^2`. The displacement of the pebble at `t = 1.0 s` is given by `y_1 = 4.9 m`. The displacement of the pebble at `t = 3.0 s` is found from. `y_3 = v_0 t + (1)/(2) at^2 = 0 + (1)/(2) (9.80 m//s^2)(3.0 s)^2 = 44 m` The distance fallen in the `2.0 -s` interval from `t = 1.0 s` to `t = 3.0 s` is then `Delta y = y_3 - y_1 = 44 m - 4.9 m = 39 m`. |
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| 82. |
A car starts from rest and moves with uniform acceleration a on a straight road from time `t=0` to `t=T`. After that, a constant deceleration brings it to rest. In this process the average speed of the car isA. `(aT)/(4)`B. `(3aT)/(2)`C. `(aT)/(2)`D. `aT` |
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Answer» Correct Answer - C For first part, `u = 0, t = T` and acceleration `= a` `:. v = 0 + aT= aT` and `S_1 = 0 + (1)/(2) aT^2 = (1)/(2) aT^2` For second part, `u = aT`, retardation `= a_1, v = 0` and time taken`= T_1` (let) `:. 0 = u - a_1 T_1 rArr aT = a_1 T_1` and from `v^2 = u^2 - 2 aS_2 rArr S_2 = (u^2)/(2 a_1) = (1)/(2) (a^2 T^2)/(a_1)` `S_1 = (1)/(2) aT xx T_1" "("As " a_1 = (a T)/(T_1))` `:. v_(av) = (S_1 + S_2)/(T + T_1) = ((1)/(2) aT^2 + (1)/(2) aT xx T_1)/(T + T_1)` =`((1)/(2) aT(T + T_1))/(T + T_1) =(1)/(2) aT`. |
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| 83. |
A ball is thrown straight up with a velocity at `t = 0` and returns to earth at `t = t_1`. Which graph shows the correct motion ?A. .B. .C. .D. . |
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Answer» Correct Answer - C `v = v_1 - g t` This is a straight line graph with negative slope `(-g)` and positive intercept `(v_1)`. |
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| 84. |
Two spheres of same size, one of mass `2 kg` and another of mass `4 kg`, are dropped simultaneously from the top of Quata Minar `(height = 72 m)`. When they are `1 m` above the ground, the two spheres have the same.A. momentumB. kinetic energyC. potential energyD. acceleration |
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Answer» Correct Answer - D When two spheres are dropped they will acquire the same acceleration which is due to gravitational effect. And also the acceleration due to gravity is independent of mass of the body. Hence, the two spheres have the same acceleration. |
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| 85. |
Particle beginning from rest, travels a distance `S` with uniform acceleration and immediately after travels a distane of `3 S` with uniform velocity followed by a distance `5 S` with uniform deceleration, and comes to rest. Then the ratio of average speed to the maximum speed of the particle is.A. `1//5`B. `2//5`C. `(3)/(5)`D. `(4)/(5)` |
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Answer» Correct Answer - C `v_0 rarr maxiumum speed` `s = (v_0 + 0)/(2) t_1 rArr t_(1) = (25)/(v_0)` `t_2 = (35)/(v_0)` `5 s = (v_0 + 0)/(2) t_2 rArr t_3 = (10 s)/(v_0)` `v_(av) = (s + 3s + 5s)/(t_1 + t_2 + t_3)` `v_(av) = (9 s)/((2s)/v_0+ (3 s)/(v_0) +(10 s)/(v_0)) rArr (v_(av))/(v_0) = (3)/(5)`. |
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| 86. |
A moving car possesses average velocities of `5 ms^-1, 10 ms^-1 and 15 ms^-1` in the first, second and third seconds respectively. What is the total distance covered by the car in these three seconds ?A. 15 mB. 30 mC. 55 mD. None of these |
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Answer» Correct Answer - B distance covered `= s = v_(av) xx time` For `1st` second : `S_1 = 5 xx 1 = 5 m` For `2nd` second : `S_(2)=10xx1=10 m` For second `3rd` second : `S_1 = 15 xx 1 = 15 m` Total distance traveled `S = S_1 + S_2 + S_3 = 5 + 10 + 15 = 30 cm`. |
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| 87. |
Each of the three graphs represents acceleration versus time for an object that already has a positive velocity at time `t_1`. Which graphs show an object whose speed is increasing for the entire time interval between `t_1 and t_2` ? A. graph `I`, onlyB. graphs `I and II`, onlyC. graphs `I and III`, onlyD. graphs `I, II and III`. |
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Answer» Correct Answer - D Acceleration is positive in all three graphs, so velocity increases in all three. |
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| 88. |
The graph shows position as a function of time for two trains running on parallel tracks. Which statement is true ? A. At time `t_B`, both trains have the same velocity.B. Both trains have the same velocity at some time after `t_B`C. Both trains have the same velocity at some time before `t_B`.D. Somewhere on the graph, both trains have the same acceleration. |
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Answer» Correct Answer - C At some time before `t_B`, slope of `B` will be equal to slope of `A` Acceleration of `A` is zero always whereas that of `B` is not zero. |
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| 89. |
A train normally travels at a uniform speed of `72 km//h` on a long stretch of straight level track. On a particular day, the train was forced to make a `2.0` minute stop at a station along this track. If the train decelerates at a uniform rate of `1.0 m//s^2` and accelerates at a rate of `0.50 m//s^2`, how much time is lost in stopping at the station ?A. 2 minB. 2 min 30 sC. 30 sD. 2 min 20 s |
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Answer» Correct Answer - B Case-I : Actual time spend in decelerating =`(v - u)/(a) = (0 - 20)/(-1.0) = 20 s` Distance travelled in decelerating =`20 xx 20 - (1)/(2) xx 1 xx 20^2 = 400 - 200 = 200 m` Time that train will have taken if it had travelled uniformly with `20 m//s` for `200 m = (200)/(20) = 10 s` Extra time spend in decelerating `= 20 - 10 = 10 s` Case-II : Actual time spend in accelerating =`(v - u)/(a) = (20 - 0)/(0.5) = 40 s` Distance travelled in these `40 s` =`0 xx 40 + (1)/(2) xx 0.5 xx 40^2 = 400 m` Time that the train will have taken if travelled uniformly with `20 m//s = (400)/(20) = 20 s` Extra time lost due to acceleration `= 40 - 20 = 20 s` Total extra time `= 10 s + 2 min + 30 s = 2 min 30 s`. |
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| 90. |
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. 2 sB. 4 sC. `2 sqrt(3)`D. `4 sqrt(3)` |
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Answer» Correct Answer - B Total length to be crossed `= 50 + 50 = 100 m` Relativity Velocity `= 10 + 15 = 25 m//s` Time taken `= (100)/(25) = 4 s`. |
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| 91. |
Acceleration versus time graph for four objects are shown below. All axes have the same scale. Which object had the greatest change in velocity during the interval ?A. .B. .C. .D. . |
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Answer» Correct Answer - D In option (b) area of graph is `(1)/(2) t_0 a_0`. But in options (a) and ( c) area is less than `(1)/(2) t_0 a_0`. In option (d), area is more than `(1)/(2) t_0 a_0`. So in option (d) change in velocity is greatest. |
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| 92. |
The graph to the right is a plot of position versus time. For which lebeled region is the velocity positive and the acceleration negative ? A. aB. bC. cD. d |
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Answer» Correct Answer - D In part (d), slope is positive and decreasing, so velocity is positive and decreasing. If velcity is decreasing, then acceleration is negative. |
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| 93. |
The plot shows the position (`x`) as a function of time (`t`) for two trains that run on a parallel track. Train `A` is next to train `B` at `t=0` and at `t=T_(0)`. A. At `T_0` both trains have the same velocityB. Both trains speed up all the timesC. Both trains have the same velocity at some time before `T_0`D. At `T_0` the trains have covered different distances. |
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Answer» Correct Answer - C Initially slope of `A` is more than `B` and finally slope of `A` is less. So speed of `A` is initially more and finally less than `B`. So somewhere in between their speeds becomes same. |
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| 94. |
Velocity versus displacement graph of a particle moving in a straight line is shown in figure. Corresponding acceleration versus velocity graph will be. A. .B. .C. .D. . |
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Answer» Correct Answer - A `a = v. (dv)/(ds)` `(dv)/(ds) = 1 s^-1` (constant). or acceleration versus velocity graph will be a straight line passing through origin with slope `1 s^-1`. |
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| 95. |
Graph of velocity versus displacement of a particle moving in a straight line is as shown in figure. The acceleration of the particle is. A. constantB. increases linearly with `x`C. increases parabolically with `x`D. none of these |
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Answer» Correct Answer - B From the graph, velocity-displacement equation can be written as : `v = v_0 + alpha x` ….(1) Here `v_0` and `alpha` are positive constants. Differentiating (1) with respect to `x` we get `(dv)/(dx) = alpha = constant` Acceleration of the particle can be written as `a = v. (dv)/(dx) = (v_0 + alpha x) . alpha` `alpha- x` equation is a linear equation. Thus acceleration increases linearly with `x`. |
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| 96. |
The velocity-time plot is shown in figure. Find the average speed in time interval `t = 0` to `t = 40 s` during the period. A. zeroB. `2.5 m//s`C. `5 m//s`D. none |
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Answer» Correct Answer - B Distance covered `D = 2[(1)/(2) xx 20 xx 5] = 100 m` Average speed `=(D)/(40) = (100)/(40) = 2.5 m//s`. |
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| 97. |
The diagram shows the displacement-time graph for a particle moving in a straight line. Find the average velocity for the interval from `t = 0` to `t = 5 s`. A. `- 2 ms^-1`B. `2 ms^-1`C. `-4 ms^-1`D. `4 ms^-1` |
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Answer» Correct Answer - A Displacement `= -10 -0 = -10` Average velocity `= (-10)/(5) = -2 m//s`. |
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| 98. |
Assertion: The average speed of a body over a given interval of time is equal to the average velocity of the body in the same interval of time if a body moves in a straight line in one direction. Reason: Because in this case distance travelled by a body is equal to the displacement of the body.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assersion is false but reason is true. |
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Answer» Correct Answer - A When a body moves on a straight path in one direction value of distance and displacement remains same so that average speed equals the average velocity for a given time interval. |
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| 99. |
The acceleration due to gravity on the planet `A` is `9` times the acceleration due to gravity on planet `B`. A man jumps to a height of `2 m` on the surface of `A`. What is the height of jump by the same person on the planet `B` ?A. 18 mB. 6 mC. `(2)/(3) m`D. `(2)/(9) m` |
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Answer» Correct Answer - A `H_(max) = (u^2)/(2 g) rArr H_(max) prop (1)/(g)` On planet `B` value of `g` is `1//9` times to that of `A`. So value of `H_(max)` will become `9` times, i.e., `2 xx 9 = 18` metre. |
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| 100. |
Assertion: In rectilinear motion when velocity is positive distance travelled increases and when velocity is negative distance travelled decreases. Reason: Distance is length of the path covered by a particle.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assersion is false but reason is true. |
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Answer» Correct Answer - D Either velocity is positive or negative distance travelled by a body is always positive, hence assertion is wrong. |
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