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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. |
a body is thrown horizontally with a velocity `sqrt(2gh)` from the top of a tower of height h. It strikes the level gound through the foot of the tower at a distance x from the tower. The value of x is :-A. hB. `(h)/(2)`C. 2hD. `(3)/(4)m` |
| Answer» Correct Answer - C | |
| 52. |
A ball is dropped from the certain height on the surface of glass. It is collide elastically the comes back to initial position. If this process it repeated then velocity time graph is:-A. B. C. D. |
| Answer» Correct Answer - C | |
| 53. |
A body is dropped from a tower with zero velocity, reaches ground in 4s. The height of the tower is about:-A. 80mB. 20mC. 160mD. 40m |
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Answer» Correct Answer - A `h=(1)/(2)"gt"^(2)=(1)/(2)xx10xx4^(2)=80m` |
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| 54. |
A stone dropped from the top of a tower of height 300 m high splashes into the water of a pong near the base of the tower. When is the splash heard at the top ? Given that the speed of sound in air is `340ms^(-1)? (g=9.8ms^(-2)`. |
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Answer» Time taken by the stone to reach the water of pond `t_(1)=sqrt((2h)/(g))=sqrt((2xx300)/(9.8))=7.825s` time taken by splash of sound `t_(2)=("distance")/(velocity")=(300)/(340)=0.882a` `therefore` Total time `t=t_(1)+t_(2)=7.825+0.882=8.707s` |
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| 55. |
A ball is thrown upwards from the top of a tower `40 m` high with a velocity of `10 m//s.` Find the time when it strikes the ground. Take `g=10 m//s^2`. |
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Answer» In the problem `u=+10m//s, a=-10 m//s^(2)` and `s=-40 m` (at the point where ball strikes the ground) Substituting in `s=ut+(1)/(2)at^(2)` `-40=10t-5t^(2)-10t-40=0` or `t^(2)-2t-8=0` Solving this we have t=4s and -2s. Taking the positive value t=4s. |
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| 56. |
A rocket is fired vertically from the ground. It moves upwards with a constant acceleration `10m//s^(2)` after 30 sec the fuel is finished. After what time from the instant of firing the rocket will attain the maximum height? `g=10m//s^(2)`:-A. 30sB. 45sC. 60sD. 75s |
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Answer» Correct Answer - C For upward motion `rArr` velocity after 30 sec. v=u+at `v=0+10xx30=300 m//s` Now fueld is finished then, time taken to reach maximum height `rArr` using v=u+at `0=300-10xt` `t=(300)/(10)=30` sec So total time `=30+30=60` sec. |
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| 57. |
Two bodies `A` (of mass `1 kg`) and `B` (of mass `3 kg`) are dropped from heights of `16 m` and `25 m`. Respectively. The ratio of the time taken to reach the ground is :A. `(5)/(2)`B. `(12)/(5)`C. `(5)/(12)`D. `(4)/(5)` |
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Answer» Correct Answer - D `(t_(1))/(t_(2))=sqrt((2h_(1))/(g))/sqrt((2h_(2))/(g))=sqrt(h_(1))/sqrt(h_(2))=sqrt(16)/(sqrt(25))=(4)/(5)` |
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| 58. |
A stone is dropped from the top of a tower of height `h`. Aftre `1 s` another stone is droppped from the balcony `20 m` below the top. Both reach the bottom simultaneously. What is the value of `h` ? Take `g=10 ms^(-2)`.A. 43125mB. 312.5mC. 31.25mD. 25.31m |
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Answer» Correct Answer - C `h=(1)/(2)"gt"^(2)` and `h-20=(1)/(2)g(t-1)^(2)` `(1)/(2)"gt"^(2)-20=(1)/(2)"gt"^(2)-20=(1)/(2)"gt"^(2)-gt+(g)/(2)` `gt=20+(g)/(2)` `10t=20+5rArr t=2.5` `h=(1)/(2)"gt"^(2)=(1)/(2)xx10xx(2.5)^(2)` `=5xx6.25=31.25m` |
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| 59. |
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. At `(H)/(2)` time after releaseB. At `(H)/(4)` metres from groundC. At `(3H)/(4)` metres from the groundD. At `(H)/(6)` metres from the ground |
| Answer» Correct Answer - C | |
| 60. |
Two balls are dropped from different heights at different instants. Seconds ball is dropped 2 sec. after the first ball. If both balls reach the ground simultaneousl after 5 sec. of dropping the first ball. then the difference of initial heights of the two balls will be :- `(g=9.8m//s^(2))`A. 58.8mB. 78.4mC. 98.0mD. 117.6m |
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Answer» Correct Answer - B `h_(1)-h_(2)=(1)/(2)g(t_(1)^(2)-t_(2)^(2))=(1)/(2)xx9.8(5)^(2)-(3)^(2)` `=8xx9.5=78.4m` |
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| 61. |
A body is freely dropped from a height h above the ground. Find the ratio of distances fallen in first one second, first two seconds, first three second first two seconds, first three seconds, also find the rato of distances fallen in `1^(st)` second, in `2^(nd)` second, in `3^(rd)` |
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Answer» From second equation of motion i.e. `h=(1)/(2)"gt"^(2)(h=ut+(1)/(2)"gt"^(2)` and u=0) `h_(1):h_(2):h_(3)......=(1)/(2)g(1)^(2):(1)/(2)g(2)^(2):(1)/(2)g(3)^(2)=1^(2):2^(2):3^(2)......=1:4:9....` Now from the of distance travelled in `n^(th)` second `S_(n)=u+(1)/(2)a(2n01)` here u=0, a=g `S_(n)=(1)/(2)g(2n-1)` or `S_(1):S_(2):S_(3):......=(1)/(2)g(2xx1-1):(1)/(2)g(2xx2-1):(1)/(2)g(2xx3-1)=1:3:5.....` |
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| 62. |
In the graph shown in fig. the time is plotted along x-axis. Which quantity associated with the projectile motion is plotted along the y -axis:- A. kinetic energyB. momentumC. horizontal velocityD. none of the above |
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Answer» Correct Answer - C Horizontal component of velocity remain same. |
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| 63. |
When a body is subjected to a uniform acceleration, it always moves in a straight line. Straight line motion is the natural tendency of the body.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 | |
| 64. |
Assertion: An object can have constant speed but variable velocity. Reason: Speed is a scalar but velocity is a vector quantity.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 | |
| 65. |
Statement-I : The speed of a body can be negative. Statement-II : If the body is moving in the opposite direction of positive motion, then its speed is negative.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 | |
| 66. |
Three particles A, B and C are projected from the same point with the same initial speeds making angles `30^(@), 45^(@) and 60^(@)` respectively with the horizontal. Which of the following statement is correct?A. A,B and C have unequal rangesB. Ranges of A and C are equal and less than that of BC. Ranges of A and C are equal and greater then that of BD. A,B and C have equal ranges |
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Answer» Correct Answer - B `because theta_(A)+theta_(B)=90^(@)` `therefore R_(A)=R_(C)=(u^(2)sin2xx30^(@))/(g_=(u^(2)sqrt(3))/(2g)` `R_(B)=(u^(2)sin2xx45^(@))/(g)=(u^(2))/(g)rArr R_(B)gtR_(A)=R_(C)` |
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| 67. |
Two particles are separated at a horizontal distance `x` as shown in (Fig. 5.57). They are projected at the same time as shown in the figure with different initial speed. Find the time after which the horizontal distance between the particles becomes zero. .A. `(x)/(u)`B. `(u)/(2x)`C. `(2u)/(x)`D. None of these |
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Answer» Correct Answer - A `u_(x_(1))t+u_(x_(2))t=x` `[(u)/sqrt(3)cos 30^(@)+u cos 60^(@)]t=x` `[(u)/sqrt(3)xxsqrt(3)/(2)+(u)/(2)]t=x rArr t=(x)/(u)` |
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| 68. |
A car moves at a velocity 2.24 `km h^(-1)` in first minute, at 3.60 `km h^(-1)` in the second minute and at 5.18 `km h^(-1)` in the third minute. Calculate the average velocity in od car in this interval of time. |
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Answer» Displacement in first minute `s_(1)=v_(1)xxt_(1)=2.24xx(1)/(60)km` Displacement in second minute `s_(2)=v_(2)xxt_(2)=3.60xx(1)/(60)km` Displacement in third minute `s_(3)=v_(3)xxt_(3)=5.18xx(1)/(60)km` Total Displacement, `s=s_(1)+s_(2)+s_(3)=(2.24)/(60)+(3.60)/(60)+(5.18)/(60)=(11.02)/(60)km`. Total time taken, `t=1+1+1=3 "min"=(1)/(20)h` `therefore "average velocity",=(s)/(t)=(11.02)/(60)xx(20)/(1)=3.67 kmh^(-1)` |
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| 69. |
Buses A and B are moving in the same direction with speed `20 ms^(-1)` and `15 ms^(-1)` respectively. Find the relative velcoity of A w.r.t. B and relative velocity of B w.r.t. A. |
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Answer» Let their direction of motion of bus A is `hati` then `vecv_(A)=(20m//s)hati` (a) Relative velocity of A w.r.t. B `=vecv_(AB)=vecV_(A)-vecv_(B)=` (actual velocity of A)-(velocity of B) `=(20 m//s)hati-(15m//s)hati=5m//shati` i.e. A is moving with speed `5m//s` w.r.t B in `hati` direction. (b) Relative velocity of B w.r.t `A=vecv_(BA)=vecv_(B)-vecv_(A)=` (Actual velocity of B)-(velocity of A) `=(15m//s)hati-(20 m//s)hati=(-5m//s)hati=(5m//s)(-hati)` i.e. B moving in `-hati` direction w.r.t A, at speed 5 m/s |
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| 70. |
In a reference frame a man A is moving with velocity `(3hati-4hatj)ms^(-1)` and another man B is moving with velocity `(hati+hatj)ms^(-1)` relative to A. Find the actual velocity of B. |
| Answer» `vecv_("actual")=vecv_(ref)+vecv_(ref)=[(hati+hatj)+(3hati-4hatj)]ms^(-1)=(4hati-3hatj)ms^(-1)` | |
| 71. |
A particle starts from rest. Its acceleration at time t=0 is `5 m//s^(2)` which varies with time as shown in the figure. The maximum speed of the particle will be A. 7.5m/sB. 15m/sC. 30m/sD. 37.5 m/s |
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Answer» Correct Answer - B Area under a-t graph gives the change in velocity during given time interval. `therefore u_("max")=(1)/(2)xx5xx6=15m//s` Since initial velocity=0 `therefore` Maximum speed of the particle =15m/s |
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| 72. |
A body is slipping from an inclined plane of height h and length l . If the angle of inclination is `theta` , the time taken by the body to come from the top to the bottom of this inclined plane isA. `sqrt((2h)/(g))`B. `sqrt((2l)/(g))`C. `(1)/(sin theta)sqrt((2h)/(g))`D. `sin theta sqrt((2h)/(g))` |
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Answer» Correct Answer - C `because (h)/(l)=sin theta rArr l=(h)/(sin theta)` component of g along l is g `sin theta` `t=sqrt((2l)/(a))=sqrt((2h)/(sintheta/(g sin theta)))=(1)/(sin theta)sqrt((2h)/(g))` |
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| 73. |
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 `v=at, (ds)/(dt)=atrArrds=at dt` `(V_("avg"))_("space")=(intVds)/(intds)=(int at at dt)/(int ad dt )=(a^(2)(t^(3))/(3))/((at^(2))/(2))=(2at)/(3)` `(V_("avg"))_("time")=(int v dt)/(int dt)=(int at dt a (t^(2))/(2))/(int dt t)=(at)/(2)` `((V_("avg"))_("space"))/((V_("avg"))_("time"))=((2at)/(3))/((at)/(2))=(4)/(3)` |
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| 74. |
In the Q.23, the angle of projection with the horizontal is:-A. `tan^(-1)((4)/(5))`B. `tan^(-1)((5)/(4))`C. `tan^(-1)((5)/(8))`D. `tan^(-1)((8)/(5))` |
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Answer» Correct Answer - C `because u_(x)=40` and `u_(y)=25` `therefore tan theta=(u_(y))/(u_(x))=(25)/(40)=(5)/(8)` |
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| 75. |
Acceleration is defined as the rate of change of velocity. Suppose we call the rate of change of acceleration as ` SLAP ?` (i) What is the unit of `SLAP` . (ii) How can we calculate instantaneous SLAP ?A. `m//s^(2)`B. `m//s^(3)`C. `m//s`D. `m^(2)//s^(3)` |
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Answer» Correct Answer - B Rate of change of acceleration `=(Deltaa)/(t)=(m)/((s^(2))/(s))=(m)/(s^(3))` |
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| 76. |
A graph between the square of the velocity of a particle and the distance(s) moved is shown in figure. The acceleration of the particle in kilometers per hour square is:- A. 2250B. 3084C. -2250D. -3084 |
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Answer» Correct Answer - D `V^(2)=U^(2)+2as` `a=(v^(2)-u^(2))/(2s)=((900)-(4600))/(2xx0.6)=-(3700)/(1.2)` `=-3084 km//hr^(2)` |
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| 77. |
The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is .A. `15^(@)`B. `30^(@)`C. `45^(@)`D. `60^(@)` |
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Answer» Correct Answer - D `u cos theta=(u)/(2)` So `theta=60^(@)` |
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| 78. |
A missile is fired for maximum range with an initial velocity of `20m//s`. If `g=10m//s^(2)`, the range of the missile isA. 40mB. 50mC. 60mD. 20m |
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Answer» Correct Answer - A Maximum range `R_("max")=(u^(2))/(g)=((20)^(2))/(10)=40m` |
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| 79. |
A force `vecF=6t^(2)hati+4thatj` is acting on a particle of mass 3 kg then what will be velocity of particle at t=3 second and if at t=0, particle is at rest:-A. `18 hati+6hatj`B. `18hati+12hatj`C. `12hati+6hatj`D. none |
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Answer» Correct Answer - A `veca=(vecF)/(m)=(1)/(3)(6t^(2)hati+4hatj)=2t^(2)hati+(4)/(3)thatj` `(dvecv)/(dt)=2t^(2)hati+(4)/(3)thatjrArrint_(0)^(v)dvecv=int_(0)^(3)(2t^(2)hati+(4)/(3)thatj)dt` `vecv=[(2t^3))/(3)hati+(4)/(3)(t^(2))/(2)hatj]_(0)^(3)` `=(2)/(3)(3)^(3)hati+(2)/(3)(3)^(2)hatj=18hati+6hatj` |
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| 80. |
A body is thrown with some velocity from the ground. Maximum height when it is thrown at `60^(@)` to horizontal is 90m. What is the height reached when it is thrown at `30^(@)` to the horizontal:-A. 90mB. 45mC. 30m/sD. 15m |
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Answer» Correct Answer - C `h=(u^(2)sin^(2)theta)/(2g)rArrh_(1)=90=((u sin 60^(@))^(2))/(2g)` `90=(u^(2))/(2g)xx[sqrt(3)/(2)]^(2)rArr(u^(2))/(2g)=120` `h_(2)=((u sin 30^(@))^(2))/(2g)=120xx(1)/(4)=30m` |
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| 81. |
Two projectiles of same mass and with same velocity are thrown at an angle `60^(@)` and `30^(@)` with the horizontal, then which quantity will remain same:-A. time of flightB. range of projectileC. max height acquiredD. all of them |
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Answer» Correct Answer - B `because theta_(1)+theta_(2)=90^(@)=90^(@) theta therefore R_(1)R_(2)` |
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| 82. |
The maximum range of a gun on horizontal terrain is 1km. If `g = 10 ms^(-2)`, what must be the muzzle velocity of the shell ?A. `1600 ms^(-1)`B. `400 ms^(-1)`C. `200sqrt(2)ms^(-1)`D. `160sqrt(10)ms^(-1)` |
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Answer» Correct Answer - B `R_("max")=(u^(2)/(g))rArr u^(2)=Rg=16000xx10` `therefore u=400m//s` |
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| 83. |
If the range of a gun which fires a shell with muzzle speed V is R , then the angle of elevation of the gun isA. `cos^(-1)((v^(2))/(Rg))`B. `cos^(-1)((Rg)/(v^(2)))`C. `(1)/(2)sin^(-1)((v^(2))/(Rg))`D. `(1)/(2)sin^(-1)((Rg)/(v^(2)))` |
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Answer» Correct Answer - D `R=(v^(2)sin 2 theta)/(g) rArr sin2 theta=(gR)/(v^(2))` `rArr 2 theta=sin ^(-1)(gR)/(v^(2))rArr theta=(1)/(2)sin^(-1)(gR)/(v^(2))` |
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| 84. |
A car runs at constant speed on a circular track of radius 10m. taking 6.28s on each lap. The average speed and average velocity on each complete lap is:-A. Velocity `10ms^(-1)`, speed `10ms^(-1)`B. Velocity zero, speed `10ms^(-1)`C. Velocity zero, speed zeroD. Velocity `10ms^(-1)` speed zero |
| Answer» Correct Answer - B | |
| 85. |
An object is projected upwards with a velocity of `100m//s`. It will strike the ground after (approximately)A. 10sB. 20sC. 5sD. 40s |
| Answer» Correct Answer - B | |
| 86. |
A monkey is sitting on the tree. A hunter fires a bullet to kill him. At the same time monkey falls downward. Will the bullet strike him?A. YesB. NoC. Depends upon luckD. Data insufficient |
| Answer» Correct Answer - A | |
| 87. |
A body is imparted motion from rest to move in a straight line. If it is then obstructed by an opposite force, thenA. the body may necessarily change directionB. the body is sure to slow donwC. the body will necessarily continue to move in the same direction at the same speedD. none of the above |
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Answer» Correct Answer - B As force is oppostie to velocity so body will be retarded and slow down. |
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| 88. |
A particle is moing with a velocity of `10m//s` towards east. After 10s its velocity changes to `10m//s` towards north. Its average acceleration is:-A. zeroB. `sqrt(2) m//s^(2)` towards N-WC. `(2)/sqrt(2) m//s^(2)` towards N-WD. `(1)/sqrt(2)m//s^(2)` towards N-W |
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Answer» Correct Answer - B Acceleration `veca=(Deltavecv)/(Deltat)=(10hatj-10hati)/(10)` `=-hati+hatj` (west north) `|veca|=sqrt(1+1)=sqrt(2)` (north-west) |
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| 89. |
When a ball is thrown up vertically with velocity `v_0`, it reaches a maximum height of h. If one wishes to triple the maximum height then the ball should be thrown with velocityA. `sqrt(3) v_(0)`B. `3v_(0)`C. `9 v_(0)`D. `3//2v_(0)` |
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Answer» Correct Answer - A `(h_(2))/(h_(1))=((u_(2)^(2))/(2g))/((u_(1)^(2))/(2g))rArr (3h_(1))/(h_(1))=(u_(2)^(2))/(v_(0))` `u_(2)^(2)=3v_(0)^(2)rArru_(2)=sqrt(3)v_(0)` |
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| 90. |
A projectile is thrown with an initial velocity of `(a hati +b hatj) ms^(-1)`. If the range of the projectile is twice the maximum height reached by it, thenA. a=2bB. b=aC. b=2aD. b=4a |
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Answer» Correct Answer - C `vecv=ahati+6hatjrArru cos theta=a` and `usin theta=b` `tantheta=(b)/(a)` also `tan theta=(4)/(n)=(4)/(2)=2` `therefore (b)/(a)=2 rArr b=2a` |
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| 91. |
A ball of mass (m) is thrown vertically up. Another ball of mass ` 2 m` is thrown at an angle ` theta` with the vertical. Both of them stay in air for the same period of time. What is the ratio of the height attained by two balls. |
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Answer» For the ball thrown vertically upwards, the time taken by the ball to come back is `T_(1)=(2u_(1))/(g)` For the ball projected at an angle `theta` with their vertical, the time of flight is `T_(2)=2u_(2) cos theta/g` Since time of flights for the balls is same, so `(2u_(1))/(g)=(2u_(2)cos theta)/(g)` or `u_(1)=u_(2) cos theta` `because h_(1)=(u_(1)^(2))/(2g)` and `h_(2)=(u_(2)^(2))/(2g) cos ^(2) theta therefore (h_(1))/(h_(2))=(u_(1)^(2))/(u_(2)^(2)cos^(2)theta)=(u_(2)^(2)cos^(2)theta)/(u_(2)^(2)cos^(2)theta)=1` |
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| 92. |
Two bodies are thrown with the same initial speed at angles `alpha and (90^(@) -alpha)` with the horizontal. What will be the ratio of (a) maximum heights attained by them and (b) horizontal ranges ? |
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Answer» Horizontal range `R=(u^(2))/(g)sin2 theta` and maximum height `H=(u^(2)sin^(2)theta)/(2g)` Case (i) when `theta=alpha, R_(1)=(u^(2))/(g)sin2alpha` and `H_(1)=(u^(2)sin^(2)alpha)/(2g)` Case(ii) when `theta=(90^(@)-alpha), R_(2)=(u^(2)sin2(90^(@)-alpha))/(g)=(u^(2)sin(180^(@)-2alpha))/(g)=(u^(2)sin 2alpha)/(g)` and `H_(2)=(u^(2)sin^(2)(90^(@)-alpha))/(g)=(u^(2)cos^(2)alpha)/(g) (H_(1))/(H_(2))=(sin^(2)alpha)/(cos^(2)alpha)-tan^(2)alpha` and `(R_(1))/(R_(2))=1` |
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| 93. |
A ball is thown at angle `theta` and another ball is thrown at angle `(90^(@)- theta)` with the horizontal direction from the same point each with speeds of 40 m/s. The second ball reaches 50 m higher than the first ball. Find the individual heights. ` g= 10 m//s^(2)` |
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Answer» For the first ball, angle of projection `=theta`, velocity of projection, `u=40 ms^(-1)` Let h be the maximum height attained by it. As maximum height `=(u^(2)sin^(2)theta)/(2g) therefore h=((40)^(2)sin^(2)theta)/(2xx10)`....(1) For second ball, Angle of projection `-(90^(@)-theta)`. velocity of projection, `u=40ms^(-1)` Maximum height reached `=(h+50)m` `therefore h+50=(u^(2)sin^(2)(90^(@)-theta))/(2g)=((40)^(2)cos^(2)theta)/(2xx10)` .....(2) By adding (1) and (2), `2h+50=((40)^(2))/(2xx10)xx(sin^(2)theta+cos^(2)theta)=((40)^(2))/(2xx10)=80rArr 2h=80-50=30rArrh=15m` Height of the first ball, h=15m `&` Height of the second ball, =h+50=15+50=65m |
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| 94. |
A cricketer can throw a ball to a maximum horizontal distance of 100 m. How high above the ground can the cricketer throw the same ball ? |
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Answer» Let u be the velocity of projection of the ball. The ball will cover maximum horizontal distance when angle of projection with horizontal, `theta=45^(@)`. Then `R_("max")=(u^(2))/(g)=100m` If ball is projected vertically upwards `(theta=90^(@))` from ground then H attains maximum value. `H_("max")=(u^(2))/(2g)=(R_("max"))/(2)` `therefore` the height to which cricketer can through the ball is `=(R_("max"))/(2)=(100)/(2)=50m`. |
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| 95. |
A ball is thrown upwards . Its height varies with time as shown in figure. If the acceleration due to gravity is `7.5 m//s^(2)`, then the height h is A. 10mB. 15mC. 20mD. 25m |
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Answer» Correct Answer - B The height h is convred in time interval t=1s to t =2s or t=5s to t=6s. Let u be the initial velocity given to the ball when projected vertically upwards. Then `h=s_(2)-s_(1)=s_(5)-s_(6)` .....(i) `s_(2)-s_(1)=(uxx2-(1)/(2)xx7.5xx2^(2))-(uxx1-(1)/(2)xx7.5xx1^(2))` `=u-(3)/(2)xx7.5` .....(ii) `s_(5)-s_(6)=(uxx5-(1)/(2)xx7.5xx5^(2))-(uxx6-(1)/(2)xx7.5xx6^(2))` `=-u+(11)/(2)xx7.5` `therefore u-(3)/(2)xx7.5=-u+(11)/(2)xx7.5` or `2u=(14)/(2)xx7.57xx7.5` or `2u=(14)/(2)xx7.5=7xx7.5` or `u=7xx(7.5)/(2)=26.25 m//s` `therefore h=26.25-(3)/(2)xx7.5=26.25-11.25=15.0m` |
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| 96. |
Three different objects of masses `m_(1) , m_(2)` and `m_(2)` are allowed to fall from rest and from the same point `O` along three different frictionless paths. The speeds of three objects on reaching the ground will be:A. `m_(1):m_(2):m_(3)`B. `m_(1):2m_(2):3m_(3)`C. `1:1:1`D. `(1)/(m_(1)):(1)/(m_(2)):(1)/(m_(3))` |
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Answer» Correct Answer - C PE=KE `mgh=(1)/(2)mv^(2)` `v=sqrt(2gh)` [does not depend upon mass] |
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| 97. |
A stone is thrown upwards and it rises to a height 0f 200m. The relative velocity of the stone with respect to the earth will be maximum at :-A. Height of 100mB. Height of 150mC. Heighest pointD. The ground |
| Answer» Correct Answer - D | |
| 98. |
The distance x of a particle moving in one dimensions, under the action of a constant force is related to time t by the equation, `t=sqrt(x)+3`, where x is in metres and t in seconds. Find the displacement of the particle when its velocity is zero.A. 2mB. 4mC. 5mD. zero |
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Answer» Correct Answer - D `t=sqrt(x)+3rArrsqrt(x)=t-3` `x=(t-4)^(2)=t^(2)-6t+9` `x=t^(2)-6t+9` `v=(dx)/(dt)=2t-6` for `v=0,2t-6=0rArrt=3` sec at t=3 sec `x=(3-3)^(2)=0` |
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| 99. |
A bird is flying with a speed of 40 km/hr. in the north direction. A train is moving with a speed of 40km/hr. in the west direction. A passenger sitting in the train will see the bird moving with velocity:-A. 40 km/hr in NE directionB. `40sqrt(2)` km/hr in NE directionC. 40 km/hr in. NW directionD. `40 sqrt(2) km//hr` in NW direction |
| Answer» Correct Answer - B | |
| 100. |
On a long horizontally moving belt, a child runs to an fro with a speed `9 kmh^(-1)` ( with respect to the belt) between his father and mother located 50m a part on the moving belt. The belt moves with a speed of 4 `kmh^(-1)`. For an observer on a stationery platform outside, what is the (i) speed of the child running in the direction of motion of the belt, (ii) speed of the child running opposited to the direction of motion of the belt , and (iii) time taken by child in (i) and (ii) ? which of the answers alter if motion is viewed by one of the parents? |
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Answer» (a) Speed of the child running in the direction of motion of the belt `=9+4 =13 kmhr^(-1)` (b) Speed of the child in opposite direction of the motion of the belt `=9-4=5 kmhr^(-1)` (c) Since both the parents and child are on the moving belt. So their relative position and speed remain unchanged. speed of child `=9xx(5)/(18)=2.5 ms^(-1)`, time taken `=("distance")/("speed")=(50)/(2.5)=20s` If the motion is =vieqed by one of the parents answer to (a) and (b) are altered while answer to (c) remain unchanged. |
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