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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.
| 1001. |
A body is dropped from a balloon meving up wigh a velocity of `4 m s^(-2) ` when the balloon is at a height of `12.5` m from the ground. The distance of separation between of separation between the body and the balloon after `5` is.A. `122.5 m`B. `100.5 m`C. `132.5 m`D. `112.5 m` |
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Answer» Correct Answer - A The distance travelled by the balloon in `5 s` upwards in `4 xx 5 =20 m`. Distance of separation `=20+ 102.5 =122.5 m`. |
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| 1002. |
A body thrown with some initial velocity wigh the horizontal direction and then allowed to move in two dimensions under the action of gravity alone is called………………………………. . |
| Answer» Correct Answer - Projectile | |
| 1003. |
A particle is projected vertically upwards from a point A on the ground. It takes `t_(1)` time to reach a point B but it still continues to move up. If it takes further `t_(2)` time to reach the ground from point B then height of point B from the ground is :-A. `1/2g(t_(1)+t_(2))^(2)`B. `"g" t_(1)t_(2)`C. `1/8g(t_(1)+t_(2))^(2)`D. `1/2"g" t_(1)t_(2)` |
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Answer» Correct Answer - D |
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| 1004. |
STATEMENT-1: Two particles starts moving with velocities `vecV_(1) & vecV_(2)` respectively in `xy`-plane. They can meet only if component of their velocity perpendicular to line joining them are equal. STATEMENT-2: Realative velocity of a body w.r.t. other body is calculated along line joining two bodies.A. Statement -1 is True, Statement-2 is Ture, Statement-2 is a correct explanantion for Statement-1.B. Statement-1 is Ture, Statement-2 is Ture, Statement-2 is Not a correct explanantion for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - C | |
| 1005. |
STATEMENT-1: A man can cross river of width `d` in minimum time `t`. On increasing river velocity, minimum time to cross the river by man will remain uncharged. STATEMENT-2: Velocity of river is perpendicular to width of river. So time to cross the river is independent of velcoity of river.A. Statement -1 is True, Statement-2 is Ture, Statement-2 is a correct explanantion for Statement-1.B. Statement-1 is Ture, Statement-2 is Ture, Statement-2 is Not a correct explanantion for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - A | |
| 1006. |
Two trains take `3s` to pass another when going in the opposite directions but only `2.5s` if the speed of one is increased by `50%`. The time one would take to pass the other when going in the same direction at their original speed isA. `10s`B. `12s`C. `15s`D. `18s` |
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Answer» Correct Answer - C |
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| 1007. |
A ball is thrown vertically down with velocity of `5m//s`. With what velocity should another ball be thrown down after 2 seconds so that it can hit the `1^(st)` ball in next 2 second.A. `40m//s`B. `10m//s`C. `15m//s`D. `20m//s` |
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Answer» Correct Answer - A |
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| 1008. |
Figure shows three blocks A, B and C connected by a cable and system of pulleys. The blocks is pulled downward with a constant velocity of `7.5 cm//s`. At t = 0 , block, A starts moving downward from rest with a constant acceleration. It is given that the velocity of block A after travelling 20 cm is 30cm/s, find the change in position, velocity and the acceleration of block C at this instant. |
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Answer» Correct Answer - `[40 cm, 45 cm//s, 2.25 cm//s^(2)]` |
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| 1009. |
An object may have :-A. Varying speed without having varying velocity,B. Varying velocity without having varying speed,C. Non zero acceleration without having varying velocity,D. Non zero acceleration without having varying speed. |
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Answer» Correct Answer - B, D |
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| 1010. |
Choose the correct statement(s):A. We can have a motion having zero displacement and nonzero average speed.B. Average velocity is half the sum of its initial and final velocity.C. Total displacement is equal to product of average velocity and time.D. Acceleration of a particle is positive if it is moving in negative direction with decreasing speed. |
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Answer» Correct Answer - A, C, D |
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| 1011. |
A particle is projected at an angle of `60^(@)` above the horizontal with a speed of `10m//s`. After some time the direction of its velocity makes an angle of `30^(@)` above the horizontal. The speed of the particle at this instant is sA. `5/(sqrt(3))m//s`B. `5sqrt(3)m//s`C. `5m//s`D. `10/(sqrt(3))m//s` |
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Answer» Correct Answer - D |
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| 1012. |
A particle starts from rest with constant acceleration. The ratio of space-average velocity to the time average velocity is :-A. `1//2`B. `3//4`C. `4//3`D. `3//2` |
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Answer» Correct Answer - C `lt v_("space") gt =(int" vds")/(int" ds")=(int sqrt(2as)ds)/(int ds)=2/3 v` `lt v_("time") gt =(int" vdt")/(int" dt")=(int" atdt")/(int" dt")=v/2 :. (lt v_(s) gt)/(lt v_(t) gt)=4:3` |
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| 1013. |
The position time graph for a particle moving along X axis has been shown in the fig. At which of the indicated points the particle has (i) negative velocity but acceleration in positive X direction. (ii) positive velocity but acceleration in negative X direction. (iii) received a sharp blow (a large force for negligible interval of time)? |
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Answer» Correct Answer - (a) E, (b) D,G (c) B,C |
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| 1014. |
If angular velocity of a disc depends an angle rotated `theta` as `omega=theta^(2)+2theta`, then its angular acceleration `alpha` at `theta=1` rad is :A. `8 rad//s^(2)`B. `10 rad//s^(2)`C. `12 rad//s^(2)`D. None of these |
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Answer» Correct Answer - C Given `omrga=theta^(2)+2theta` `(domega)/(dtheta)=2theta+2` `alpha=omega(domega)/(dtheta)=(theta^(2)+2theta)(2theta+2)` at `theta=1` `alpha=12 rad//sec^(2)` |
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| 1015. |
If the radii of circular path of two particles are in the ratio of `1 :2`, then in order to have same centripatal acceleration, their speeds should be in the ratio of :A. `1:4`B. `4:1`C. `1:sqrt(2)`D. `sqrt(2):1` |
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Answer» Correct Answer - C Centripetal acceleration `=v^(2)/R` `v_(1)^(2)/R_(1)=v_(2)^(2)/R_(2)=v_(1)/v_(2)=sqrt(R_(1)/R_(2))=sqrt(1/2)` |
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| 1016. |
Displacement time graph of a particle moving in a straight line is a shown in figure. A. in region `A` acceleration is positiveB. in region `B` acceleration is negativeC. in region `C` acceleration is positiveD. in region `D` acceleration is negative |
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Answer» Correct Answer - A::B |
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| 1017. |
A particle `P` lying on a smooth horizontal `x-y` plane starts from `(3hati+4hatj)m` with velocity `(2hati)m//s`. Another particle `Q` is projected (horizontally from origin with velocity `(xhati+yhatj)` so that is strikes `P` after `2s`. ThenA. `x=2.0`B. `x=3.5`C. `y=2.0`D. `y=3.5` |
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Answer» Correct Answer - B::C |
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| 1018. |
Two tourist A and B who are at a distance of 40 km from their camp must reach it together in the shortest possible time. They have one bicycle and they decide to use it in turn. ‘A’ started walking at a speed of `5 km hr^(1)` and B moved on the bicycle at a speed of `15 km hr^(1)`. After moving certain distance B left the bicycle and walked the remaining distance. A, on reaching near the bicycle, picks it up and covers the remaining distance riding it. Both reached the camp together. (a) Find the average speed of each tourist. (b) How long was the bicycle left unused? |
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Answer» Correct Answer - (a) `7.5 km//hr^(–1)` (b) `2 hr 40 min` |
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| 1019. |
A body is projected upwards with a velocity `u`. It passes through a certain point above the grond after `t_(1)`, Find the time after which the body posses thoruth the same point during the journey. |
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Answer» Suppose `v` be the velcity attained by the body after time (i) `t_(1)`. The `v=u-g t_(1)` Let the body reaches the same point at time `t_(2)`. Now velocity will be be donward with same magitude `v`, then …(ii) `-v=u-g t_(2)` (i)-(ii) `rArr2v=g (t_(2)-t_(1))` or `t_(2)-t_(1)=(2v)/(g)=(2)/(g) (u-gt_(1))=2((u)/(g) -t_(1))`. |
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| 1020. |
The dispacement of a body is given by `4s=M +2Nt^(4)`, where `M` and `N` are constants. The velocity of the body at the end of `1 s` from the start is .A. `2N`B. `(M+2N)/(4)`C. `2(M_N)`D. `(2M+N)/(4)` |
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Answer» Correct Answer - A `s=(M+2Nt^(4))/(4) rArr v=(ds)/(dt) =2Nt^(3) ` Putting `1 s`, we get `v=2 N` . |
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| 1021. |
A time-velocity graph of two vehicles A and B starting from rest at the same time is given in the figure-1.107. The statement that can be deduced correctly from the graph is : A. Acceleration of A is greater thanthat of BB. Acceleration of B is greater than that of AC. Acceleration of A is increasing at a slower rate than that of BD. Velocity of B is greaterthan that of A. |
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Answer» Correct Answer - A |
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| 1022. |
Mark the correct statements :A. The magnitude of the instantaneous velocity of a particle is equal to its instantaneous speed.B. The magnitude of average velocity in an interval is equal to its average speed in that interval.C. It is possible to have a situation in which the speed ofa particle is always zero but the average speed is not zero.D. It is possible to have a situation in which the speed of a particle is never zero but the average speed in an interval is zero. |
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Answer» Correct Answer - A |
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| 1023. |
A juggler keeps (n) balls going with one hand so that at any instant, ( n-1) blls in air in air and one ball in the hand, If each ball rises to a height of 9x) metre, find the time for which each ball stays the hand. |
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Answer» Let (u) the linitial velocity of the ball while goig upwards. The final velocity of the ball at height (x) is, ` v=0`. Using the relation , ` v^2 = u ^2 = 2 a S, we have ` o= u^2 - 2 gx ` or ` u =sqrt 2 gx` If (t) is the time taken by ball in going up through distance` (x), when ` 0= u + ( - g0 t ` or t = u/g`. Total time after which the ball comes into the hand is , ` T= 2 t = ( 2u0/ g 2/g sqrt( 2 gt) = 2 sqrt ( 2x)/g` During time (T), (n- 1) balls will be in air and ne ball will be in hand. Time for one ball in hand `= T/(n-1) = ( 2 aqrt 2 x //g)/ (( n-1)) = 2/( ( n-1)) sqrt ( 2x)/g`. |
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| 1024. |
Man A is sitting in a car moving with a speed of 54 `(km)/(hr)` observes a man B in front of the car crossing perpendicularly a road of width 15 m in three seconds. Then the velocity of man B (in `(m)/(s)`) will be:A. Speed of man `B` is `5sqrt(10)m//s`B. Speed of man `B` is `5ms^(-1)`C. Actual direction of motion of `B` is at an angle of `tan^(-1)(1/3)` with direction of motion of carD. Actual direction of motion of `B` is at an angle of `tan^(-1)(3)` with direction opposite to the direction of motion of car. |
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Answer» Correct Answer - A::C |
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| 1025. |
Man A is sitting in a car moving with a speed of 54 `(km)/(hr)` observes a man B in front of the car crossing perpendicularly a road of width 15 m in three seconds. Then the velocity of man B (in `(m)/(s)`) will be:A. Speed of man `B` is `5sqrt(10)` m/sB. Speed of man `B` is `5 ms^(-1)`C. Actual direction of motion of B is at an angle of `tan^(-1) (1/3)` with direction of motion of carD. Actual direction of motion of B si at angle of `tan^(-1) (3)` with direction opposite to the direction of motion of car |
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Answer» Correct Answer - A, C |
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| 1026. |
From a building two balls `A` and `B` are thrown such that `A` is thrown upwards and `B` downwards ( both vertically with the same speed ). If `v_(A) and v_(B)` are their respective velocities on reaching the ground , thenA. `v_(B) gt v_(B)`B. `v_(A) = v_(B)`C. `v_(A) gt v_(B)`D. their velocities depends on their masses |
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Answer» Correct Answer - B From conservaiton of enegry, potential enegry at height `h` `= K.E.` at ground Therefore, at hieght `h, P.E.` of ball `A` `P.E. = m_(A) gh` `V_(A) = sqrt(2gh)` Similarly, `V_(B) - sqrt(2gh)` Therefore, `V_(A) = V_(B)` |
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| 1027. |
At time `t=0`, particle `A` is at `(1m,2m)` and `B` is at `(5m,5m)`. Velociyt of `B` is `(2hati+4hatj)m//s` Velocity of particle `A` is `sqrt(2)v)` at `45^(@)` with `x`-axis. A collides with `B` . Value of `v` is ………..m/sA. 5B. 15C. 25D. 10 |
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Answer» Correct Answer - D |
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| 1028. |
At time `t=0`, particle `A` is at `(1m,2m)` and `B` is at `(5m,5m)`. Velocity of `B` is `(2hati+4hatj)m//s` Velocity of particle `A` is `sqrt(2)v)` at `45^(@)` with `x`-axis. A collides with `B` . Time when `A` will collide with `B` is …….second.A. `0.5s`B. `1.5s`C. `4s`D. `3s` |
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Answer» Correct Answer - A |
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| 1029. |
The position of a particle is given by `x=2(t-t^(2))` where `t` is expressed in seconds and `x` is in metre. The particleA. never does to negative `x`-axisB. never goes to positive `x`-axisC. starts from the origin goes up to `x=1/2m` in the positive `x` -axis and then moves in opposites directionD. has zero initial velocity |
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Answer» Correct Answer - C |
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| 1030. |
The position of a particle is given by `x=2(t-t^(2))` where `t` is expressed in seconds and `x` is in metre. The acceleration of the particle is |
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Answer» Correct Answer - C |
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| 1031. |
The position of a particle is given by `x=2(t-t^(2))` where `t` is expressed in seconds and `x` is in metre. When does the object return to its initial velocity?A. Ate `t=4s`B. At `t=7s`C. At `t=8s`D. Impossible to determine the given information |
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Answer» Correct Answer - B |
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| 1032. |
The position of a particle is given by `x=2(t-t^(2))` where `t` is expressed in seconds and `x` is in metre. The total distance travelled by the paticle between `t=0` to `t=1s` isA. `0m`B. `1m`C. `2m`D. `1/2m` |
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Answer» Correct Answer - B |
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| 1033. |
The position of a particle is given by `x=2(t-t^(2))` where `t` is expressed in seconds and `x` is in metre. The maximum value of position co-ordinate of particle on positive `x`-axis isA. `1m`B. `2m`C. `1/2m`D. `4m` |
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Answer» Correct Answer - C |
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| 1034. |
Which of the following statements `is//are` correct ?A. If the velcity of a body changes, it must have some acceleration.B. If the speed of a body change, it must have some acceleration.C. If the body has acceleration, its speed must change.D. 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, definitely there will be acceleration. If speed changes, then velocity also changes. So definitely there will be acceleration. Acceleration, may be due to change in the direction of velocity only and not magnitude. If body has acceleration ofvelocity. |
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| 1035. |
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. `2/(n^(2))-1/n`B. `2/(n^(2))+1/n`C. `2/n-1/(n^(2))`D. `2/n+1/(n^(2))` |
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Answer» Correct Answer - C distance covered `n^(th)` second is `(a)/(2)(2n-1)` distance covered in n second `=1/2 an^(2)` this implies ratio `=2/n-1/(n^(2))` |
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| 1036. |
A particle starts from rest and moves with an acceleration of `a={2+|t-2|}m//s^(2)` The velocity of the particle at `t=4` sec isA. `16m//s`B. `20 m//s`C. `8 m//s`D. `12 m//s` |
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Answer» Correct Answer - D |
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| 1037. |
Two bikes `A and B` start from a point. A moves with uniform speed `40 m//s and B` starts from rest with uniform acceleration `2 m//s^2`. If `B` starts at `t = 10` and `A` starts from the same point at `t = 10 s`, then the time during the journey in which `A` was ahead of `B` is :A. `20 s`B. `8s`C. `10 s`D. `A` is never ahead of `B` |
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Answer» Correct Answer - D A will be ahead of B when `x_(A) gt X_(B)` 40(t-10) gt (0) t +`1/2(2) t^(2)` as A is 10 sec. late than B. `rArr t^(2)-40 t +400 lt 0` `rArr (t-20)^(2) lt 0` Which is not possible. So A will never be ahead at B. |
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| 1038. |
A body starts from rest and then moves with uniform acceleration. Then.A. Its displacement is directly proportional to square of timeB. Its displacement is inversely proportion to the square of the time.C. It may move along a circle.D. It always moves in a straight line. |
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Answer» Correct Answer - A::D From `s=(1)/(2) st^(2) , u=0` `s prop t^(2) `, since the particle starts from rest and accelration is constant., so there is no change in the direction of velocity and the particle moves in a straight line always. |
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| 1039. |
The figure shows velocity-time graph of a particle moving along a straight line. Identify the correct statement. A. The particle starts from the originB. The particle crosses it initial position at `t = 2 s`C. The average speed of the particle in the time interval, `0letle2s` is zeroD. All of the above |
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Answer» Correct Answer - D s=net area of v-t graph At `2s`, net area=0 `:. s=0` and the particle crosses its initial position. |
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| 1040. |
A cart is moving horizontally along a straight line with constant speed `30 ms^-1.` A particle is to be fired vertically upwards from the moving cart in such a way that it returns to the cart at the same point from where it was projected after the cart has moved `80 m.` At what speed (relative to the cart) must the projectile be fired? (Take `g = 10 ms^-2`)A. `10 ms^-1`B. `10 sqrt8 ms^-1`C. `40/3 ms^-1`D. None of these |
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Answer» Correct Answer - B `t=(80m)/(30m//s)=8/3s.` Now in vertical direction `t=(2u_r)/(a_r)` or `u_r=(ta_r)/2 (a_r=g=10m//s^2)` `=((8//3)(10))/2` `=40/3 m//s` |
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| 1041. |
At `t=0`, an arrow is fired vertically upwards with a speed of `100 m s^(-1)`. A second arrow is fired vertically upwads with the same speed at `t=5 s`. Then .A. The two arrows will be at the same height above the `t=20 s`,B. The two arrows will reach back their starting points at `t=20 s` and at `t=25 s`.C. The ratio of the speeds of the first and second arrow at`t= 20 s` will be `2 :1`.D. The maximum height attained by either arrow will be `1000 m`, |
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Answer» Correct Answer - A::B::C Let they meet at height `h` after time `t`. `h=100 t-(1)/(2)g t^(2) rarr` for first aarow `=100 (t-5) -(1)/(2) g (t-5)^(2) rarr` for decond arrow `rArr t =- 12.5 s` (after solving ). So (a) is correct. Time of flight of first arrow: `T=(2u)/(g) =(2xx 100)/(10) =20 s` Second arrow will reach after `5 s` of reaching first. So `b` is rossect. `{:(v_(1) =100 -10 xx 20 =- 100 m s^(-1))`,`(v_(2) =100 -10 xx 15 =-50 m s^(-1)):}` Ratio: `(v_(1))/(v_(2))=2: 1. `So (c) is correct. Maximum deight attained `H=(u^(2))/(2g)=((100)^(2))/(2xx10)=500 m`. Hence, `d` incorrect. |
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| 1042. |
Two bodies of masses M and m are allowed to fall from the same height . If air resistance for each body be same , will the two bodies reach the ground simultaneously ? |
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Answer» Let (F) be thet air resistance on each body. Then acceleration of the body of mass(M) is ltbRgt ` a_1 = (Mg -F)/M =g - F/M` The accleration of the body of mass (M) is ` `a_2 = ( mg -F)/m = g F/M` As, ` Mgtm., so a_1 gta_2`. Therefore, the body of mass (M) ( i.e. heavier body ) will reahc ground first, . |
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| 1043. |
From the top of a tower of height `200 m`, a ball `A` is projected up with `10 m s^(-1)`. And `2 s` later another ball `B` is projected verticall down with the same speed. Then .A. Both `A` and `B` will reach the ground simultaneouslyB. Ball `A` will hit the ground `2 s` later than B hitting the ground.C. Both the balls will ground with same velocity.D. Both the balls will hit the ground with different velocity. |
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Answer» Correct Answer - A::C Ball `A` will return to the top of tower after with speed of `10 m s^(-1)` downwards . And this time, `B` also projected downwards with `10 m s^(-10)`. So both reach ground simultaneously. Also they will hit the ground with the same speed. |
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| 1044. |
Two bodies of masses `(m_(1))` and `(m_(2))` are droppded from heithts `h_(1)` and `h_(2)`, respectively. They reach the ground after time `t_(1)` and `t_(2)` and strike the ground with `v_(1)` and `v_(2)`, respectively Choose the correct relations from the following.A. `(t_(1))/(t_(2)) =sqrt((h_(1))/(h_(2))`B. `(t_(1))/(t_(2)) =sqrt((h_(2))/(h_(1))`C. `(v_(1))/(v_(2)) =sqrt((h_(1))/(h_(2))`D. `(v_(1))/(v_(2)) =(h_(2))/(h_(1))` |
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Answer» Correct Answer - A::C `t=sqrt ((2h)/(g)), v=sqrt 2gh`. |
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| 1045. |
A disc is rotating with constant angular velocity `omega` in anticlockwise direction. An insect sitting at the centre (which is origin of our co-ordinate system) begins to crawl along a radius at time `t = 0` with a constant speed V relative to the disc. At time `t = 0` the velocity of the insect is along the X direction. (a) Write the position vector `(vec(r))` of the insect at time ‘t’. (b) Write the velocity vector `(vec(v))` of the insect at time ‘t’. (c) Show that the X component of the velocity of the insect become zero when the disc has rotated through an angle `theta` given by `tan theta = (1)/(theta)`. |
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Answer» Correct Answer - (a) `vec(r) = vt [cos (omegat)hati + sin (omegat)] hatj` (b) `vec(V_(p)) = V [cos (omegat) - omegat sin (omegat)]` `hati + V [sin (omegat) + omegat cos (omegat)] hatj` |
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| 1046. |
A lift is moving up with acceleration a `A` person inside the ligt throws the ball upwards with a velocity `u` relativeto hand. a. What is the time of flight of the ball? b. What is the maximum height reached by the ball in the lift? |
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Answer» a.`vec a_(BL)=vec a_(B)-vec a_(L) =g+a` `vec s=vec u t+(1)/(2) vec a_(BL)t^(2), 0=uT -(1)/(2)(g+a)T_(1)^(2)` b. `v^(2)-u^(2)=2` as `0-u^(2)=-2 (g+a)H rArr H=u^(2)/(2 (g+a))`. |
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| 1047. |
The acceleration due to gravity near the surface of the earth is `vec(g)` . A ball is projected with velocity `vec(u)` from the ground. (a) Express the time of flight of the ball. (b) Write the expression of average velocity of the ball for its entire duration of flight. Express both answers in terms of `vec(u)` and `vec(g)`. |
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Answer» Correct Answer - (a) `t = - (2vec(u).vec(g))/(|vec(g)|^(2))` (b) `vec(V_(av)) = vec(u) - (vec(g)(vecu.vec(g)))/(|vec(g)|^(2))` |
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| 1048. |
On a smooth horizontal platform a mass of 2 kg is dragged with a horizontal force of 10 Nt. On platform there are so many holes spreaded on its surface below which there is an air blower which exerts a force on block in upward direction depending on its height above the platformas `vec(F) = 20 (2 - h)N`, where h is the height of the block above the platform. Let at `t = 0` block starts from rest from origin of coordinate system shown. Find the equation of trajectory of the block during its motion. Consider x-axis along the motion ofparticle and y-axis in vertical up direction. |
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Answer» Correct Answer - `[2 sqrtx = sin^(-1) (y - 1) - pi]` |
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| 1049. |
A ball is thrown straight up in air with an initial velocity u. Air exerts a force on it in horizontal direction which produces an acceleration depending on its height from ground as `a_(x) = ah^(2)`. Find the displacement of ball from the projection point as a function of time. |
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Answer» Correct Answer - [`r = sqrt(x_(t)^(2) + y_(t)^(2))` where `x_(t) = (au^(2)t^(4))/(12) + (ag^(2) t^(6))/(120) - (aug t^(5))/(20), y_(1) ut - (1)/(2) g t^(2)]` |
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| 1050. |
If ` vec a` and vec b` are two non collinerar unit vectors and if ` | vec a+ vec b | = sqrt 3`, then find the value of ` (vec a-vec b). (2 vec a+ vec b)`. |
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Answer» Given ` a= b= 1` and ` | vec a + vec b| = sqrt 2` :. ` a^2 +b^1+ 2 ab cos theta =3 ` or ` cos thete = 1//2` Now ( `vec a-vec b`). (` 2 vec a+vec b`)= `2 a^2 -b^2 - ab cos theta` = 2- 1 1/2 1/2` . |
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