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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. |
Fig. 2 (b). 12` shows thae distance (S) -time (t) graphs of two trains, which start moving simultaneously in the same direetion. From the graphs, find: . (a) How much (B) is ahead of (A) when motion starts (b) What is the speed of (B) ? ( c) What and where (A) will catch (B) ? ` (d) What is the differnce in speeds of (A) and (B) ? lt |
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Answer» (a) When motion starts, (B) is ahead of (A) by the distance `OC= 80 km`. (b) Speed of `B =(DE)/(CE) =(120-80)/(2-0) =20 km h^(-1)` (c ) Since the two graphs at point `D, so A` woll catch (B) after 2 ours and at a destance of `120 km` from the origh. (d) Speed of `A =(DF)/(FO) =(120-0)/(2-0) =60 km h^(-1)` Differnce in speed `60 -20 40 km h^(-1)`. |
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| 52. |
Jet plane A is moving towards east at a speed of `900 km//hr`. Another plane B has its nose pointed towards `45° N` of E but appears to be moving in direction `60° N` of W to the pilot in A. Find the true velocity of B. `[sin 60° = 0.866 , sin 75° = 0.966]` |
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Answer» Correct Answer - `807 kph` |
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| 53. |
In the previous question, the height attained by the rocket before deceleration is :A. 1 KmB. 10 KmC. 20 KmD. 60 Km |
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Answer» Correct Answer - B |
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| 54. |
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» Lrt boy standing at (A) throw a ball with a velocity (u) at an angle `theta` with the building and the ball just enters window (W), As the boy is at `78.4 m` from the building and the ball just enters the window `39.2 m` above the ground therefore Max , height, `u^(2) sin^(2) theta)/(2 g) =39.2 m` ....(i) and horizontal range, `(u^(2) sin 2 theta)/g =2 xx 78.4 m` ...(ii) Dividing (i) by (ii), we get `(u^(2) sin^(2) theta)/ 2 g) g/(u^(2) 2 sin theta cos theta) =(39.2)/( 2 xx 78.4)` or ` (sin ^(2) theta)/(2 xx 2 sin theta cos theta) =1/4` or ` tan theta =1 or theta =45^(@)` Substituting in (ii), we get `(u^(2) sin 90^(@))/(9.8) =2 xx 78 .4` or `u=zsqrt (2 xx 78.4 xx 9.8 ) =3 9 .2 ms^(-1)`. |
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| 55. |
Which is greater , the angular velicty of the hour hand of a watch or angular velocity of earth around its own axis ? |
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Answer» For hour hand of watch, time period, ` T_h =12 h, For earth, `T_e= 24 h`. Angular velocity, ` omega=(2 pi)/ T` :. ` (omega _h)/(omega_e) = T_e/T_h = (24)/(12) =2` or ` omega_h=2 omega_e` Thus angular velocity of hour hand of a watch is greater than the angular velocity of earth around its own axis. |
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| 56. |
What are the angular velocities of a second hand, minute hand and hour hand of a clock ? |
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Answer» Second hand of a clocj completes one rotation in ` 60 seconds` . So ` T =60 s, theta = 2 pi ` rad. ` omega = (theta)/T = (2 pi) /(60) = (pi) /(30) rad//s ` For minute hand of a clock, ` T =60 min =60 xx 60 s = 3600 s` ` omega = (2 pi)/(3600) = (pi)/(1800) rad//s` For hour hand of a clock, ` T = 12 h= 12 xx 60 xx 60 s` ` omega = (2 pi)/(12 xx 60 xx 60) = (pi) /(21600) rad//s`. |
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| 57. |
Choose the correct statement:A. The magnitudes of sped and velocity are the same when a body travels in a straight line path.B. The average speed of a moving body can be equal to zero, but its average velocity cannot be equal to zero .C. To describe the velocity , direction is necessary.D. Both (a) and ( c) |
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Answer» Correct Answer - D (1) When a body travels in a strianght line path, its distance and displacment are equal , and hence, in a given time , its velocity is equal to its speed. (2) . The average velocity of a body may be equal to zerom but the average speed will not be equal to zero uncles and unit the distance covered is zero (i.e the body is at rest (3) . To describe velocity, direaction is necessary as velocity is speed in a specific direaction. |
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| 58. |
A smooth ring A of mass m can slide on a fixed horizontal rod. A string tied to the ring passes over a fixed pulley B and carries a block C of mass `M(=2m)` as shown in figure. At an instant the string between the ring and the pulley makes an angle `theta` with the rod. (a). Show that, if the ring slides with a speed v, the block descends with speed `v cos theta`, (b). With what acceleration will the ring starts moving if the system is released from rest with `theta= 30^0` |
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Answer» Correct Answer - `[v sec theta]` |
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| 59. |
Wind is blowing west to east along two parallel tracks. Two train moving with the same speed in opposite directions on these tracks have the steam tracks. If one is double than the other, what is the speed of each train ?A. equal to that of wind`B. double that of windC. three times that of windD. half that of wind |
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Answer» Correct Answer - C Let (u) and (v) be the speed of train and wind respectively. The speed of steam track of train moving in the direction of wind ` = u-v` The speed of steamtrack of train moving in the oppsite direction of wind ` =u +v`. As per question, (u+v) = 2 (u-v) ` or u= 3 v`. |
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| 60. |
A particle is resting over a smooth horizontal floor. At `t = 0,` a horizontal force starts acting on it. Magnitude of the force increases with time according to law `F = alpha t,` Where alpha is a positive constant. From figure, which of the following statements are correct ? A. Curve `1` shows acceleration against timeB. Curve `2` shows velocity against timeC. Curve `2` shows velocity against accelerationD. none of these |
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Answer» Correct Answer - D `F=alphat` `ma=alphat` `rArr aalphat` curve 1 is against time `(mdv)/(dt) = prop t`. Integrating this we get `v prop t` this implise parabola. also `v prop a^(2)rArr ` curve 3 shows velocity against acceleration. |
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| 61. |
A particle is resting over a smooth horizontal floor. At `t = 0,` a horizontal force starts acting on it. Magnitude of the force increases with time according to law `F = alpha t,` Where alpha is a positive constant. From figure, which of the following statements are correct ? A. Curve 1 can be the plot of acceleration against timeB. Curve 2 can be the plot of velocity against timeC. Curve 2 can be the plot of velocity against accelerationD. Curve 1 can be the plot of displacement against time |
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Answer» Correct Answer - A::B `a=F/m=alpha/mt……(i)` or `apropt` i.e. a-t graph is a straight line passing through origin. If `u=0,` then integration of Eq. (i) gives, `v=(alphat^2)/(2m)` or `vpropt^2` Hence in this situation (when `u=0`) v-t graph is a parabola passing through origin. |
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| 62. |
In projectile motion, kinetec eneragy is…………………………at the point of frpjection and is………………………………………….at the highest point. |
| Answer» Correct Answer - maximum , minimum | |
| 63. |
The maximum height attanide by a projectile is equal to…………………………..of its maximum range. |
| Answer» Correct Answer - one fourth | |
| 64. |
If `vec A =3 hat I + 4 hat j ` and `vec B =7hat I + 14 hat j`, find a vector having the some magnitude as `vec A` and parallel to `vec B`. |
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Answer» Here,` | vec A| =sxqart 3^(2) + 4^(2) =5` , `|vec B| =sqrt 7^(2) + 24 ^(2) =25` Unit vector in the direction of `vec B` is `hat B =(vec B)/B = (7 hat I + 24 hat j)/(25) =1/(25) (7 hat i+ 24 hat j)` and pparallel to `vec B` is `=|vec A| hat B = 5 xx 1/(25) (7 hat i + 24 hat j)` `=1/5 (7 hat + 24 hat j)`1. |
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| 65. |
For a moving particle, which of the following options may be correct? Here, `V_(av)` is average velocity and `v_(av)` the average speed.A. `|V_(av)|ltv_(av)`B. `|V_(av)|gtv_(av)`C. `V_(av)=0` but `v_(av) !=0`D. `V_(av)!=0 but v_(av) =0` |
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Answer» Correct Answer - A::C `v_(av)=s/t and v_(av)=d/t` Now, `dge|s|` `:. v_(av) ge|v_(av)|` |
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| 66. |
A point mass starts moving in straight line with constant acceleration `a` from rest at `t=0`. At time `t=2s`, the acceleration changes the sign remaining the same in magnitude. The mass returns to the initial position at time `t=t_(0)` after start of motion. Here `t_(0)` isA. `4s`B. `(4+2sqrt(2))s`C. `(2+2sqrt(2))s`D. `(4+4sqrt(2))s` |
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Answer» Correct Answer - B |
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| 67. |
In a car race car `A` takes `t_(0)` time less to finish than car `B` and pases 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 toA. `(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 |
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| 68. |
In a car race, `A` takes a time of `t` s, less than car `B` at the finish and passes the finishing point with a velocity `v` more than car `B`. Assuming that the cars start from rest and travel with constant accelerations `a_(1)` and `a_(2)`. Respectively, show that `v=sqrt(a_(1) a_(2)t)`. |
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Answer» Let (S) be the distance to be covered by each car. Let ` t_(1), t_(2)` be the times taken by cars ` A and B` to complete their journey and ` v_(1), v_(2)` be their velocitues at the finishing pont. According to given problem ` ` v_(1) -v_(2) =v and ` t_(2)-t_(1) =t` As, distance travelled =average velcity x time interval when ` u=0, so` ` S= (0+ v_(1))/2 t_(1) =(0+v_(2))/2 t_(2) or S=(v_(1) t_(1)/2 =(v_(2) t_(2)/2` or ` v_(1) =2 S//t_(1) and v_(2) a_(2) t_(2)^(2)` or ` t_(1)=sqrt(2 S)/a_(1) and t_(2) =sqrt (2 S)/a_(2)` Now `v/t =(v_(1)-v_(2))/(t_(2)-t_(1))=((2 S//t_(1)) -(2 S//t_(2)))/(t)(2)-t_(1) =(2 S)/ t_(1) t_(2)` ` (2 S)/(sqrt(2 S)/a_(1) xx sqrt92 S)/a_(2) = sqrta_(1) a_(2)` or `v=t sqer a_(1) a_(2)`. |
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| 69. |
The velocity of a particle moving in a straight line is directly proportional to `3//4th` power of time elapsed. How does its displacement and acceleration depend on time? |
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Answer» Correct Answer - A::D `v prop t^(3/4)` `a=(dv)/(dt)rArr a prop t^(-1/4)` `s=int vdt rArr spropt^(7/4)` |
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| 70. |
A body is projected with a veocity of `40 ms^(-1)`. After `2 s` it fcrosses a vertical pole of height ` 20 .4 m` Find the angle of projection and horizontal range of projectile. (g = 9.8 ms^(-2)`. |
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Answer» Let ` theta`be the angle of projection of projectile with horizontal. Taking vertical upward motion of the projectile from point projection up to the top of vertical pole, we have ` u_y = u sin theta = 40 sin theta, a_y =- 9.8 m//s^2 ,` ` t= 2 s , y= 20 .4 m` As ` y= u-y t + 1/2 a-y t^2` :. ` 20. 4 = 40 sin theta xx 2 + 1/2 (- 9.8 0 xx 2^2` or ` is theta = ( 20 . 4 + 19 .6) //80 = 1//2 or theta = 30^2` Horizontal range, ` R = (u^2 sin 2 theta) /g` `= ( 40^2)/( 9.8) sin ( 2 xx 30^@) = 141.4 m`. |
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| 71. |
A car is moving along a straight (OP). It moves from `O "to" P` in `18 seconds` and returns from `P "to" Q` in 6 seconds, where `OP=360 m` and `OQ=240 m` What are the car the average velocity and average speed of the car in going (a) from `O "to" P` and back to `Q`? |
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Answer» (a) For a motion from `O to P`, Displacement `=360 m` and path length `=360 m` Average velocity`= (displacement) /(time) =(360 m)/(18s)` `=20 ms^(-1)` along `vec (OP)` Average speed `=(path length)/(time)` `=(360 m)/(18 s) = 20 ms^(-1)` (b) For a motion from `O to P` and back to `Q , displacement`=OQ =24 0 m` path length `=OP +PQ =360 + (360 +(360 -240)` `=360 +120 =480 m` Average velocity `=(240)/(18+6)` `= 10 ms^(-1)` along `vec (OP)` Average speed `=(480)/(18 +6) =20 ms^(-1)`. |
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| 72. |
A body travelling along a straight line , one thired of the total distance with a velocity ` 4 ms^(-1)`. The remaining part of the distance was covered with a velocity ` 2 ms^(-1)` for half the time and with velocity ` 6 ms^(-1)` for the other half of time . What is the mean velocity averaged over te whle time of motin ?A. `5m//s`B. `4m//s`C. `4.5m//s`D. `3.5m//s` |
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Answer» Correct Answer - B |
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| 73. |
A particle travels along a straight line. It covers halg the distance with a speed (v). The remaining part of the distance was covere with speed `v_(1)` for half the time and with speed `v_(2)` for the other half the time . Find the average speed of the particle over the entire motion. |
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Answer» Let (S) be the total distance to betravelled by the particle. Time taken by particle to travel halgf the distance `(S//2)` is `t_(1) =(S//2)/v =S/(2v)` Let `2t_(2)` be the time takenby particle to cover the remainig distance `S//2` Distance travelled by particle in time `t_(2)` while travelling with speed `v_(1)` is Distance travelled by particle in time `t_(2)` while travelling with speed `v_(2` is `S_(2) =v_(1) t_(2)` `S_(1) +S_(2) =v_(1) t_(2) + v_(2) t_(2) ` But `S_(1) +S_(2) =S/2 =v_(1) t_(2) +v_(2) t_(2) + (v_(1) +v_(2)) t_(2)` or `2t_(2) =S/((v_(1) +v_(2))` Average speed `=(total distance travelled)/(total time talke)` `=S/(t_(1) +2t_(2)) =S/((S//2 v) + S//(v_(1) +v_(2))` `=(2 v(v_(1) + v_(2))/(v_(1) +v_(2) +2v)`. |
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| 74. |
A person travels along a straight road for the first half length with a velocity `v_(1) ` and the second half length with velocity ` v_(2)`. What is the mean velocity of the person ? |
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Answer» Time take by person to travel first half length, ` t_(1) =((d//2))/v_(1) =d/(2 v_(1)` Time taken by person to travel second half length , `t_(2) =((d //2))/v_(2) =d/(2 v_(1))` :. Total time ` =t_(1) + t_(2) d/2 [1/v_(1) + 1/v_(2)]` `=(d (v_(1) + v_(2))/(2 v_91) v_(2))` :. Mean velocity or average velocity ` =d/(t_(1) + t_(2)) =(2 u _(1) v_(2))/((v_(1) + v_(2))`. |
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| 75. |
An athelete completes one round of a circular track of radius ` R in 40 seconds`. What will be the displacement at the end of ` 2 min. 20 second ? |
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Answer» In `2 minutes 20 seconds`, and athelete will complete =(2 xx 60 + 20 )/(40) =3.5` frounds, i.e. three complete rotations and one half rotation, i.e.., finally it will be at the oppoiste end of the diamere from the strating point. So displacement` =2 R`. |
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| 76. |
An athelete moves along a path PQRSTUVP, in 44 seconds as shown in the figure, then the average speed of the athlete is ______ `m s^(-1)` A. zeroB. ` 5/7`C. `5/11`D. none of these |
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Answer» Correct Answer - B Total distance traced by the athlete= sum of perimeter of two circles ` = 2 pi r_(1) + 2pir_(2) = 2pi (r_(1) + r_(2)` ` = 2pi (2+3) = 10 pi ` time taken to trace the path , (t) = 44s Average speed = `(" total distance")/(" total time taken")` ` (10pi)/44` ` = (10xx 22)/(7 xx 44) = 5//7 ms ^(-1)` |
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| 77. |
An athelete takes 2.0 s to reach his maximum speed of 18.0 km/h. What is the magnitude of his average acceleration ? |
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Answer» Initial velocity u = 0 ( starts from rest) |
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| 78. |
A truck driver, starting with zero speed at time zero, drove in such a way that the speed time graph is approximately an isosceles triangle with the base along the time axis. The maximum speed was `30 m//s`, and the total elapsed time was 50.0 s. What distance did he travel. |
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Answer» Correct Answer - [570 m] |
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| 79. |
A large rectangular box ABCD falls vertically with an acceleration a. A toy gun fixed at A and aimed towards C fires a particle P.(a) P will hit C if a = g. (b) P will hit the roof BC if a > g. (c) P will hit the wall CD or the floor AD if a < g. (d) May be either (a), (b) or (c), depending on the speed of projection of P. |
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Answer» Correct Answer is: (a, b, & c) Superimpose an upward acceleration a on the system. The box becomes stationary. The particle has an upward acceleration a and a downward acceleration g. If a = g, the particle has no acceleration and will hit C. If a > g, the particle has a net upward acceleration, and if a < g, the particle has a net downward acceleration. |
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| 80. |
A large box moves vertically downwards with an acceleration a. A toy gun fixed at `A` aimed towards `C` fires a particle `P`. A. `P` will hit `C` if `a = g`B. `P` will hit the roof `BC`, if `fa gt g`C. `P` will hit the wall `CD` if ` alt g`D. May be either `(A),(B)` or `(C )`, depending on the speed of projection of `P`. |
| Answer» Correct Answer - A::B | |
| 81. |
While starting from a station, a train driver was instructed to stop his train after time T and to cover maximum possible distance in that time. (a) If the maximum acceleration and retardation for the train are both equal to ‘a’, find the maximum distance it can cover. (b) Will the train travel more distance if maximum acceleration is ‘a’ but the maximum retardation caused by the brakes is ‘2a’? Find this distance. |
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Answer» Correct Answer - (a) `(1)/(4) aT^(2)` (b) Yes, `(1)/(3) aT^(2)` |
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| 82. |
Two particles projected vertically upward from points `(0,0)` and `(1,0)m` with uniform velocity `10m//s` and `vm//s` respectively, as shown in the figure. It is found that they collide after time `t` in space. Time `t` is `1/(5(sqrt(x)-1))`. Calculate `x` A. `2`B. `3`C. `5`D. `8` |
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Answer» Correct Answer - B |
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| 83. |
A large rectangular box moves vertically downward with an acceleration `a`. A toy gun fixed at `A` and aimed towards `C` fires a particle `P`. A. `P` will hit `C` if `a=g`B. `P` will hit the roof `BC`, if `agt g`C. `P` will hit the `CD` if `alt g`D. May be either (a), (b) or (c) depending on the speed of projection of `P`. |
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Answer» Correct Answer - D |
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| 84. |
A stone is throun with an initial speed of ` 4.9 ms^(-1)~ from a bridge in vertical upward direction. If falls down in water after `2` seconds. The height of the bridge is. |
| Answer» Correct Answer - (b) | |
| 85. |
Between two stations a train starting from rest first accelerates uniformly, then moves with constant velocity and finally retarts uniformly to come to rest. If the ratio of the time taken be `1 : 8 : 1` and the maximum speed attained be `60 km//h`, then what is the average speed over the whole journey ? |
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Answer» Correct Answer - [54 kph] |
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| 86. |
A lift accelerates downwards from rest at rate of `2 m//s^(2)`, starting `100 m` above the ground. After 3 sec, an object falls out of the lift. Which will reach the ground first? What is the time interval between their striking the ground? |
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Answer» Correct Answer - object, 3.3 s After 3 sec distance covered `=1//2xx2xx9=9m` velocity of life`=2xx3=6 m//s darr :. U_(p)=6m//s darr, a=g darr` height `=(100-9)=91 m` `:.` Time to reach the ground `=91=6t+1/2xxgxxt^(2) t=3.7` sec Total time taken by object to reach the ground `=3+3.7=6.7` sec Time to reach on the ground by lift `=1/2xx2xxt^(2)=100 rArr t=10` sec So interval `=10-6.7=3.3`sec |
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| 87. |
A particle starts from rest at time `t=0` and moves on a straight line with acceleration a (ms^-2) as plotted in Fig. 2 (b) .17. Find the time at which the speed og the particle is maximum. Also calculate the displacement of theparticle from starting point after `4 s`. . |
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Answer» Taking motion from `0 to 2 seconds` , we have `u=0, a=- `10 m//s^(2), t=2 s, v=? `v=u + at =0 + (-10) xx 2= =- 20 ms^(-1)` Taking motion from `2 to 4 swconsa ` we have ltBrgt `u==- 20 ms^(-1) , a=10 m//s^(2) , t=2 s, v=? Theregore, the speed is maximum at time `t=2 seconds` Taking motion from `0t0 2swons, it `S_(1) is the distance coverd, then` `S_91) =(v^(2)-u^(2))/(2a) =((-20)^(2) -(0)^(2))/(2(-10)) =-20m` Taking motion from `2 to 4` seconds it `S_(2) is the distance covered then `S_(2) =(v^(2)-u^(2))/(2a) =(-(-20)^(2))/(2 xx 10) =- 20 m :. Total displacement `=S_(1) +S_(2)` `=- 20 +(-20)` `=- 20 + (-20)` =-40 m` lt. |
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| 88. |
A lift starts from rest. Its acceleration is plotted against time. When it comes to rest its height above its starting point is A. `20 m`B. `64 m`C. `32 m`D. `36 m` |
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Answer» Correct Answer - B At 4s `u=at=8m//s` `s_1=1/2at^2=1/2xx2xx4^2=16m` From 4s to 8s `a=0, v=constant=8m//s` `:. s_2=vt=(8)(4)=32m` From 8s to 12s `s_3=s_1=16m` `:. s_(Total)=s_1+s_2+s_3=64m` |
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| 89. |
You are on the roof of the phusics building, `46,0 m` above the ground (Fig.4.40). Your physics professor, who is `1.80 m` tall, is walking alongside the bulding at a constant speed of `1.10 m s^(-1)`. If you wish to drop a flower on your professor`s head, where should the professor be when you release the flower? Assume that the flower is in free fall. . |
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Answer» The time needed for the egg to fall is `tsqrt((2Deltah)/(9)) =sqrt((2(46.0m-1.89 m))/((9.80 ms^(-2))))=3.00 s` and so the professor should be at a distance `v_(y)t=(1.20 ms^(-1))(3.00 s)=3.60 m`. |
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| 90. |
If `vec(A)=vec(B)+vec(C )`, and the magnitudes of `vec(A)`,`vec(B)`,`vec(C )` are 5,4, and 3 units, then the angle between `vec(A)` and `vec(C )` is |
| Answer» Since ` vec A^(2) + vec B^(@) = vec c^(@), therfore angle between ` vec A + vec B ` is 90^(@).Let `theta` be the angle between ` vec A and vec C`, then , ` cos theta =3//5 or theta = cos^(-1) (3//5)` | |
| 91. |
If vectors ` vec A , vec` B` and vec C`have magnitudes `5, 12 and 13 units and ` vec A + vec B =vec C`, find theangle between ` vec B and vec C`. |
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Answer» As ` vec A + vec B = vec C` :. ` vec A = vec C - vec B` Then ` vec A . Vec A =( vec C - vec B) .(vec C - vec B)` `=vec C. vec C -2 vec C . Vec B + vev B . Vec B` or `A^(2) =C^(2) -2 CB cos theta + B^(2)` or `cos theta = (C^(32) + B^(2) -A%(2))/(2 C B)` `=(13^(2) + 12^(2) 12^(2) -5^(2))/(2xx 13 xx 12)= 9288)/(312) =(12) /(13)` `theta = cos ^(-1) (12 // 13)`. |
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| 92. |
If `vec(A)=vec(B)+vec(C )`, and the magnitudes of `vec(A)`,`vec(B)`,`vec(C )` are 5,4, and 3 units, then the angle between `vec(A)` and `vec(C )` isA. cos^(-1) ( 3/5)`B. cos^(-1) (4/5)`C. ` pi //2`D. sin ^(-1) (3/5)` |
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Answer» Correct Answer - A Herem ` | vec A| = 5 , | vec B|= 4, |vec C | = 3` As ` vec A = vec B+ vec C. so vec B = vec A - vec C` As ` vec B` is now resultant of ` vec A` and ` vec C`. So it is given by ` `B^2 = A^2 + C^2 -2 AC cos theta` ` (4)^2= (5)^2 + (3)^2 - 2 xx 5 xx 3 cos theta` ` - 16 - 25 - 9 =- 30 cos theta` ` -18 =- 30 cos theta ` ` cos theta = (-18)/(-30) = 3/2 theta = cos^(-1) ( 3/5)`. |
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| 93. |
Two particles, A and B are moving in concentric circles in anticlockwise sense in the same plane with radii of the circles being `gamma_(A) = 1.0 m` and `gamma_(B) = 2.0 m` respectively. The particles move with same angular speed of `omega = 4 rad//s`. Find the angular velocity of B as observed by A if (a) Particles lie on a line passing through the centre of the circle. (b) Particles lie on two perpendicular lines passing through the centre. |
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Answer» Correct Answer - (a) `omega = 4 rad//s` (b) `omega = 4 rad//s` |
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| 94. |
A man is on straight road AC, standing at A. He wants to get to a point P which is in field at a distance ‘d’ off the road (see figure). Distance AB is `l = 50`. The man can run on the road at a speed `v_(1) = 5 m//s` and his speed in the field is `v_(2) = 3 m//s`. (a) Find the minimum value of ‘d’ for which man can reach point P in least possible time by travelling only in the field along the straight line AP. (b) If value of ‘d’ is half the value found in (a), what length the man must run on the road before entering the field, in order to reach ‘P’ in least possible time. |
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Answer» Correct Answer - (a) `d_(min) = (200)/(3)` (b) `25 m` |
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| 95. |
The retardation fo a moving particle if the relation between time and position is ` t= A x^3 + Bx^2 ` where ` A` and ` B` are appropriate constants will be . |
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Answer» Give, ` t= Ax^3 = Bx^2` ltbgt :. ` (dt)/(dx) =3 Ax^2 = 2 2 Bx` or ` v= (dx)/(dt) = ( 3 Ax^2 + 2 Bx)^(-10` Now , (dv) /dx) = (-1) (3 Ax^2 + 2 Bx)^(-1)` Acceleration , ` a = 9dv)/(dt0 = (dv)/(dx) xx (dx)/(dt)` ` = (-(6 Ax^2 + 2 B))/((3 A x^2 + 2 Bx)^20 xx 1/( (3 Ax^2 + 2 Bx0` `= (-(6 Ax+ 2 B))/( (2 Ax^2 = 2 Bx)^3)` Retardation ` =- a = ( 6 Ax+ 2 B)/((3 Ax^2 + 2 Bx)^3)`. |
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| 96. |
The relation between time t and displacement x is `t = alpha x^2 + beta x,` where `alpha and beta` are constants. The retardation isA. `2 alpha v^3`B. `2 beta v^3`C. `2 alpha beta v^3`D. `2 beta^2 v^3` |
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Answer» Correct Answer - A `(dt)/(dx)=(2 alphax+beta)` `:. (dx)/(dt)=v=(1/(2 alphax+beta))` `a=(dv)/(dt)=-2 alpha(1/(2 alphax+beta))^2.(dx)/(dt)` `=-2alpha(v)^2(v)=-2 alpha(v)^3` |
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| 97. |
A truck travelling due North at ` 50 km h^(-1)` turns Westamd travels at the same speed. What is the change in velocity ?A. ` 50 km h^(-1) ` North -WestB. ` 50 sqrt 2 km h^(-1)` South- WestC. ` 50 sqrt 2 km h^(-1)` South -West ` ` 50 sqrt 2 km h^(-1)south -West.D. |
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Answer» Correct Answer - D Here, ` v_N= 50 km h^(-1)`, change in veloicty ` = vec _w - vec v_N = vec v_w+ (- vecv_N0` :. Magnitude of change in velocity ` = [ | v_w |^2 + |v_N|^2]^(1//2)` `= (50^2 + 50^2)^(1//2) = 50 sqrt 2 ` South West . |
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| 98. |
The motion of a particle along a straight line is described by equation : `x = 8 + 12 t - t^3` where `x` is in metre and `t` in second. The retardation of the particle when its velocity becomes zero is.A. (a) ` 24 m//s^2`B. (b) ` zero`C. (c ) ` 6 m//s`D. (d) ` 12 m//s^2` |
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Answer» Correct Answer - (d) Here, ` x = 8 + 12 t-t^3` :. ` When velocity is zero (v=0) . The ` 12 -3 t^3 =0 ` or ` 12 =3 t^2 ` or t^2 =4` :. ` t=2 s` Also, ` a= (dv)/(dt) = d/(dt) (12 -3 t^2) =- 6t` At ` t= 2 s a =- 6 t =- xx 2 =- 12 m//s^2` So retardation is ` 12 m//s^(2)`. |
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| 99. |
The retardation fo a moving particle if the relation between time and position is ` t= A x^3 + Bx^2 ` where ` A` and ` B` are appropriate constants will be .A. (a)` (6 A x + 2 B)/((3 A x^2 + 2 B x)^3)`B. (b) ` (6 B x + 6 A)`/((3 Ax^2 + 2 B x)^3)`C. (c ) `( 6 A+ 2 B x)/((3 A x+2 B x^2)^3)`D. (d) (6 A+2 B x)/( (3 A+ 2 B x^3)^2)` |
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Answer» Correct Answer - A Here, ` t= A x^3 + B x^2 , (dt)/(dx) (Ax^3 + B x^2)` ` (dt)/(dx) = 3 A x^2 + 2 B x`, ` v = (dx)/(dt) = ( Ax^2 + 2 B x)^1` Now ` (dv)/(dx) = (-1) [3 A x^2 + 2 B x ]^(-2)` xx (6 A + 2 B0` Acceleration ` a = (dv)/(dt) = (dv)/(dx) xx (dx)/(dt) = v (dv)/(dx)` ` a = 1/(3 A x^2+2 B) xx ((- 10 ( 6 A x + 2 B))/((3 A x^2 +2 B x)^2 )` ` =(- (6 A x + 2 B))/((3 A x^2 + 2 B x)^3)` Retardation `=- a= (6 a x + 2 B)/((3 A x^2 + 2 B x )^3)`. |
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| 100. |
(a) A boy on a skateboard is sliding down on a smooth incline having inclination angle q. He throws a ball such that he catches it back after time T. With what velocity was the ball thrown by the boy relative to himself ? (b) Barrel of an anti aircraft gun is rotating in vertical plane (it is rotating up from the horizontal position towards vertical orientation in the plane of the fig). The length of the barrel is `L = sqrt(2) m` and barrel is rotating with angular velocity `omega = 2 rad//s`. At the instant angle `theta` is `45°` a shell is fired with a velocity `2 sqrt(2) m//s` with respect to the exit point of the barrel. The tank recoils with speed `4 m//s`. What is the launch speed of the shell as seen from the ground? |
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Answer» Correct Answer - (a) `(1)/(2)Tg cos theta` Perpendicualr to the incline (b) `4sqrt(2) ms^(-1)` |
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