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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.
| 1. |
Speeds of two identical cars are u and 4u at at specific instant. The ratio of the respective distances in which the two cars are stopped from that instant isA. `1:1`B. `1:4`C. `1:8`D. `1:16` |
| Answer» Correct Answer - D | |
| 2. |
A train starts from rest with constant acceleration `a=1 m//s^(2)` A passenger at a distance S from the train runs at this maximum velocity of `10 m//s` to catch the train at the same moment at which the train starts. Find the speed of the train when the passenger catches it for the critical distance:A. `8 m//s`B. `10 m//s`C. `12 m//s`D. `15 m//s` |
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Answer» Correct Answer - B For critical distance, passenger catches the train in time `t=(V_(P))/(a_(t))` So, required velocity of train `a_(t).t` `=a_(t)((v_(P))/(a_(t)))=V_(P)//2 =10 m//sec` |
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| 3. |
A train starts from rest with constant acceleration `a=1 m//s^(2)` A passenger at a distance S from the train runs at this maximum velocity of `10 m//s` to catch the train at the same moment at which the train starts. If `S=25.5` m and passenger keeps running, find the time in which he will catch the train:A. 5 secB. 4 secC. 3 secD. `2sqrt(2)` sec |
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Answer» Correct Answer - C At time t, `X_(t)` and `X_(p)` are coordinates of train and passenger respectively. `X_(t)=(1)/(2)a_(1)t^(2)` and `X_(p)=v_(P)t-S` If passenger catches the train, ,brgt `X_(t)=X_(p)` or `(1)/(2)a_(1)t^(2)=v_(P)t-S or t=(V_(p)-sqrt(V_(P)^(2)-2a_(1)S))/(a_(1))` `=(10-sqrt((10)^(2)-2(1)(25.5)))/(1)=3` seconds |
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| 4. |
A train starts from rest with constant acceleration `a=1 m//s^(2)` A passenger at a distance S from the train runs at this maximum velocity of `10 m//s` to catch the train at the same moment at which the train starts. Find the critical distance `S_(c)` for which passenger will take the ten seconds time to catch the trainA. 50 mB. 35 mC. 30 mD. 25 m |
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Answer» Correct Answer - A The critical distance `S_(c)` for which passenger will take the ten seconds time to catch the train is given by `S_(C)=(v_(P)^(2))/(2a_(1))` The time is 10 seconds if `V_(P)^(2)-2a_(1)S=0` `S_(C)=(V_(P)^(2))/(2a_(t))=((10)^(2))/(2(1))=50m` |
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| 5. |
A particle is moving along X-axis under a force such that its position -time graph is as shown in figure. As the particle passes position (6) :A. it is moving along positive X-direction with a speed that is increasing with timeB. it is moving along positive X-direction with a speed that is decreasing with timeC. it is moving along negative X-direction with a speed that is increasing with timeD. it is moving along negative X-direction with a speed that is decreasing with time |
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Answer» Correct Answer - D If `(dx)/(dt)` slope of `x - t` curve is zero. Then particle will be at rest. If `(dx)/(dt) gt 0` velocity is `+ve` If `(dx)/(dt) lt 0` velocity is `-ve` |
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| 6. |
A particle is moving along X-axis under a force such that its position -time graph is as shown in figure. As the particle passes position (5) :A. it is instantaneously at rest and will now move along positive X-directionB. it is moving along positive X-direction with a speed that is decreasing with timeC. it is moving along negative X-direction with a maximum speedD. it is moving along negative X-direction with a minimum speed |
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Answer» Correct Answer - C If `(dx)/(dt)` slope of `x - t` curve is zero. Then particle will be at rest. If `(dx)/(dt) gt 0` velocity is `+ve` If `(dx)/(dt) lt 0` velocity is `-ve` |
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| 7. |
A particle is moving along X-axis under a force such that its position -time graph is as shown in figure. As the particle passes position (1):A. it is moving along negative X-direction with a speed that is increasing with timeB. it is moving along positive X-direction with a speed that is decreasing with timeC. it is moving along negative X-direction with a speed that is decreasing with timeD. it is moving along positive X-direction with a speed that is increasing with time |
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Answer» Correct Answer - D If `(dx)/(dt)` slope of `x - t` curve is zero. Then particle will be at rest. If `(dx)/(dt) gt 0` velocity is `+ve` If `(dx)/(dt) lt 0` velocity is `-ve` |
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| 8. |
A particle is moving along X-axis under a force such that its position -time graph is as shown in figure. As the particle passes position (2):A. it is moving along negative X-direction with a maximum speedB. it is moving along positive X-direction with a minimum speedC. it is moving along negative X-direction and its speed is zero hereD. it is moving along negative X-direction with a minimum speed |
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Answer» Correct Answer - A If `(dx)/(dt)` slope of `x - t` curve is zero. Then particle will be at rest. If `(dx)/(dt) gt 0` velocity is `+ve` If `(dx)/(dt) lt 0` velocity is `-ve` |
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| 9. |
A body is released from the top of a tower of height h. It takes t sec to reach the ground. Where will be the ball after time `(t)/(2)` sec?A. `(h)/(2)`B. `(h)/(4)`C. `(3h)/(4)`D. `(3h)/(2)` |
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Answer» Correct Answer - C `h=(1)/(2)g t^(2)` |
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| 10. |
In the question number 67, the time taken by the ball to reach the ground isA. 2 sB. 3 sC. 5 sD. 7 s |
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Answer» Correct Answer - C Let `t_1` be the time taken by the ball to reach the highest point. here, v = 0, u = `20 m s^(-1), a = - g = - 10 m s^(-2), t = t_1` As `v = u + at` `therefore 0 = 20 + (-10)t_1` or `t_1 = 2 s` Taking vertical downward motion of the ball from the highest point to ground. Here, u = 0, `a = +g = 10 m s^(-2)`, S = 20 m + 25 m = 45 m, `t = t_2` As `S = ut + 1/2 at^(2)` `therefore` 45 = `0 + 1/2(10)t^(2)_2` `t^(2)_2 = 45 xx 2/10 = 90/10 = 9` or `t_2 = 3 s` Total time taken by the ball to reach the ground = `t_1 + t_2 = 2 s + 3 s = 5 s` |
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| 11. |
The velocity of a particle at an instant is `10 m s^(-1)`. After 3 s its velocity will becomes `16 m s^(-1)`. The velocity at 2 s, before the given instant will beA. `6 m s^(-1)`B. `4 m s^(-1)`C. `2 m s^(-1)`D. `1 m s^(-1)` |
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Answer» Correct Answer - A Here, u = `10 m s^(-1), t = 3 s, v = 16 m s^(-1)` `a = v - u/t = 16 - 10/3 = 2 m s^(-2)`. Now velocity at 2 s, before the given instant `10 = u + 2 xx 2` `therefore` u = 6 m `s^(-1)` (since v = u + at) |
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| 12. |
Two trains ` A and B` of length `400 m` each are moving on two parallel tracks with a uniform speed of ` 72 km h^(-1)` in the same direction with ` A` ahead of `B` .The driver of B decides to overtake Aand accelerates by 1m/s².if after 50s ,the guard of B just passes the driver of A , what was the original distance between them ?A. 750 mB. 1000 mC. 1250 mD. 2250 m |
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Answer» Correct Answer - C For train A, `u_A = 72 km h^(-1) = 72 xx 5/18 m s^(-1) = 20 m s^(-1) , a_A = 0`, t = 50 s, `therefore S_A = u_(A)t - (20)(50) = 1000 m` For train B, `U_b = 72 km h^(-1) = 72 xx 5/18 m s^(-1) = 20 m s^(-1)` `a_B = 1 m s^(-2)` , t = 50 s `therefore S_B = u_(B)t + 1/2 a_(B)t^(2)` `= (20 xx 50) + 1/2 xx 1 xx (50)^(2) = 2250` m Original distance between A and `B = S_B - S_A = 2250 m - 1000 m = 1250 m` |
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| 13. |
Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of `72 km h^(-1)` in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by `1 ms^(-2)`. If after 50s, the guard of B just brushed past the driver of A, what was the original distance between them ? |
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Answer» For train A: Initial velocity, u = 72 km/h = 20 m/s Time, t = 50 s Acceleration, `a_I = 0` (Since it is moving with a uniform velocity) From second equation of motion, distance `(s_|)`covered by train A can be obtained as: `s_|=ut+1/2a_|t^2`= 20 × 50 + 0 = 1000 m For train B: Initial velocity, u = 72 km/h = 20 m/s Acceleration, `a = 1 m//s^2` Time, t = 50 s From second equation of motion, distance `(s_(||))`covered by train A can be obtained as: `S_(||)=ut+1/2at^2` `=20xx50+1/2xx1xx(50)^2=2250 m` Hence, the original distance between the driver of train A and the guard of train B is 2250 – 1000 = 1250 m |
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| 14. |
The `v - t` graph for a particle is shown. The distance travelled in the first four seconds is: A. 12 mB. 16 mC. 20 mD. 24 m |
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Answer» Correct Answer - B Area below `v - t` curve provides displacement Hence, `s = int_(0)^(u) dt` |
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| 15. |
The displacement-time graph of moving particle is shown below The instantaneous velocity of the particle in negative at the pointA. CB. DC. ED. F |
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Answer» Correct Answer - C The slope of the tangent at any point on the displacement-time graph gives instantaneous velocity at that instant. In the given graph, the slope is negative at point E. |
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| 16. |
The distance moved by a freely falling body (starting from rest) during ` st, 2nd, 3rd,……nth` second of its motion are proportional to .A. even numbersB. odd numbersC. all integral numbersD. squares of integral numbers |
| Answer» Correct Answer - B | |
| 17. |
The distance moved by a freely falling body (starting from rest) during ` st, 2nd, 3rd,……nth` second of its motion are proportional to .A. `(n-1)`B. `(2n-1)`C. `(n^(2)-1)`D. `(2n-1)//n^(2)` |
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Answer» Correct Answer - B `S_(n)=g(n-(1)/(2))`, ratio `=(g)/(2):(3g)/(2):(5g)/(2)` …. `(2n-1)(g)/(2)` `S_(n)prop(2n-1)` |
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| 18. |
In the given v-t graph the distance travelled by the body in 5 seconds will be A. 100 mB. 80 mC. 40 mD. 20 m |
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Answer» Correct Answer - A The distance is equal to total area under v-t graph = `20 xx 2 / 2 + 20 xx 2 + 20 xx 1 + 20 xx 1 /2 + 20 xx 1/2` = 100 m |
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| 19. |
Prove that the distances traversed during equal intervals of time by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely 1: 3: 5: …………….]. |
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Answer» Let us divide the time interval of motion of an object under free fall into many equal intervals `tau` and find out the distance traversed during successive intervals of time. Since initial velocity is zero. We have `y = - (1)/(2) g t^(2)` Using this equation. we can calculate the position of the object after different time intervals, `0 , tau, 2 tau, 3 tau`.... which are given in second column of Table 3.2. If we take `(-1//2) g tau^(2)` as `y_(0)-`the position coordinate after first time interval `tau`, then third column gives the positions in the unit of `y_(0)`. The fourth column gives the distance traversed in successive `tau s`. we find that the distances are in the simple ratio `1 : 3 : 5 : 7 : 9 : 11`.... as shown in the last column. This law was established by Galileo Galilei (1564 - 1642) who was the first to make quantitative studies of free fall. |
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| 20. |
The given acceleration-time graph represents which of the following physical situation? A. A cricket ball moving with a uniform speed is hit with a bat for a very short time interval.B. A ball is falling freely from the top of a tower.C. A car moving with constant velocity on a straight road.D. A football is kicked into the air vertically upwards. |
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Answer» Correct Answer - A The a-t graph shows that initially the body moves with uniform velocity. Its acceleration increases for a short time and then fails to zero and there after the body moves with a constant velocity. Such as physical situation arises when a cricket ball moving with a uniform speed is hit with a bat for a very short time interval. |
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| 21. |
A particle moving with a velocity equal to `0.4 m//s` is subjected to an acceleration of `0.15 m//s^(2)` for `2 s`. in a direction at the right angle to its direction of motion. The resultant velocity is |
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Answer» In vector form, 1 st equation of motion is `vec(v) = vec(u) + vec(a)t` So, `v = sqrt(u^(2) + (at)^(2) + 2u (at) cos theta)` Here `u = 0.4 m//s, a = 0.15 m//s^(2), t = 2s` and `theta = 90^(@)` So, `u = sqrt([(0.4)^(2) + (0.15 xx 2)^(2) + 0]) = 0.5 m//s` |
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| 22. |
The jet plane starts from rest at `s =0` and is subjected to the acceleration shown. Determine the speed of the plane when it has travelled 60 m. A. `46.47 m//s`B. `36.47 m//s`C. `26.47 m//s`D. `16.47 m//s` |
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Answer» Correct Answer - A `a=(dv)/(ds)v implies int_(0)^(S)a.ds=int_(u)^(v)v.dv=v^(2)-u^(2)` `implies"Area under" a~s "curve"=v^(2)-u^(2)` `implies((1)/(2)xx150xx22.5)-(1)/(2)(90xx13.5)=v^(2)-0` `impliesv^(2)=75xx22.5-45xx13.5` `impliesv=sqrt(75xx22.5-45xx13.5)implies46.47` |
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| 23. |
The figure shows a particle moving along x - axis subjected to three particles of acceleration (a). Rank the periods according to the increase they produce in the particle velocity, greatest first. A. B. C. D. |
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Answer» Correct Answer - D Area of acceleration-time graph gives velocity. Area of region gives velocity of particles `m s^(-1)` `v_1 = 2s xx (5 m)/s^(2) = 10 m s^(-1)` Similarly, `v_2 = 4s xx (2)m//s^(2) = 8 m//s` and `v_3 = 1s xx 7 m//s^(2)` = 7m/s `therefore v_1 gt v_2 gt v_3` |
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| 24. |
A body of mass `m`, moving along the positive x direction is subjected to a resistive force `F = Kv^(2)` (where K is a constant and `v` the particle n velocity). If `m = 10 kg v = 10 m//s` at `t = 0, " and " K = 2N (m//s)^(-2)` the velocity when `t = 2s` is:A. `(10)/(3) m//s`B. `2 m//s`C. `- (10)/(3) m//s`D. `(3)/(10) m//s` |
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Answer» Correct Answer - B `F = - k v^(2)` `m (dv)/(dt) = - kv^(2)` `underset(10)overset(v)int (dv)/(v^(2)) = - (k)/(m) underset(0)overset(t)int dt` `- |(1)/(v)|_(10)^(v) = (2)/(10) |t|_(0)^(t)` `(1)/(v) - (1)/(10) = (1)/(5) (2)` `v = 2 m//s` |
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| 25. |
A particle is thrown vertically upward from the ground with some velocity and it strikes the ground again in time 2 s. The maximum height achieved by the particle is : `(g=10 m//s^(2))`A. 2.50 mB. 1.25 mC. 6.25 mD. 5 m |
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Answer» Correct Answer - D `A` to `B`, then `B` to `A`, time`=2 s` `A` to `B`, time`=1 s` `0=u-gxx1impliesu=10 m//s` `0=u^(2)-2gHimpliesH=u^(2)/(2g)=5 m` |
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| 26. |
A body is thrown vertically upward from a point `A` 125 m above the ground. It goes up to a maximum height of 250 m above the ground and passes through `A` on its downward journey. The velocity of the body when it is at a height of 70 m above the ground is `(g=10 m//s^(2))`A. `20 m//s`B. `50 m//s`C. `60 m//s`D. `80 m//s` |
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Answer» Correct Answer - C The body is freely falling from a height of 250 m its velocity at a height of 70 m from the ground means velocity of freely falling body after travelling 180 m. `v=sqrt(2gh)` |
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| 27. |
A boy throws a ball in air in such a manner that when the ball is at its maximum height he throws another ball. If the balls are thrown with the time difference 1 second, the maximum height attained by each ball isA. 9.8 mB. 19.6 mC. 4.9 mD. 2.45 m |
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Answer» Correct Answer - C `h=(g)/(2n^(2))` |
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| 28. |
A boy throws n balls per second at regular time intervals. When the first ball reaches the maximum height he throws the second one vertically up. The maximum height reached by each bass isA. `(g)/(2(n-1)^(2))`B. `(g)/(2n^(2))`C. `(g)/(n^(2))`D. `(g)/(n)` |
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Answer» Correct Answer - B Time interval between two balls`=` Time of ascend `=(1)/(n)=(u)/(g)u=(g)/(n),h=(u^(2))/(2g)` |
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| 29. |
A boy throws balls into air.He throws one, whenver the previous one is at its highest point. How high do the balls rise if he throws one ball each sec ?A. 19.6 mB. 9.8 mC. 4.9 mD. 2.45 m |
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Answer» Correct Answer - C `v = u - g t` `0 = u - 9.8 xx 1 " " u = 9.8` `s = 9.8 xx 1 - (1)/(2) xx 9.8 xx (1)^(2)` `s = 4.9 m` |
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| 30. |
In which of the following exmples of motion, can the body be considered approxinmately a point object : (a) a railway carriage moving without jerks between two two stations. (b) a mondey sistting on top of a man cycling smoothly on a circulat track. (c ) a spinning cricket ball that turns sharply on hitting the round . (d) a tumbling beake theat has slopped off the edge of a table ?A. a,bB. b,cC. a,cD. b,d |
| Answer» Correct Answer - A | |
| 31. |
Speedometer measures the speed of the car inA. `m s^(-1)`B. `km h^(-1)`C. `cm s^(-1)`D. `km min^(-1)` |
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Answer» Correct Answer - B Speedometer measures the speed of the car n `km h^(-1)`. |
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| 32. |
Speedometer of a car measuresA. average speedB. average velocityC. instantaneous speedD. instantaneous velocity |
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Answer» Correct Answer - C Speedometer of the car measures the instantaneous speed of the car. |
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| 33. |
Assertion : The speedometer of an automobile measure the average speed of the automobile. Reason : Average velocity is equal to total displacement per total time taken.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 |
| Answer» Correct Answer - (b) | |
| 34. |
A carrom striker is given velocity on carrom based has always. Friction cause constant retardation. Striker hits boundary of carrom and come to rest at point from where it started. Take initial velocity direction as positive, choose correct graph describing motion : (`v-`velocity, s-displacement, t-time)A. B. C. D. |
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Answer» Correct Answer - B::C Position 1 st increase but velocity decrease and later position decreases but velocity increase. |
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| 35. |
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. 45 m |
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Answer» Correct Answer - B `S=u_(1)t_(1)+u_(2)t_(2)+u_(3)t_(3)` |
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| 36. |
The average velocity of a body moving with uniform acceleration after travelling a distance of `3.06 m` is `0.34 m s^(-1)`. If the change in velocity of the body is `0.18 ms^(-1)` during this time, its uniform acceleration is .A. 0.01B. 0.2C. 0.03D. 0.04 |
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Answer» Correct Answer - B `t=("distance")/(V_(ave)),a=("change in velocity")/("total time")=(v-u)/(t)` |
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| 37. |
Stopping distance of a moving vehicle is directly proportional toA. square of the initial velocityB. square of the initial accelerationC. the initial velocityD. the initial acceleration |
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Answer» Correct Answer - A Let `d_s` is the distance travelled by the vehicle before it stops. here, final velocity v = 0, initial velocity = u, S = `d_s` Using equation of motion `v^(2) = u^(2) + 2aS` `therefore` `(0)^(2) = u^(2) + 2ad_s`, `d_s = -u^(2)/2a` or `d_s propto u^(2)` |
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| 38. |
The distance covered by a moving body is directly proportional to the square to the time. The acceleration of the body isA. increasingB. decreasingC. zeroD. constant |
| Answer» Correct Answer - D | |
| 39. |
When the speed of a car is `u`, the minimum distance over which it can be stopped is `a`, If speed becomes nu, what will be the minimum distance over which it can be stopped during the same time?A. `s//n`B. `ns`C. `s//n^(2)`D. `n^(2)s` |
| Answer» Correct Answer - D | |
| 40. |
Car `A` starts from a point `O` and moves with constant velocity `9 m//s`. After `2 s`, another car `B` begins its journey from `O` and follows car `A`. If car `B` starts from rest and moves under constant acceleration `4 m//s^(2)`, after how much time and at what distance from `O` the cars meet? |
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Answer» Here the cars starts at different timings. Let the cars meet after time t (from start of A) and at distance d from `O`. Car `B` will take time `(t-2)` sec. Car `A` (uniform motion): `d=vt=9t …(i)` Car `B` (accelerated motion): `d=0+1/2a(t-2)^(2)` `=(1)/(2).4(t-2)^(2) ...(ii)` From (i) and (ii), we get `9t=2(t-2)^(2)` `2t^(2)-17t+8=0` Solving we get `t =8 s,1/2` s `t=1/2 s` is not possible, because car `B` starts `2 s` later `d=9t=9xx8=72m` |
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| 41. |
The position `x` of a particle with respect to time `t` along the x-axis is given by `x=9t^(2)-t^(3)` where `x` is in meter and `t` in second. What will be the position of this particle when it achieves maximum speed along the positive `x` directionA. 32 mB. 54 mC. 81 mD. 24 m |
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Answer» Correct Answer - B `x=9t^(2)-t^(3)` `v=(dx)/(dt)=18t-3t^(2)` `a=(dv)/(dt)=18-6t` When` v` is maximum, `a=(dv)/(dt)=0` `18-6t=0impliest=3s` At `t=3 s, x=9(3)^(2)-(3)^(3)=54 m` |
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| 42. |
In `1.0 s`, a particle goes from point `A` to point `B` , moving in a semicircle of radius `1.0 m ` (see figure ). The magnitude of the average velocity A. `3.14 m//s`B. `2.0 m//s`C. `1.0 m//s`D. zero |
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Answer» Correct Answer - B `|"average velocity"|=|("displacement")/("time")|` `=(AB)/("time")=(2)/(1)=2 m//s`. |
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| 43. |
In a car race, car `A` takes `4 s` less than can `B` at the finish and passes the finishing point with a velocity `v` more than the car `B` . Assuming that the cars start form rest and travel with constant acceleration `a_(1)=4 m s^(-2)` and `a_(2) =1 m s^(-2)` respectively, find the velocity of `v` in m `s^(-1)`. |
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Answer» Correct Answer - 8 `t_(1)=t_(2)-t,v_(1)v_(2)=v,S=(1)/(2)a_(1)t_(1)^(2),S=(1)/(2)a_(2)t_(2)^(2)` `v_(1)=a_(1)t_(1),v_(2)=a_(2)t_(2)impliesv_(2)+v=a_(1)t_(1)` `impliesa_(2)t_(2)+v=a_(1)t_(1)=a_(1)t_(2)impliest_(2)=(v+a_(1)t)/(a_(1)-a_(2))` `sqrt((a_(2))/(a_(1)))=(t_(1))/(t_(2))=1-(t)/(t_(2))rArr sqrt((a_(2))/(a_(1)))=1-(t(a_(1)-a_(2)))/((v+a_(1)t))` `implies(sqrt(a_(2)))/(sqrt(a_(1)))=(v+a_(2)t)/(v+a_(1)t)impliessqrt(a_(2))v+a_(1)sqrt(a_(2)t)=vsqrt(a_(1))+a_(2)sqrt(a_(1)t)` `impliesv=(sqrt(a_(1)a_(2)))t=8ms^(-1)` |
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| 44. |
A motorist droves north for 35.0 minutes at `85.0 km//h` and then stops for 15.0 minutes He next continues north, travelling 130 km is 2.00 hours. What is his total displacementA. 85 kmB. 179.6 kmC. 20 kmD. 140 km |
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Answer» Correct Answer - B `overline(S_(1))=overline(V_(1))t,overline(S)=overline(S_(1))+overline(S_(2))` |
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| 45. |
The variation of velocity of a particle with time moving along a straight line is illustrated in the following figure. The distance travelled by the particle in four seconds is. A. 60 mB. 55 mC. 25 mD. 30 m |
| Answer» Correct Answer - B | |
| 46. |
In Q.118, total distance travelled by the particle from `t = 0` to `t = t_(0)` is:A. `(A_(0))/(2 alpha)`B. `(A_(0))/(alpha)`C. `(2A_(0))/(alpha)`D. `(4 A_(0))/(alpha)` |
| Answer» Correct Answer - A | |
| 47. |
In relation to a velocity-time graphA. the curve can be a circleB. the area under the curve and above the time axis between any two instants gives the average acceleration.C. the slope at any instant gives the rate of change of acceleration at that instantD. the area under the curve and above the time axis gives the displacement. |
| Answer» Correct Answer - D | |
| 48. |
If the distance travelled by a particle and corresponding time be laid off along y and x axes respectively, then the correct statement of the following isA. the curve may lie in fourth quadrantB. the curve lies in first quadrantC. the curve exhibits peaks corresponding to maximumD. the curve may drop as time passes |
| Answer» Correct Answer - B | |
| 49. |
The numerical ratio of displacement to the distance covered is alwaysA. always less than 1B. always greater than 1C. always equal to 1D. may be less than 1 or equal to the |
| Answer» Correct Answer - D | |
| 50. |
The numerical ratio of displacement to the distance covered is alwaysA. always equal to oneB. always less than oneC. always greater than oneD. equal to or more than one |
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Answer» Correct Answer - D (d) : Distances `ge` [Displacement] |
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