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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. |
In the above question, the acceleration of the rocket is :A. `50ms^(-2)`B. `100ms^(-2)`C. `10 ms^(-2)`D. `1000ms^(-2)` |
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Answer» Correct Answer - A Acceleration =slope of OA line `tan theta=(1000)/(20)=50 m//s^(2)` |
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| 2. |
In above question the retardation of rocket is:-A. `50ms^(-2)`B. `100ms^(-2)`C. `500ms^(-2)`D. `10ms^(-2)` |
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Answer» Correct Answer - D Retardation=Slope of AB line `=tan theta=(1000)/((120-20))=(1000)/(100)=10 m//s^(2)` |
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| 3. |
In the previous question, the mean velocity of the rocket reaching the maximum height is :A. `100ms^(-1)`B. `50ms^(-1)`C. `500ms^(-1)`D. `25//3ms^(-1)` |
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Answer» Correct Answer - C `"Mean velocity"=("Displacement")/("Total time")=("Total height")/("Total time")` `=(10,000m)/(20sec)=500 m//s` |
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| 4. |
In the previous question, the height attained by the rocket before deceleration is :A. 1 kmB. 10kmC. 20kmD. 60 km |
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Answer» Correct Answer - B Height before retardation `(1)/(2)xx(20)xx1000=10,000=10 km` |
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| 5. |
A ball is projected to attain the maximum range. If the height attained is H, the range isA. HB. 2HC. 4HD. H/2 |
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Answer» Correct Answer - C Maximum range for a given velocity is attained at `theta=45^(@)` `R_("max")=(u^(2)sin2theta)/(g)rArrR_("max")=(u^(2)sin90^(@))/(g)=(u^(2))/(g)` `therefore R_("max")=4.H_("max")` |
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| 6. |
In the Q.23, the maximum height attained by the arrow is:-A. 25mB. 40mC. 31.25mD. 12.5m |
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Answer» Correct Answer - C `H=(1)/(2)"gt"^(2)=(1)/(2)xx10xx[(5)/(2)]^(2)=31.25m` [time for downward journey `=(T)/(2)=(5)/(2)`] In this time it will fall a distance equal to maximum height. |
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| 7. |
A body stans at `78.4 m` from a building and throws a ball which just enters a window `39.2 m` above the ground. Calculate the velocity of projection of the ball. Fig. 2 (d) . 22. . |
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Answer» Maximum height `=(u^(2)sin^(2)theta)/(2g)=39.2m`...(i) Range `=(u^(2)sin2 theta)/(g)=(2u^(2)sin thetacostheta)/(g)=2xx78.4` ....(ii) from equation (i) divided by equation (ii) `tan theta=1rArrtheta=45^(@)` from equation (ii) range `=(u^(2)sin90^(@))/(g)=2xx78.4rArru=sqrt(2xx78.4xx9.8)=39.2 m//s`. |
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| 8. |
The equation of projectile is `y=16x-(x^(2))/(4)` the horizontal range is:-A. 16mB. 8mC. 64mD. 12.8m |
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Answer» Correct Answer - C `y=16x-(x^(2))/(y)` `y=16x(1-(x)/(64))` Comparing with `y=x tan theta(1-(x)/(R))` R=64m |
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| 9. |
If R is the maximum horizontal range of a particle, then the greatest height attained by it is :A. RB. 2RC. `(R)/(2)`D. `(R)/(4)` |
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Answer» Correct Answer - D For `theta=45^(@)` range will be maximum `R_("max")=(u^(2))/(g)` `rArr H=(u^(2)sin^(2)545^(@))/(2g)=(u^(2))/(4g)=(R)/(4)` |
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| 10. |
What is the ratio of P.E. w.r.t. ground and K.E. at the top most point of the projectile motion:-A. `cos^(2)theta`B. `sin^(2)theta`C. `tan^(2)theta`D. `cot^(2)theta` |
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Answer» Correct Answer - C `(K.E.)_("TOP")=(1)/(2)mu^(2)cos^(2)theta` vertical part of initial `K.E.=(P.E.)_("top")` `(P.E)_("TOP")=mgH=mg(u^(2)sin^(2)sin^(2)theta)/(2g)=(1)/(2)mu^(2) sin^(2)theta` `rArr((P.E.)_("TOP"))/((K.E.)_("TOP"))=` |
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| 11. |
In the above, the change in speed is:-A. ` u cos theta`B. uC. `u sin theta`D. `(u cos theta-u)` |
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Answer» Correct Answer - D `Deltavecv=vecv_("top")-vecv_(1)` `(u cos theta)hati-[(u cos theta)hati+(u sin theta)hatj]=-(u sin theta) hatj` `|Deltavecv|=u sin theta` change in speed `=(u cos theta)-v` |
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| 12. |
If a projectile is fired at an angle `theta` with the vertical with velocity u, then maximum height attained is given by:-A. `(u^(2)cos theta)/(2g)`B. `(u^(2)sin^(2)theta)/(2g)`C. `(u^(2)sin^(@)theta)/(g)`D. `(u^(2)cos^(2)theta)/(2g)` |
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Answer» Correct Answer - D `H=(u_(y)^(2))/(2g)=(u^(2)cos^(2)theta)/(2g)` |
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| 13. |
In case of a projectile fired at an angle equally inclined to the horizontal and vertical with velocity (u). The horizontal range is:-A. `(u^(2))/(g)`B. `(u^(2))/(2g)`C. `(2u^(2))/(g)`D. `(u^(2))/(g^(2))` |
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Answer» Correct Answer - A `theta=45^(@), R=(u^(2)sin 2xx45^(@))/(g)=(u^(2))/(g)` |
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| 14. |
A particle is fired with velocity `u` making angle `theta` with the horizontal.What is the change in velocity when it is at the highest point?A. `u cos theta`B. uniform retardationC. `u sin theta`D. `(u cos theta-u)` |
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Answer» Correct Answer - C `Deltavecv=vecv_("top")-vecv_(1)` `(u cos theta)hati-[(u cos theta)hati+(u sin theta)hatj]=-(u sin theta) hatj` `|Deltavecv|=u sin theta` change in speed `=(u cos theta)-v` |
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| 15. |
Assertion: In projectile motion, when horizontal range is n times the maximum height, the angle of projection is given by `tan theta=(4)/(n)` Reason: In the case of horizontal projection the vertical increases with time.A. If both assertion `&` Reason are True `&` the Reason is a corrrect explanation of the Asserion.B. If both Assertion `&` Reason are True but Reason is not correct explanation of the Assertion.C. If Assertion is Trie but the Reason is False.D. If both Assertion `&` Reason are false |
| Answer» Correct Answer - B | |
| 16. |
An object travels 10km at a speed of 100 m/s and another 10km at 50 k/s. The average speed over the whole distance is:-A. mgdB. `sqrt(2gd)`C. `2sqrt(g//d)`D. `2sqrt((mg)/(d))` |
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Answer» Correct Answer - B Initial velocity that is thrown down =v=`v_("initial")` Distance travelled =d, Acceleration=g `v^(2)-u^(2)=2as`. i.e., `v_("final")^(2)=v_("initial")^(2)+2gd` `therefore_("final")=sqrt(v_("initial")^(2)+2gd)` Option is not given. Only if `v_("initial")` has been zero, it would have been `sqrt(2gd)` |
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| 17. |
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. 24mB. 32mC. 54mD. 81m |
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Answer» Correct Answer - C `x=9t^(2)-t^(3) therefore v=18t-3t^(2)` `rArr=(dv)/(dt)=18-6t` `rArr(dv)/(dt)=18-6t` for maximum speed `(dv)/(dt)=0`, and `(d^(2)v)/(dt^(2))=` negative so `18-6t=0 rArrt=3s` `x_(t=3s)=9(3)^(2)-(3)^(3)=81-27=54m` |
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| 18. |
A projectile is thrown from a point in a horizontal plane such that the horizontal and vertical velocities are `9.8 ms^(-1)` and `19.6 ms^(-1)`. It will strike the plane after covering distance ofA. 4.9mB. 9.8mC. 19.6mD. 39.2m |
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Answer» Correct Answer - D `R=u_(x)T=u_(x)xx(2u_(y))/(g)=(2xx9.8xx19.6)/(9.8)=39.2m` |
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| 19. |
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. 2sB. 4sC. `2sqrt(3)s`D. `4sqrt(3)s` |
| Answer» Correct Answer - B | |
| 20. |
The velocity of a particle moving with constant acceleration at an instant `t_(0)` is `10m//s` After 5 seconds of that instant the velocity of the particle is 20m/s. The velocity at 3 second before `t_(0)` is:-A. 8m/sB. 4m/sC. 6m/sD. 7m/s |
| Answer» Correct Answer - B | |
| 21. |
Assertion: The magnitude of average velocity of the object over an interval of time is either smaller than or equal to the average speed of the object over the same interval. Reason: Path length (distance) is either equal or greater than the magnitude of displacement.A. If both assertion `&` Reason are True `&` the Reason is a corrrect explanation of the Asserion.B. If both Assertion `&` Reason are True but Reason is not correct explanation of the Assertion.C. If Assertion is Trie but the Reason is False.D. If both Assertion `&` Reason are false |
| Answer» Correct Answer - A | |
| 22. |
Velocity-time graph for a moving object is shown in the figure. Total displacement of the object during the time interval when there is non-zero acceleration and retardation is. A. 60mB. 50mC. 30mD. 40m |
| Answer» Correct Answer - B | |
| 23. |
Which of the following statements is incorrect? (i) Average velocity is path length divided by time interval. (ii) In general, speed is greater than the magnitude of the velocity. (iii) A particle moving in a given direction with a non-zero velocity can have zero speed. (iv) The magnitude of average velocity is the average speed.A. S2 and S3B. S1 and S4C. S1. S3 and S4D. All four statements |
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Answer» Correct Answer - C Average velocity `=("Displacement")/("Time interval")` A particle moving in a given direction with non-zero velocity cannot have zero speed. In general, average speed is not equal to magnitude of average velocity. However, it can be so if the motion is along a straight line without change in direction. |
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| 24. |
A person walks up a stalled escalator in 90 s. When standingon the same escalator, now moving, he is carried in 60 s.The time it would take him to walk up the moving escalator will be:A. 27sB. 72sC. 18 sD. 36s |
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Answer» Correct Answer - D speed of man `(v_(m))=(d)/(90)` speed of escaltor `(v_(s))=(d)/(60)` `t=(d)/(v_(m)+v_(s))=(d)/((d)/(900)+(d)/(90))=(60xx90)/(150)=36` sec |
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| 25. |
Velocity-time graph of a particle moving in a straight line is shown in figure. Plot the corresponding displacement-time graph of the particle if at time `t=0,` displacement `s=0.` |
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Answer» Displacement=area under velocity-time graph. Hence, `s_(DA)=(1)/(2)xxxx10=10m` `s_(AB)=2xx10=20m` or `s_(OAB)=10+20=30m` `s_(BC)=2xx((10+20)/(2))=30m` or `s_(OABC)=30+30=60m` and `s_(CD)=2xx((20+0)/(2))=20m` or `s_(OABCD)=60+20=80m` Between 0 to 2 s and 4 to 6s motion is accelerated, hence displacement-time graph is a parabola. Between 2 to 4 s motion is uniform, so displacement-time will be a straight line. Between 6 to 8 s motion is decelerated hence displacement-time graph is again a parabola but inverted in shape. At the end of 8s velocity is zero. therefore, slope of displacement time graph should be zero. The corresponding graph is shown in figure. |
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| 26. |
Assertion: If a body moves on a straight line, magnitude of its displacement and distance covered by it will be same. Reason: Along the straight line body can move only in one direction.A. If both assertion `&` Reason are True `&` the Reason is a corrrect explanation of the Asserion.B. If both Assertion `&` Reason are True but Reason is not correct explanation of the Assertion.C. If Assertion is Trie but the Reason is False.D. If both Assertion `&` Reason are false |
| Answer» Correct Answer - D | |
| 27. |
If the distance covered is zero, the displacement:-A. must be zeroB. may or may not be zeroC. cannot be zeroD. depends upon the particle |
| Answer» Correct Answer - A | |
| 28. |
The location of a particle is changed. What can we say about the displacement and distance covered by the particle?A. Both cannot be zeroB. One of the two may be zeroC. Both must be zeroD. If one is positive, the other is negative and vice-versa |
| Answer» Correct Answer - A | |
| 29. |
If the displacement of a body is zero is the distance covered by it necessarily zero ? Explain with suitable illustration.A. must be zeroB. may or may not be zeroC. cannot be zeroD. depends upon the particle |
| Answer» Correct Answer - B | |
| 30. |
The numerical ratio of displacement to the distance covered is alwaysA. Less than oneB. Equal to oneC. Equal to or less than oneD. Equal to or greater than one |
| Answer» Correct Answer - C | |
| 31. |
A particle moves in straight line in same direction for 20 seconds with velocity `3 m//s` and the moves with velocity `4 m//s` for another 20 sec and finally moves with velocity `5 m//s` for next 20 seconds. What is the average velocity of the particle?A. 3 m/sB. 4m/sC. 5m/sD. zero |
| Answer» Correct Answer - B | |
| 32. |
Which of the following velocity-time graphs shows a realistic situation for a body in motion ?A. B. C. D. |
| Answer» Correct Answer - B | |
| 33. |
A particle has an initial velocity `(6hati+8hatj) ms^(-1)` and an acceleration of `(0.8hati+0.6hatj)ms^(-2)`. Its speed after 10s is |
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Answer» Velcocity of particle after `10s vecv=(6hati+8hatj)+10(3hati+4hatj)=36hati+48hatj` speed `=sqrt((36)^(2)+(48)^(2))=+sqrt(6^(2)+8^(2))=60ms^(-1)` |
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| 34. |
A jet airplance travelling at the speed of ` 500 km ^(-1)` ejects its products of combustion at the speed of ` 1500 km h^(-1)` relative to the jet plane. What is the speed of the burnt gases with respect to observer on the ground ?A. 1500 km/hB. 2000km/hC. 1000km/hD. 500km/h |
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Answer» Correct Answer - C `V_("gas,plane")=V_("gas")+V_("plane")` `V_("gas")=V_("gas,plane")-V_("plane")` `=1500-500=1000lkm//h` |
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| 35. |
A boat man can row with a speed of 10 km/hr. in still water. The river flow steadily at 5 km/hr. and the width of the river is 2 km. if the boat man cross the river with reference to minimum distance of approach then time elapsed in rowing the boat will be:-A. `(2sqrt(3))/(5)` hourB. `(2)/(5)sqrt(3)` hourC. `(3sqrt(2))/(5)`hourD. `(5sqrt(2))/(3)` hour |
| Answer» Correct Answer - B | |
| 36. |
Assertion: In successive time intervals if the average velocities of a particle are equal then the particle must be moving with constant velocity. Reason: When a particle moves with uniform velocity, its displacement always increases with time.A. If both assertion `&` Reason are True `&` the Reason is a corrrect explanation of the Asserion.B. If both Assertion `&` Reason are True but Reason is not correct explanation of the Assertion.C. If Assertion is Trie but the Reason is False.D. If both Assertion `&` Reason are false |
| Answer» Correct Answer - D | |
| 37. |
A particle moving in a straight line covers half the distance with speed of `3 m//s`. The half of the distance is covered in two equal intervals with speed of `4.5 m//s and 7.5 m//s` respectively. The average speed of the particle during this motion is :A. 4.0 m/sB. 5.0 m/sC. 5.5 m/sD. 4.8 m/s |
| Answer» Correct Answer - A | |
| 38. |
Assertion: In projectile motion, the acceleration is constant in both magnitude and direction but the velocity changes in both magnitude and direction. Reason: When a force or accelration is acting in an oblique direction to the direction of velocity then both magnitude and direction of the velocity may be changed.A. If both assertion `&` Reason are True `&` the Reason is a corrrect explanation of the Asserion.B. If both Assertion `&` Reason are True but Reason is not correct explanation of the Assertion.C. If Assertion is Trie but the Reason is False.D. If both Assertion `&` Reason are false |
| Answer» Correct Answer - A | |
| 39. |
Assertion: A body moving with constant acceleration always travels equal distance in equal time intervals. Reason: Motion of the body with constant acceleration is a uniform motion.A. If both assertion `&` Reason are True `&` the Reason is a corrrect explanation of the Asserion.B. If both Assertion `&` Reason are True but Reason is not correct explanation of the Assertion.C. If Assertion is Trie but the Reason is False.D. If both Assertion `&` Reason are false |
| Answer» Correct Answer - D | |
| 40. |
If a ball is thrown vertically upwards with speed `u`, the distance covered during the last `t` second of its ascent isA. utB. `(1)/(2)"gt"^(2)`C. `ut-(1)/(2)"gt"^(2)`D. `(u+gt)t` |
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Answer» Correct Answer - B Distance convered in last t second of ascent = distance covered in first 1 second of descent `=0+(1)/(2)"gt"^(2)` |
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| 41. |
Which of the following equation represents the motion of a body moving with constant finite acceleration ? In these equation, y denotes the displacement in time t and p.q and r the arbitary constants ?A. `y=(p+qt)(r+pt)B. y=p+tqrC. y=(p+t)(q+t)(r+1)D. y=(p+qt)r |
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Answer» Correct Answer - C Acceleration `=(d^(2)y)/(dt^(2))` It is the second derivative where `t^(2)` terms can only survive and yield a constant value. Hence (i) represent the motion with constant finite acceleration. |
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| 42. |
A stone falls freely such that the distance covered by it in the last second of its motion is equal to the distance covered by it in the first 5 seconds. It remained in air for :-A. 12sB. 13sC. 25sD. 26s |
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Answer» Correct Answer - B Let total time of fall =n,u=0 `S_("last")=S_(5)` `(g)/(2)(2n-1)=(1)/(2)g(5)^(2)` `2n-1=25 rArr t=13 sec` |
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| 43. |
An elevator is accelerating upward at a rate of `6ft(sec^(2)` when a bolt from its celling falls to the floor of the lift (Distance=9.5feet). The time taken (in seconds) by the falling bolt to hit the floor is (take `g=32ft//sec^(2)`)A. `sqrt(2)`B. `(1)/sqrt(2)`C. `2sqrt(2)`D. `(1)/(2sqrt(2))` |
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Answer» Correct Answer - B Effective acceleration in ascending lift =(g_f) `therefores=ut+(1)/(2)at^(2)` `9.5=0+(1)/(2)(g+f)t^(2)` or `9.5=(1)/(2)(32+6)t^(2)` or `t^(2)=(9.5xx2)/(38)=(1)/(2)` or `t=(1)/sqrt(2)` sec |
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| 44. |
A stone falls from a ballon that id descending at a uniform rate of 12m/s. The displacement of the stone from the point of release after 10 sec isA. 490mB. 510mC. 610mD. 725m |
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Answer» Correct Answer - C Initial velocity of stone u=12m/sec downwards `therefore` displacement `h=ut+(1)/(2)"gt"^(2)` `=12xx10+(1)/(2)xx9.8xx(10)^(2)` `=120+400=610 m` |
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| 45. |
The velocity-time relation of an electron starting from rest is given by u=id, where `k=2m//s^(2)`. The distance traversed in 3 sec is:-A. 9mB. 16mC. 27mD. 36m |
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Answer» Correct Answer - A `u=ktrArr (ds)/(dt)=kt (because k= 2m//s^(2))` `rArr int_(0)^(x)ds=2int_(0)^(3)tdt` `rArr s=2|(t^(2))/(2)|_(0)^(2)rArr s=9m` |
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| 46. |
The distance travelled by a particle starting from rest and moving with an acceleration `(4)/(3) ms^-2`, in the third second is.A. `(10)/(3)m`B. `(19)/(3)m`C. `6m`D. 4m |
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Answer» Correct Answer - A `S_(n)=u+(a)/(2)(2n-1)=0+(4)/(3xx2)(2xx3-1)=(10)/(3)m`. |
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| 47. |
A boy standing at the top of a tower of 20m of height drops a stone. Assuming `g=10ms^(-2)`, the velocity with which it hits the ground is :-A. 10.0 m/sB. 20.0m/sC. 40.0m/sD. 5.0m/s |
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Answer» Correct Answer - B `v=|(Deltavecv)/(Deltat)|=sqrt(2xx10xx20)=20 m//s` |
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| 48. |
A particle experiences a constant acceleration for 20 sec after starting from rest. If it travels distance `S_1` in the first 10 sec and a distance `S_2` in the next 10 sec, ThenA. `S_(1)`B. `2S_(1)`C. `3S_(1)`D. `4S_(1)` |
| Answer» Correct Answer - C | |
| 49. |
If a body starts from rest, the time in which it covers a particular displacement with uniform acceleration is :A. inversely proportional to the square root of the displacementB. inversely proportional to the displacmentC. directily proportional to the displacementD. directly proportional to the square root of the displacement. |
| Answer» Correct Answer - D | |
| 50. |
A stone is thrown vertically upwards with an initial speed `u` from the top of a tower, reaches the ground with a speed `3 u`. The height of the tower is :A. `(3u^(2))/(g)`B. `(4u^(2))/(g)`C. `(6u^(2))/(g)`D. `(9mu^(2))/(g)` |
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Answer» Correct Answer - B `v^(2)=u^(2)+2gh` `h=(v^(2)-u^(2))/(2g)=(9u^(2)-u^(2))/(2g)=(4u^(2))/(g)` |
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